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

Optimal. Leaf size=131 \[ -\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 \sqrt {a} c^{5/2}}-\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)}{8 c^2 x^2}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4} \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {446, 94, 93, 208} \begin {gather*} -\frac {3 \sqrt {a+b x^2} \sqrt {c+d x^2} (b c-a d)}{8 c^2 x^2}-\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}}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 c x^4} \end {gather*}

Antiderivative was successfully verified.

[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 93

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 94

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[p, 1] &&  !SumSimplerQ[m, 1])

Rule 208

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

Rule 446

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*} \int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {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)) \operatorname {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 ) \operatorname {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 ) \operatorname {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]  time = 0.07, size = 110, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (-2 a c+3 a d x^2-5 b c x^2\right )}{8 c^2 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 {gather*}

Antiderivative was successfully verified.

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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 2.57, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \sqrt {c+d x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 2.13, size = 360, normalized size = 2.75 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

________________________________________________________________________________________

giac [B]  time = 2.29, size = 1101, normalized size = 8.40 \begin {gather*} -\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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 352, normalized size = 2.69 \begin {gather*} -\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (3 a^{2} d^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-6 a b c d \,x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )+3 b^{2} c^{2} x^{4} \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 a c +2 \sqrt {a c}\, \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}}{x^{2}}\right )-6 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a d \,x^{2}+10 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, b c \,x^{2}+4 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, a c \right )}{16 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {a c}\, c^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/c^2*(3*a^2*d^2*x^4*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*
x^2+b*c*x^2+a*c)^(1/2))/x^2)-6*a*b*c*d*x^4*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2))/x^2)+3*b^2*c^2*x^4*ln((a*d*x^2+b*c*x^2+2*a*c+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2))/x^2)
-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*a*d*x^2+10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*
b*c*x^2+4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*a*c)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/x^4/(a*c)^(
1/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is a*d-b*c zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^{3/2}}{x^5\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x^{2}\right )^{\frac {3}{2}}}{x^{5} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________