3.2.9 \(\int \frac {x^6}{(a+b x^3) (c+d x^3)} \, dx\)

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

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Rubi [A]  time = 0.27, antiderivative size = 296, 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 = {479, 522, 200, 31, 634, 617, 204, 628} \begin {gather*} -\frac {a^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{4/3} (b c-a d)}+\frac {a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{4/3} (b c-a d)}-\frac {a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{4/3} (b c-a d)}+\frac {c^{4/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 d^{4/3} (b c-a d)}-\frac {c^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 d^{4/3} (b c-a d)}+\frac {c^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} d^{4/3} (b c-a d)}+\frac {x}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/((a + b*x^3)*(c + d*x^3)),x]

[Out]

x/(b*d) - (a^(4/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(4/3)*(b*c - a*d)) + (c^(4/3)
*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/(Sqrt[3]*d^(4/3)*(b*c - a*d)) + (a^(4/3)*Log[a^(1/3) + b^(
1/3)*x])/(3*b^(4/3)*(b*c - a*d)) - (c^(4/3)*Log[c^(1/3) + d^(1/3)*x])/(3*d^(4/3)*(b*c - a*d)) - (a^(4/3)*Log[a
^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(4/3)*(b*c - a*d)) + (c^(4/3)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x
+ d^(2/3)*x^2])/(6*d^(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 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 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[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 {x^6}{\left (a+b x^3\right ) \left (c+d x^3\right )} \, dx &=\frac {x}{b d}-\frac {\int \frac {a c+(b c+a d) x^3}{\left (a+b x^3\right ) \left (c+d x^3\right )} \, dx}{b d}\\ &=\frac {x}{b d}+\frac {a^2 \int \frac {1}{a+b x^3} \, dx}{b (b c-a d)}-\frac {c^2 \int \frac {1}{c+d x^3} \, dx}{d (b c-a d)}\\ &=\frac {x}{b d}+\frac {a^{4/3} \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b (b c-a d)}+\frac {a^{4/3} \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 b (b c-a d)}-\frac {c^{4/3} \int \frac {1}{\sqrt [3]{c}+\sqrt [3]{d} x} \, dx}{3 d (b c-a d)}-\frac {c^{4/3} \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 d (b c-a d)}\\ &=\frac {x}{b d}+\frac {a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{4/3} (b c-a d)}-\frac {c^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 d^{4/3} (b c-a d)}-\frac {a^{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 b^{4/3} (b c-a d)}+\frac {a^{5/3} \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 b (b c-a d)}+\frac {c^{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 d^{4/3} (b c-a d)}-\frac {c^{5/3} \int \frac {1}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{2 d (b c-a d)}\\ &=\frac {x}{b d}+\frac {a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{4/3} (b c-a d)}-\frac {c^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 d^{4/3} (b c-a d)}-\frac {a^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{4/3} (b c-a d)}+\frac {c^{4/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 d^{4/3} (b c-a d)}+\frac {a^{4/3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{b^{4/3} (b c-a d)}-\frac {c^{4/3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}\right )}{d^{4/3} (b c-a d)}\\ &=\frac {x}{b d}-\frac {a^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{4/3} (b c-a d)}+\frac {c^{4/3} \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} d^{4/3} (b c-a d)}+\frac {a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{4/3} (b c-a d)}-\frac {c^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 d^{4/3} (b c-a d)}-\frac {a^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{4/3} (b c-a d)}+\frac {c^{4/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 d^{4/3} (b c-a d)}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 238, normalized size = 0.80 \begin {gather*} \frac {-\frac {a^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{4/3}}+\frac {2 a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{4/3}}-\frac {2 \sqrt {3} a^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{4/3}}-\frac {6 a x}{b}+\frac {c^{4/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{d^{4/3}}-\frac {2 c^{4/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{d^{4/3}}+\frac {2 \sqrt {3} c^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )}{d^{4/3}}+\frac {6 c x}{d}}{6 b c-6 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/((a + b*x^3)*(c + d*x^3)),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^6}{\left (a+b x^3\right ) \left (c+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^6/((a + b*x^3)*(c + d*x^3)),x]

[Out]

IntegrateAlgebraic[x^6/((a + b*x^3)*(c + d*x^3)), x]

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fricas [A]  time = 0.94, size = 228, normalized size = 0.77 \begin {gather*} -\frac {2 \, \sqrt {3} a d \left (-\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) + 2 \, \sqrt {3} b c \left (\frac {c}{d}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} d x \left (\frac {c}{d}\right )^{\frac {2}{3}} - \sqrt {3} c}{3 \, c}\right ) - a d \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) - b c \left (\frac {c}{d}\right )^{\frac {1}{3}} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right ) + 2 \, a d \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x - \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right ) + 2 \, b c \left (\frac {c}{d}\right )^{\frac {1}{3}} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right ) - 6 \, {\left (b c - a d\right )} x}{6 \, {\left (b^{2} c d - a b d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)/(d*x^3+c),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 308, normalized size = 1.04 \begin {gather*} -\frac {a^{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 b^{2} c - a^{2} b d\right )}} + \frac {c^{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^{2} d - a c d^{2}\right )}} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} a \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} b^{3} c - \sqrt {3} a b^{2} d} - \frac {\left (-c d^{2}\right )^{\frac {1}{3}} c \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 d^{2} - \sqrt {3} a d^{3}} + \frac {\left (-a b^{2}\right )^{\frac {1}{3}} a \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b^{3} c - a b^{2} d\right )}} - \frac {\left (-c d^{2}\right )^{\frac {1}{3}} c \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 d^{2} - a d^{3}\right )}} + \frac {x}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)/(d*x^3+c),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 266, normalized size = 0.90 \begin {gather*} -\frac {\sqrt {3}\, a^{2} \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}} b^{2}}-\frac {a^{2} \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}} b^{2}}+\frac {a^{2} \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}} b^{2}}+\frac {\sqrt {3}\, c^{2} \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}} d^{2}}+\frac {c^{2} \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}} d^{2}}-\frac {c^{2} \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}} d^{2}}+\frac {x}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^3+a)/(d*x^3+c),x)

