3.5.29 \(\int \frac {(a+b x^2)^p}{x^2 (d+e x)^3} \, dx\) [429]

3.5.29.1 Optimal result

Integrand size = 20, antiderivative size = 754 \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)^3} \, dx=-\frac {e^3 \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 {3 e^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,1,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}+\frac {2 e^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 {e^2 x^2}{d^2}\right )}{d^5}+\frac {e^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,3,\frac {3}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^5}+\frac {2 e^4 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,2,\frac {5}{2},-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{3 d^7}+\frac {e^4 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^7}-\frac {\left (a+b x^2\right )^p \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 )}{d^3 x}-\frac {3 e^3 \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 d^4 \left (b d^2+a e^2\right ) (1+p)}+\frac {3 e \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 d^4 (1+p)}-\frac {2 b e^3 \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 )}{d^2 \left (b d^2+a e^2\right )^2 (1+p)}+\frac {b e^3 \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 d^2 \left (b d^2+a e^2\right )^3 (1+p)}-\frac {3 b^2 e^3 \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)} \]

output

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

3.5.29.2 Mathematica [A] (warning: unable to verify)

Time = 0.77 (sec) , antiderivative size = 478, normalized size of antiderivative = 0.63 \[ \int \frac {\left (a+b x^2\right )^p}{x^2 (d+e x)^3} \, dx=\frac {\left (a+b x^2\right )^p \left (\frac {4 d e \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+2 p) (d+e x)}+\frac {d^2 e \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (2-2 p,-p,-p,3-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(-1+p) (d+e x)^2}+\frac {3 e \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}-\frac {2 d \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )}{x}-\frac {3 e \left (1+\frac {a}{b x^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^2}\right )}{p}\right )}{2 d^4} \]

input

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

output

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

3.5.29.3 Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 460, normalized size of antiderivative = 0.61, 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.

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

\(\Big \downarrow \) 622

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

\(\Big \downarrow \) 2009

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

input

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

output

(-3*e^3*(a + b*x^2)^(1 + p))/(4*(b*d^2 + a*e^2)*(d^2 - e^2*x^2)^2) - (3*e^ 
3*(2*a*e^2 + b*d^2*(3 - p))*(a + b*x^2)^(1 + p))/(4*d^2*(b*d^2 + a*e^2)^2* 
(d^2 - e^2*x^2)) - ((a + b*x^2)^p*AppellF1[-1/2, -p, 3, 1/2, -((b*x^2)/a), 
 (e^2*x^2)/d^2])/(d^3*x*(1 + (b*x^2)/a)^p) + (3*e^2*x*(a + b*x^2)^p*Appell 
F1[1/2, -p, 3, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^5*(1 + (b*x^2)/a)^p) 
- (3*e^3*(2*a^2*e^4 + 2*a*b*d^2*e^2*(2 - p) + b^2*d^4*(2 - 3*p + p^2))*(a 
+ b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d 
^2 + a*e^2)])/(4*d^4*(b*d^2 + a*e^2)^3*(1 + p)) + (3*e*(a + b*x^2)^(1 + p) 
*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(2*a*d^4*(1 + p)) - (b 
^2*e^3*(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))
 

3.5.29.3.1 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]
 

3.5.29.4 Maple [F]

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

input

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

output

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

3.5.29.5 Fricas [F]

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

input

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

output

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

3.5.29.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.5.29.7 Maxima [F]

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

input

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

output

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

3.5.29.8 Giac [F]

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

input

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

output

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

3.5.29.9 Mupad [F(-1)]

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

input

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

output

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