3.96 \(\int \frac{x^2}{(1-x^3) \sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=96 \[ \frac{\log \left (1-x^3\right )}{6 \sqrt [3]{a+b}}-\frac{\log \left (\sqrt [3]{a+b}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a+b}}-\frac{\tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a+b}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{a+b}} \]

[Out]

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

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Rubi [A]  time = 0.0781431, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {444, 55, 617, 204, 31} \[ \frac{\log \left (1-x^3\right )}{6 \sqrt [3]{a+b}}-\frac{\log \left (\sqrt [3]{a+b}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a+b}}-\frac{\tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a+b}}+1}{\sqrt{3}}\right )}{\sqrt{3} \sqrt [3]{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 444

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

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 31

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

Rubi steps

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

Mathematica [A]  time = 0.0808377, size = 80, normalized size = 0.83 \[ \frac{-3 \log \left (\sqrt [3]{a+b}-\sqrt [3]{a+b x^3}\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a+b}}+1}{\sqrt{3}}\right )+\log \left (1-x^3\right )}{6 \sqrt [3]{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.044, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{-{x}^{3}+1}{\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^3+1)/(b*x^3+a)^(1/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.35095, size = 983, normalized size = 10.24 \begin{align*} \left [\frac{3 \, \sqrt{\frac{1}{3}}{\left (a + b\right )} \sqrt{\frac{{\left (-a - b\right )}^{\frac{1}{3}}}{a + b}} \log \left (\frac{2 \, b x^{3} + 3 \, \sqrt{\frac{1}{3}}{\left ({\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (a + b\right )} -{\left (a + b\right )}{\left (-a - b\right )}^{\frac{1}{3}} - 2 \,{\left (b x^{3} + a\right )}^{\frac{2}{3}}{\left (-a - b\right )}^{\frac{2}{3}}\right )} \sqrt{\frac{{\left (-a - b\right )}^{\frac{1}{3}}}{a + b}} + 3 \, a - 3 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (-a - b\right )}^{\frac{2}{3}} + b}{x^{3} - 1}\right ) +{\left (-a - b\right )}^{\frac{2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} -{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (-a - b\right )}^{\frac{1}{3}} +{\left (-a - b\right )}^{\frac{2}{3}}\right ) - 2 \,{\left (-a - b\right )}^{\frac{2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{1}{3}} +{\left (-a - b\right )}^{\frac{1}{3}}\right )}{6 \,{\left (a + b\right )}}, -\frac{6 \, \sqrt{\frac{1}{3}}{\left (a + b\right )} \sqrt{-\frac{{\left (-a - b\right )}^{\frac{1}{3}}}{a + b}} \arctan \left (\sqrt{\frac{1}{3}}{\left (2 \,{\left (b x^{3} + a\right )}^{\frac{1}{3}} -{\left (-a - b\right )}^{\frac{1}{3}}\right )} \sqrt{-\frac{{\left (-a - b\right )}^{\frac{1}{3}}}{a + b}}\right ) -{\left (-a - b\right )}^{\frac{2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{2}{3}} -{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (-a - b\right )}^{\frac{1}{3}} +{\left (-a - b\right )}^{\frac{2}{3}}\right ) + 2 \,{\left (-a - b\right )}^{\frac{2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac{1}{3}} +{\left (-a - b\right )}^{\frac{1}{3}}\right )}{6 \,{\left (a + b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{2}}{x^{3} \sqrt [3]{a + b x^{3}} - \sqrt [3]{a + b x^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(-x**3+1)/(b*x**3+a)**(1/3),x)

[Out]

-Integral(x**2/(x**3*(a + b*x**3)**(1/3) - (a + b*x**3)**(1/3)), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError