Integrand size = 27, antiderivative size = 392 \[ \int \frac {(f x)^m \left (d+e x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^2\right )}{2 a \left (b^2-4 a c\right ) f \left (a+b x^2+c x^4\right )}+\frac {c \left (b \left (4 a e+\sqrt {b^2-4 a c} d (1-m)\right )-2 a \left (\sqrt {b^2-4 a c} e (1-m)+2 c d (3-m)\right )+b^2 (d-d m)\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) f (1+m)}-\frac {c \left (b \left (4 a e-\sqrt {b^2-4 a c} d (1-m)\right )+2 a \left (\sqrt {b^2-4 a c} e (1-m)-2 c d (3-m)\right )+b^2 d (1-m)\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) f (1+m)} \]
1/2*(f*x)^(1+m)*(b^2*d-2*a*c*d-a*b*e+c*(-2*a*e+b*d)*x^2)/a/(-4*a*c+b^2)/f/ (c*x^4+b*x^2+a)-1/2*c*(f*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-2* c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(b^2*d*(1-m)+b*(4*a*e-d*(1-m)*(-4*a*c+b^2)^( 1/2))+2*a*(-2*c*d*(3-m)+e*(1-m)*(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/ f/(1+m)/(b+(-4*a*c+b^2)^(1/2))+1/2*c*(f*x)^(1+m)*hypergeom([1, 1/2+1/2*m], [3/2+1/2*m],-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))*(b^2*(-d*m+d)+b*(4*a*e+d*(1-m )*(-4*a*c+b^2)^(1/2))-2*a*(2*c*d*(3-m)+e*(1-m)*(-4*a*c+b^2)^(1/2)))/a/(-4* a*c+b^2)^(3/2)/f/(1+m)/(b-(-4*a*c+b^2)^(1/2))
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.53 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.41 \[ \int \frac {(f x)^m \left (d+e x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {x (f x)^m \left (d (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},2,2,\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+e (1+m) x^2 \operatorname {AppellF1}\left (\frac {3+m}{2},2,2,\frac {5+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )}{a^2 (1+m) (3+m)} \]
(x*(f*x)^m*(d*(3 + m)*AppellF1[(1 + m)/2, 2, 2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + e*(1 + m)*x^2*A ppellF1[(3 + m)/2, 2, 2, (5 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2 *c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(a^2*(1 + m)*(3 + m))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (d+e x^2\right ) (f x)^m}{\left (a+b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 1600 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {(f x)^m \left (d (1-m) b^2+a e (m+1) b+c (b d-2 a e) (1-m) x^2-2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int -\frac {(f x)^m \left (-\left ((d-d m) b^2\right )-a e (m+1) b-c (b d-2 a e) (1-m) x^2+2 a c d (3-m)\right )}{c x^4+b x^2+a}dx}{2 a \left (b^2-4 a c\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\) |
3.3.24.3.1 Defintions of rubi rules used
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(a + b*x^2 + c*x^4)^(p + 1) *((d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^2)/(2*a*f*(p + 1)*(b^2 - 4*a *c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) Int[(f*x)^m*(a + b*x^2 + c *x^4)^(p + 1)*Simp[d*(b^2*(m + 2*(p + 1) + 1) - 2*a*c*(m + 4*(p + 1) + 1)) - a*b*e*(m + 1) + c*(m + 2*(2*p + 3) + 1)*(b*d - 2*a*e)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && Int egerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
\[\int \frac {\left (f x \right )^{m} \left (e \,x^{2}+d \right )}{\left (c \,x^{4}+b \,x^{2}+a \right )^{2}}d x\]
\[ \int \frac {(f x)^m \left (d+e x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}} \,d x } \]
integral((e*x^2 + d)*(f*x)^m/(c^2*x^8 + 2*b*c*x^6 + (b^2 + 2*a*c)*x^4 + 2* a*b*x^2 + a^2), x)
Timed out. \[ \int \frac {(f x)^m \left (d+e x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {(f x)^m \left (d+e x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}} \,d x } \]
\[ \int \frac {(f x)^m \left (d+e x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(f x)^m \left (d+e x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int \frac {{\left (f\,x\right )}^m\,\left (e\,x^2+d\right )}{{\left (c\,x^4+b\,x^2+a\right )}^2} \,d x \]