3.2.8 \(\int \frac {x^7}{(a+b x^3) (c+d x^3)} \, dx\)

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

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Rubi [A]  time = 0.31, antiderivative size = 301, 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 = {479, 584, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {a^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3} (b c-a d)}-\frac {a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} (b c-a d)}-\frac {a^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} (b c-a d)}-\frac {c^{5/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 d^{5/3} (b c-a d)}+\frac {c^{5/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 d^{5/3} (b c-a d)}+\frac {c^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} d^{5/3} (b c-a d)}+\frac {x^2}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/((a + b*x^3)*(c + d*x^3)),x]

[Out]

x^2/(2*b*d) - (a^(5/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(5/3)*(b*c - a*d)) + (c^(
5/3)*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/(Sqrt[3]*d^(5/3)*(b*c - a*d)) - (a^(5/3)*Log[a^(1/3) +
b^(1/3)*x])/(3*b^(5/3)*(b*c - a*d)) + (c^(5/3)*Log[c^(1/3) + d^(1/3)*x])/(3*d^(5/3)*(b*c - a*d)) + (a^(5/3)*L
og[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*b^(5/3)*(b*c - a*d)) - (c^(5/3)*Log[c^(2/3) - c^(1/3)*d^(1/3
)*x + d^(2/3)*x^2])/(6*d^(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 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 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 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 {x^7}{\left (a+b x^3\right ) \left (c+d x^3\right )} \, dx &=\frac {x^2}{2 b d}-\frac {\int \frac {x \left (2 a c+2 (b c+a d) x^3\right )}{\left (a+b x^3\right ) \left (c+d x^3\right )} \, dx}{2 b d}\\ &=\frac {x^2}{2 b d}-\frac {\int \left (\frac {2 a^2 d x}{(-b c+a d) \left (a+b x^3\right )}+\frac {2 b c^2 x}{(b c-a d) \left (c+d x^3\right )}\right ) \, dx}{2 b d}\\ &=\frac {x^2}{2 b d}+\frac {a^2 \int \frac {x}{a+b x^3} \, dx}{b (b c-a d)}-\frac {c^2 \int \frac {x}{c+d x^3} \, dx}{d (b c-a d)}\\ &=\frac {x^2}{2 b d}-\frac {a^{5/3} \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 b^{4/3} (b c-a d)}+\frac {a^{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 b^{4/3} (b c-a d)}+\frac {c^{5/3} \int \frac {1}{\sqrt [3]{c}+\sqrt [3]{d} x} \, dx}{3 d^{4/3} (b c-a d)}-\frac {c^{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 d^{4/3} (b c-a d)}\\ &=\frac {x^2}{2 b d}-\frac {a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} (b c-a d)}+\frac {c^{5/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 d^{5/3} (b c-a d)}+\frac {a^{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 b^{5/3} (b c-a d)}+\frac {a^2 \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 b^{4/3} (b c-a d)}-\frac {c^{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 d^{5/3} (b c-a d)}-\frac {c^2 \int \frac {1}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{2 d^{4/3} (b c-a d)}\\ &=\frac {x^2}{2 b d}-\frac {a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} (b c-a d)}+\frac {c^{5/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 d^{5/3} (b c-a d)}+\frac {a^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3} (b c-a d)}-\frac {c^{5/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 d^{5/3} (b c-a d)}+\frac {a^{5/3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{b^{5/3} (b c-a d)}-\frac {c^{5/3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}\right )}{d^{5/3} (b c-a d)}\\ &=\frac {x^2}{2 b d}-\frac {a^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} (b c-a d)}+\frac {c^{5/3} \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} d^{5/3} (b c-a d)}-\frac {a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} (b c-a d)}+\frac {c^{5/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 d^{5/3} (b c-a d)}+\frac {a^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3} (b c-a d)}-\frac {c^{5/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 d^{5/3} (b c-a d)}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 242, normalized size = 0.80 \begin {gather*} \frac {\frac {a^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{5/3}}-\frac {2 a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{5/3}}-\frac {2 \sqrt {3} a^{5/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{5/3}}-\frac {3 a x^2}{b}-\frac {c^{5/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{d^{5/3}}+\frac {2 c^{5/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{d^{5/3}}+\frac {2 \sqrt {3} c^{5/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )}{d^{5/3}}+\frac {3 c x^2}{d}}{6 b c-6 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7/((a + b*x^3)*(c + d*x^3)),x]

