\(\int \frac {(a+b x^2)^p}{x^3 (c+d x)^3} \, dx\) [53]

Optimal result

Integrand size = 20, antiderivative size = 473 \[ \int \frac {\left (a+b x^2\right )^p}{x^3 (c+d x)^3} \, dx=\frac {d^2 \left (b c^2+3 a d^2\right ) \left (a+b x^2\right )^{1+p}}{2 a c \left (b c^2+a d^2\right ) \left (c^2-d^2 x^2\right )^2}-\frac {c \left (a+b x^2\right )^{1+p}}{2 a x^2 \left (c^2-d^2 x^2\right )^2}+\frac {d^2 \left (b^2 c^4+6 a^2 d^4+a b c^2 d^2 (9-2 p)\right ) \left (a+b x^2\right )^{1+p}}{2 a c^3 \left (b c^2+a d^2\right )^2 \left (c^2-d^2 x^2\right )}+\frac {3 d \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{2},-p,3,\frac {1}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 x}-\frac {d^3 x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^6}+\frac {d^4 \left (6 a^2 d^4+a b c^2 d^2 (12-5 p)+b^2 c^4 \left (6-7 p+2 p^2\right )\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 c^5 \left (b c^2+a d^2\right )^3 (1+p)}-\frac {\left (6 a d^2+b c^2 p\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a^2 c^5 (1+p)} \] Output:

1/2*d^2*(3*a*d^2+b*c^2)*(b*x^2+a)^(p+1)/a/c/(a*d^2+b*c^2)/(-d^2*x^2+c^2)^2 
-1/2*c*(b*x^2+a)^(p+1)/a/x^2/(-d^2*x^2+c^2)^2+1/2*d^2*(b^2*c^4+6*a^2*d^4+a 
*b*c^2*d^2*(9-2*p))*(b*x^2+a)^(p+1)/a/c^3/(a*d^2+b*c^2)^2/(-d^2*x^2+c^2)+3 
*d*(b*x^2+a)^p*AppellF1(-1/2,3,-p,1/2,d^2*x^2/c^2,-b*x^2/a)/c^4/x/((1+b*x^ 
2/a)^p)-d^3*x*(b*x^2+a)^p*AppellF1(1/2,3,-p,3/2,d^2*x^2/c^2,-b*x^2/a)/c^6/ 
((1+b*x^2/a)^p)+1/2*d^4*(6*a^2*d^4+a*b*c^2*d^2*(12-5*p)+b^2*c^4*(2*p^2-7*p 
+6))*(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^5/(a*d^2+b*c^2)^3/(p+1)-1/2*(b*c^2*p+6*a*d^2)*(b*x^2+a)^(p+1)*hypergeom 
([1, p+1],[2+p],1+b*x^2/a)/a^2/c^5/(p+1)
 

Mathematica [A] (warning: unable to verify)

Time = 1.39 (sec) , antiderivative size = 537, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right )^p}{x^3 (c+d x)^3} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {6 c d^2 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+2 p) (c+d x)}-\frac {c^2 d^2 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+p) (c+d x)^2}-\frac {6 d^2 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{p}+\frac {6 c d \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}+\frac {c^2 \left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (1-p,-p,2-p,-\frac {a}{b x^2}\right )}{(-1+p) x^2}+\frac {6 d^2 \left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^2}\right )}{p}\right )}{2 c^5} \] Input:

Integrate[(a + b*x^2)^p/(x^3*(c + d*x)^3),x]
 

Output:

((a + b*x^2)^p*((-6*c*d^2*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) 
) - (c^2*d^2*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)])/((-1 + p)*((d*(-Sqrt[-(a/b)] + x))/ 
(c + d*x))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p*(c + d*x)^2) - (6*d^2*Ap 
pellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x), (c + Sqrt[-( 
a/b)]*d)/(c + d*x)])/(p*((d*(-Sqrt[-(a/b)] + x))/(c + d*x))^p*((d*(Sqrt[-( 
a/b)] + x))/(c + d*x))^p) + (6*c*d*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x 
^2)/a)])/(x*(1 + (b*x^2)/a)^p) + (c^2*Hypergeometric2F1[1 - p, -p, 2 - p, 
-(a/(b*x^2))])/((-1 + p)*(1 + a/(b*x^2))^p*x^2) + (6*d^2*Hypergeometric2F1 
[-p, -p, 1 - p, -(a/(b*x^2))])/(p*(1 + a/(b*x^2))^p)))/(2*c^5)
 

Rubi [A] (verified)

Time = 2.24 (sec) , antiderivative size = 744, normalized size of antiderivative = 1.57, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^p}{x^3 (c+d x)^3} \, dx\)

