Integrand size = 27, antiderivative size = 210 \[ \int \frac {x^3 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+q)}-\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+q)} \]
-1/2*(e*x^2+d)^(1+q)*hypergeom([1, 1+q],[2+q],2*c*(e*x^2+d)/(2*c*d-e*(b-(- 4*a*c+b^2)^(1/2))))*(1-b/(-4*a*c+b^2)^(1/2))/(1+q)/(2*c*d-e*(b-(-4*a*c+b^2 )^(1/2)))-1/2*(e*x^2+d)^(1+q)*hypergeom([1, 1+q],[2+q],2*c*(e*x^2+d)/(2*c* d-e*(b+(-4*a*c+b^2)^(1/2))))*(1+b/(-4*a*c+b^2)^(1/2))/(1+q)/(2*c*d-e*(b+(- 4*a*c+b^2)^(1/2)))
Time = 0.35 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=-\frac {\left (d+e x^2\right )^{1+q} \left (\left (-b d+\sqrt {b^2-4 a c} d+2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^2\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )+\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \operatorname {Hypergeometric2F1}\left (1,1+q,2+q,\frac {2 c \left (d+e x^2\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )\right )}{4 \sqrt {b^2-4 a c} \left (c d^2+e (-b d+a e)\right ) (1+q)} \]
-1/4*((d + e*x^2)^(1 + q)*((-(b*d) + Sqrt[b^2 - 4*a*c]*d + 2*a*e)*Hypergeo metric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d + (-b + Sqrt[b^2 - 4*a *c])*e)] + (b*d + Sqrt[b^2 - 4*a*c]*d - 2*a*e)*Hypergeometric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]))/(Sqrt[b^2 - 4*a*c]*(c*d^2 + e*(-(b*d) + a*e))*(1 + q))
Time = 0.38 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1578, 1200, 2009}
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.
\(\displaystyle \int \frac {x^3 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx\) |
\(\Big \downarrow \) 1578 |
\(\displaystyle \frac {1}{2} \int \frac {x^2 \left (e x^2+d\right )^q}{c x^4+b x^2+a}dx^2\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \frac {1}{2} \int \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (e x^2+d\right )^q}{2 c x^2+b-\sqrt {b^2-4 a c}}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (e x^2+d\right )^q}{2 c x^2+b+\sqrt {b^2-4 a c}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (d+e x^2\right )^{q+1} \operatorname {Hypergeometric2F1}\left (1,q+1,q+2,\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(q+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (d+e x^2\right )^{q+1} \operatorname {Hypergeometric2F1}\left (1,q+1,q+2,\frac {2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(q+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )\) |
(-(((1 - b/Sqrt[b^2 - 4*a*c])*(d + e*x^2)^(1 + q)*Hypergeometric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)])/((2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(1 + q))) - ((1 + b/Sqrt[b^2 - 4*a*c])*(d + e*x^2)^(1 + q)*Hypergeometric2F1[1, 1 + q, 2 + q, (2*c*(d + e*x^2))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(1 + q)))/2
3.5.3.3.1 Defintions of rubi rules used
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
\[\int \frac {x^{3} \left (e \,x^{2}+d \right )^{q}}{c \,x^{4}+b \,x^{2}+a}d x\]
\[ \int \frac {x^3 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{3}}{c x^{4} + b x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{3}}{c x^{4} + b x^{2} + a} \,d x } \]
\[ \int \frac {x^3 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q} x^{3}}{c x^{4} + b x^{2} + a} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx=\int \frac {x^3\,{\left (e\,x^2+d\right )}^q}{c\,x^4+b\,x^2+a} \,d x \]