3.5.73 \(\int \frac {1}{x^4 (a+b x^3)^{2/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=299 \[ -\frac {(3 a d+2 b c) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{5/3} c^2}+\frac {(3 a d+2 b c) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} c^2}+\frac {\log (x) (3 a d+2 b c)}{6 a^{5/3} c^2}-\frac {d^{5/3} \log \left (c+d x^3\right )}{6 c^2 (b c-a d)^{2/3}}+\frac {d^{5/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2 (b c-a d)^{2/3}}-\frac {d^{5/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} c^2 (b c-a d)^{2/3}}-\frac {\sqrt [3]{a+b x^3}}{3 a c x^3} \]

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Rubi [A]  time = 0.32, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {446, 103, 156, 57, 617, 204, 31, 58} \begin {gather*} -\frac {(3 a d+2 b c) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{5/3} c^2}+\frac {(3 a d+2 b c) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} c^2}+\frac {\log (x) (3 a d+2 b c)}{6 a^{5/3} c^2}-\frac {d^{5/3} \log \left (c+d x^3\right )}{6 c^2 (b c-a d)^{2/3}}+\frac {d^{5/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2 (b c-a d)^{2/3}}-\frac {d^{5/3} \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^2 (b c-a d)^{2/3}}-\frac {\sqrt [3]{a+b x^3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

-(a + b*x^3)^(1/3)/(3*a*c*x^3) + ((2*b*c + 3*a*d)*ArcTan[(a^(1/3) + 2*(a + b*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])/(
3*Sqrt[3]*a^(5/3)*c^2) - (d^(5/3)*ArcTan[(1 - (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]])/(Sqrt
[3]*c^2*(b*c - a*d)^(2/3)) + ((2*b*c + 3*a*d)*Log[x])/(6*a^(5/3)*c^2) - (d^(5/3)*Log[c + d*x^3])/(6*c^2*(b*c -
a*d)^(2/3)) - ((2*b*c + 3*a*d)*Log[a^(1/3) - (a + b*x^3)^(1/3)])/(6*a^(5/3)*c^2) + (d^(5/3)*Log[(b*c - a*d)^(
1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(2*c^2*(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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

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

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^4 \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^2 (a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a c x^3}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{3} (2 b c+3 a d)+\frac {2 b d x}{3}}{x (a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 a c}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a c x^3}+\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{2/3} (c+d x)} \, dx,x,x^3\right )}{3 c^2}-\frac {(2 b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{x (a+b x)^{2/3}} \, dx,x,x^3\right )}{9 a c^2}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a c x^3}+\frac {(2 b c+3 a d) \log (x)}{6 a^{5/3} c^2}-\frac {d^{5/3} \log \left (c+d x^3\right )}{6 c^2 (b c-a d)^{2/3}}+\frac {d^{5/3} \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^2 (b c-a d)^{2/3}}+\frac {d^{4/3} \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^2 \sqrt [3]{b c-a d}}+\frac {(2 b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{6 a^{5/3} c^2}+\frac {(2 b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{a^{2/3}+\sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{6 a^{4/3} c^2}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a c x^3}+\frac {(2 b c+3 a d) \log (x)}{6 a^{5/3} c^2}-\frac {d^{5/3} \log \left (c+d x^3\right )}{6 c^2 (b c-a d)^{2/3}}-\frac {(2 b c+3 a d) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{5/3} c^2}+\frac {d^{5/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2 (b c-a d)^{2/3}}+\frac {d^{5/3} \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^2 (b c-a d)^{2/3}}-\frac {(2 b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{3 a^{5/3} c^2}\\ &=-\frac {\sqrt [3]{a+b x^3}}{3 a c x^3}+\frac {(2 b c+3 a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{3 \sqrt {3} a^{5/3} c^2}-\frac {d^{5/3} \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^2 (b c-a d)^{2/3}}+\frac {(2 b c+3 a d) \log (x)}{6 a^{5/3} c^2}-\frac {d^{5/3} \log \left (c+d x^3\right )}{6 c^2 (b c-a d)^{2/3}}-\frac {(2 b c+3 a d) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{5/3} c^2}+\frac {d^{5/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2 (b c-a d)^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 303, normalized size = 1.01 \begin {gather*} \frac {\frac {(3 a d+2 b c) \left (\log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )\right )}{a^{2/3} c}+\frac {3 a d^{5/3} \left (-\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 \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )+2 \sqrt {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 )\right )}{c (b c-a d)^{2/3}}-\frac {6 \sqrt [3]{a+b x^3}}{x^3}}{18 a c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

