\(\int \frac {(a+b x^2)^2}{x^5 (c+d x^2)^{3/2}} \, dx\) [898]

Optimal result

Integrand size = 24, antiderivative size = 131 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {(b c-a d)^2}{c^3 \sqrt {c+d x^2}}-\frac {a^2 \sqrt {c+d x^2}}{4 c^2 x^4}-\frac {a (8 b c-7 a d) \sqrt {c+d x^2}}{8 c^3 x^2}-\frac {\left (8 b^2 c^2-24 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{7/2}} \] Output:

(-a*d+b*c)^2/c^3/(d*x^2+c)^(1/2)-1/4*a^2*(d*x^2+c)^(1/2)/c^2/x^4-1/8*a*(-7 
*a*d+8*b*c)*(d*x^2+c)^(1/2)/c^3/x^2-1/8*(15*a^2*d^2-24*a*b*c*d+8*b^2*c^2)* 
arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {c} \left (8 b^2 c^2 x^4-8 a b c x^2 \left (c+3 d x^2\right )+a^2 \left (-2 c^2+5 c d x^2+15 d^2 x^4\right )\right )}{x^4 \sqrt {c+d x^2}}+\left (-8 b^2 c^2+24 a b c d-15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 c^{7/2}} \] Input:

Integrate[(a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x]
 

Output:

((Sqrt[c]*(8*b^2*c^2*x^4 - 8*a*b*c*x^2*(c + 3*d*x^2) + a^2*(-2*c^2 + 5*c*d 
*x^2 + 15*d^2*x^4)))/(x^4*Sqrt[c + d*x^2]) + (-8*b^2*c^2 + 24*a*b*c*d - 15 
*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*c^(7/2))
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {354, 100, 27, 87, 61, 73, 221}

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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^2}{x^6 \left (d x^2+c\right )^{3/2}}dx^2\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {4 b^2 c x^2+a (8 b c-5 a d)}{2 x^4 \left (d x^2+c\right )^{3/2}}dx^2}{2 c}-\frac {a^2}{2 c x^4 \sqrt {c+d x^2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {4 b^2 c x^2+a (8 b c-5 a d)}{x^4 \left (d x^2+c\right )^{3/2}}dx^2}{4 c}-\frac {a^2}{2 c x^4 \sqrt {c+d x^2}}\right )\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \int \frac {1}{x^2 \left (d x^2+c\right )^{3/2}}dx^2}{2 c}-\frac {a (8 b c-5 a d)}{c x^2 \sqrt {c+d x^2}}}{4 c}-\frac {a^2}{2 c x^4 \sqrt {c+d x^2}}\right )\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \left (\frac {\int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2}{c}+\frac {2}{c \sqrt {c+d x^2}}\right )}{2 c}-\frac {a (8 b c-5 a d)}{c x^2 \sqrt {c+d x^2}}}{4 c}-\frac {a^2}{2 c x^4 \sqrt {c+d x^2}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \left (\frac {2 \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{c d}+\frac {2}{c \sqrt {c+d x^2}}\right )}{2 c}-\frac {a (8 b c-5 a d)}{c x^2 \sqrt {c+d x^2}}}{4 c}-\frac {a^2}{2 c x^4 \sqrt {c+d x^2}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\left (8 b^2 c^2-3 a d (8 b c-5 a d)\right ) \left (\frac {2}{c \sqrt {c+d x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{c^{3/2}}\right )}{2 c}-\frac {a (8 b c-5 a d)}{c x^2 \sqrt {c+d x^2}}}{4 c}-\frac {a^2}{2 c x^4 \sqrt {c+d x^2}}\right )\)

Input:

Int[(a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x]
 

Output:

