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

Optimal. Leaf size=181 \[ -\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}+\frac {\left (c+d x^2\right )^{3/2} \left (3 a d (a d+8 b c)+8 b^2 c^2\right )}{24 c^2}+\frac {\sqrt {c+d x^2} \left (3 a d (a d+8 b c)+8 b^2 c^2\right )}{8 c}-\frac {\left (3 a d (a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}}-\frac {a \left (c+d x^2\right )^{5/2} (a d+8 b c)}{8 c^2 x^2} \]

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Rubi [A]  time = 0.21, antiderivative size = 178, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 89, 78, 50, 63, 208} \begin {gather*} -\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}+\frac {1}{24} \left (c+d x^2\right )^{3/2} \left (\frac {3 a d (a d+8 b c)}{c^2}+8 b^2\right )+\frac {\sqrt {c+d x^2} \left (3 a d (a d+8 b c)+8 b^2 c^2\right )}{8 c}-\frac {\left (3 a d (a d+8 b c)+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}}-\frac {a \left (c+d x^2\right )^{5/2} (a d+8 b c)}{8 c^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(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 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 )^2 \left (c+d x^2\right )^{3/2}}{x^5} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^2 (c+d x)^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}+\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {1}{2} a (8 b c+a d)+2 b^2 c x\right ) (c+d x)^{3/2}}{x^2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac {1}{16} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{24} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac {1}{16} \left (c \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right )\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{8} c \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {1}{24} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac {1}{16} \left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=\frac {1}{8} c \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {1}{24} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}+\frac {\left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{8 d}\\ &=\frac {1}{8} c \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \sqrt {c+d x^2}+\frac {1}{24} \left (8 b^2+\frac {3 a d (8 b c+a d)}{c^2}\right ) \left (c+d x^2\right )^{3/2}-\frac {a^2 \left (c+d x^2\right )^{5/2}}{4 c x^4}-\frac {a (8 b c+a d) \left (c+d x^2\right )^{5/2}}{8 c^2 x^2}-\frac {\left (8 b^2 c^2+24 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 116, normalized size = 0.64 \begin {gather*} \frac {1}{24} \left (\frac {\sqrt {c+d x^2} \left (-3 a^2 \left (2 c+5 d x^2\right )-24 a b x^2 \left (c-2 d x^2\right )+8 b^2 x^4 \left (4 c+d x^2\right )\right )}{x^4}-\frac {3 \left (3 a^2 d^2+24 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.19, size = 119, normalized size = 0.66 \begin {gather*} \frac {\left (-3 a^2 d^2-24 a b c d-8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{8 \sqrt {c}}+\frac {\sqrt {c+d x^2} \left (-6 a^2 c-15 a^2 d x^2-24 a b c x^2+48 a b d x^4+32 b^2 c x^4+8 b^2 d x^6\right )}{24 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-6*a^2*c - 24*a*b*c*x^2 - 15*a^2*d*x^2 + 32*b^2*c*x^4 + 48*a*b*d*x^4 + 8*b^2*d*x^6))/(24*x^4
) + ((-8*b^2*c^2 - 24*a*b*c*d - 3*a^2*d^2)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/(8*Sqrt[c])

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fricas [A]  time = 1.47, size = 267, normalized size = 1.48 \begin {gather*} \left [\frac {3 \, {\left (8 \, b^{2} c^{2} + 24 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {c} x^{4} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (8 \, b^{2} c d x^{6} + 16 \, {\left (2 \, b^{2} c^{2} + 3 \, a b c d\right )} x^{4} - 6 \, a^{2} c^{2} - 3 \, {\left (8 \, a b c^{2} + 5 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c x^{4}}, \frac {3 \, {\left (8 \, b^{2} c^{2} + 24 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} c d x^{6} + 16 \, {\left (2 \, b^{2} c^{2} + 3 \, a b c d\right )} x^{4} - 6 \, a^{2} c^{2} - 3 \, {\left (8 \, a b c^{2} + 5 \, a^{2} c d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{24 \, c x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*(8*b^2*c^2 + 24*a*b*c*d + 3*a^2*d^2)*sqrt(c)*x^4*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2)
+ 2*(8*b^2*c*d*x^6 + 16*(2*b^2*c^2 + 3*a*b*c*d)*x^4 - 6*a^2*c^2 - 3*(8*a*b*c^2 + 5*a^2*c*d)*x^2)*sqrt(d*x^2 +
c))/(c*x^4), 1/24*(3*(8*b^2*c^2 + 24*a*b*c*d + 3*a^2*d^2)*sqrt(-c)*x^4*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (8*b
^2*c*d*x^6 + 16*(2*b^2*c^2 + 3*a*b*c*d)*x^4 - 6*a^2*c^2 - 3*(8*a*b*c^2 + 5*a^2*c*d)*x^2)*sqrt(d*x^2 + c))/(c*x
^4)]