((-6*(a + b*x^3)^(1/3))/x^3 + ((2*b*c + 3*a*d)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]]
- 2*Log[a^(1/3) - (a + b*x^3)^(1/3)] + Log[a^(2/3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)]))/(a^(2/3)
*c) + (3*a*d^(5/3)*(2*Sqrt[3]*ArcTan[(-1 + (2*d^(1/3)*(a + b*x^3)^(1/3))/(b*c - a*d)^(1/3))/Sqrt[3]] + 2*Log[(
b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(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)]))/(c*(b*c - a*d)^(2/3)))/(18*a*c)

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IntegrateAlgebraic [A]  time = 0.73, size = 388, normalized size = 1.30 \begin {gather*} \frac {(-3 a d-2 b c) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{a}\right )}{9 a^{5/3} c^2}+\frac {(3 a d+2 b c) \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{18 a^{5/3} c^2}+\frac {(3 a d+2 b c) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} c^2}+\frac {d^{5/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 c^2 (b c-a d)^{2/3}}-\frac {d^{5/3} \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^2 (b c-a d)^{2/3}}-\frac {d^{5/3} \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^2 (b c-a d)^{2/3}}-\frac {\sqrt [3]{a+b x^3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^4*(a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

-1/3*(a + b*x^3)^(1/3)/(a*c*x^3) + ((2*b*c + 3*a*d)*ArcTan[(a^(1/3) + 2*(a + b*x^3)^(1/3))/(Sqrt[3]*a^(1/3))])
/(3*Sqrt[3]*a^(5/3)*c^2) - (d^(5/3)*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^2*(b*c - a*d)^(2/3)) + ((-2*b*c - 3*a*d)*Log[-a^(1/3) + (a + b*x^3)^(1/3)])/(9*a^(5/3)*c^2) +
(d^(5/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*c^2*(b*c - a*d)^(2/3)) + ((2*b*c + 3*a*d)*Log[
a^(2/3) + a^(1/3)*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(18*a^(5/3)*c^2) - (d^(5/3)*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^2*(b*c - a*d)^(2/3))

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fricas [B]  time = 1.25, size = 562, normalized size = 1.88 \begin {gather*} -\frac {6 \, \sqrt {3} a^{3} d \left (\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x^{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 ) + 3 \, a^{3} d \left (\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x^{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 ) - 6 \, a^{3} d \left (\frac {d^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x^{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 (2 \, a b c + 3 \, a^{2} d\right )} x^{3} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\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 )} \sqrt {-\left (-a^{2}\right )^{\frac {1}{3}}}}{3 \, a^{2}}\right ) - \left (-a^{2}\right )^{\frac {2}{3}} {\left (2 \, b c + 3 \, a d\right )} x^{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}} {\left (2 \, b c + 3 \, a d\right )} x^{3} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} a - \left (-a^{2}\right )^{\frac {2}{3}}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} a^{2} c}{18 \, a^{3} c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

-1/18*(6*sqrt(3)*a^3*d*(d^2/(b^2*c^2 - 2*a*b*c*d + a^2*d^2))^(1/3)*x^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) + 3*a^3*d*(d^2/(b^2*c^2 - 2*a*b*c*d
+ a^2*d^2))^(1/3)*x^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)) - 6*a^3*d*(d^
2/(b^2*c^2 - 2*a*b*c*d + a^2*d^2))^(1/3)*x^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)*(2*a*b*c + 3*a^2*d)*x^3*sqrt(-(-a^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-a^2)^(1/3)*a
- 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-a^2)^(2/3))*sqrt(-(-a^2)^(1/3))/a^2) - (-a^2)^(2/3)*(2*b*c + 3*a*d)*x^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)*(2*b*c + 3*a*d)*x^3*log(
(b*x^3 + a)^(1/3)*a - (-a^2)^(2/3)) + 6*(b*x^3 + a)^(1/3)*a^2*c)/(a^3*c^2*x^3)

