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

Optimal. Leaf size=301 \[ \frac {b^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{5/3} (b c-a d)}-\frac {b^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{5/3} (b c-a d)}+\frac {b^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3} (b c-a d)}-\frac {d^{5/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{5/3} (b c-a d)}+\frac {d^{5/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{5/3} (b c-a d)}-\frac {d^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{5/3} (b c-a d)}-\frac {1}{2 a c x^2} \]

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Rubi [A]  time = 0.25, antiderivative size = 301, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {480, 522, 200, 31, 634, 617, 204, 628} \begin {gather*} \frac {b^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{5/3} (b c-a d)}-\frac {b^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{5/3} (b c-a d)}+\frac {b^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3} (b c-a d)}-\frac {d^{5/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{5/3} (b c-a d)}+\frac {d^{5/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{5/3} (b c-a d)}-\frac {d^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{5/3} (b c-a d)}-\frac {1}{2 a c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, 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 480

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

Rule 522

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

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]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 259, normalized size = 0.86 \begin {gather*} \frac {\frac {2 b^{5/3} x^2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{5/3}}-\frac {b^{5/3} x^2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{5/3}}-\frac {2 \sqrt {3} b^{5/3} x^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{5/3}}+\frac {3 b}{a}-\frac {2 d^{5/3} x^2 \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{c^{5/3}}+\frac {d^{5/3} x^2 \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{c^{5/3}}+\frac {2 \sqrt {3} d^{5/3} x^2 \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )}{c^{5/3}}-\frac {3 d}{c}}{6 x^2 (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 2.62, size = 301, normalized size = 1.00 \begin {gather*} -\frac {2 \, \sqrt {3} b c x^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) + 2 \, \sqrt {3} a d x^{2} \left (-\frac {d^{2}}{c^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} c x \left (-\frac {d^{2}}{c^{2}}\right )^{\frac {2}{3}} - \sqrt {3} d}{3 \, d}\right ) - b c x^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{2} - a b x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) - a d x^{2} \left (-\frac {d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (d^{2} x^{2} + c d x \left (-\frac {d^{2}}{c^{2}}\right )^{\frac {1}{3}} + c^{2} \left (-\frac {d^{2}}{c^{2}}\right )^{\frac {2}{3}}\right ) + 2 \, b c x^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, a d x^{2} \left (-\frac {d^{2}}{c^{2}}\right )^{\frac {1}{3}} \log \left (d x - c \left (-\frac {d^{2}}{c^{2}}\right )^{\frac {1}{3}}\right ) + 3 \, b c - 3 \, a d}{6 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 309, normalized size = 1.03 \begin {gather*} \frac {b^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (a^{2} b c - a^{3} d\right )}} - \frac {d^{2} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b c^{3} - a c^{2} d\right )}} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} b \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} a^{2} b c - \sqrt {3} a^{3} d} + \frac {\left (-c d^{2}\right )^{\frac {1}{3}} d \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c^{3} - \sqrt {3} a c^{2} d} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} b \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{2} b c - a^{3} d\right )}} + \frac {\left (-c d^{2}\right )^{\frac {1}{3}} d \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c^{3} - a c^{2} d\right )}} - \frac {1}{2 \, a c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 257, normalized size = 0.85 \begin {gather*} \frac {\sqrt {3}\, b \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (a d -b c \right ) \left (\frac {a}{b}\right )^{\frac {2}{3}} a}+\frac {b \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (a d -b c \right ) \left (\frac {a}{b}\right )^{\frac {2}{3}} a}-\frac {b \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (a d -b c \right ) \left (\frac {a}{b}\right )^{\frac {2}{3}} a}-\frac {\sqrt {3}\, d \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (a d -b c \right ) \left (\frac {c}{d}\right )^{\frac {2}{3}} c}-\frac {d \ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (a d -b c \right ) \left (\frac {c}{d}\right )^{\frac {2}{3}} c}+\frac {d \ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \left (a d -b c \right ) \left (\frac {c}{d}\right )^{\frac {2}{3}} c}-\frac {1}{2 a c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.14, size = 328, normalized size = 1.09 \begin {gather*} -\frac {\sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, {\left (a b c \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} d \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {\sqrt {3} d \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, {\left (b c^{2} \left (\frac {c}{d}\right )^{\frac {1}{3}} - a c d \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )} \left (\frac {c}{d}\right )^{\frac {1}{3}}} + \frac {b \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (a b c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}} - \frac {d \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}} - a c d \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}} - \frac {b \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, {\left (a b c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}} + \frac {d \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \, {\left (b c^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}} - a c d \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}} - \frac {1}{2 \, a c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 11.83, size = 1829, normalized size = 6.08

