3.697 \(\int \frac {x^5 (a+b x^3)^{4/3}}{c+d x^3} \, dx\)

Optimal. Leaf size=211 \[ \frac {c (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}+\frac {c (b c-a d)^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{10/3}}+\frac {c \sqrt [3]{a+b x^3} (b c-a d)}{d^3}-\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d} \]

[Out]

c*(-a*d+b*c)*(b*x^3+a)^(1/3)/d^3-1/4*c*(b*x^3+a)^(4/3)/d^2+1/7*(b*x^3+a)^(7/3)/b/d+1/6*c*(-a*d+b*c)^(4/3)*ln(d
*x^3+c)/d^(10/3)-1/2*c*(-a*d+b*c)^(4/3)*ln((-a*d+b*c)^(1/3)+d^(1/3)*(b*x^3+a)^(1/3))/d^(10/3)+1/3*c*(-a*d+b*c)
^(4/3)*arctan(1/3*(1-2*d^(1/3)*(b*x^3+a)^(1/3)/(-a*d+b*c)^(1/3))*3^(1/2))/d^(10/3)*3^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {446, 80, 50, 58, 617, 204, 31} \[ -\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {c \sqrt [3]{a+b x^3} (b c-a d)}{d^3}+\frac {c (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}+\frac {c (b c-a d)^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{10/3}}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

(c*(b*c - a*d)*(a + b*x^3)^(1/3))/d^3 - (c*(a + b*x^3)^(4/3))/(4*d^2) + (a + b*x^3)^(7/3)/(7*b*d) + (c*(b*c -
a*d)^(4/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt[3]*d^(10/3)) + (c*(b*c
- a*d)^(4/3)*Log[c + d*x^3])/(6*d^(10/3)) - (c*(b*c - a*d)^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^
(1/3)])/(2*d^(10/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d
*x)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x
] && NegQ[(b*c - a*d)/b]

Rule 80

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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]]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x (a+b x)^{4/3}}{c+d x} \, dx,x,x^3\right )\\ &=\frac {\left (a+b x^3\right )^{7/3}}{7 b d}-\frac {c \operatorname {Subst}\left (\int \frac {(a+b x)^{4/3}}{c+d x} \, dx,x,x^3\right )}{3 d}\\ &=-\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d}+\frac {(c (b c-a d)) \operatorname {Subst}\left (\int \frac {\sqrt [3]{a+b x}}{c+d x} \, dx,x,x^3\right )}{3 d^2}\\ &=\frac {c (b c-a d) \sqrt [3]{a+b x^3}}{d^3}-\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d}-\frac {\left (c (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 d^3}\\ &=\frac {c (b c-a d) \sqrt [3]{a+b x^3}}{d^3}-\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d}+\frac {c (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {\left (c (b c-a d)^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}-\frac {\left (c (b c-a d)^{5/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{d}}+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 d^{11/3}}\\ &=\frac {c (b c-a d) \sqrt [3]{a+b x^3}}{d^3}-\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d}+\frac {c (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}-\frac {\left (c (b c-a d)^{4/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}\right )}{d^{10/3}}\\ &=\frac {c (b c-a d) \sqrt [3]{a+b x^3}}{d^3}-\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d}+\frac {c (b c-a d)^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{10/3}}+\frac {c (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^{10/3}}-\frac {c (b c-a d)^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{10/3}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 255, normalized size = 1.21 \[ \frac {c (b c-a d) \left (\sqrt [3]{b c-a d} \log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )-2 \sqrt [3]{b c-a d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )-2 \sqrt {3} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}-1}{\sqrt {3}}\right )+6 \sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{6 d^{10/3}}-\frac {c \left (a+b x^3\right )^{4/3}}{4 d^2}+\frac {\left (a+b x^3\right )^{7/3}}{7 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

-1/4*(c*(a + b*x^3)^(4/3))/d^2 + (a + b*x^3)^(7/3)/(7*b*d) + (c*(b*c - a*d)*(6*d^(1/3)*(a + b*x^3)^(1/3) - 2*S
qrt[3]*(b*c - a*d)^(1/3)*ArcTan[(-1 + (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] - 2*(b*c - a*d
)^(1/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)] + (b*c - a*d)^(1/3)*Log[(b*c - a*d)^(2/3) - d^(1/3)
*(b*c - a*d)^(1/3)*(a + b*x^3)^(1/3) + d^(2/3)*(a + b*x^3)^(2/3)]))/(6*d^(10/3))

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fricas [A]  time = 1.10, size = 298, normalized size = 1.41 \[ \frac {28 \, \sqrt {3} {\left (b^{2} c^{2} - a b c d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}} - \sqrt {3} {\left (b c - a d\right )}}{3 \, {\left (b c - a d\right )}}\right ) + 14 \, {\left (b^{2} c^{2} - a b c d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right ) - 28 \, {\left (b^{2} c^{2} - a b c d\right )} \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right ) + 3 \, {\left (4 \, b^{2} d^{2} x^{6} + 28 \, b^{2} c^{2} - 35 \, a b c d + 4 \, a^{2} d^{2} - {\left (7 \, b^{2} c d - 8 \, a b d^{2}\right )} x^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{84 \, b d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