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giac [A]  time = 0.76, size = 377, normalized size = 1.26 \begin {gather*} -\frac {d^{2} \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^{3} - a c^{2} d\right )}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} d \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^{3} - \sqrt {3} a c^{2} d} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {1}{3}} d \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^{3} - a c^{2} d\right )}} + \frac {\sqrt {3} {\left (2 \, b c + 3 \, a d\right )} \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 )}{9 \, a^{\frac {5}{3}} c^{2}} + \frac {{\left (2 \, b c + 3 \, a d\right )} \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 )}{18 \, a^{\frac {5}{3}} c^{2}} - \frac {{\left (2 \, b c + 3 \, a d\right )} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{9 \, a^{\frac {5}{3}} c^{2}} - \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{3 \, a c x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*d^2*(-(b*c - a*d)/d)^(1/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b*c^3 - a*c^2*d) + (-b*c
*d^2 + a*d^3)^(1/3)*d*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^3 - sqrt(3)*a*c^2*d) + 1/6*(-b*c*d^2 + a*d^3)^(1/3)*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))/(b*c^3 - a*c^2*d) + 1/9*sqrt(3)*(2*b*c + 3*a*d)*arctan(1/3*s
qrt(3)*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/(a^(5/3)*c^2) + 1/18*(2*b*c + 3*a*d)*log((b*x^3 + a)^(2/3) + (
b*x^3 + a)^(1/3)*a^(1/3) + a^(2/3))/(a^(5/3)*c^2) - 1/9*(2*b*c + 3*a*d)*log(abs((b*x^3 + a)^(1/3) - a^(1/3)))/
(a^(5/3)*c^2) - 1/3*(b*x^3 + a)^(1/3)/(a*c*x^3)

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maple [F]  time = 0.64, 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^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^3+a)^(2/3)/(d*x^3+c),x)

[Out]

int(1/x^4/(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^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(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^4), x)

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mupad [B]  time = 11.14, size = 1959, normalized size = 6.55

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^3)^(2/3)*(c + d*x^3)),x)

[Out]

