3.5.61 \(\int \frac {1}{x^4 \sqrt [3]{a+b x^3} (c+d x^3)} \, dx\)

Optimal. Leaf size=296 \[ -\frac {(3 a d+b c) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{4/3} c^2}-\frac {(3 a d+b c) \tan ^{-1}\left (\frac {2 \sqrt [3]{a+b x^3}+\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{4/3} c^2}+\frac {\log (x) (3 a d+b c)}{6 a^{4/3} c^2}+\frac {d^{4/3} \log \left (c+d x^3\right )}{6 c^2 \sqrt [3]{b c-a d}}-\frac {d^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2 \sqrt [3]{b c-a d}}-\frac {d^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} c^2 \sqrt [3]{b c-a d}}-\frac {\left (a+b x^3\right )^{2/3}}{3 a c x^3} \]

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Rubi [A]  time = 0.31, antiderivative size = 296, 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, 55, 617, 204, 31, 56} \begin {gather*} -\frac {(3 a d+b c) \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x^3}\right )}{6 a^{4/3} c^2}-\frac {(3 a d+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^{4/3} c^2}+\frac {\log (x) (3 a d+b c)}{6 a^{4/3} c^2}+\frac {d^{4/3} \log \left (c+d x^3\right )}{6 c^2 \sqrt [3]{b c-a d}}-\frac {d^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 c^2 \sqrt [3]{b c-a d}}-\frac {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} c^2 \sqrt [3]{b c-a d}}-\frac {\left (a+b x^3\right )^{2/3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 55

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

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

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Mathematica [C]  time = 0.61, size = 156, normalized size = 0.53 \begin {gather*} -\frac {\frac {(3 a d+b c) \left (3 \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 )-3 \log (x)\right )}{a^{4/3}}-\frac {9 d^2 \left (a+b x^3\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )}{b c-a d}+\frac {6 c \left (a+b x^3\right )^{2/3}}{a x^3}}{18 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/18*((6*c*(a + b*x^3)^(2/3))/(a*x^3) - (9*d^2*(a + b*x^3)^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, (d*(a + b*x^3
))/(-(b*c) + a*d)])/(b*c - a*d) + ((b*c + 3*a*d)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*x^3)^(1/3))/a^(1/3))/Sqrt[3]
] - 3*Log[x] + 3*Log[a^(1/3) - (a + b*x^3)^(1/3)]))/a^(4/3))/c^2

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IntegrateAlgebraic [A]  time = 0.71, size = 386, normalized size = 1.30 \begin {gather*} \frac {(-3 a d-b c) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{a}\right )}{9 a^{4/3} c^2}+\frac {(3 a d+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^{4/3} c^2}-\frac {(3 a d+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^{4/3} c^2}-\frac {d^{4/3} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 c^2 \sqrt [3]{b c-a d}}+\frac {d^{4/3} \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 \sqrt [3]{b c-a d}}-\frac {d^{4/3} \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 \sqrt [3]{b c-a d}}-\frac {\left (a+b x^3\right )^{2/3}}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a + b*x^3)^(2/3)/(a*c*x^3) - ((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^(4/3)*c^2) - (d^(4/3)*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)^(1/3)) + ((-(b*c) - 3*a*d)*Log[-a^(1/3) + (a + b*x^3)^(1/3)])/(9*a^(4/3)*c^2) - (d
^(4/3)*Log[(b*c - a*d)^(1/3) + d^(1/3)*(a + b*x^3)^(1/3)])/(3*c^2*(b*c - a*d)^(1/3)) + ((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^(4/3)*c^2) + (d^(4/3)*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)^(1/3))

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fricas [A]  time = 0.62, size = 837, normalized size = 2.83

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.88, size = 383, normalized size = 1.29 \begin {gather*} -\frac {d^{2} \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 )}{3 \, {\left (b c^{3} - a c^{2} d\right )}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{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^{3} - \sqrt {3} a c^{2} d} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{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^{3} - a c^{2} d\right )}} + \frac {{\left (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 {4}{3}} c^{2}} - \frac {\sqrt {3} {\left (a^{\frac {2}{3}} b c + 3 \, a^{\frac {5}{3}} 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^{2} c^{2}} - \frac {{\left (a^{\frac {1}{3}} b c + 3 \, a^{\frac {4}{3}} 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 {2}{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)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