\(\Big \downarrow \) 622

\(\displaystyle \int \left (\frac {3 c d^2 \left (a+b x^2\right )^p}{x \left (c^2-d^2 x^2\right )^3}-\frac {3 c^2 d \left (a+b x^2\right )^p}{x^2 \left (c^2-d^2 x^2\right )^3}+\frac {d^3 \left (a+b x^2\right )^p}{\left (d^2 x^2-c^2\right )^3}+\frac {c^3 \left (a+b x^2\right )^p}{x^3 \left (c^2-d^2 x^2\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d^4 \left (a+b x^2\right )^{p+1} \left (6 a^2 d^4+4 a b c^2 d^2 (3-p)+b^2 c^4 \left (p^2-5 p+6\right )\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{4 c^5 (p+1) \left (a d^2+b c^2\right )^3}+\frac {3 d^4 \left (a+b x^2\right )^{p+1} \left (2 a^2 d^4+2 a b c^2 d^2 (2-p)+b^2 c^4 \left (p^2-3 p+2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{4 c^5 (p+1) \left (a d^2+b c^2\right )^3}+\frac {d^2 \left (a+b x^2\right )^{p+1} \left (6 a^2 d^4+a b c^2 d^2 (9-p)+2 b^2 c^4\right )}{4 a c^3 \left (c^2-d^2 x^2\right ) \left (a d^2+b c^2\right )^2}-\frac {\left (a+b x^2\right )^{p+1} \left (3 a d^2+b c^2 p\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{2 a^2 c^5 (p+1)}-\frac {d^3 x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^6}+\frac {3 d \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{2},-p,3,\frac {1}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 x}-\frac {3 d^2 \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{2 a c^5 (p+1)}+\frac {d^2 \left (3 a d^2+2 b c^2\right ) \left (a+b x^2\right )^{p+1}}{4 a c \left (c^2-d^2 x^2\right )^2 \left (a d^2+b c^2\right )}-\frac {c \left (a+b x^2\right )^{p+1}}{2 a x^2 \left (c^2-d^2 x^2\right )^2}+\frac {3 d^4 \left (a+b x^2\right )^{p+1}}{4 c \left (c^2-d^2 x^2\right )^2 \left (a d^2+b c^2\right )}+\frac {3 d^4 \left (a+b x^2\right )^{p+1} \left (2 a d^2+b c^2 (3-p)\right )}{4 c^3 \left (c^2-d^2 x^2\right ) \left (a d^2+b c^2\right )^2}\)

Input:

Int[(a + b*x^2)^p/(x^3*(c + d*x)^3),x]
 

Output:

(3*d^4*(a + b*x^2)^(1 + p))/(4*c*(b*c^2 + a*d^2)*(c^2 - d^2*x^2)^2) + (d^2 
*(2*b*c^2 + 3*a*d^2)*(a + b*x^2)^(1 + p))/(4*a*c*(b*c^2 + a*d^2)*(c^2 - d^ 
2*x^2)^2) - (c*(a + b*x^2)^(1 + p))/(2*a*x^2*(c^2 - d^2*x^2)^2) + (3*d^4*( 
2*a*d^2 + b*c^2*(3 - p))*(a + b*x^2)^(1 + p))/(4*c^3*(b*c^2 + a*d^2)^2*(c^ 
2 - d^2*x^2)) + (d^2*(2*b^2*c^4 + 6*a^2*d^4 + a*b*c^2*d^2*(9 - p))*(a + b* 
x^2)^(1 + p))/(4*a*c^3*(b*c^2 + a*d^2)^2*(c^2 - d^2*x^2)) + (3*d*(a + b*x^ 
2)^p*AppellF1[-1/2, -p, 3, 1/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^4*x*(1 + 
(b*x^2)/a)^p) - (d^3*x*(a + b*x^2)^p*AppellF1[1/2, -p, 3, 3/2, -((b*x^2)/a 
), (d^2*x^2)/c^2])/(c^6*(1 + (b*x^2)/a)^p) + (d^4*(6*a^2*d^4 + 4*a*b*c^2*d 
^2*(3 - p) + b^2*c^4*(6 - 5*p + p^2))*(a + b*x^2)^(1 + p)*Hypergeometric2F 
1[1, 1 + p, 2 + p, (d^2*(a + b*x^2))/(b*c^2 + a*d^2)])/(4*c^5*(b*c^2 + a*d 
^2)^3*(1 + p)) + (3*d^4*(2*a^2*d^4 + 2*a*b*c^2*d^2*(2 - p) + b^2*c^4*(2 - 
3*p + p^2))*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (d^2*(a 
 + b*x^2))/(b*c^2 + a*d^2)])/(4*c^5*(b*c^2 + a*d^2)^3*(1 + p)) - (3*d^2*(a 
 + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(2*a* 
c^5*(1 + p)) - ((3*a*d^2 + b*c^2*p)*(a + b*x^2)^(1 + p)*Hypergeometric2F1[ 
1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(2*a^2*c^5*(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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((b*x^2+a)^p/x^3/(d*x+c)^3,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^p/(d^3*x^6 + 3*c*d^2*x^5 + 3*c^2*d*x^4 + c^3*x^3), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((b*x^2+a)^p/x^3/(d*x+c)^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((b*x^2+a)^p/x^3/(d*x+c)^3,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^p}{x^3 (c+d x)^3} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{p}}{d^{3} x^{6}+3 c \,d^{2} x^{5}+3 c^{2} d \,x^{4}+c^{3} x^{3}}d x \] Input:

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

Output:

int((a + b*x**2)**p/(c**3*x**3 + 3*c**2*d*x**4 + 3*c*d**2*x**5 + d**3*x**6 
),x)