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

Optimal. Leaf size=299 \[ -\frac {b^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} (b c-a d)}+\frac {b^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} (b c-a d)}+\frac {b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3} (b c-a d)}+\frac {d^{4/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{4/3} (b c-a d)}-\frac {d^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{4/3} (b c-a d)}-\frac {d^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{4/3} (b c-a d)}-\frac {1}{a c x} \]

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Rubi [A]  time = 0.27, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {480, 584, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {b^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} (b c-a d)}+\frac {b^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} (b c-a d)}+\frac {b^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3} (b c-a d)}+\frac {d^{4/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{4/3} (b c-a d)}-\frac {d^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{4/3} (b c-a d)}-\frac {d^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{4/3} (b c-a d)}-\frac {1}{a c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(a*c*x)) + (b^(4/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(4/3)*(b*c - a*d)) - (d^
(4/3)*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/(Sqrt[3]*c^(4/3)*(b*c - a*d)) + (b^(4/3)*Log[a^(1/3)
+ b^(1/3)*x])/(3*a^(4/3)*(b*c - a*d)) - (d^(4/3)*Log[c^(1/3) + d^(1/3)*x])/(3*c^(4/3)*(b*c - a*d)) - (b^(4/3)*
Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(4/3)*(b*c - a*d)) + (d^(4/3)*Log[c^(2/3) - c^(1/3)*d^(1/
3)*x + d^(2/3)*x^2])/(6*c^(4/3)*(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 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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 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 584

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[((g*x)^m*(a + b*x^n)^p*(e + f*x^n))/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

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

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Mathematica [A]  time = 0.21, size = 244, normalized size = 0.82 \begin {gather*} \frac {\frac {b^{4/3} x \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{4/3}}-\frac {2 b^{4/3} x \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{4/3}}-\frac {2 \sqrt {3} b^{4/3} x \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{4/3}}+\frac {6 b}{a}-\frac {d^{4/3} x \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{c^{4/3}}+\frac {2 d^{4/3} x \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{c^{4/3}}+\frac {2 \sqrt {3} d^{4/3} x \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )}{c^{4/3}}-\frac {6 d}{c}}{6 a d x-6 b c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

