Integrand size = 27, antiderivative size = 194 \[ \int \frac {(f x)^m \left (d+e x^2\right )}{a+b x^2+c x^4} \, dx=\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f (1+m)}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f (1+m)} \]
(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)))*(e+(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))/f/(1+m)/(b-(-4*a*c+b^2)^(1/2)) +(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)))*(e+(b*e-2*c*d)/(-4*a*c+b^2)^(1/2))/f/(1+m)/(b+(-4*a*c+b^2)^(1/2))
Time = 0.35 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.80 \[ \int \frac {(f x)^m \left (d+e x^2\right )}{a+b x^2+c x^4} \, dx=\frac {x (f x)^m \left (\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+\left (-b d+\sqrt {b^2-4 a c} d+2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \sqrt {b^2-4 a c} (1+m)} \]
(x*(f*x)^m*((b*d + Sqrt[b^2 - 4*a*c]*d - 2*a*e)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + (-(b*d) + Sqrt[b^2 - 4*a*c]*d + 2*a*e)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (-2*c*x^2)/ (b + Sqrt[b^2 - 4*a*c])]))/(2*a*Sqrt[b^2 - 4*a*c]*(1 + m))
Time = 0.39 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1608, 27, 278}
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.
\(\displaystyle \int \frac {\left (d+e x^2\right ) (f x)^m}{a+b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 1608 |
\(\displaystyle \frac {1}{2} \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \int \frac {2 (f x)^m}{2 c x^2+b-\sqrt {b^2-4 a c}}dx+\frac {1}{2} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {2 (f x)^m}{2 c x^2+b+\sqrt {b^2-4 a c}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \int \frac {(f x)^m}{2 c x^2+b-\sqrt {b^2-4 a c}}dx+\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {(f x)^m}{2 c x^2+b+\sqrt {b^2-4 a c}}dx\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {(f x)^{m+1} \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(f x)^{m+1} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \left (\sqrt {b^2-4 a c}+b\right )}\) |
((e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*(f*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])])/((b - Sqrt[b^2 - 4*a*c])*f*(1 + m)) + ((e - (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*(f*x)^(1 + m )*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4* a*c])])/((b + Sqrt[b^2 - 4*a*c])*f*(1 + m))
3.3.23.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2))/((a_) + (b_.)*(x_)^2 + (c_.) *(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0]
\[\int \frac {\left (f x \right )^{m} \left (e \,x^{2}+d \right )}{c \,x^{4}+b \,x^{2}+a}d x\]
\[ \int \frac {(f x)^m \left (d+e x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
\[ \int \frac {(f x)^m \left (d+e x^2\right )}{a+b x^2+c x^4} \, dx=\int \frac {\left (f x\right )^{m} \left (d + e x^{2}\right )}{a + b x^{2} + c x^{4}}\, dx \]
\[ \int \frac {(f x)^m \left (d+e x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
\[ \int \frac {(f x)^m \left (d+e x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{c x^{4} + b x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {(f x)^m \left (d+e x^2\right )}{a+b x^2+c x^4} \, dx=\int \frac {{\left (f\,x\right )}^m\,\left (e\,x^2+d\right )}{c\,x^4+b\,x^2+a} \,d x \]