\(\int \frac {(a+b x^2)^p}{x^2 (c+d x)^2} \, dx\) [47]

Optimal result

Integrand size = 20, antiderivative size = 296 \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx=-\frac {d^3 \left (a+b x^2\right )^{1+p}}{c \left (b c^2+a d^2\right ) \left (c^2-d^2 x^2\right )}-\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{2},-p,2,\frac {1}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 x}+\frac {d^2 x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4}-\frac {d^3 \left (a d^2+b c^2 (1-p)\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{c^3 \left (b c^2+a d^2\right )^2 (1+p)}+\frac {d \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{a c^3 (1+p)} \] Output:

-d^3*(b*x^2+a)^(p+1)/c/(a*d^2+b*c^2)/(-d^2*x^2+c^2)-(b*x^2+a)^p*AppellF1(- 
1/2,2,-p,1/2,d^2*x^2/c^2,-b*x^2/a)/c^2/x/((1+b*x^2/a)^p)+d^2*x*(b*x^2+a)^p 
*AppellF1(1/2,2,-p,3/2,d^2*x^2/c^2,-b*x^2/a)/c^4/((1+b*x^2/a)^p)-d^3*(a*d^ 
2+b*c^2*(1-p))*(b*x^2+a)^(p+1)*hypergeom([1, p+1],[2+p],d^2*(b*x^2+a)/(a*d 
^2+b*c^2))/c^3/(a*d^2+b*c^2)^2/(p+1)+d*(b*x^2+a)^(p+1)*hypergeom([1, p+1], 
[2+p],1+b*x^2/a)/a/c^3/(p+1)
 

Mathematica [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.16 \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx=\frac {\left (a+b x^2\right )^p \left (\frac {c d \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+2 p) (c+d x)}+\frac {d \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{p}-\frac {c \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}-\frac {d \left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^2}\right )}{p}\right )}{c^3} \] Input:

Integrate[(a + b*x^2)^p/(x^2*(c + d*x)^2),x]
 

Output:

((a + b*x^2)^p*((c*d*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (c - Sqrt[-(a/b)]* 
d)/(c + d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)])/((-1 + 2*p)*((d*(-Sqrt[-(a/ 
b)] + x))/(c + d*x))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p*(c + d*x)) + ( 
d*AppellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x), (c + Sqr 
t[-(a/b)]*d)/(c + d*x)])/(p*((d*(-Sqrt[-(a/b)] + x))/(c + d*x))^p*((d*(Sqr 
t[-(a/b)] + x))/(c + d*x))^p) - (c*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x 
^2)/a)])/(x*(1 + (b*x^2)/a)^p) - (d*Hypergeometric2F1[-p, -p, 1 - p, -(a/( 
b*x^2))])/(p*(1 + a/(b*x^2))^p)))/c^3
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {622, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx\)

\(\Big \downarrow \) 622

\(\displaystyle \int \left (-\frac {2 c d \left (a+b x^2\right )^p}{x \left (c^2-d^2 x^2\right )^2}+\frac {c^2 \left (a+b x^2\right )^p}{x^2 \left (c^2-d^2 x^2\right )^2}+\frac {d^2 \left (a+b x^2\right )^p}{\left (d^2 x^2-c^2\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{2},-p,2,\frac {1}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 x}+\frac {d^2 x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4}+\frac {d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a c^3 (p+1)}-\frac {d^3 \left (a+b x^2\right )^{p+1}}{c \left (c^2-d^2 x^2\right ) \left (a d^2+b c^2\right )}-\frac {d^3 \left (a+b x^2\right )^{p+1} \left (a d^2+b c^2 (1-p)\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{c^3 (p+1) \left (a d^2+b c^2\right )^2}\)

Input:

Int[(a + b*x^2)^p/(x^2*(c + d*x)^2),x]
 

Output:

-((d^3*(a + b*x^2)^(1 + p))/(c*(b*c^2 + a*d^2)*(c^2 - d^2*x^2))) - ((a + b 
*x^2)^p*AppellF1[-1/2, -p, 2, 1/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^2*x*(1 
 + (b*x^2)/a)^p) + (d^2*x*(a + b*x^2)^p*AppellF1[1/2, -p, 2, 3/2, -((b*x^2 
)/a), (d^2*x^2)/c^2])/(c^4*(1 + (b*x^2)/a)^p) - (d^3*(a*d^2 + b*c^2*(1 - p 
))*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (d^2*(a + b*x^2) 
)/(b*c^2 + a*d^2)])/(c^3*(b*c^2 + a*d^2)^2*(1 + p)) + (d*(a + b*x^2)^(1 + 
p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(a*c^3*(1 + p))
 

Defintions of rubi rules used

rule 622

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Int[ExpandIntegrand[x^m*(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, m, p}, x] && ILtQ[n, -1]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2} \left (d x +c \right )^{2}}d x\]

Input:

int((b*x^2+a)^p/x^2/(d*x+c)^2,x)
 

Output:

int((b*x^2+a)^p/x^2/(d*x+c)^2,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x + c\right )}^{2} x^{2}} \,d x } \] Input:

integrate((b*x^2+a)^p/x^2/(d*x+c)^2,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^p/(d^2*x^4 + 2*c*d*x^3 + c^2*x^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)**p/x**2/(d*x+c)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x + c\right )}^{2} x^{2}} \,d x } \] Input:

integrate((b*x^2+a)^p/x^2/(d*x+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^p/((d*x + c)^2*x^2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((b*x^2+a)^p/x^2/(d*x+c)^2,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[0,1,2,0]%%%} / %%%{1,[0,0,0,2]%%%} Error: Bad Argument 
Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{x^2\,{\left (c+d\,x\right )}^2} \,d x \] Input:

int((a + b*x^2)^p/(x^2*(c + d*x)^2),x)
 

Output:

int((a + b*x^2)^p/(x^2*(c + d*x)^2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^p}{x^2 (c+d x)^2} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{p}}{d^{2} x^{4}+2 c d \,x^{3}+c^{2} x^{2}}d x \] Input:

int((b*x^2+a)^p/x^2/(d*x+c)^2,x)
 

Output:

int((a + b*x**2)**p/(c**2*x**2 + 2*c*d*x**3 + d**2*x**4),x)