3.4.16 \(\int \frac {x^8}{(a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=150 \[ \frac {-a^2 d^2-2 b^2 c^2}{3 b^2 d \sqrt {c+d x^3} (b c-a d)^2}-\frac {a^2}{3 b^2 \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}+\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{5/2}} \]

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Rubi [A]  time = 0.19, antiderivative size = 149, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 89, 78, 63, 208} \begin {gather*} -\frac {a^2 d^2+2 b^2 c^2}{3 b^2 d \sqrt {c+d x^3} (b c-a d)^2}-\frac {a^2}{3 b^2 \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)}+\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

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

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 {x^8}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(a+b x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac {a^2}{3 b^2 (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2} a (2 b c+a d)+b (b c-a d) x}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{3 b^2 (b c-a d)}\\ &=-\frac {2 b^2 c^2+a^2 d^2}{3 b^2 d (b c-a d)^2 \sqrt {c+d x^3}}-\frac {a^2}{3 b^2 (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}-\frac {(a (4 b c-a d)) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{6 b (b c-a d)^2}\\ &=-\frac {2 b^2 c^2+a^2 d^2}{3 b^2 d (b c-a d)^2 \sqrt {c+d x^3}}-\frac {a^2}{3 b^2 (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}-\frac {(a (4 b c-a d)) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{3 b d (b c-a d)^2}\\ &=-\frac {2 b^2 c^2+a^2 d^2}{3 b^2 d (b c-a d)^2 \sqrt {c+d x^3}}-\frac {a^2}{3 b^2 (b c-a d) \left (a+b x^3\right ) \sqrt {c+d x^3}}+\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 134, normalized size = 0.89 \begin {gather*} \frac {\frac {a (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{5/2}}-\frac {\sqrt {b} \left (a^2 d \left (c+d x^3\right )+2 a b c^2+2 b^2 c^2 x^3\right )}{d \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)^2}}{3 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 151, normalized size = 1.01 \begin {gather*} \frac {\left (4 a b c-a^2 d\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 b^{3/2} (a d-b c)^{5/2}}+\frac {-a^2 c d-a^2 d^2 x^3-2 a b c^2-2 b^2 c^2 x^3}{3 b d \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^8/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

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

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fricas [B]  time = 0.92, size = 746, normalized size = 4.97 \begin {gather*} \left [-\frac {{\left ({\left (4 \, a b^{2} c d^{2} - a^{2} b d^{3}\right )} x^{6} + 4 \, a^{2} b c^{2} d - a^{3} c d^{2} + {\left (4 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) + 2 \, {\left (2 \, a b^{3} c^{3} - a^{2} b^{2} c^{2} d - a^{3} b c d^{2} + {\left (2 \, b^{4} c^{3} - 2 \, a b^{3} c^{2} d + a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a b^{5} c^{4} d - 3 \, a^{2} b^{4} c^{3} d^{2} + 3 \, a^{3} b^{3} c^{2} d^{3} - a^{4} b^{2} c d^{4} + {\left (b^{6} c^{3} d^{2} - 3 \, a b^{5} c^{2} d^{3} + 3 \, a^{2} b^{4} c d^{4} - a^{3} b^{3} d^{5}\right )} x^{6} + {\left (b^{6} c^{4} d - 2 \, a b^{5} c^{3} d^{2} + 2 \, a^{3} b^{3} c d^{4} - a^{4} b^{2} d^{5}\right )} x^{3}\right )}}, -\frac {{\left ({\left (4 \, a b^{2} c d^{2} - a^{2} b d^{3}\right )} x^{6} + 4 \, a^{2} b c^{2} d - a^{3} c d^{2} + {\left (4 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) + {\left (2 \, a b^{3} c^{3} - a^{2} b^{2} c^{2} d - a^{3} b c d^{2} + {\left (2 \, b^{4} c^{3} - 2 \, a b^{3} c^{2} d + a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a b^{5} c^{4} d - 3 \, a^{2} b^{4} c^{3} d^{2} + 3 \, a^{3} b^{3} c^{2} d^{3} - a^{4} b^{2} c d^{4} + {\left (b^{6} c^{3} d^{2} - 3 \, a b^{5} c^{2} d^{3} + 3 \, a^{2} b^{4} c d^{4} - a^{3} b^{3} d^{5}\right )} x^{6} + {\left (b^{6} c^{4} d - 2 \, a b^{5} c^{3} d^{2} + 2 \, a^{3} b^{3} c d^{4} - a^{4} b^{2} d^{5}\right )} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 195, normalized size = 1.30 \begin {gather*} -\frac {{\left (4 \, a b c - a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (d x^{3} + c\right )} b^{2} c^{2} - 2 \, b^{2} c^{3} + 2 \, a b c^{2} d + {\left (d x^{3} + c\right )} a^{2} d^{2}}{3 \, {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b - \sqrt {d x^{3} + c} b c + \sqrt {d x^{3} + c} a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

