Integrand size = 17, antiderivative size = 216 \[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx=-\frac {\left (a+b x^2\right )^p}{2 d (c+d x)^2}+\frac {p \left (a+b x^2\right )^p}{2 d (1-p) \left (c^2-d^2 x^2\right )}-\frac {2 b p x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},1-p,2,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 a c^3}-\frac {b \left (a d^2-b c^2 (1-2 p)\right ) \left (a+b x^2\right )^p \operatorname {Hypergeometric2F1}\left (2,p,1+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 d \left (b c^2+a d^2\right )^2 (1-p)} \] Output:
-1/2*(b*x^2+a)^p/d/(d*x+c)^2+1/2*p*(b*x^2+a)^p/d/(1-p)/(-d^2*x^2+c^2)-2/3* b*p*x^3*(b*x^2+a)^p*AppellF1(3/2,2,1-p,5/2,d^2*x^2/c^2,-b*x^2/a)/a/c^3/((1 +b*x^2/a)^p)-1/2*b*(a*d^2-b*c^2*(1-2*p))*(b*x^2+a)^p*hypergeom([2, p],[p+1 ],d^2*(b*x^2+a)/(a*d^2+b*c^2))/d/(a*d^2+b*c^2)^2/(1-p)
Time = 0.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.66 \[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\frac {\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (a+b x^2\right )^p \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{2 d (-1+p) (c+d x)^2} \] Input:
Integrate[(a + b*x^2)^p/(c + d*x)^3,x]
Output:
((a + b*x^2)^p*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)])/(2*d*(-1 + p)*((d*(-Sqrt[-(a/b)] + x))/(c + d*x))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p*(c + d*x)^2)
Time = 0.61 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.49, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx\) |
| \(\Big \downarrow \) 505 |
| \(\displaystyle \int \left (\frac {3 c d^2 x^2 \left (a+b x^2\right )^p}{\left (c^2-d^2 x^2\right )^3}-\frac {3 c^2 d x \left (a+b x^2\right )^p}{\left (c^2-d^2 x^2\right )^3}+\frac {d^3 x^3 \left (a+b x^2\right )^p}{\left (d^2 x^2-c^2\right )^3}+\frac {c^3 \left (a+b x^2\right )^p}{\left (c^2-d^2 x^2\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,3,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^5}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^3}-\frac {3 b^2 c^2 d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (3,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{2 (p+1) \left (a d^2+b c^2\right )^3}+\frac {b d \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (2,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{4 (p+1) \left (a d^2+b c^2\right )^3}-\frac {c^2 d \left (a+b x^2\right )^{p+1}}{4 \left (c^2-d^2 x^2\right )^2 \left (a d^2+b c^2\right )}\) |
Input:
Int[(a + b*x^2)^p/(c + d*x)^3,x]
Output:
-1/4*(c^2*d*(a + b*x^2)^(1 + p))/((b*c^2 + a*d^2)*(c^2 - d^2*x^2)^2) + (x* (a + b*x^2)^p*AppellF1[1/2, -p, 3, 3/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^3 *(1 + (b*x^2)/a)^p) + (d^2*x^3*(a + b*x^2)^p*AppellF1[3/2, -p, 3, 5/2, -(( b*x^2)/a), (d^2*x^2)/c^2])/(c^5*(1 + (b*x^2)/a)^p) + (b*d*(2*a*d^2 + b*c^2 *(1 + p))*(a + b*x^2)^(1 + p)*Hypergeometric2F1[2, 1 + p, 2 + p, (d^2*(a + b*x^2))/(b*c^2 + a*d^2)])/(4*(b*c^2 + a*d^2)^3*(1 + p)) - (3*b^2*c^2*d*(a + b*x^2)^(1 + p)*Hypergeometric2F1[3, 1 + p, 2 + p, (d^2*(a + b*x^2))/(b* c^2 + a*d^2)])/(2*(b*c^2 + a*d^2)^3*(1 + p))
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]
\[\int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{3}}d x\]
Input:
int((b*x^2+a)^p/(d*x+c)^3,x)
Output:
int((b*x^2+a)^p/(d*x+c)^3,x)
\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((b*x^2+a)^p/(d*x+c)^3,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^p/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)
\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {\left (a + b x^{2}\right )^{p}}{\left (c + d x\right )^{3}}\, dx \] Input:
integrate((b*x**2+a)**p/(d*x+c)**3,x)
Output:
Integral((a + b*x**2)**p/(c + d*x)**3, x)
\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((b*x^2+a)^p/(d*x+c)^3,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^p/(d*x + c)^3, x)
\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((b*x^2+a)^p/(d*x+c)^3,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^p/(d*x + c)^3, x)
Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{{\left (c+d\,x\right )}^3} \,d x \] Input:
int((a + b*x^2)^p/(c + d*x)^3,x)
Output:
int((a + b*x^2)^p/(c + d*x)^3, x)
\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{p}}{d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}}d x \] Input:
int((b*x^2+a)^p/(d*x+c)^3,x)
Output:
int((a + b*x**2)**p/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3),x)