\(\int \frac {(a+b x^2)^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx\) [949]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 26, antiderivative size = 131 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx=-\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}-\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 \sqrt {a} c^{5/2}} \]

[Out]

-3/8*(-a*d+b*c)^2*arctanh(c^(1/2)*(b*x^2+a)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))/c^(5/2)/a^(1/2)-1/4*(b*x^2+a)^(3/2)
*(d*x^2+c)^(1/2)/c/x^4-3/8*(-a*d+b*c)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/c^2/x^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {457, 96, 95, 214} \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx=-\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 \sqrt {a} c^{5/2}}-\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)}{8 c^2 x^2}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4} \]

[In]

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

[Out]

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

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

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

Rule 214

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 457

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^3 \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = -\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}+\frac {(3 (b c-a d)) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2 \sqrt {c+d x}} \, dx,x,x^2\right )}{8 c} \\ & = -\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}+\frac {\left (3 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{16 c^2} \\ & = -\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}+\frac {\left (3 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{8 c^2} \\ & = -\frac {3 (b c-a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 c^2 x^2}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4}-\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 \sqrt {a} c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (-2 a c-5 b c x^2+3 a d x^2\right )}{8 c^2 x^4}-\frac {3 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x^2}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{8 \sqrt {a} c^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.16 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.25

method result size
risch \(-\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (-3 a d \,x^{2}+5 c b \,x^{2}+2 a c \right )}{8 c^{2} x^{4}}-\frac {3 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{16 c^{2} \sqrt {a c}\, \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\) \(164\)
default \(-\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (3 \ln \left (\frac {a d \,x^{2}+c b \,x^{2}+2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}+2 a c}{x^{2}}\right ) a^{2} d^{2} x^{4}-6 \ln \left (\frac {a d \,x^{2}+c b \,x^{2}+2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}+2 a c}{x^{2}}\right ) a b c d \,x^{4}+3 \ln \left (\frac {a d \,x^{2}+c b \,x^{2}+2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}+2 a c}{x^{2}}\right ) b^{2} c^{2} x^{4}-6 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, d a \,x^{2} \sqrt {a c}+10 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b c \,x^{2} \sqrt {a c}+4 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, a c \sqrt {a c}\right )}{16 c^{2} \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, x^{4} \sqrt {a c}}\) \(303\)
elliptic \(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (-\frac {a \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{4 c \,x^{4}}+\frac {3 a \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, d}{8 c^{2} x^{2}}-\frac {5 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, b}{8 c \,x^{2}}-\frac {3 a^{2} \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) d^{2}}{16 c^{2} \sqrt {a c}}+\frac {3 a \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) d b}{8 c \sqrt {a c}}-\frac {3 b^{2} \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right )}{16 \sqrt {a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\) \(327\)

[In]

int((b*x^2+a)^(3/2)/x^5/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-3*a*d*x^2+5*b*c*x^2+2*a*c)/c^2/x^4-3/16*(a^2*d^2-2*a*b*c*d+b^2*c^2)/c^2
/(a*c)^(1/2)*ln((2*a*c+(a*d+b*c)*x^2+2*(a*c)^(1/2)*(b*d*x^4+(a*d+b*c)*x^2+a*c)^(1/2))/x^2)*((b*x^2+a)*(d*x^2+c
))^(1/2)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.75 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx=\left [\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {a c} x^{4} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {a c}}{x^{4}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} + {\left (5 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{32 \, a c^{3} x^{4}}, \frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-a c} x^{4} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-a c}}{2 \, {\left (a b c d x^{4} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} + {\left (5 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{16 \, a c^{3} x^{4}}\right ] \]

[In]

integrate((b*x^2+a)^(3/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/32*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a*c)*x^4*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 +
8*(a*b*c^2 + a^2*c*d)*x^2 - 4*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a*c))/x^4) - 4*(2
*a^2*c^2 + (5*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a*c^3*x^4), 1/16*(3*(b^2*c^2 - 2*a*b
*c*d + a^2*d^2)*sqrt(-a*c)*x^4*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a*c)
/(a*b*c*d*x^4 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x^2)) - 2*(2*a^2*c^2 + (5*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(b*x^2 +
 a)*sqrt(d*x^2 + c))/(a*c^3*x^4)]

Sympy [F]

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

[In]

integrate((b*x**2+a)**(3/2)/x**5/(d*x**2+c)**(1/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((b*x^2+a)^(3/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
 more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1101 vs. \(2 (105) = 210\).