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(((b^5/(a^5*(a*d - b*c)^3))^(1/3)*(((81*a^10*b^3*c^10*d^3*(a*d + b*c)*(a*d - b*c)^4*(b^5/(a^5*(a*d - b*c)^3
))^(1/3) - 81*a^8*b^3*c^8*d^3*x*(a*d - b*c)^4*(a^2*d^2 + b^2*c^2 + a*b*c*d))*(b^5/(a^5*(a*d - b*c)^3))^(2/3))/
9 + 9*a^6*b^9*c^11*d^4 - 9*a^7*b^8*c^10*d^5 - 9*a^10*b^5*c^7*d^8 + 9*a^11*b^4*c^6*d^9))/3 + 3*a^6*b^6*c^6*d^6*
x*(a^2*d^2 + b^2*c^2))*(b^5/(27*a^8*d^3 - 27*a^5*b^3*c^3 + 81*a^6*b^2*c^2*d - 81*a^7*b*c*d^2))^(1/3) + log(((-
d^5/(c^5*(a*d - b*c)^3))^(1/3)*(((81*a^10*b^3*c^10*d^3*(a*d + b*c)*(a*d - b*c)^4*(-d^5/(c^5*(a*d - b*c)^3))^(1
/3) - 81*a^8*b^3*c^8*d^3*x*(a*d - b*c)^4*(a^2*d^2 + b^2*c^2 + a*b*c*d))*(-d^5/(c^5*(a*d - b*c)^3))^(2/3))/9 +
9*a^6*b^9*c^11*d^4 - 9*a^7*b^8*c^10*d^5 - 9*a^10*b^5*c^7*d^8 + 9*a^11*b^4*c^6*d^9))/3 + 3*a^6*b^6*c^6*d^6*x*(a
^2*d^2 + b^2*c^2))*(d^5/(27*b^3*c^8 - 27*a^3*c^5*d^3 + 81*a^2*b*c^6*d^2 - 81*a*b^2*c^7*d))^(1/3) + (log(((3^(1
/2)*1i - 1)*(b^5/(a^5*(a*d - b*c)^3))^(1/3)*(((3^(1/2)*1i - 1)^2*(81*a^8*b^3*c^8*d^3*x*(a*d - b*c)^4*(a^2*d^2
+ b^2*c^2 + a*b*c*d) - (81*a^10*b^3*c^10*d^3*(3^(1/2)*1i - 1)*(a*d + b*c)*(a*d - b*c)^4*(b^5/(a^5*(a*d - b*c)^
3))^(1/3))/2)*(b^5/(a^5*(a*d - b*c)^3))^(2/3))/36 - 9*a^6*b^9*c^11*d^4 + 9*a^7*b^8*c^10*d^5 + 9*a^10*b^5*c^7*d
^8 - 9*a^11*b^4*c^6*d^9))/6 - 3*a^6*b^6*c^6*d^6*x*(a^2*d^2 + b^2*c^2))*(b^5/(27*a^8*d^3 - 27*a^5*b^3*c^3 + 81*
a^6*b^2*c^2*d - 81*a^7*b*c*d^2))^(1/3)*(3^(1/2)*1i - 1))/2 - (log(((3^(1/2)*1i + 1)*(b^5/(a^5*(a*d - b*c)^3))^
(1/3)*(((3^(1/2)*1i + 1)^2*(81*a^8*b^3*c^8*d^3*x*(a*d - b*c)^4*(a^2*d^2 + b^2*c^2 + a*b*c*d) + (81*a^10*b^3*c^
10*d^3*(3^(1/2)*1i + 1)*(a*d + b*c)*(a*d - b*c)^4*(b^5/(a^5*(a*d - b*c)^3))^(1/3))/2)*(b^5/(a^5*(a*d - b*c)^3)
)^(2/3))/36 - 9*a^6*b^9*c^11*d^4 + 9*a^7*b^8*c^10*d^5 + 9*a^10*b^5*c^7*d^8 - 9*a^11*b^4*c^6*d^9))/6 + 3*a^6*b^
6*c^6*d^6*x*(a^2*d^2 + b^2*c^2))*(b^5/(27*a^8*d^3 - 27*a^5*b^3*c^3 + 81*a^6*b^2*c^2*d - 81*a^7*b*c*d^2))^(1/3)
*(3^(1/2)*1i + 1))/2 + (log(((3^(1/2)*1i - 1)*(-d^5/(c^5*(a*d - b*c)^3))^(1/3)*(((3^(1/2)*1i - 1)^2*(81*a^8*b^
3*c^8*d^3*x*(a*d - b*c)^4*(a^2*d^2 + b^2*c^2 + a*b*c*d) - (81*a^10*b^3*c^10*d^3*(3^(1/2)*1i - 1)*(a*d + b*c)*(
a*d - b*c)^4*(-d^5/(c^5*(a*d - b*c)^3))^(1/3))/2)*(-d^5/(c^5*(a*d - b*c)^3))^(2/3))/36 - 9*a^6*b^9*c^11*d^4 +
9*a^7*b^8*c^10*d^5 + 9*a^10*b^5*c^7*d^8 - 9*a^11*b^4*c^6*d^9))/6 - 3*a^6*b^6*c^6*d^6*x*(a^2*d^2 + b^2*c^2))*(d
^5/(27*b^3*c^8 - 27*a^3*c^5*d^3 + 81*a^2*b*c^6*d^2 - 81*a*b^2*c^7*d))^(1/3)*(3^(1/2)*1i - 1))/2 - (log(((3^(1/
2)*1i + 1)*(-d^5/(c^5*(a*d - b*c)^3))^(1/3)*(((3^(1/2)*1i + 1)^2*(81*a^8*b^3*c^8*d^3*x*(a*d - b*c)^4*(a^2*d^2
+ b^2*c^2 + a*b*c*d) + (81*a^10*b^3*c^10*d^3*(3^(1/2)*1i + 1)*(a*d + b*c)*(a*d - b*c)^4*(-d^5/(c^5*(a*d - b*c)
^3))^(1/3))/2)*(-d^5/(c^5*(a*d - b*c)^3))^(2/3))/36 - 9*a^6*b^9*c^11*d^4 + 9*a^7*b^8*c^10*d^5 + 9*a^10*b^5*c^7
*d^8 - 9*a^11*b^4*c^6*d^9))/6 + 3*a^6*b^6*c^6*d^6*x*(a^2*d^2 + b^2*c^2))*(d^5/(27*b^3*c^8 - 27*a^3*c^5*d^3 + 8
1*a^2*b*c^6*d^2 - 81*a*b^2*c^7*d))^(1/3)*(3^(1/2)*1i + 1))/2 - 1/(2*a*c*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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