3.5.72 \(\int \frac {1}{x (a+b x^3)^{2/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=245 \[ \frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3} c}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} c}-\frac {\log (x)}{2 a^{2/3} c}+\frac {d^{2/3} \log \left (c+d x^3\right )}{6 c (b c-a d)^{2/3}}-\frac {d^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c (b c-a d)^{2/3}}+\frac {d^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} c (b c-a d)^{2/3}} \]

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Rubi [A]  time = 0.21, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {446, 86, 57, 617, 204, 31, 58} \begin {gather*} \frac {\log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{2 a^{2/3} c}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} c}-\frac {\log (x)}{2 a^{2/3} c}+\frac {d^{2/3} \log \left (c+d x^3\right )}{6 c (b c-a d)^{2/3}}-\frac {d^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c (b c-a d)^{2/3}}+\frac {d^{2/3} \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} c (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[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]
&& PosQ[(b*c - a*d)/b]

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 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

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

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Mathematica [A]  time = 0.45, size = 308, normalized size = 1.26 \begin {gather*} -\frac {-\frac {2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{a^{2/3}}+\frac {\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{a^{2/3}}+\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {2 d^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{2/3}}-\frac {d^{2/3} \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 )}{(b c-a d)^{2/3}}+\frac {2 \sqrt {3} d^{2/3} \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 )}{(b c-a d)^{2/3}}}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.59, size = 333, normalized size = 1.36 \begin {gather*} \frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{a}\right )}{3 a^{2/3} c}-\frac {\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{6 a^{2/3} c}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{a}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} a^{2/3} c}-\frac {d^{2/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 c (b c-a d)^{2/3}}+\frac {d^{2/3} \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 c (b c-a d)^{2/3}}+\frac {d^{2/3} \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} c (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.50, size = 472, normalized size = 1.93 \begin {gather*} -\frac {2 \, \sqrt {3} a^{2} \left (-\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \left (-\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {2}{3}} - \sqrt {3} d}{3 \, d}\right ) + a^{2} \left (-\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} d^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c d - a d^{2}\right )} \left (-\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {2}{3}}\right ) - 2 \, a^{2} \left (-\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} d - {\left (b c - a d\right )} \left (-\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, \sqrt {3} {\left (a^{2}\right )}^{\frac {1}{6}} a \arctan \left (\frac {{\left (a^{2}\right )}^{\frac {1}{6}} {\left (\sqrt {3} {\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, a^{2}}\right ) + {\left (a^{2}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (a^{2}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right )}{6 \, a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.76, size = 321, normalized size = 1.31 \begin {gather*} \frac {d \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 c^{2} - a c d\right )}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} \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 c^{2} - \sqrt {3} a c d} - \frac {{\left (-b c d^{2} + a d^{3}\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 )}{6 \, {\left (b c^{2} - a c d\right )}} - \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{3 \, a^{\frac {2}{3}} c} - \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{6 \, a^{\frac {2}{3}} c} + \frac {\log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{3 \, a^{\frac {2}{3}} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right ) x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.94, size = 1413, normalized size = 5.77