-1/3*d^2*(-(b*c - a*d)/d)^(2/3)*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)^(2/3)*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^3 - sqrt(3)*a*c^2*d) + 1/6*(-b*c*d^2 + a*d^3)^(2/3)*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^3 - a*c^2*d) + 1/18*(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^(4/3)*c^2) - 1/9*sqrt(3)*(a^(2/3)*b*c + 3*a^(5/3)*d)*arctan(1/3*sqrt(3)
*(2*(b*x^3 + a)^(1/3) + a^(1/3))/a^(1/3))/(a^2*c^2) - 1/9*(a^(1/3)*b*c + 3*a^(4/3)*d)*log(abs((b*x^3 + a)^(1/3
) - a^(1/3)))/(a^(5/3)*c^2) - 1/3*(b*x^3 + a)^(2/3)/(a*c*x^3)

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maple [F]  time = 0.59, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{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)^(1/3)/(d*x^3+c),x)

[Out]

int(1/x^4/(b*x^3+a)^(1/3)/(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 {1}{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)^(1/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 11.36, size = 1929, normalized size = 6.52

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(- (((((3*b^4*d^3*(a + b*x^3)^(1/3)*(18*a^4*d^4 + b^4*c^4 - 2*a^2*b^2*c^2*d^2 + 4*a*b^3*c^3*d - 12*a^3*b*c*
d^3))/a^2 - 3*a*b^4*c^4*d^3*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d)*(-(3*a*d + b*c)^3/(a^4*c^6))^(2/3))*(-(3*a*d + b
*c)^3/(a^4*c^6))^(1/3))/9 + (b^5*d^4*(b^3*c^3 - 27*a^3*d^3 + 8*a*b^2*c^2*d + 18*a^2*b*c*d^2))/(3*a^2*c))*(-(3*
a*d + b*c)^3/(a^4*c^6))^(2/3))/81 - (4*b^5*d^7*(a + b*x^3)^(1/3)*(3*a*d + b*c)^2)/(27*a^2*c^5))*(-(27*a^3*d^3
+ b^3*c^3 + 9*a*b^2*c^2*d + 27*a^2*b*c*d^2)/(729*a^4*c^6))^(1/3) + log(- (-d^4/(27*b*c^7 - 27*a*c^6*d))^(2/3)*
((-d^4/(27*b*c^7 - 27*a*c^6*d))^(1/3)*((3*b^4*d^3*(a + b*x^3)^(1/3)*(18*a^4*d^4 + b^4*c^4 - 2*a^2*b^2*c^2*d^2
+ 4*a*b^3*c^3*d - 12*a^3*b*c*d^3))/a^2 - 243*a*b^4*c^4*d^3*(-d^4/(27*b*c^7 - 27*a*c^6*d))^(2/3)*(2*a^2*d^2 + b
^2*c^2 - 3*a*b*c*d)) + (b^5*d^4*(b^3*c^3 - 27*a^3*d^3 + 8*a*b^2*c^2*d + 18*a^2*b*c*d^2))/(3*a^2*c)) - (4*b^5*d
^7*(a + b*x^3)^(1/3)*(3*a*d + b*c)^2)/(27*a^2*c^5))*(-d^4/(27*b*c^7 - 27*a*c^6*d))^(1/3) - log((((((3*b^4*d^3*
(a + b*x^3)^(1/3)*(18*a^4*d^4 + b^4*c^4 - 2*a^2*b^2*c^2*d^2 + 4*a*b^3*c^3*d - 12*a^3*b*c*d^3))/a^2 - 3*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 + b*c)^3/(a^4*c^6))^(2/3))*((3^(1/2)*
