Integrand size = 17, antiderivative size = 322 \[ \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx=-\frac {d^2 e \left (a+b x^2\right )^{1+p}}{4 \left (b d^2+a e^2\right ) \left (d^2-e^2 x^2\right )^2}+\frac {x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}+\frac {e^2 x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,3,\frac {5}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}+\frac {b e \left (2 a e^2+b d^2 (1+p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{4 \left (b d^2+a e^2\right )^3 (1+p)}-\frac {3 b^2 d^2 e \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (3,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 \left (b d^2+a e^2\right )^3 (1+p)} \]
-1/4*d^2*e*(b*x^2+a)^(p+1)/(a*e^2+b*d^2)/(-e^2*x^2+d^2)^2+x*(b*x^2+a)^p*Ap pellF1(1/2,3,-p,3/2,e^2*x^2/d^2,-b*x^2/a)/d^3/((1+b*x^2/a)^p)+e^2*x^3*(b*x ^2+a)^p*AppellF1(3/2,3,-p,5/2,e^2*x^2/d^2,-b*x^2/a)/d^5/((1+b*x^2/a)^p)+1/ 4*b*e*(2*a*e^2+b*d^2*(p+1))*(b*x^2+a)^(p+1)*hypergeom([2, p+1],[2+p],e^2*( b*x^2+a)/(a*e^2+b*d^2))/(a*e^2+b*d^2)^3/(p+1)-3/2*b^2*d^2*e*(b*x^2+a)^(p+1 )*hypergeom([3, p+1],[2+p],e^2*(b*x^2+a)/(a*e^2+b*d^2))/(a*e^2+b*d^2)^3/(p +1)
Time = 0.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.44 \[ \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\frac {\left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (a+b x^2\right )^p \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{2 e (-1+p) (d+e x)^2} \]
((a + b*x^2)^p*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)])/(2*e*(-1 + p)*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p*(d + e*x)^2)
Time = 0.56 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {505, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx\) |
\(\Big \downarrow \) 505 |
\(\displaystyle \int \left (\frac {3 d e^2 x^2 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}-\frac {3 d^2 e x \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac {e^3 x^3 \left (a+b x^2\right )^p}{\left (e^2 x^2-d^2\right )^3}+\frac {d^3 \left (a+b x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^2 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,3,\frac {5}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {3 b^2 d^2 e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (3,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}+\frac {b e \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (2,p+1,p+2,\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{4 (p+1) \left (a e^2+b d^2\right )^3}-\frac {d^2 e \left (a+b x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+b d^2\right )}\) |
-1/4*(d^2*e*(a + b*x^2)^(1 + p))/((b*d^2 + a*e^2)*(d^2 - e^2*x^2)^2) + (x* (a + b*x^2)^p*AppellF1[1/2, -p, 3, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^3 *(1 + (b*x^2)/a)^p) + (e^2*x^3*(a + b*x^2)^p*AppellF1[3/2, -p, 3, 5/2, -(( b*x^2)/a), (e^2*x^2)/d^2])/(d^5*(1 + (b*x^2)/a)^p) + (b*e*(2*a*e^2 + b*d^2 *(1 + p))*(a + b*x^2)^(1 + p)*Hypergeometric2F1[2, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(4*(b*d^2 + a*e^2)^3*(1 + p)) - (3*b^2*d^2*e*(a + b*x^2)^(1 + p)*Hypergeometric2F1[3, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b* d^2 + a*e^2)])/(2*(b*d^2 + a*e^2)^3*(1 + p))
3.5.27.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E xpandIntegrand[(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^( -n), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && PosQ[a/b]
\[\int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{3}}d x\]
\[ \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3}} \,d x } \]
\[ \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{(d+e x)^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \]