3.5.74 \(\int \frac {x^7}{(a+b x^3)^{2/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=279 \[ \frac {(2 a d+3 b c) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{5/3} d^2}+\frac {(2 a d+3 b c) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3} d^2}+\frac {c^{5/3} \log \left (c+d x^3\right )}{6 d^2 (b c-a d)^{2/3}}-\frac {c^{5/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2 (b c-a d)^{2/3}}-\frac {c^{5/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^2 (b c-a d)^{2/3}}+\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d} \]

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Rubi [A]  time = 0.50, antiderivative size = 400, normalized size of antiderivative = 1.43, number of steps used = 16, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {494, 470, 584, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {(2 a d+3 b c) \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3} d^2}-\frac {(2 a d+3 b c) \log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{18 b^{5/3} d^2}+\frac {(2 a d+3 b c) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3} d^2}-\frac {c^{5/3} \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{3 d^2 (b c-a d)^{2/3}}+\frac {c^{5/3} \log \left (\frac {x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{6 d^2 (b c-a d)^{2/3}}-\frac {c^{5/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} d^2 (b c-a d)^{2/3}}+\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

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

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 470

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

Rule 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

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 )^{2/3} \left (c+d x^3\right )} \, dx &=a^2 \operatorname {Subst}\left (\int \frac {x^7}{\left (1-b x^3\right )^2 \left (c-(b c-a d) x^3\right )} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )\\ &=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}-\frac {a \operatorname {Subst}\left (\int \frac {x \left (2 c+(b c+2 a d) x^3\right )}{\left (1-b x^3\right ) \left (c+(-b c+a d) x^3\right )} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 b d}\\ &=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}-\frac {a \operatorname {Subst}\left (\int \left (\frac {(3 b c+2 a d) x}{a d \left (1-b x^3\right )}+\frac {3 b c^2 x}{a d \left (-c+(b c-a d) x^3\right )}\right ) \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 b d}\\ &=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}-\frac {c^2 \operatorname {Subst}\left (\int \frac {x}{-c+(b c-a d) x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{d^2}-\frac {(3 b c+2 a d) \operatorname {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 b d^2}\\ &=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}-\frac {c^{5/3} \operatorname {Subst}\left (\int \frac {1}{-\sqrt [3]{c}+\sqrt [3]{b c-a d} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 d^2 \sqrt [3]{b c-a d}}+\frac {c^{5/3} \operatorname {Subst}\left (\int \frac {-\sqrt [3]{c}+\sqrt [3]{b c-a d} x}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 d^2 \sqrt [3]{b c-a d}}-\frac {(3 b c+2 a d) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt [3]{b} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{4/3} d^2}+\frac {(3 b c+2 a d) \operatorname {Subst}\left (\int \frac {1-\sqrt [3]{b} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{4/3} d^2}\\ &=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}+\frac {(3 b c+2 a d) \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3} d^2}-\frac {c^{5/3} \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 d^2 (b c-a d)^{2/3}}+\frac {c^{5/3} \operatorname {Subst}\left (\int \frac {\sqrt [3]{c} \sqrt [3]{b c-a d}+2 (b c-a d)^{2/3} x}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{6 d^2 (b c-a d)^{2/3}}-\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{2 d^2 \sqrt [3]{b c-a d}}-\frac {(3 b c+2 a d) \operatorname {Subst}\left (\int \frac {\sqrt [3]{b}+2 b^{2/3} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{18 b^{5/3} d^2}+\frac {(3 b c+2 a d) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{6 b^{4/3} d^2}\\ &=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}+\frac {(3 b c+2 a d) \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3} d^2}-\frac {(3 b c+2 a d) \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{18 b^{5/3} d^2}-\frac {c^{5/3} \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 d^2 (b c-a d)^{2/3}}+\frac {c^{5/3} \log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{6 d^2 (b c-a d)^{2/3}}+\frac {c^{5/3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{d^2 (b c-a d)^{2/3}}-\frac {(3 b c+2 a d) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{5/3} d^2}\\ &=\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d}+\frac {(3 b c+2 a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{5/3} d^2}-\frac {c^{5/3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2 (b c-a d)^{2/3}}+\frac {(3 b c+2 a d) \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{5/3} d^2}-\frac {(3 b c+2 a d) \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{18 b^{5/3} d^2}-\frac {c^{5/3} \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 d^2 (b c-a d)^{2/3}}+\frac {c^{5/3} \log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{6 d^2 (b c-a d)^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 190, normalized size = 0.68 \begin {gather*} \frac {5 c x^2 \left (\left (a+b x^3\right ) \left (\frac {d x^3}{c}+1\right )^{2/3}-a \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {(a d-b c) x^3}{a \left (d x^3+c\right )}\right )\right )-x^5 \left (\frac {b x^3}{a}+1\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3} (2 a d+3 b c) F_1\left (\frac {5}{3};\frac {2}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{15 b c d \left (a+b x^3\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^7/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