1i)/2 + 1/2)*(-(3*a*d + b*c)^3/(a^4*c^6))^(1/3))/9 - (b^5*d^4*(b^3*c^3 - 27*a^3*d^3 + 8*a*b^2*c^2*d + 18*a^2*b
*c*d^2))/(3*a^2*c))*((3^(1/2)*1i)/2 - 1/2)*(-(3*a*d + b*c)^3/(a^4*c^6))^(2/3))/81 - (4*b^5*d^7*(a + b*x^3)^(1/
3)*(3*a*d + b*c)^2)/(27*a^2*c^5))*((3^(1/2)*1i)/2 + 1/2)*(-(27*a^3*d^3 + b^3*c^3 + 9*a*b^2*c^2*d + 27*a^2*b*c*
d^2)/(729*a^4*c^6))^(1/3) + log((((((3*b^4*d^3*(a + b*x^3)^(1/3)*(18*a^4*d^4 + b^4*c^4 - 2*a^2*b^2*c^2*d^2 + 4
*a*b^3*c^3*d - 12*a^3*b*c*d^3))/a^2 + 3*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 + b*c)^3/(a^4*c^6))^(2/3))*((3^(1/2)*1i)/2 - 1/2)*(-(3*a*d + b*c)^3/(a^4*c^6))^(1/3))/9 + (b^5*d^4*(
b^3*c^3 - 27*a^3*d^3 + 8*a*b^2*c^2*d + 18*a^2*b*c*d^2))/(3*a^2*c))*((3^(1/2)*1i)/2 + 1/2)*(-(3*a*d + b*c)^3/(a
^4*c^6))^(2/3))/81 - (4*b^5*d^7*(a + b*x^3)^(1/3)*(3*a*d + b*c)^2)/(27*a^2*c^5))*((3^(1/2)*1i)/2 - 1/2)*(-(27*
a^3*d^3 + b^3*c^3 + 9*a*b^2*c^2*d + 27*a^2*b*c*d^2)/(729*a^4*c^6))^(1/3) + (log(- ((3^(1/2)*1i - 1)^2*(((3^(1/
2)*1i - 1)*(-d^4/(27*b*c^7 - 27*a*c^6*d))^(1/3)*((3*b^4*d^3*(a + b*x^3)^(1/3)*(18*a^4*d^4 + b^4*c^4 - 2*a^2*b^
2*c^2*d^2 + 4*a*b^3*c^3*d - 12*a^3*b*c*d^3))/a^2 - (243*a*b^4*c^4*d^3*(3^(1/2)*1i - 1)^2*(-d^4/(27*b*c^7 - 27*
a*c^6*d))^(2/3)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d))/4))/2 + (b^5*d^4*(b^3*c^3 - 27*a^3*d^3 + 8*a*b^2*c^2*d + 18
*a^2*b*c*d^2))/(3*a^2*c))*(-d^4/(27*b*c^7 - 27*a*c^6*d))^(2/3))/4 - (4*b^5*d^7*(a + b*x^3)^(1/3)*(3*a*d + b*c)
^2)/(27*a^2*c^5))*(3^(1/2)*1i - 1)*(-d^4/(27*b*c^7 - 27*a*c^6*d))^(1/3))/2 - (log(((3^(1/2)*1i + 1)^2*(((3^(1/
2)*1i + 1)*(-d^4/(27*b*c^7 - 27*a*c^6*d))^(1/3)*((3*b^4*d^3*(a + b*x^3)^(1/3)*(18*a^4*d^4 + b^4*c^4 - 2*a^2*b^
2*c^2*d^2 + 4*a*b^3*c^3*d - 12*a^3*b*c*d^3))/a^2 - (243*a*b^4*c^4*d^3*(3^(1/2)*1i + 1)^2*(-d^4/(27*b*c^7 - 27*
a*c^6*d))^(2/3)*(2*a^2*d^2 + b^2*c^2 - 3*a*b*c*d))/4))/2 - (b^5*d^4*(b^3*c^3 - 27*a^3*d^3 + 8*a*b^2*c^2*d + 18
*a^2*b*c*d^2))/(3*a^2*c))*(-d^4/(27*b*c^7 - 27*a*c^6*d))^(2/3))/4 - (4*b^5*d^7*(a + b*x^3)^(1/3)*(3*a*d + b*c)
^2)/(27*a^2*c^5))*(3^(1/2)*1i + 1)*(-d^4/(27*b*c^7 - 27*a*c^6*d))^(1/3))/2 - (a + b*x^3)^(2/3)/(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} \sqrt [3]{a + b x^{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)**(1/3)/(d*x**3+c),x)

[Out]

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

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