(-1/2*a^2/(c*x^4*Sqrt[c + d*x^2]) + (-((a*(8*b*c - 5*a*d))/(c*x^2*Sqrt[c + 
 d*x^2])) + ((8*b^2*c^2 - 3*a*d*(8*b*c - 5*a*d))*(2/(c*Sqrt[c + d*x^2]) - 
(2*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/c^(3/2)))/(2*c))/(4*c))/2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(\frac {-\frac {\sqrt {x^{2} d +c}\, a \left (-7 a d \,x^{2}+8 x^{2} b c +2 a c \right )}{8 x^{4}}-\frac {\left (15 a^{2} d^{2}-24 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}+\frac {a^{2} d^{2}-2 a b c d +b^{2} c^{2}}{\sqrt {x^{2} d +c}}}{c^{3}}\)	\(113\)
risch	\(-\frac {\sqrt {x^{2} d +c}\, a \left (-7 a d \,x^{2}+8 x^{2} b c +2 a c \right )}{8 c^{3} x^{4}}+\frac {c \left (15 a^{2} d^{2}-24 a b c d +8 b^{2} c^{2}\right ) \left (\frac {1}{c \sqrt {x^{2} d +c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )}{c^{\frac {3}{2}}}\right )-\frac {a d \left (7 a d -8 b c \right )}{\sqrt {x^{2} d +c}}}{8 c^{3}}\)	\(134\)
default	\(a^{2} \left (-\frac {1}{4 c \,x^{4} \sqrt {x^{2} d +c}}-\frac {5 d \left (-\frac {1}{2 c \,x^{2} \sqrt {x^{2} d +c}}-\frac {3 d \left (\frac {1}{c \sqrt {x^{2} d +c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\right )+b^{2} \left (\frac {1}{c \sqrt {x^{2} d +c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )}{c^{\frac {3}{2}}}\right )+2 a b \left (-\frac {1}{2 c \,x^{2} \sqrt {x^{2} d +c}}-\frac {3 d \left (\frac {1}{c \sqrt {x^{2} d +c}}-\frac {\ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )\)	\(212\)

Input:

int((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/c^3*(-1/8*(d*x^2+c)^(1/2)*a*(-7*a*d*x^2+8*b*c*x^2+2*a*c)/x^4-1/8*(15*a^2 
*d^2-24*a*b*c*d+8*b^2*c^2)/c^(1/2)*arctanh((d*x^2+c)^(1/2)/c^(1/2))+(a^2*d 
^2-2*a*b*c*d+b^2*c^2)/(d*x^2+c)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.80 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (8 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} + {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4}\right )} \sqrt {c} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (2 \, a^{2} c^{3} - {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4} + {\left (8 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{16 \, {\left (c^{4} d x^{6} + c^{5} x^{4}\right )}}, \frac {{\left ({\left (8 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{6} + {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{2} + c} \sqrt {-c}}{c}\right ) - {\left (2 \, a^{2} c^{3} - {\left (8 \, b^{2} c^{3} - 24 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} x^{4} + {\left (8 \, a b c^{3} - 5 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{8 \, {\left (c^{4} d x^{6} + c^{5} x^{4}\right )}}\right ] \] Input:

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x, algorithm="fricas")
 

Output:

[1/16*(((8*b^2*c^2*d - 24*a*b*c*d^2 + 15*a^2*d^3)*x^6 + (8*b^2*c^3 - 24*a* 
b*c^2*d + 15*a^2*c*d^2)*x^4)*sqrt(c)*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt( 
c) + 2*c)/x^2) - 2*(2*a^2*c^3 - (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2)* 
x^4 + (8*a*b*c^3 - 5*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c^4*d*x^6 + c^5*x^4 
), 1/8*(((8*b^2*c^2*d - 24*a*b*c*d^2 + 15*a^2*d^3)*x^6 + (8*b^2*c^3 - 24*a 
*b*c^2*d + 15*a^2*c*d^2)*x^4)*sqrt(-c)*arctan(sqrt(d*x^2 + c)*sqrt(-c)/c) 
- (2*a^2*c^3 - (8*b^2*c^3 - 24*a*b*c^2*d + 15*a^2*c*d^2)*x^4 + (8*a*b*c^3 
- 5*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c^4*d*x^6 + c^5*x^4)]
 

Sympy [F]