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fricas [A]  time = 0.89, size = 238, normalized size = 0.80 \begin {gather*} -\frac {2 \, \sqrt {3} b c x \left (-\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - 2 \, \sqrt {3} a d x \left (\frac {d}{c}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (\frac {d}{c}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) - b c x \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (-\frac {b}{a}\right )^{\frac {2}{3}} - a \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right ) - a d x \left (\frac {d}{c}\right )^{\frac {1}{3}} \log \left (d x^{2} - c x \left (\frac {d}{c}\right )^{\frac {2}{3}} + c \left (\frac {d}{c}\right )^{\frac {1}{3}}\right ) + 2 \, b c x \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x + a \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 2 \, a d x \left (\frac {d}{c}\right )^{\frac {1}{3}} \log \left (d x + c \left (\frac {d}{c}\right )^{\frac {2}{3}}\right ) + 6 \, b c - 6 \, a d}{6 \, {\left (a b c^{2} - a^{2} c d\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 305, normalized size = 1.02 \begin {gather*} \frac {b^{2} \left (-\frac {a}{b}\right )^{\frac {2}{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 {2}{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 {2}{3}} \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 {2}{3}} \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 {2}{3}} \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 {2}{3}} \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}{a c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 257, normalized size = 0.86 \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 {1}{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 {1}{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 {1}{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 {1}{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 {1}{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 {1}{3}} c}-\frac {1}{a c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.20, size = 300, normalized size = 1.00 \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 - a^{2} d\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} - a c d\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 {1}{3}} - a^{2} d \left (\frac {a}{b}\right )^{\frac {1}{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 {1}{3}} - a c d \left (\frac {c}{d}\right )^{\frac {1}{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 {1}{3}} - a^{2} d \left (\frac {a}{b}\right )^{\frac {1}{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 {1}{3}} - a c d \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}} - \frac {1}{a c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 3.85, size = 716, normalized size = 2.39 \begin {gather*} \ln \left (b-a^2\,d\,x\,{\left (-\frac {b^4}{a^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}+a\,b\,c\,x\,{\left (-\frac {b^4}{a^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}\right )\,{\left (-\frac {b^4}{27\,a^7\,d^3-81\,a^6\,b\,c\,d^2+81\,a^5\,b^2\,c^2\,d-27\,a^4\,b^3\,c^3}\right )}^{1/3}+\ln \left (d-b\,c^2\,x\,{\left (\frac {d^4}{c^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}+a\,c\,d\,x\,{\left (\frac {d^4}{c^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}\right )\,{\left (-\frac {d^4}{-27\,a^3\,c^4\,d^3+81\,a^2\,b\,c^5\,d^2-81\,a\,b^2\,c^6\,d+27\,b^3\,c^7}\right )}^{1/3}-\frac {1}{a\,c\,x}-\frac {\ln \left (b+2\,a^2\,d\,x\,{\left (-\frac {b^4}{a^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}-2\,a\,b\,c\,x\,{\left (-\frac {b^4}{a^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}-\sqrt {3}\,b\,1{}\mathrm {i}\right )\,{\left (-\frac {b^4}{27\,a^7\,d^3-81\,a^6\,b\,c\,d^2+81\,a^5\,b^2\,c^2\,d-27\,a^4\,b^3\,c^3}\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}+\frac {\ln \left (b+2\,a^2\,d\,x\,{\left (-\frac {b^4}{a^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}-2\,a\,b\,c\,x\,{\left (-\frac {b^4}{a^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}+\sqrt {3}\,b\,1{}\mathrm {i}\right )\,{\left (-\frac {b^4}{27\,a^7\,d^3-81\,a^6\,b\,c\,d^2+81\,a^5\,b^2\,c^2\,d-27\,a^4\,b^3\,c^3}\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}-\frac {\ln \left (d+2\,b\,c^2\,x\,{\left (\frac {d^4}{c^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}-2\,a\,c\,d\,x\,{\left (\frac {d^4}{c^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}-\sqrt {3}\,d\,1{}\mathrm {i}\right )\,{\left (-\frac {d^4}{-27\,a^3\,c^4\,d^3+81\,a^2\,b\,c^5\,d^2-81\,a\,b^2\,c^6\,d+27\,b^3\,c^7}\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}+\frac {\ln \left (d+2\,b\,c^2\,x\,{\left (\frac {d^4}{c^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}-2\,a\,c\,d\,x\,{\left (\frac {d^4}{c^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}+\sqrt {3}\,d\,1{}\mathrm {i}\right )\,{\left (-\frac {d^4}{-27\,a^3\,c^4\,d^3+81\,a^2\,b\,c^5\,d^2-81\,a\,b^2\,c^6\,d+27\,b^3\,c^7}\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b - a^2*d*x*(-b^4/(a^4*(a*d - b*c)^3))^(1/3) + a*b*c*x*(-b^4/(a^4*(a*d - b*c)^3))^(1/3))*(-b^4/(27*a^7*d^3
- 27*a^4*b^3*c^3 + 81*a^5*b^2*c^2*d - 81*a^6*b*c*d^2))^(1/3) + log(d - b*c^2*x*(d^4/(c^4*(a*d - b*c)^3))^(1/3
) + a*c*d*x*(d^4/(c^4*(a*d - b*c)^3))^(1/3))*(-d^4/(27*b^3*c^7 - 27*a^3*c^4*d^3 + 81*a^2*b*c^5*d^2 - 81*a*b^2*
c^6*d))^(1/3) - 1/(a*c*x) - (log(b - 3^(1/2)*b*1i + 2*a^2*d*x*(-b^4/(a^4*(a*d - b*c)^3))^(1/3) - 2*a*b*c*x*(-b
^4/(a^4*(a*d - b*c)^3))^(1/3))*(-b^4/(27*a^7*d^3 - 27*a^4*b^3*c^3 + 81*a^5*b^2*c^2*d - 81*a^6*b*c*d^2))^(1/3)*
(3^(1/2)*1i + 1))/2 + (log(b + 3^(1/2)*b*1i + 2*a^2*d*x*(-b^4/(a^4*(a*d - b*c)^3))^(1/3) - 2*a*b*c*x*(-b^4/(a^
4*(a*d - b*c)^3))^(1/3))*(-b^4/(27*a^7*d^3 - 27*a^4*b^3*c^3 + 81*a^5*b^2*c^2*d - 81*a^6*b*c*d^2))^(1/3)*(3^(1/
2)*1i - 1))/2 - (log(d - 3^(1/2)*d*1i + 2*b*c^2*x*(d^4/(c^4*(a*d - b*c)^3))^(1/3) - 2*a*c*d*x*(d^4/(c^4*(a*d -
b*c)^3))^(1/3))*(-d^4/(27*b^3*c^7 - 27*a^3*c^4*d^3 + 81*a^2*b*c^5*d^2 - 81*a*b^2*c^6*d))^(1/3)*(3^(1/2)*1i +
1))/2 + (log(d + 3^(1/2)*d*1i + 2*b*c^2*x*(d^4/(c^4*(a*d - b*c)^3))^(1/3) - 2*a*c*d*x*(d^4/(c^4*(a*d - b*c)^3)
)^(1/3))*(-d^4/(27*b^3*c^7 - 27*a^3*c^4*d^3 + 81*a^2*b*c^5*d^2 - 81*a*b^2*c^6*d))^(1/3)*(3^(1/2)*1i - 1))/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**2/(b*x**3+a)/(d*x**3+c),x)

[Out]

Timed out

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