3.721 \(\int \frac{x^6}{\sqrt [3]{a+b x^3} (c+d x^3)} \, dx\)

Optimal. Leaf size=273 \[ \frac{(a d+3 b c) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{4/3} d^2}-\frac{(a d+3 b c) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3} d^2}+\frac{c^{4/3} \log \left (c+d x^3\right )}{6 d^2 \sqrt [3]{b c-a d}}-\frac{c^{4/3} \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2 \sqrt [3]{b c-a d}}+\frac{c^{4/3} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} d^2 \sqrt [3]{b c-a d}}+\frac{x \left (a+b x^3\right )^{2/3}}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.509068, antiderivative size = 394, normalized size of antiderivative = 1.44, number of steps used = 15, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {494, 470, 522, 200, 31, 634, 617, 204, 628} \[ \frac{(a d+3 b c) \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{9 b^{4/3} d^2}-\frac{(a d+3 b c) \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{18 b^{4/3} d^2}-\frac{(a d+3 b c) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3} d^2}-\frac{c^{4/3} \log \left (\sqrt [3]{c}-\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{3 d^2 \sqrt [3]{b c-a d}}+\frac{c^{4/3} \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{6 d^2 \sqrt [3]{b c-a d}}+\frac{c^{4/3} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt{3} \sqrt [3]{c}}\right )}{\sqrt{3} d^2 \sqrt [3]{b c-a d}}+\frac{x \left (a+b x^3\right )^{2/3}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^6/((a + b*x^3)^(1/3)*(c + d*x^3)),x]

[Out]

(x*(a + b*x^3)^(2/3))/(3*b*d) - ((3*b*c + a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[
3]*b^(4/3)*d^2) + (c^(4/3)*ArcTan[(c^(1/3) + (2*(b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3))/(Sqrt[3]*c^(1/3))])/(S
qrt[3]*d^2*(b*c - a*d)^(1/3)) + ((3*b*c + a*d)*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(9*b^(4/3)*d^2) - ((3*b
*c + 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^(4/3)*d^2) - (c^(4/3
)*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3)])/(3*d^2*(b*c - a*d)^(1/3)) + (c^(4/3)*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
)^(1/3))

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 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 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 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 31

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

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]

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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.521259, size = 288, normalized size = 1.05 \[ \frac{\frac{2 \left (-a \sqrt [3]{c} \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a x^3+b\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}+c^{2/3}\right )+6 x \left (a+b x^3\right )^{2/3} \sqrt [3]{b c-a d}+2 a \sqrt [3]{c} \log \left (\sqrt [3]{c}-\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{a x^3+b}}\right )-2 \sqrt{3} a \sqrt [3]{c} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a x^3+b}}+1}{\sqrt{3}}\right )\right )}{\sqrt [3]{b c-a d}}-\frac{3 x^4 \sqrt [3]{\frac{b x^3}{a}+1} (a d+3 b c) F_1\left (\frac{4}{3};\frac{1}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c \sqrt [3]{a+b x^3}}}{36 b d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^6/((a + b*x^3)^(1/3)*(c + d*x^3)),x]

[Out]

((-3*(3*b*c + a*d)*x^4*(1 + (b*x^3)/a)^(1/3)*AppellF1[4/3, 1/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(c*(a + b
*x^3)^(1/3)) + (2*(6*(b*c - a*d)^(1/3)*x*(a + b*x^3)^(2/3) - 2*Sqrt[3]*a*c^(1/3)*ArcTan[(1 + (2*(b*c - a*d)^(1
/3)*x)/(c^(1/3)*(b + a*x^3)^(1/3)))/Sqrt[3]] + 2*a*c^(1/3)*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(b + a*x^3)^(1/
3)] - a*c^(1/3)*Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(b + a*x^3)^(2/3) + (c^(1/3)*(b*c - a*d)^(1/3)*x)/(b + a
*x^3)^(1/3)]))/(b*c - a*d)^(1/3))/(36*b*d)

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Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{6}}{d{x}^{3}+c}{\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x)

[Out]

int(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(x^6/((b*x^3 + a)^(1/3)*(d*x^3 + c)), x)

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Fricas [A]  time = 2.02761, size = 2026, normalized size = 7.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

[-1/18*(6*sqrt(3)*b^2*c*(-c/(b*c - a*d))^(1/3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-c/(b*c -
a*d))^(1/3))/x) - 6*b^2*c*(-c/(b*c - a*d))^(1/3)*log(-((b*c - a*d)*x*(-c/(b*c - a*d))^(2/3) - (b*x^3 + a)^(1/
3)*c)/x) + 3*b^2*c*(-c/(b*c - a*d))^(1/3)*log(-((b*c - a*d)*x^2*(-c/(b*c - a*d))^(1/3) - (b*x^3 + a)^(1/3)*(b*
c - a*d)*x*(-c/(b*c - a*d))^(2/3) - (b*x^3 + a)^(2/3)*c)/x^2) - 6*(b*x^3 + a)^(2/3)*b*d*x - 3*sqrt(1/3)*(3*b^2
*c + a*b*d)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3)*x^2 - 3*sqrt(1/3)*(b^(4/3)*x^3 + (b*x^3
+ a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^(2/3)*x)*sqrt(-1/b^(2/3)) + 2*a) - 2*(3*b*c + a*d)*b^(2/3)*log(-(b^(
1/3)*x - (b*x^3 + a)^(1/3))/x) + (3*b*c + a*d)*b^(2/3)*log((b^(2/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3
+ a)^(2/3))/x^2))/(b^2*d^2), -1/18*(6*sqrt(3)*b^2*c*(-c/(b*c - a*d))^(1/3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)
*(b*x^3 + a)^(1/3)*(-c/(b*c - a*d))^(1/3))/x) - 6*b^2*c*(-c/(b*c - a*d))^(1/3)*log(-((b*c - a*d)*x*(-c/(b*c -
a*d))^(2/3) - (b*x^3 + a)^(1/3)*c)/x) + 3*b^2*c*(-c/(b*c - a*d))^(1/3)*log(-((b*c - a*d)*x^2*(-c/(b*c - a*d))^
(1/3) - (b*x^3 + a)^(1/3)*(b*c - a*d)*x*(-c/(b*c - a*d))^(2/3) - (b*x^3 + a)^(2/3)*c)/x^2) - 6*(b*x^3 + a)^(2/
3)*b*d*x - 2*(3*b*c + a*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) + (3*b*c + a*d)*b^(2/3)*log((b^(2/3
)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) - 6*sqrt(1/3)*(3*b^2*c + a*b*d)*arctan(sqrt(1/3)
*(b^(1/3)*x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3))/(b^2*d^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**3+a)**(1/3)/(d*x**3+c),x)

[Out]

Integral(x**6/((a + b*x**3)**(1/3)*(c + d*x**3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(x^6/((b*x^3 + a)^(1/3)*(d*x^3 + c)), x)