Time = 0.59 (sec) , antiderivative size = 1101, normalized size of antiderivative = 8.40 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx=-\frac {b {\left (\frac {3 \, {\left (\sqrt {b d} b^{3} c^{2} - 2 \, \sqrt {b d} a b^{2} c d + \sqrt {b d} a^{2} b d^{2}\right )} \arctan \left (-\frac {b^{2} c + a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} b c^{2}} + \frac {2 \, {\left (5 \, \sqrt {b d} b^{9} c^{5} - 23 \, \sqrt {b d} a b^{8} c^{4} d + 42 \, \sqrt {b d} a^{2} b^{7} c^{3} d^{2} - 38 \, \sqrt {b d} a^{3} b^{6} c^{2} d^{3} + 17 \, \sqrt {b d} a^{4} b^{5} c d^{4} - 3 \, \sqrt {b d} a^{5} b^{4} d^{5} - 15 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{7} c^{4} + 28 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b^{6} c^{3} d - 2 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{2} b^{5} c^{2} d^{2} - 20 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{3} b^{4} c d^{3} + 9 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a^{4} b^{3} d^{4} + 15 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} b^{5} c^{3} + \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a b^{4} c^{2} d + 9 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a^{2} b^{3} c d^{2} - 9 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4} a^{3} b^{2} d^{3} - 5 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} b^{3} c^{2} - 6 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} a b^{2} c d + 3 \, \sqrt {b d} {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{6} a^{2} b d^{2}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4}\right )}^{2} c^{2}}\right )}}{8 \, {\left | b \right |}} \]

[In]

integrate((b*x^2+a)^(3/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-1/8*b*(3*(sqrt(b*d)*b^3*c^2 - 2*sqrt(b*d)*a*b^2*c*d + sqrt(b*d)*a^2*b*d^2)*arctan(-1/2*(b^2*c + a*b*d - (sqrt
(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c^2) +
 2*(5*sqrt(b*d)*b^9*c^5 - 23*sqrt(b*d)*a*b^8*c^4*d + 42*sqrt(b*d)*a^2*b^7*c^3*d^2 - 38*sqrt(b*d)*a^3*b^6*c^2*d
^3 + 17*sqrt(b*d)*a^4*b^5*c*d^4 - 3*sqrt(b*d)*a^5*b^4*d^5 - 15*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2
*c + (b*x^2 + a)*b*d - a*b*d))^2*b^7*c^4 + 28*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*
b*d - a*b*d))^2*a*b^6*c^3*d - 2*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^
2*a^2*b^5*c^2*d^2 - 20*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^3*b^4
*c*d^3 + 9*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^4*b^3*d^4 + 15*sq
rt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*b^5*c^3 + sqrt(b*d)*(sqrt(b*x^2
+ a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a*b^4*c^2*d + 9*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d)
 - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a^2*b^3*c*d^2 - 9*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*
c + (b*x^2 + a)*b*d - a*b*d))^4*a^3*b^2*d^3 - 5*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a
)*b*d - a*b*d))^6*b^3*c^2 - 6*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^6*
a*b^2*c*d + 3*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^6*a^2*b*d^2)/((b^4
*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*b^2
*c - 2*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*x^2 + a)*sqrt(b*d
) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4)^2*c^2))/abs(b)

Mupad [F(-1)]

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

[In]

int((a + b*x^2)^(3/2)/(x^5*(c + d*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^(3/2)/(x^5*(c + d*x^2)^(1/2)), x)