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

integrate((b*x**2+a)**2/x**5/(d*x**2+c)**(3/2),x)
 

Output:

Integral((a + b*x**2)**2/(x**5*(c + d*x**2)**(3/2)), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=-\frac {b^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {3}{2}}} + \frac {3 \, a b d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{c^{\frac {5}{2}}} - \frac {15 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, c^{\frac {7}{2}}} + \frac {b^{2}}{\sqrt {d x^{2} + c} c} - \frac {3 \, a b d}{\sqrt {d x^{2} + c} c^{2}} + \frac {15 \, a^{2} d^{2}}{8 \, \sqrt {d x^{2} + c} c^{3}} - \frac {a b}{\sqrt {d x^{2} + c} c x^{2}} + \frac {5 \, a^{2} d}{8 \, \sqrt {d x^{2} + c} c^{2} x^{2}} - \frac {a^{2}}{4 \, \sqrt {d x^{2} + c} c x^{4}} \] Input:

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x, algorithm="maxima")
 

Output:

-b^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(3/2) + 3*a*b*d*arcsinh(c/(sqrt(c*d)* 
abs(x)))/c^(5/2) - 15/8*a^2*d^2*arcsinh(c/(sqrt(c*d)*abs(x)))/c^(7/2) + b^ 
2/(sqrt(d*x^2 + c)*c) - 3*a*b*d/(sqrt(d*x^2 + c)*c^2) + 15/8*a^2*d^2/(sqrt 
(d*x^2 + c)*c^3) - a*b/(sqrt(d*x^2 + c)*c*x^2) + 5/8*a^2*d/(sqrt(d*x^2 + c 
)*c^2*x^2) - 1/4*a^2/(sqrt(d*x^2 + c)*c*x^4)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left (8 \, b^{2} c^{2} - 24 \, a b c d + 15 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{8 \, \sqrt {-c} c^{3}} + \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{\sqrt {d x^{2} + c} c^{3}} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d - 8 \, \sqrt {d x^{2} + c} a b c^{2} d - 7 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2} + 9 \, \sqrt {d x^{2} + c} a^{2} c d^{2}}{8 \, c^{3} d^{2} x^{4}} \] Input:

integrate((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x, algorithm="giac")
 

Output:

1/8*(8*b^2*c^2 - 24*a*b*c*d + 15*a^2*d^2)*arctan(sqrt(d*x^2 + c)/sqrt(-c)) 
/(sqrt(-c)*c^3) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)/(sqrt(d*x^2 + c)*c^3) - 
1/8*(8*(d*x^2 + c)^(3/2)*a*b*c*d - 8*sqrt(d*x^2 + c)*a*b*c^2*d - 7*(d*x^2 
+ c)^(3/2)*a^2*d^2 + 9*sqrt(d*x^2 + c)*a^2*c*d^2)/(c^3*d^2*x^4)
 

Mupad [B] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.37 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{c}-\frac {\left (d\,x^2+c\right )\,\left (25\,a^2\,d^2-40\,a\,b\,c\,d+16\,b^2\,c^2\right )}{8\,c^2}+\frac {{\left (d\,x^2+c\right )}^2\,\left (15\,a^2\,d^2-24\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^3}}{{\left (d\,x^2+c\right )}^{5/2}-2\,c\,{\left (d\,x^2+c\right )}^{3/2}+c^2\,\sqrt {d\,x^2+c}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )\,\left (15\,a^2\,d^2-24\,a\,b\,c\,d+8\,b^2\,c^2\right )}{8\,c^{7/2}} \] Input:

int((a + b*x^2)^2/(x^5*(c + d*x^2)^(3/2)),x)
                                                                                    
                                                                                    
 

Output:

