Integrating Inverse square root of a polynomial

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I'm Looking for integrals of the following kind



$intlimits_0^M left( -x^3 +bx^2 -omega right)^-frac12 ,dx , ,$



for positive constants $b, omega$, and $M$ is on of the roots of the polynomial ($x=0$ isn't a root). I used the trigonometric method to find all three real roots of this polynomial.



I haven't found any good way of calculating this integral.




definite-integrals cubic-equations elliptic-integrals




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  • The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^-frac12mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
    – J. M. is not a mathematician
    Jun 15 '16 at 18:13











  • Thanks, I've changed the terminology
    – Amir Sagiv
    Jun 15 '16 at 19:31














up vote
0
down vote

favorite
1












I'm Looking for integrals of the following kind



$intlimits_0^M left( -x^3 +bx^2 -omega right)^-frac12 ,dx , ,$



for positive constants $b, omega$, and $M$ is on of the roots of the polynomial ($x=0$ isn't a root). I used the trigonometric method to find all three real roots of this polynomial.



I haven't found any good way of calculating this integral.




definite-integrals cubic-equations elliptic-integrals




share|cite|improve this question






















  • The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^-frac12mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
    – J. M. is not a mathematician
    Jun 15 '16 at 18:13











  • Thanks, I've changed the terminology
    – Amir Sagiv
    Jun 15 '16 at 19:31












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I'm Looking for integrals of the following kind



$intlimits_0^M left( -x^3 +bx^2 -omega right)^-frac12 ,dx , ,$



for positive constants $b, omega$, and $M$ is on of the roots of the polynomial ($x=0$ isn't a root). I used the trigonometric method to find all three real roots of this polynomial.



I haven't found any good way of calculating this integral.




definite-integrals cubic-equations elliptic-integrals




share|cite|improve this question














I'm Looking for integrals of the following kind



$intlimits_0^M left( -x^3 +bx^2 -omega right)^-frac12 ,dx , ,$



for positive constants $b, omega$, and $M$ is on of the roots of the polynomial ($x=0$ isn't a root). I used the trigonometric method to find all three real roots of this polynomial.



I haven't found any good way of calculating this integral.





definite-integrals cubic-equations elliptic-integrals





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 17 '16 at 7:54

























asked Jun 15 '16 at 17:45









Amir Sagiv

1708




1708











  • The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^-frac12mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
    – J. M. is not a mathematician
    Jun 15 '16 at 18:13











  • Thanks, I've changed the terminology
    – Amir Sagiv
    Jun 15 '16 at 19:31
















  • The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^-frac12mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
    – J. M. is not a mathematician
    Jun 15 '16 at 18:13











  • Thanks, I've changed the terminology
    – Amir Sagiv
    Jun 15 '16 at 19:31















The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^-frac12mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
– J. M. is not a mathematician
Jun 15 '16 at 18:13





The result will be in terms of elliptic integrals and not elliptic functions; similarly, integrals like $intleft(a+bx+cx^2right)^-frac12mathrm dx$ are expressible in terms of inverse trigonometric functions and logarithms, not trigonometric functions and exponentials. In any event, there is a standard reduction algorithm in Abramowitz and Stegun that you might want to look into.
– J. M. is not a mathematician
Jun 15 '16 at 18:13













Thanks, I've changed the terminology
– Amir Sagiv
Jun 15 '16 at 19:31




Thanks, I've changed the terminology
– Amir Sagiv
Jun 15 '16 at 19:31










1 Answer
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0
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I think you'll find everything you need and more here:



http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf



You're welcome.






share|cite|improve this answer
















  • 1




    I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
    – Amir Sagiv
    Jun 15 '16 at 19:53










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













I think you'll find everything you need and more here:



http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf



You're welcome.






share|cite|improve this answer
















  • 1




    I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
    – Amir Sagiv
    Jun 15 '16 at 19:53














up vote
0
down vote













I think you'll find everything you need and more here:



http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf



You're welcome.






share|cite|improve this answer
















  • 1




    I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
    – Amir Sagiv
    Jun 15 '16 at 19:53












up vote
0
down vote










up vote
0
down vote









I think you'll find everything you need and more here:



http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf



You're welcome.






share|cite|improve this answer












I think you'll find everything you need and more here:



http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf



You're welcome.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 15 '16 at 17:54









Mathemagician1234

13.6k24054




13.6k24054







  • 1




    I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
    – Amir Sagiv
    Jun 15 '16 at 19:53












  • 1




    I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
    – Amir Sagiv
    Jun 15 '16 at 19:53







1




1




I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53




I don't see anything there which looks like my integral. All the polynomials under the root are either of 2nd or 4th degree, but I can't see how to transform it to a cubic,
– Amir Sagiv
Jun 15 '16 at 19:53












 

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