1/84*(28*sqrt(3)*(b^2*c^2 - a*b*c*d)*((b*c - a*d)/d)^(1/3)*arctan(-1/3*(2*sqrt(3)*(b*x^3 + a)^(1/3)*d*((b*c -
a*d)/d)^(2/3) - sqrt(3)*(b*c - a*d))/(b*c - a*d)) + 14*(b^2*c^2 - a*b*c*d)*((b*c - a*d)/d)^(1/3)*log((b*x^3 +
a)^(2/3) - (b*x^3 + a)^(1/3)*((b*c - a*d)/d)^(1/3) + ((b*c - a*d)/d)^(2/3)) - 28*(b^2*c^2 - a*b*c*d)*((b*c - a
*d)/d)^(1/3)*log((b*x^3 + a)^(1/3) + ((b*c - a*d)/d)^(1/3)) + 3*(4*b^2*d^2*x^6 + 28*b^2*c^2 - 35*a*b*c*d + 4*a
^2*d^2 - (7*b^2*c*d - 8*a*b*d^2)*x^3)*(b*x^3 + a)^(1/3))/(b*d^3)

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giac [B]  time = 0.29, size = 348, normalized size = 1.65 \[ \frac {{\left (b^{10} c^{3} d^{4} - 2 \, a b^{9} c^{2} d^{5} + a^{2} b^{8} c d^{6}\right )} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{9} c d^{7} - a b^{8} d^{8}\right )}} - \frac {\sqrt {3} {\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c^{2} - a c d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{4}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} {\left (b c^{2} - a c d\right )} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{4}} + \frac {28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{8} c^{2} d^{4} - 7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{7} c d^{5} - 28 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a b^{7} c d^{5} + 4 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} b^{6} d^{6}}{28 \, b^{7} d^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

1/3*(b^10*c^3*d^4 - 2*a*b^9*c^2*d^5 + a^2*b^8*c*d^6)*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c
- a*d)/d)^(1/3)))/(b^9*c*d^7 - a*b^8*d^8) - 1/3*sqrt(3)*(-b*c*d^2 + a*d^3)^(1/3)*(b*c^2 - a*c*d)*arctan(1/3*s
qrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)/d)^(1/3))/d^4 - 1/6*(-b*c*d^2 + a*d^3)^(1/
3)*(b*c^2 - a*c*d)*log((b*x^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/
d^4 + 1/28*(28*(b*x^3 + a)^(1/3)*b^8*c^2*d^4 - 7*(b*x^3 + a)^(4/3)*b^7*c*d^5 - 28*(b*x^3 + a)^(1/3)*a*b^7*c*d^
5 + 4*(b*x^3 + a)^(7/3)*b^6*d^6)/(b^7*d^7)

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} x^{5}}{d \,x^{3}+c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^3+a)^(4/3)/(d*x^3+c),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 = 5.06, size = 348, normalized size = 1.65 \[ \frac {{\left (b\,x^3+a\right )}^{7/3}}{7\,b\,d}-{\left (b\,x^3+a\right )}^{4/3}\,\left (\frac {a}{4\,b\,d}+\frac {b^2\,c-a\,b\,d}{4\,b^2\,d^2}\right )-\frac {c\,\ln \left (\frac {3\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a^2\,c\,d^2-2\,a\,b\,c^2\,d+b^2\,c^3\right )}{d}-\frac {c\,{\left (a\,d-b\,c\right )}^{4/3}\,\left (9\,a\,d^3-9\,b\,c\,d^2\right )}{3\,d^{10/3}}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{10/3}}-\frac {c\,\ln \left (\frac {3\,c\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d}-\frac {3\,c\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{10/3}}+\frac {c\,\ln \left (\frac {3\,c\,{\left (b\,x^3+a\right )}^{1/3}\,{\left (a\,d-b\,c\right )}^2}{d}+\frac {3\,c\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{7/3}}{d^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (a\,d-b\,c\right )}^{4/3}}{3\,d^{10/3}}+\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (b^2\,c-a\,b\,d\right )\,\left (\frac {a}{b\,d}+\frac {b^2\,c-a\,b\,d}{b^2\,d^2}\right )}{b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*x^3)^(4/3))/(c + d*x^3),x)

[Out]

(a + b*x^3)^(7/3)/(7*b*d) - (a + b*x^3)^(4/3)*(a/(4*b*d) + (b^2*c - a*b*d)/(4*b^2*d^2)) - (c*log((3*(a + b*x^3
)^(1/3)*(b^2*c^3 + a^2*c*d^2 - 2*a*b*c^2*d))/d - (c*(a*d - b*c)^(4/3)*(9*a*d^3 - 9*b*c*d^2))/(3*d^(10/3)))*(a*
d - b*c)^(4/3))/(3*d^(10/3)) - (c*log((3*c*(a + b*x^3)^(1/3)*(a*d - b*c)^2)/d - (3*c*((3^(1/2)*1i)/2 - 1/2)*(a
*d - b*c)^(7/3))/d^(4/3))*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c)^(4/3))/(3*d^(10/3)) + (c*log((3*c*(a + b*x^3)^(1/
3)*(a*d - b*c)^2)/d + (3*c*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c)^(7/3))/d^(4/3))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*
c)^(4/3))/(3*d^(10/3)) + ((a + b*x^3)^(1/3)*(b^2*c - a*b*d)*(a/(b*d) + (b^2*c - a*b*d)/(b^2*d^2)))/(b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Timed out

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