log(- (((((27*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(4*a^2*d^2 - 2*b^2*c^2 + a*b*c*d))/a - 81*a*b^4*c^4*d^3*(2*a^2*d^2
+ b^2*c^2 - 3*a*b*c*d)*(d^5/(c^6*(a*d - b*c)^2))^(1/3))*(d^5/(c^6*(a*d - b*c)^2))^(2/3))/9 + (b^5*d^4*(8*b^3*
c^3 - 27*a^3*d^3 + 28*a*b^2*c^2*d + 18*a^2*b*c*d^2))/(3*a^3*c))*(d^5/(c^6*(a*d - b*c)^2))^(1/3))/3 - (2*b^4*d^
6*(a + b*x^3)^(1/3)*(27*a^3*d^3 + 4*b^3*c^3 + 18*a*b^2*c^2*d + 36*a^2*b*c*d^2))/(9*a^3*c^4))*(d^5/(27*b^2*c^8
+ 27*a^2*c^6*d^2 - 54*a*b*c^7*d))^(1/3) + log(- (((((27*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(4*a^2*d^2 - 2*b^2*c^2 +
a*b*c*d))/a - 27*a*b^4*c^4*d^3*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(-(3*a*d + 2*b*c)^3/(a^5*c^6))^(1/3))*(-(3*a
*d + 2*b*c)^3/(a^5*c^6))^(2/3))/81 + (b^5*d^4*(8*b^3*c^3 - 27*a^3*d^3 + 28*a*b^2*c^2*d + 18*a^2*b*c*d^2))/(3*a
^3*c))*(-(3*a*d + 2*b*c)^3/(a^5*c^6))^(1/3))/9 - (2*b^4*d^6*(a + b*x^3)^(1/3)*(27*a^3*d^3 + 4*b^3*c^3 + 18*a*b
^2*c^2*d + 36*a^2*b*c*d^2))/(9*a^3*c^4))*(-(27*a^3*d^3 + 8*b^3*c^3 + 36*a*b^2*c^2*d + 54*a^2*b*c*d^2)/(729*a^5
*c^6))^(1/3) + (log(((3^(1/2)*1i - 1)*((((27*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(4*a^2*d^2 - 2*b^2*c^2 + a*b*c*d))/
a - (81*a*b^4*c^4*d^3*(3^(1/2)*1i - 1)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(d^5/(c^6*(a*d - b*c)^2))^(1/3))/2)*(
3^(1/2)*1i - 1)^2*(d^5/(c^6*(a*d - b*c)^2))^(2/3))/36 + (b^5*d^4*(8*b^3*c^3 - 27*a^3*d^3 + 28*a*b^2*c^2*d + 18
*a^2*b*c*d^2))/(3*a^3*c))*(d^5/(c^6*(a*d - b*c)^2))^(1/3))/6 + (2*b^4*d^6*(a + b*x^3)^(1/3)*(27*a^3*d^3 + 4*b^
3*c^3 + 18*a*b^2*c^2*d + 36*a^2*b*c*d^2))/(9*a^3*c^4))*(3^(1/2)*1i - 1)*(d^5/(27*b^2*c^8 + 27*a^2*c^6*d^2 - 54
*a*b*c^7*d))^(1/3))/2 - (log(((3^(1/2)*1i + 1)*((((27*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(4*a^2*d^2 - 2*b^2*c^2 + a
*b*c*d))/a + (81*a*b^4*c^4*d^3*(3^(1/2)*1i + 1)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(d^5/(c^6*(a*d - b*c)^2))^(1
/3))/2)*(3^(1/2)*1i + 1)^2*(d^5/(c^6*(a*d - b*c)^2))^(2/3))/36 + (b^5*d^4*(8*b^3*c^3 - 27*a^3*d^3 + 28*a*b^2*c
^2*d + 18*a^2*b*c*d^2))/(3*a^3*c))*(d^5/(c^6*(a*d - b*c)^2))^(1/3))/6 - (2*b^4*d^6*(a + b*x^3)^(1/3)*(27*a^3*d
^3 + 4*b^3*c^3 + 18*a*b^2*c^2*d + 36*a^2*b*c*d^2))/(9*a^3*c^4))*(3^(1/2)*1i + 1)*(d^5/(27*b^2*c^8 + 27*a^2*c^6
*d^2 - 54*a*b*c^7*d))^(1/3))/2 + log((2*b^4*d^6*(a + b*x^3)^(1/3)*(27*a^3*d^3 + 4*b^3*c^3 + 18*a*b^2*c^2*d + 3
6*a^2*b*c*d^2))/(9*a^3*c^4) - (((((27*b^5*c^3*d^3*(a + b*x^3)^(1/3)*(4*a^2*d^2 - 2*b^2*c^2 + a*b*c*d))/a - 27*
a*b^4*c^4*d^3*((3^(1/2)*1i)/2 - 1/2)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(-(3*a*d + 2*b*c)^3/(a^5*c^6))^(1/3))*(
(3^(1/2)*1i)/2 + 1/2)*(-(3*a*d + 2*b*c)^3/(a^5*c^6))^(2/3))/81 - (b^5*d^4*(8*b^3*c^3 - 27*a^3*d^3 + 28*a*b^2*c
^2*d + 18*a^2*b*c*d^2))/(3*a^3*c))*((3^(1/2)*1i)/2 - 1/2)*(-(3*a*d + 2*b*c)^3/(a^5*c^6))^(1/3))/9)*((3^(1/2)*1
i)/2 - 1/2)*(-(27*a^3*d^3 + 8*b^3*c^3 + 36*a*b^2*c^2*d + 54*a^2*b*c*d^2)/(729*a^5*c^6))^(1/3) - log((((((27*b^
5*c^3*d^3*(a + b*x^3)^(1/3)*(4*a^2*d^2 - 2*b^2*c^2 + a*b*c*d))/a + 27*a*b^4*c^4*d^3*((3^(1/2)*1i)/2 + 1/2)*(2*
a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(-(3*a*d + 2*b*c)^3/(a^5*c^6))^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(-(3*a*d + 2*b*c)^
3/(a^5*c^6))^(2/3))/81 + (b^5*d^4*(8*b^3*c^3 - 27*a^3*d^3 + 28*a*b^2*c^2*d + 18*a^2*b*c*d^2))/(3*a^3*c))*((3^(
1/2)*1i)/2 + 1/2)*(-(3*a*d + 2*b*c)^3/(a^5*c^6))^(1/3))/9 - (2*b^4*d^6*(a + b*x^3)^(1/3)*(27*a^3*d^3 + 4*b^3*c
^3 + 18*a*b^2*c^2*d + 36*a^2*b*c*d^2))/(9*a^3*c^4))*((3^(1/2)*1i)/2 + 1/2)*(-(27*a^3*d^3 + 8*b^3*c^3 + 36*a*b^
2*c^2*d + 54*a^2*b*c*d^2)/(729*a^5*c^6))^(1/3) - (a + b*x^3)^(1/3)/(3*a*c*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \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**4/(b*x**3+a)**(2/3)/(d*x**3+c),x)

[Out]

Integral(1/(x**4*(a + b*x**3)**(2/3)*(c + d*x**3)), x)

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