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giac [A]  time = 0.35, size = 182, normalized size = 1.01 \begin {gather*} \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d + 24 \, \sqrt {d x^{2} + c} b^{2} c d + 48 \, \sqrt {d x^{2} + c} a b d^{2} + \frac {3 \, {\left (8 \, b^{2} c^{2} d + 24 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {3 \, {\left (8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} - 8 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} - 3 \, \sqrt {d x^{2} + c} a^{2} c d^{3}\right )}}{d^{2} x^{4}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*(d*x^2 + c)^(3/2)*b^2*d + 24*sqrt(d*x^2 + c)*b^2*c*d + 48*sqrt(d*x^2 + c)*a*b*d^2 + 3*(8*b^2*c^2*d + 2
4*a*b*c*d^2 + 3*a^2*d^3)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) - 3*(8*(d*x^2 + c)^(3/2)*a*b*c*d^2 - 8*sqrt
(d*x^2 + c)*a*b*c^2*d^2 + 5*(d*x^2 + c)^(3/2)*a^2*d^3 - 3*sqrt(d*x^2 + c)*a^2*c*d^3)/(d^2*x^4))/d

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maple [A]  time = 0.01, size = 256, normalized size = 1.41 \begin {gather*} -\frac {3 a^{2} d^{2} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )}{8 \sqrt {c}}-3 a b \sqrt {c}\, d \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )-b^{2} c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\frac {3 \sqrt {d \,x^{2}+c}\, a^{2} d^{2}}{8 c}+3 \sqrt {d \,x^{2}+c}\, a b d +\sqrt {d \,x^{2}+c}\, b^{2} c +\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a^{2} d^{2}}{8 c^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} a b d}{c}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} b^{2}}{3}-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2} d}{8 c^{2} x^{2}}-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a b}{c \,x^{2}}-\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}} a^{2}}{4 c \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.13, size = 222, normalized size = 1.23 \begin {gather*} -b^{2} c^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - 3 \, a b \sqrt {c} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {3 \, a^{2} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{8 \, \sqrt {c}} + \frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} + \sqrt {d x^{2} + c} b^{2} c + 3 \, \sqrt {d x^{2} + c} a b d + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d}{c} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{8 \, c^{2}} + \frac {3 \, \sqrt {d x^{2} + c} a^{2} d^{2}}{8 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b}{c x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d}{8 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{4 \, c x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.65, size = 208, normalized size = 1.15 \begin {gather*} \frac {\sqrt {d\,x^2+c}\,\left (\frac {3\,a^2\,c\,d^2}{8}+b\,a\,c^2\,d\right )-\left (\frac {5\,a^2\,d^2}{8}+b\,c\,a\,d\right )\,{\left (d\,x^2+c\right )}^{3/2}}{{\left (d\,x^2+c\right )}^2-2\,c\,\left (d\,x^2+c\right )+c^2}+\sqrt {d\,x^2+c}\,\left (c\,b^2+2\,a\,d\,b\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{3/2}}{3}+\frac {\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,\left (3\,a^2\,d^2+24\,a\,b\,c\,d+8\,b^2\,c^2\right )\,1{}\mathrm {i}}{4\,\sqrt {c}\,\left (\frac {3\,a^2\,d^2}{4}+6\,a\,b\,c\,d+2\,b^2\,c^2\right )}\right )\,\left (3\,a^2\,d^2+24\,a\,b\,c\,d+8\,b^2\,c^2\right )\,1{}\mathrm {i}}{8\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c + d*x^2)^(1/2)*((3*a^2*c*d^2)/8 + a*b*c^2*d) - ((5*a^2*d^2)/8 + a*b*c*d)*(c + d*x^2)^(3/2))/((c + d*x^2)^2
- 2*c*(c + d*x^2) + c^2) + (c + d*x^2)^(1/2)*(b^2*c + 2*a*b*d) + (b^2*(c + d*x^2)^(3/2))/3 + (atan(((c + d*x^
2)^(1/2)*(3*a^2*d^2 + 8*b^2*c^2 + 24*a*b*c*d)*1i)/(4*c^(1/2)*((3*a^2*d^2)/4 + 2*b^2*c^2 + 6*a*b*c*d)))*(3*a^2*
d^2 + 8*b^2*c^2 + 24*a*b*c*d)*1i)/(8*c^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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