((a^2*d^2 + b^2*c^2 - 2*a*b*c*d)/c - ((c + d*x^2)*(25*a^2*d^2 + 16*b^2*c^2 
 - 40*a*b*c*d))/(8*c^2) + ((c + d*x^2)^2*(15*a^2*d^2 + 8*b^2*c^2 - 24*a*b* 
c*d))/(8*c^3))/((c + d*x^2)^(5/2) - 2*c*(c + d*x^2)^(3/2) + c^2*(c + d*x^2 
)^(1/2)) - (atanh((c + d*x^2)^(1/2)/c^(1/2))*(15*a^2*d^2 + 8*b^2*c^2 - 24* 
a*b*c*d))/(8*c^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 554, normalized size of antiderivative = 4.23 \[ \int \frac {\left (a+b x^2\right )^2}{x^5 \left (c+d x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {d \,x^{2}+c}\, a^{2} c^{3}+5 \sqrt {d \,x^{2}+c}\, a^{2} c^{2} d \,x^{2}+15 \sqrt {d \,x^{2}+c}\, a^{2} c \,d^{2} x^{4}-8 \sqrt {d \,x^{2}+c}\, a b \,c^{3} x^{2}-24 \sqrt {d \,x^{2}+c}\, a b \,c^{2} d \,x^{4}+8 \sqrt {d \,x^{2}+c}\, b^{2} c^{3} x^{4}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} c \,d^{2} x^{4}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} d^{3} x^{6}-24 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b \,c^{2} d \,x^{4}-24 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b c \,d^{2} x^{6}+8 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{3} x^{4}+8 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{2} d \,x^{6}-15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} c \,d^{2} x^{4}-15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} d^{3} x^{6}+24 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b \,c^{2} d \,x^{4}+24 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b c \,d^{2} x^{6}-8 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{3} x^{4}-8 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{2} d \,x^{6}}{8 c^{4} x^{4} \left (d \,x^{2}+c \right )} \] Input:

int((b*x^2+a)^2/x^5/(d*x^2+c)^(3/2),x)
 

Output:

( - 2*sqrt(c + d*x**2)*a**2*c**3 + 5*sqrt(c + d*x**2)*a**2*c**2*d*x**2 + 1 
5*sqrt(c + d*x**2)*a**2*c*d**2*x**4 - 8*sqrt(c + d*x**2)*a*b*c**3*x**2 - 2 
4*sqrt(c + d*x**2)*a*b*c**2*d*x**4 + 8*sqrt(c + d*x**2)*b**2*c**3*x**4 + 1 
5*sqrt(c)*log((sqrt(c + d*x**2) - sqrt(c) + sqrt(d)*x)/sqrt(c))*a**2*c*d** 
2*x**4 + 15*sqrt(c)*log((sqrt(c + d*x**2) - sqrt(c) + sqrt(d)*x)/sqrt(c))* 
a**2*d**3*x**6 - 24*sqrt(c)*log((sqrt(c + d*x**2) - sqrt(c) + sqrt(d)*x)/s 
qrt(c))*a*b*c**2*d*x**4 - 24*sqrt(c)*log((sqrt(c + d*x**2) - sqrt(c) + sqr 
t(d)*x)/sqrt(c))*a*b*c*d**2*x**6 + 8*sqrt(c)*log((sqrt(c + d*x**2) - sqrt( 
c) + sqrt(d)*x)/sqrt(c))*b**2*c**3*x**4 + 8*sqrt(c)*log((sqrt(c + d*x**2) 
- sqrt(c) + sqrt(d)*x)/sqrt(c))*b**2*c**2*d*x**6 - 15*sqrt(c)*log((sqrt(c 
+ d*x**2) + sqrt(c) + sqrt(d)*x)/sqrt(c))*a**2*c*d**2*x**4 - 15*sqrt(c)*lo 
g((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)*x)/sqrt(c))*a**2*d**3*x**6 + 24*sq 
rt(c)*log((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)*x)/sqrt(c))*a*b*c**2*d*x** 
4 + 24*sqrt(c)*log((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)*x)/sqrt(c))*a*b*c 
*d**2*x**6 - 8*sqrt(c)*log((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)*x)/sqrt(c 
))*b**2*c**3*x**4 - 8*sqrt(c)*log((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)*x) 
/sqrt(c))*b**2*c**2*d*x**6)/(8*c**4*x**4*(c + d*x**2))