[Out]

((-3*a*x^2)/b + (3*c*x^2)/d - (2*Sqrt[3]*a^(5/3)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(5/3) + (2*Sqr
t[3]*c^(5/3)*ArcTan[(1 - (2*d^(1/3)*x)/c^(1/3))/Sqrt[3]])/d^(5/3) - (2*a^(5/3)*Log[a^(1/3) + b^(1/3)*x])/b^(5/
3) + (2*c^(5/3)*Log[c^(1/3) + d^(1/3)*x])/d^(5/3) + (a^(5/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b
^(5/3) - (c^(5/3)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/d^(5/3))/(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^7}{\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^7/((a + b*x^3)*(c + d*x^3)),x]

[Out]

IntegrateAlgebraic[x^7/((a + b*x^3)*(c + d*x^3)), x]

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fricas [A]  time = 1.36, size = 273, normalized size = 0.91 \begin {gather*} \frac {2 \, \sqrt {3} a d \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) - 2 \, \sqrt {3} b c \left (-\frac {c^{2}}{d^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} d x \left (-\frac {c^{2}}{d^{2}}\right )^{\frac {1}{3}} + \sqrt {3} c}{3 \, c}\right ) + a d \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + b c \left (-\frac {c^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left (c x^{2} - d x \left (-\frac {c^{2}}{d^{2}}\right )^{\frac {2}{3}} - c \left (-\frac {c^{2}}{d^{2}}\right )^{\frac {1}{3}}\right ) - 2 \, a d \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right ) - 2 \, b c \left (-\frac {c^{2}}{d^{2}}\right )^{\frac {1}{3}} \log \left (c x + d \left (-\frac {c^{2}}{d^{2}}\right )^{\frac {2}{3}}\right ) + 3 \, {\left (b c - a d\right )} x^{2}}{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^7/(b*x^3+a)/(d*x^3+c),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 311, normalized size = 1.03 \begin {gather*} -\frac {a^{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 b^{2} c - a^{2} b d\right )}} + \frac {c^{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^{2} d - a c d^{2}\right )}} - \frac {\left (-a b^{2}\right )^{\frac {2}{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^{4} c - \sqrt {3} a b^{3} d} + \frac {\left (-c d^{2}\right )^{\frac {2}{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^{3} - \sqrt {3} a d^{4}} + \frac {\left (-a b^{2}\right )^{\frac {2}{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^{4} c - a b^{3} d\right )}} - \frac {\left (-c d^{2}\right )^{\frac {2}{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^{3} - a d^{4}\right )}} + \frac {x^{2}}{2 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)/(d*x^3+c),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 269, normalized size = 0.89 \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 {1}{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 {1}{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 {1}{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 {1}{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 {1}{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 {1}{3}} d^{2}}+\frac {x^{2}}{2 b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^3+a)/(d*x^3+c),x)

[Out]

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

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maxima [A]  time = 1.27, size = 324, normalized size = 1.08 \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 - a b^{2} d\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} - a d^{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 {1}{3}} - a b^{2} d \left (\frac {a}{b}\right )^{\frac {1}{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 {1}{3}} - a d^{3} \left (\frac {c}{d}\right )^{\frac {1}{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 {1}{3}} - a b^{2} d \left (\frac {a}{b}\right )^{\frac {1}{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 {1}{3}} - a d^{3} \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}} + \frac {x^{2}}{2 \, b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(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^2*d)*(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 - a*d^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)^(1/3) - a*b^2*d*(a/b)^(1/3)) - 1/6*c^2*log(x^2 - x*(c/d)^(1/3) + (
c/d)^(2/3))/(b*c*d^2*(c/d)^(1/3) - a*d^3*(c/d)^(1/3)) - 1/3*a^2*log(x + (a/b)^(1/3))/(b^3*c*(a/b)^(1/3) - a*b^
2*d*(a/b)^(1/3)) + 1/3*c^2*log(x + (c/d)^(1/3))/(b*c*d^2*(c/d)^(1/3) - a*d^3*(c/d)^(1/3)) + 1/2*x^2/(b*d)

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mupad [B]  time = 11.36, size = 1751, normalized size = 5.82

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/((a + b*x^3)*(c + d*x^3)),x)

[Out]

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

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

[Out]

Timed out

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