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

Optimal. Leaf size=174 \[ \frac {c \log \left (c+d x^3\right )}{6 d^{2/3} (b c-a d)^{4/3}}-\frac {c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} (b c-a d)^{4/3}}-\frac {c \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} d^{2/3} (b c-a d)^{4/3}}+\frac {a}{b \sqrt [3]{a+b x^3} (b c-a d)} \]

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Rubi [A]  time = 0.17, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 78, 56, 617, 204, 31} \begin {gather*} \frac {c \log \left (c+d x^3\right )}{6 d^{2/3} (b c-a d)^{4/3}}-\frac {c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 d^{2/3} (b c-a d)^{4/3}}-\frac {c \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^{2/3} (b c-a d)^{4/3}}+\frac {a}{b \sqrt [3]{a+b x^3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 56

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

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

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Mathematica [C]  time = 0.03, size = 77, normalized size = 0.44 \begin {gather*} \frac {b c \left (a+b x^3\right ) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )+2 a (b c-a d)}{2 b \sqrt [3]{a+b x^3} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(b*c - a*d) + b*c*(a + b*x^3)*Hypergeometric2F1[2/3, 1, 5/3, (d*(a + b*x^3))/(-(b*c) + a*d)])/(2*b*(b*c -
a*d)^2*(a + b*x^3)^(1/3))

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IntegrateAlgebraic [A]  time = 0.26, size = 230, normalized size = 1.32 \begin {gather*} -\frac {c \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 d^{2/3} (b c-a d)^{4/3}}+\frac {c \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 )}{6 d^{2/3} (b c-a d)^{4/3}}-\frac {c \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{b c-a d}}\right )}{\sqrt {3} d^{2/3} (b c-a d)^{4/3}}+\frac {a}{b \sqrt [3]{a+b x^3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.82, size = 872, normalized size = 5.01 \begin {gather*} \left [-\frac {3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} c^{2} d - a^{2} b c d^{2} + {\left (b^{3} c^{2} d - a b^{2} c d^{2}\right )} x^{3}\right )} \sqrt {-\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \log \left (\frac {2 \, b d^{2} x^{3} - b c d + 3 \, a d^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c d - a d^{2}\right )} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b c - a d\right )}\right )} \sqrt {-\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} - 3 \, {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{d x^{3} + c}\right ) - {\left (b^{2} c x^{3} + a b c\right )} {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}}\right ) + 2 \, {\left (b^{2} c x^{3} + a b c\right )} {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right ) - 6 \, {\left (a b c d^{2} - a^{2} d^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{6 \, {\left (a b^{3} c^{2} d^{2} - 2 \, a^{2} b^{2} c d^{3} + a^{3} b d^{4} + {\left (b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x^{3}\right )}}, \frac {6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} c^{2} d - a^{2} b c d^{2} + {\left (b^{3} c^{2} d - a b^{2} c d^{2}\right )} x^{3}\right )} \sqrt {\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} d - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right )} \sqrt {\frac {{\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}}{b c - a d}}}{d}\right ) + {\left (b^{2} c x^{3} + a b c\right )} {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} - {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (b^{2} c x^{3} + a b c\right )} {\left (b c d^{2} - a d^{3}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d + {\left (b c d^{2} - a d^{3}\right )}^{\frac {1}{3}}\right ) + 6 \, {\left (a b c d^{2} - a^{2} d^{3}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{6 \, {\left (a b^{3} c^{2} d^{2} - 2 \, a^{2} b^{2} c d^{3} + a^{3} b d^{4} + {\left (b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x^{3}\right )}}\right ] \end {gather*}

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]

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

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giac [B]  time = 0.25, size = 301, normalized size = 1.73 \begin {gather*} -\frac {\frac {6 \, {\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b c \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 )}{\sqrt {3} b^{2} c^{2} d^{2} - 2 \, \sqrt {3} a b c d^{3} + \sqrt {3} a^{2} d^{4}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} b c \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 )}{b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}} + \frac {2 \, b c \left (-\frac {b c - a d}{d}\right )^{\frac {2}{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 )}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}} - \frac {6 \, a}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}}}{6 \, b} \end {gather*}

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]

-1/6*(6*(-b*c*d^2 + a*d^3)^(2/3)*b*c*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c
- a*d)/d)^(1/3))/(sqrt(3)*b^2*c^2*d^2 - 2*sqrt(3)*a*b*c*d^3 + sqrt(3)*a^2*d^4) - (-b*c*d^2 + a*d^3)^(2/3)*b*c*
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))/(b^2*c^2*d^2 - 2*a*
b*c*d^3 + a^2*d^4) + 2*b*c*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^2*c^
2 - 2*a*b*c*d + a^2*d^2) - 6*a/((b*x^3 + a)^(1/3)*(b*c - a*d)))/b

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maple [F]  time = 0.67, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 = 4.98, size = 412, normalized size = 2.37 \begin {gather*} -\frac {a}{b\,{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}-\frac {c\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^2\,d^2-b\,c^3\,d\right )-\frac {c^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{9\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^2\,d^2-b\,c^3\,d\right )-\frac {{\left (c-\sqrt {3}\,c\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{36\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c-\sqrt {3}\,c\,1{}\mathrm {i}\right )}{6\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,c^2\,d^2-b\,c^3\,d\right )-\frac {{\left (c+\sqrt {3}\,c\,1{}\mathrm {i}\right )}^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{36\,d^{4/3}\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (c+\sqrt {3}\,c\,1{}\mathrm {i}\right )}{6\,d^{2/3}\,{\left (a\,d-b\,c\right )}^{4/3}} \end {gather*}

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]

(log((a + b*x^3)^(1/3)*(a*c^2*d^2 - b*c^3*d) - ((c - 3^(1/2)*c*1i)^2*(9*a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3*c^3
*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(36*d^(4/3)*(a*d - b*c)^(8/3)))*(c - 3^(1/2)*c*1i))/(6*d^(2/3)*(a
*d - b*c)^(4/3)) - (c*log((a + b*x^3)^(1/3)*(a*c^2*d^2 - b*c^3*d) - (c^2*(9*a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3
*c^3*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(9*d^(4/3)*(a*d - b*c)^(8/3))))/(3*d^(2/3)*(a*d - b*c)^(4/3))
- a/(b*(a + b*x^3)^(1/3)*(a*d - b*c)) + (log((a + b*x^3)^(1/3)*(a*c^2*d^2 - b*c^3*d) - ((c + 3^(1/2)*c*1i)^2*
(9*a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3*c^3*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(36*d^(4/3)*(a*d - b*c)^
(8/3)))*(c + 3^(1/2)*c*1i))/(6*d^(2/3)*(a*d - b*c)^(4/3))

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

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]

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

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