-1/3*(4*a*b*c - a^2*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*sqr
t(-b^2*c + a*b*d)) - 1/3*(2*(d*x^3 + c)*b^2*c^2 - 2*b^2*c^3 + 2*a*b*c^2*d + (d*x^3 + c)*a^2*d^2)/((b^3*c^2*d -
2*a*b^2*c*d^2 + a^2*b*d^3)*((d*x^3 + c)^(3/2)*b - sqrt(d*x^3 + c)*b*c + sqrt(d*x^3 + c)*a*d))

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maple [C]  time = 0.36, size = 978, normalized size = 6.52

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a)^2/(d*x^3+c)^(3/2),x)

[Out]

-2/3/b^2/d/(d*x^3+c)^(1/2)-2*a/b^2*(-2/3/(a*d-b*c)/((x^3+c/d)*d)^(1/2)-1/3*I*b/d^2*2^(1/2)*sum(1/(-a*d+b*c)/(a
*d-b*c)*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(
-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+
(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(
-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3
)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d
+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2
)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*b+a)))+a^2/b^2*(-1
/3*b/(a*d-b*c)^2*(d*x^3+c)^(1/2)/(b*x^3+a)-2/3*d/(a*d-b*c)^2/((x^3+c/d)*d)^(1/2)+1/2*I*b/d*2^(1/2)*sum(1/(a*d-
b*c)^3*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-
c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(
-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-
c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)
/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+
I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)
^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*b+a)))

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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(x^8/(b*x^3+a)^2/(d*x^3+c)^(3/2),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 positive or negative?

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mupad [B]  time = 7.90, size = 367, normalized size = 2.45 \begin {gather*} \frac {\sqrt {d\,x^3+c}\,\left (x^3\,\left (\frac {\left (\frac {3\,b\,d\,\left (a\,d+b\,c\right )-b\,d\,\left (a\,d+2\,b\,c\right )}{3\,\left (a^2\,b\,d^3-2\,a\,b^2\,c\,d^2+b^3\,c^2\,d\right )}-\frac {b\,d\,\left (a\,d+b\,c\right )}{a^2\,b\,d^3-2\,a\,b^2\,c\,d^2+b^3\,c^2\,d}\right )\,\left (a\,d+b\,c\right )}{b\,d}+\frac {a\,b\,c\,d}{a^2\,b\,d^3-2\,a\,b^2\,c\,d^2+b^3\,c^2\,d}\right )+\frac {a\,c\,\left (\frac {3\,b\,d\,\left (a\,d+b\,c\right )-b\,d\,\left (a\,d+2\,b\,c\right )}{3\,\left (a^2\,b\,d^3-2\,a\,b^2\,c\,d^2+b^3\,c^2\,d\right )}-\frac {b\,d\,\left (a\,d+b\,c\right )}{a^2\,b\,d^3-2\,a\,b^2\,c\,d^2+b^3\,c^2\,d}\right )}{b\,d}\right )}{b\,d\,x^6+\left (a\,d+b\,c\right )\,x^3+a\,c}+\frac {a\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x)

[Out]

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

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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(x**8/(b*x**3+a)**2/(d*x**3+c)**(3/2),x)

[Out]

Timed out

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