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((((81*b^6*c^5*d^3 - 162*a*b^5*c^4*d^4)*(a + b*x^3)^(1/3) - (243*a*b^6*c^6*d^3 - 729*a^2*b^5*c^5*d^4 + 486*
a^3*b^4*c^4*d^5)*(1/(27*a^2*c^3))^(1/3))*(1/(27*a^2*c^3))^(2/3) - 9*b^5*c^2*d^4)*(1/(27*a^2*c^3))^(1/3) + 6*b^
4*d^5*(a + b*x^3)^(1/3))*(1/(27*a^2*c^3))^(1/3) + log(- (((81*b^6*c^5*d^3 - 162*a*b^5*c^4*d^4)*(a + b*x^3)^(1/
3) - (-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2 - 54*a*b*c^4*d))^(1/3)*(243*a*b^6*c^6*d^3 - 729*a^2*b^5*c^5*d^4 + 486*
a^3*b^4*c^4*d^5))*(-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2 - 54*a*b*c^4*d))^(2/3) - 9*b^5*c^2*d^4)*(-d^2/(27*b^2*c^5
+ 27*a^2*c^3*d^2 - 54*a*b*c^4*d))^(1/3) - 6*b^4*d^5*(a + b*x^3)^(1/3))*(-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2 - 5
4*a*b*c^4*d))^(1/3) + log(((3^(1/2)*1i)/2 - 1/2)*(((3^(1/2)*1i)/2 - 1/2)^2*((81*b^6*c^5*d^3 - 162*a*b^5*c^4*d^
4)*(a + b*x^3)^(1/3) - ((3^(1/2)*1i)/2 - 1/2)*(243*a*b^6*c^6*d^3 - 729*a^2*b^5*c^5*d^4 + 486*a^3*b^4*c^4*d^5)*
(1/(27*a^2*c^3))^(1/3))*(1/(27*a^2*c^3))^(2/3) - 9*b^5*c^2*d^4)*(1/(27*a^2*c^3))^(1/3) + 6*b^4*d^5*(a + b*x^3)
^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(1/(27*a^2*c^3))^(1/3) - log(6*b^4*d^5*(a + b*x^3)^(1/3) - ((3^(1/2)*1i)/2 + 1/
2)*(((3^(1/2)*1i)/2 + 1/2)^2*((81*b^6*c^5*d^3 - 162*a*b^5*c^4*d^4)*(a + b*x^3)^(1/3) + ((3^(1/2)*1i)/2 + 1/2)*
(243*a*b^6*c^6*d^3 - 729*a^2*b^5*c^5*d^4 + 486*a^3*b^4*c^4*d^5)*(1/(27*a^2*c^3))^(1/3))*(1/(27*a^2*c^3))^(2/3)
- 9*b^5*c^2*d^4)*(1/(27*a^2*c^3))^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(1/(27*a^2*c^3))^(1/3) + (log(6*b^4*d^5*(a +
b*x^3)^(1/3) + ((3^(1/2)*1i - 1)*(((3^(1/2)*1i - 1)^2*((81*b^6*c^5*d^3 - 162*a*b^5*c^4*d^4)*(a + b*x^3)^(1/3)
- ((3^(1/2)*1i - 1)*(-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2 - 54*a*b*c^4*d))^(1/3)*(243*a*b^6*c^6*d^3 - 729*a^2*b^5
*c^5*d^4 + 486*a^3*b^4*c^4*d^5))/2)*(-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2 - 54*a*b*c^4*d))^(2/3))/4 - 9*b^5*c^2*d
^4)*(-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2 - 54*a*b*c^4*d))^(1/3))/2)*(3^(1/2)*1i - 1)*(-d^2/(27*b^2*c^5 + 27*a^2*
c^3*d^2 - 54*a*b*c^4*d))^(1/3))/2 - (log(6*b^4*d^5*(a + b*x^3)^(1/3) - ((3^(1/2)*1i + 1)*(((3^(1/2)*1i + 1)^2*
((81*b^6*c^5*d^3 - 162*a*b^5*c^4*d^4)*(a + b*x^3)^(1/3) + ((3^(1/2)*1i + 1)*(-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2
- 54*a*b*c^4*d))^(1/3)*(243*a*b^6*c^6*d^3 - 729*a^2*b^5*c^5*d^4 + 486*a^3*b^4*c^4*d^5))/2)*(-d^2/(27*b^2*c^5
+ 27*a^2*c^3*d^2 - 54*a*b*c^4*d))^(2/3))/4 - 9*b^5*c^2*d^4)*(-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2 - 54*a*b*c^4*d)
)^(1/3))/2)*(3^(1/2)*1i + 1)*(-d^2/(27*b^2*c^5 + 27*a^2*c^3*d^2 - 54*a*b*c^4*d))^(1/3))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x**3+a)**(2/3)/(d*x**3+c),x)

[Out]

Integral(1/(x*(a + b*x**3)**(2/3)*(c + d*x**3)), x)

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