\(\int \frac {(a+b x^2)^p}{x (c+d x)^3} \, dx\) [52]

Optimal result

Integrand size = 20, antiderivative size = 388 \[ \int \frac {\left (a+b x^2\right )^p}{x (c+d x)^3} \, dx=\frac {c d^2 \left (a+b x^2\right )^{1+p}}{\left (b c^2+a d^2\right ) \left (c^2-d^2 x^2\right )^2}+\frac {d^2 \left (a d^2+b c^2 (3-2 p)\right ) \left (a+b x^2\right )^{1+p}}{2 c \left (b c^2+a d^2\right )^2 \left (c^2-d^2 x^2\right )}-\frac {3 d x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4}-\frac {d^3 x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,3,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 c^6}+\frac {d^2 \left (a^2 d^4+a b c^2 d^2 (2-p)+b^2 c^4 \left (1-3 p+2 p^2\right )\right ) \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{2 c^3 \left (b c^2+a d^2\right )^3 (1+p)}-\frac {\left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{2 a c^3 (1+p)} \] Output:

c*d^2*(b*x^2+a)^(p+1)/(a*d^2+b*c^2)/(-d^2*x^2+c^2)^2+1/2*d^2*(a*d^2+b*c^2* 
(3-2*p))*(b*x^2+a)^(p+1)/c/(a*d^2+b*c^2)^2/(-d^2*x^2+c^2)-3*d*x*(b*x^2+a)^ 
p*AppellF1(1/2,3,-p,3/2,d^2*x^2/c^2,-b*x^2/a)/c^4/((1+b*x^2/a)^p)-1/3*d^3* 
x^3*(b*x^2+a)^p*AppellF1(3/2,3,-p,5/2,d^2*x^2/c^2,-b*x^2/a)/c^6/((1+b*x^2/ 
a)^p)+1/2*d^2*(a^2*d^4+a*b*c^2*d^2*(2-p)+b^2*c^4*(2*p^2-3*p+1))*(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)^3/(p+1)-1/2*(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 = 0.61 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^p}{x (c+d x)^3} \, dx=\frac {\left (a+b x^2\right )^p \left (-\frac {2 c \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+2 p) (c+d x)}-\frac {c^2 \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{(-1+p) (c+d x)^2}+\frac {-\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )+\left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^2}\right )}{p}\right )}{2 c^3} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.50 (sec) , antiderivative size = 464, normalized size of antiderivative = 1.20, 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.

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

\(\Big \downarrow \) 622

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c*d^2*(a + b*x^2)^(1 + p))/(4*(b*c^2 + a*d^2)*(c^2 - d^2*x^2)^2) + (d^2*( 
2*a*d^2 + b*c^2*(3 - p))*(a + b*x^2)^(1 + p))/(4*c*(b*c^2 + a*d^2)^2*(c^2 
- d^2*x^2)) - (3*d*x*(a + b*x^2)^p*AppellF1[1/2, -p, 3, 3/2, -((b*x^2)/a), 
 (d^2*x^2)/c^2])/(c^4*(1 + (b*x^2)/a)^p) - (d^3*x^3*(a + b*x^2)^p*AppellF1 
[3/2, -p, 3, 5/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(3*c^6*(1 + (b*x^2)/a)^p) 
+ (d^2*(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^3*(b*c^2 + a*d^2)^3*(1 + p)) - ((a + b*x^2)^(1 + p)*Hyper 
geometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(2*a*c^3*(1 + p)) + (3*b^2*c 
*d^2*(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))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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