[Out]

x/b/d-1/3/b^2*a^2/(a*d-b*c)/(a/b)^(2/3)*ln(x+(a/b)^(1/3))+1/6/b^2*a^2/(a*d-b*c)/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)
*x+(a/b)^(2/3))-1/3/b^2*a^2/(a*d-b*c)/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))+1/3/d^2*c^2/
(a*d-b*c)/(c/d)^(2/3)*ln(x+(c/d)^(1/3))-1/6/d^2*c^2/(a*d-b*c)/(c/d)^(2/3)*ln(x^2-(c/d)^(1/3)*x+(c/d)^(2/3))+1/
3/d^2*c^2/(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 = 1.35, size = 349, normalized size = 1.18 \begin {gather*} \frac {\sqrt {3} a^{2} \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 (b^{3} c \left (\frac {a}{b}\right )^{\frac {1}{3}} - a b^{2} d \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {\sqrt {3} c^{2} \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 d^{2} \left (\frac {c}{d}\right )^{\frac {1}{3}} - a d^{3} \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )} \left (\frac {c}{d}\right )^{\frac {1}{3}}} - \frac {a^{2} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b^{3} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}} + \frac {c^{2} \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 d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}} - a d^{3} \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}} + \frac {a^{2} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, {\left (b^{3} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b^{2} d \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}} - \frac {c^{2} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \, {\left (b c d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}} - a d^{3} \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}} + \frac {x}{b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)/(d*x^3+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.83, size = 873, normalized size = 2.95 \begin {gather*} \ln \left (a\,x+b^2\,c\,{\left (-\frac {a^4}{b^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}-a\,b\,d\,{\left (-\frac {a^4}{b^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}\right )\,{\left (\frac {a^4}{-27\,a^3\,b^4\,d^3+81\,a^2\,b^5\,c\,d^2-81\,a\,b^6\,c^2\,d+27\,b^7\,c^3}\right )}^{1/3}+\ln \left (c\,x+a\,d^2\,{\left (\frac {c^4}{d^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}-b\,c\,d\,{\left (\frac {c^4}{d^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}\right )\,{\left (\frac {c^4}{27\,a^3\,d^7-81\,a^2\,b\,c\,d^6+81\,a\,b^2\,c^2\,d^5-27\,b^3\,c^3\,d^4}\right )}^{1/3}+\frac {x}{b\,d}+\frac {\ln \left (\frac {3\,x\,\left (a^6\,c^2\,d^4+a^2\,b^4\,c^6\right )}{b\,d}-\frac {3\,a\,c^2\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {a^4}{b^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}\,\left (a^5\,d^5-a^4\,b\,c\,d^4+a\,b^4\,c^4\,d-b^5\,c^5\right )}{2\,d}\right )\,{\left (\frac {a^4}{-27\,a^3\,b^4\,d^3+81\,a^2\,b^5\,c\,d^2-81\,a\,b^6\,c^2\,d+27\,b^7\,c^3}\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}-\frac {\ln \left (\frac {3\,x\,\left (a^6\,c^2\,d^4+a^2\,b^4\,c^6\right )}{b\,d}+\frac {3\,a\,c^2\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {a^4}{b^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}\,\left (a^5\,d^5-a^4\,b\,c\,d^4+a\,b^4\,c^4\,d-b^5\,c^5\right )}{2\,d}\right )\,{\left (\frac {a^4}{-27\,a^3\,b^4\,d^3+81\,a^2\,b^5\,c\,d^2-81\,a\,b^6\,c^2\,d+27\,b^7\,c^3}\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}+\frac {\ln \left (\frac {3\,x\,\left (a^6\,c^2\,d^4+a^2\,b^4\,c^6\right )}{b\,d}+\frac {3\,a^2\,c\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {c^4}{d^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}\,\left (a^5\,d^5-a^4\,b\,c\,d^4+a\,b^4\,c^4\,d-b^5\,c^5\right )}{2\,b}\right )\,{\left (\frac {c^4}{27\,a^3\,d^7-81\,a^2\,b\,c\,d^6+81\,a\,b^2\,c^2\,d^5-27\,b^3\,c^3\,d^4}\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}-\frac {\ln \left (\frac {3\,x\,\left (a^6\,c^2\,d^4+a^2\,b^4\,c^6\right )}{b\,d}-\frac {3\,a^2\,c\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {c^4}{d^4\,{\left (a\,d-b\,c\right )}^3}\right )}^{1/3}\,\left (a^5\,d^5-a^4\,b\,c\,d^4+a\,b^4\,c^4\,d-b^5\,c^5\right )}{2\,b}\right )\,{\left (\frac {c^4}{27\,a^3\,d^7-81\,a^2\,b\,c\,d^6+81\,a\,b^2\,c^2\,d^5-27\,b^3\,c^3\,d^4}\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(x^6/((a + b*x^3)*(c + d*x^3)),x)

[Out]

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

[Out]

Timed out

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