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]

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

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Rubi [A]  time = 0.08, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [B]  time = 0.93, size = 387, normalized size = 4.03 \[ \left [\frac {3 \, \sqrt {\frac {1}{3}} {\left (a + b\right )} \sqrt {\frac {{\left (-a - b\right )}^{\frac {1}{3}}}{a + b}} \log \left (\frac {2 \, b x^{3} + 3 \, \sqrt {\frac {1}{3}} {\left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a + b\right )} - {\left (a + b\right )} {\left (-a - b\right )}^{\frac {1}{3}} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (-a - b\right )}^{\frac {2}{3}}\right )} \sqrt {\frac {{\left (-a - b\right )}^{\frac {1}{3}}}{a + b}} + 3 \, a - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (-a - b\right )}^{\frac {2}{3}} + b}{x^{3} - 1}\right ) + {\left (-a - b\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (-a - b\right )}^{\frac {1}{3}} + {\left (-a - b\right )}^{\frac {2}{3}}\right ) - 2 \, {\left (-a - b\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} + {\left (-a - b\right )}^{\frac {1}{3}}\right )}{6 \, {\left (a + b\right )}}, -\frac {6 \, \sqrt {\frac {1}{3}} {\left (a + b\right )} \sqrt {-\frac {{\left (-a - b\right )}^{\frac {1}{3}}}{a + b}} \arctan \left (\sqrt {\frac {1}{3}} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} - {\left (-a - b\right )}^{\frac {1}{3}}\right )} \sqrt {-\frac {{\left (-a - b\right )}^{\frac {1}{3}}}{a + b}}\right ) - {\left (-a - b\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (-a - b\right )}^{\frac {1}{3}} + {\left (-a - b\right )}^{\frac {2}{3}}\right ) + 2 \, {\left (-a - b\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} + {\left (-a - b\right )}^{\frac {1}{3}}\right )}{6 \, {\left (a + b\right )}}\right ] \]

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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giac [A]  time = 20.99, size = 113, normalized size = 1.18 \[ -\frac {{\left (a + b\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + {\left (a + b\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (a + b\right )}^{\frac {1}{3}}}\right )}{\sqrt {3} a + \sqrt {3} b} + \frac {\log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a + b\right )}^{\frac {1}{3}} + {\left (a + b\right )}^{\frac {2}{3}}\right )}{6 \, {\left (a + b\right )}^{\frac {1}{3}}} - \frac {\log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - {\left (a + b\right )}^{\frac {1}{3}} \right |}\right )}{3 \, {\left (a + b\right )}^{\frac {1}{3}}} \]

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]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\left (-x^{3}+1\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}\, dx \]

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 [A]  time = 1.47, size = 110, normalized size = 1.15 \[ -\frac {\frac {2 \, \sqrt {3} b \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + {\left (a + b\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (a + b\right )}^{\frac {1}{3}}}\right )}{{\left (a + b\right )}^{\frac {1}{3}}} - \frac {b \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (a + b\right )}^{\frac {1}{3}} + {\left (a + b\right )}^{\frac {2}{3}}\right )}{{\left (a + b\right )}^{\frac {1}{3}}} + \frac {2 \, b \log \left ({\left (b x^{3} + a\right )}^{\frac {1}{3}} - {\left (a + b\right )}^{\frac {1}{3}}\right )}{{\left (a + b\right )}^{\frac {1}{3}}}}{6 \, b} \]

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]

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

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mupad [B]  time = 0.59, size = 157, normalized size = 1.64 \[ \frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}-\frac {9\,a+9\,b}{9\,{\left (-a-b\right )}^{2/3}}\right )}{3\,{\left (-a-b\right )}^{1/3}}+\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}-\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,a+9\,b\right )}{36\,{\left (-a-b\right )}^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,{\left (-a-b\right )}^{1/3}}-\frac {\ln \left ({\left (b\,x^3+a\right )}^{1/3}-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,a+9\,b\right )}{36\,{\left (-a-b\right )}^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{6\,{\left (-a-b\right )}^{1/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((a + b*x^3)^(1/3) - (9*a + 9*b)/(9*(- a - b)^(2/3)))/(3*(- a - b)^(1/3)) + (log((a + b*x^3)^(1/3) - ((3^(1
/2)*1i - 1)^2*(9*a + 9*b))/(36*(- a - b)^(2/3)))*(3^(1/2)*1i - 1))/(6*(- a - b)^(1/3)) - (log((a + b*x^3)^(1/3
) - ((3^(1/2)*1i + 1)^2*(9*a + 9*b))/(36*(- a - b)^(2/3)))*(3^(1/2)*1i + 1))/(6*(- a - b)^(1/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{2}}{x^{3} \sqrt [3]{a + b x^{3}} - \sqrt [3]{a + b x^{3}}}\, dx \]

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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