(-((3*b*c + 2*a*d)*x^5*(1 + (b*x^3)/a)^(2/3)*(1 + (d*x^3)/c)^(2/3)*AppellF1[5/3, 2/3, 1, 8/3, -((b*x^3)/a), -(
(d*x^3)/c)]) + 5*c*x^2*((a + b*x^3)*(1 + (d*x^3)/c)^(2/3) - a*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[2/3, 2/3
, 5/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))]))/(15*b*c*d*(a + b*x^3)^(2/3)*(1 + (d*x^3)/c)^(2/3))

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IntegrateAlgebraic [C]  time = 4.57, size = 527, normalized size = 1.89 \begin {gather*} \frac {(2 a d+3 b c) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{9 b^{5/3} d^2}+\frac {(2 a d+3 b c) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{2 \sqrt [3]{a+b x^3}+\sqrt [3]{b} x}\right )}{3 \sqrt {3} b^{5/3} d^2}+\frac {(-2 a d-3 b c) \log \left (\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}+b^{2/3} x^2\right )}{18 b^{5/3} d^2}+\frac {\left (c^{5/3}-i \sqrt {3} c^{5/3}\right ) \log \left (2 x \sqrt [3]{b c-a d}+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{6 d^2 (b c-a d)^{2/3}}+\frac {\sqrt {\frac {1}{6} \left (-1-i \sqrt {3}\right )} c^{5/3} \tan ^{-1}\left (\frac {3 x \sqrt [3]{b c-a d}}{\sqrt {3} x \sqrt [3]{b c-a d}-\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x^3}-3 i \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{d^2 (b c-a d)^{2/3}}+\frac {i \left (\sqrt {3} c^{5/3}+i c^{5/3}\right ) \log \left (\left (\sqrt {3}+i\right ) c^{2/3} \left (a+b x^3\right )^{2/3}+\sqrt [3]{c} \left (-\sqrt {3} x+i x\right ) \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}-2 i x^2 (b c-a d)^{2/3}\right )}{12 d^2 (b c-a d)^{2/3}}+\frac {x^2 \sqrt [3]{a+b x^3}}{3 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^7/((a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

(x^2*(a + b*x^3)^(1/3))/(3*b*d) + ((3*b*c + 2*a*d)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3)
)])/(3*Sqrt[3]*b^(5/3)*d^2) + (Sqrt[(-1 - I*Sqrt[3])/6]*c^(5/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c -
a*d)^(1/3)*x - (3*I)*c^(1/3)*(a + b*x^3)^(1/3) - Sqrt[3]*c^(1/3)*(a + b*x^3)^(1/3))])/(d^2*(b*c - a*d)^(2/3))
+ ((3*b*c + 2*a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(9*b^(5/3)*d^2) + ((c^(5/3) - I*Sqrt[3]*c^(5/3))*Lo
g[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(6*d^2*(b*c - a*d)^(2/3)) + ((-3*b*c - 2
*a*d)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(18*b^(5/3)*d^2) + ((I/12)*(I*c^(5/3
) + Sqrt[3]*c^(5/3))*Log[(-2*I)*(b*c - a*d)^(2/3)*x^2 + c^(1/3)*(b*c - a*d)^(1/3)*(I*x - Sqrt[3]*x)*(a + b*x^3
)^(1/3) + (I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(d^2*(b*c - a*d)^(2/3))

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fricas [B]  time = 1.28, size = 558, normalized size = 2.00 \begin {gather*} \frac {6 \, \sqrt {3} b^{3} c \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {2}{3}} + \sqrt {3} c x}{3 \, c x}\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} d x^{2} + 6 \, b^{3} c \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \log \left (\frac {{\left (b c - a d\right )} \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}} c}{x}\right ) - 3 \, b^{3} c \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \log \left (\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} c^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c^{2} - a c d\right )} \left (-\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x}{x^{2}}\right ) - 2 \, \sqrt {3} {\left (3 \, b^{2} c + 2 \, a b d\right )} {\left (b^{2}\right )}^{\frac {1}{6}} \arctan \left (\frac {{\left (\sqrt {3} {\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{3 \, b^{2} x}\right ) + 2 \, {\left (b^{2}\right )}^{\frac {2}{3}} {\left (3 \, b c + 2 \, a d\right )} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) - {\left (b^{2}\right )}^{\frac {2}{3}} {\left (3 \, b c + 2 \, a d\right )} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{3} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(x^7/((b*x^3 + a)^(2/3)*(d*x^3 + c)), x)

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maple [F]  time = 0.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x)

[Out]

int(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(x^7/((b*x^3 + a)^(2/3)*(d*x^3 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^7}{{\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/((a + b*x^3)^(2/3)*(c + d*x^3)),x)

[Out]

int(x^7/((a + b*x^3)^(2/3)*(c + d*x^3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{\left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(b*x**3+a)**(2/3)/(d*x**3+c),x)

[Out]

Integral(x**7/((a + b*x**3)**(2/3)*(c + d*x**3)), x)

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