\(\int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx\) [2569]

Optimal result

Integrand size = 20, antiderivative size = 219 \[ \int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x^2+x^3}}\right )-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-x^2+x^3}}\right )}{\sqrt [3]{2}}-\log \left (-x+\sqrt [3]{-x^2+x^3}\right )+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{-x^2+x^3}\right )}{\sqrt [3]{2}}+\frac {1}{2} \log \left (x^2+x \sqrt [3]{-x^2+x^3}+\left (-x^2+x^3\right )^{2/3}\right )-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{-x^2+x^3}+\sqrt [3]{2} \left (-x^2+x^3\right )^{2/3}\right )}{2 \sqrt [3]{2}} \]

[Out]

3^(1/2)*arctan(3^(1/2)*x/(x+2*(x^3-x^2)^(1/3)))-1/2*3^(1/2)*arctan(3^(1/2)*x/(x+2^(2/3)*(x^3-x^2)^(1/3)))*2^(2
/3)-ln(-x+(x^3-x^2)^(1/3))+1/2*ln(-2*x+2^(2/3)*(x^3-x^2)^(1/3))*2^(2/3)+1/2*ln(x^2+x*(x^3-x^2)^(1/3)+(x^3-x^2)
^(2/3))-1/4*ln(2*x^2+2^(2/3)*x*(x^3-x^2)^(1/3)+2^(1/3)*(x^3-x^2)^(2/3))*2^(2/3)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2067, 129, 494, 245, 384} \[ \int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx=\frac {\sqrt {3} \sqrt [3]{x-1} x^{2/3} \arctan \left (\frac {\frac {2 \sqrt [3]{x}}{\sqrt [3]{x-1}}+1}{\sqrt {3}}\right )}{\sqrt [3]{x^3-x^2}}-\frac {\sqrt {3} \sqrt [3]{x-1} x^{2/3} \arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{x}}{\sqrt [3]{x-1}}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{x^3-x^2}}-\frac {3 \sqrt [3]{x-1} x^{2/3} \log \left (\sqrt [3]{x-1}-\sqrt [3]{x}\right )}{2 \sqrt [3]{x^3-x^2}}+\frac {3 \sqrt [3]{x-1} x^{2/3} \log \left (\sqrt [3]{x-1}-\sqrt [3]{2} \sqrt [3]{x}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^3-x^2}}-\frac {\sqrt [3]{x-1} x^{2/3} \log (x+1)}{2 \sqrt [3]{2} \sqrt [3]{x^3-x^2}} \]

[In]

Int[x/((1 + x)*(-x^2 + x^3)^(1/3)),x]

[Out]

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

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 494

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

Rule 2067

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.))^(q_.), x_Symbol]
:> Dist[e^IntPart[m]*(e*x)^FracPart[m]*((a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a
+ b*x^n)^FracPart[p])), Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, j, m, n,
p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [3]{-1+x} x^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\sqrt [3]{-1+x} (1+x)} \, dx}{\sqrt [3]{-x^2+x^3}} \\ & = \frac {\left (3 \sqrt [3]{-1+x} x^{2/3}\right ) \text {Subst}\left (\int \frac {x^3}{\sqrt [3]{-1+x^3} \left (1+x^3\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{-x^2+x^3}} \\ & = \frac {\left (3 \sqrt [3]{-1+x} x^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{-1+x^3}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{-x^2+x^3}}-\frac {\left (3 \sqrt [3]{-1+x} x^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{-1+x^3} \left (1+x^3\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{-x^2+x^3}} \\ & = \frac {\sqrt {3} \sqrt [3]{-1+x} x^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{x}}{\sqrt [3]{-1+x}}}{\sqrt {3}}\right )}{\sqrt [3]{-x^2+x^3}}-\frac {\sqrt {3} \sqrt [3]{-1+x} x^{2/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{x}}{\sqrt [3]{-1+x}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{-x^2+x^3}}-\frac {3 \sqrt [3]{-1+x} x^{2/3} \log \left (\sqrt [3]{-1+x}-\sqrt [3]{x}\right )}{2 \sqrt [3]{-x^2+x^3}}+\frac {3 \sqrt [3]{-1+x} x^{2/3} \log \left (\sqrt [3]{-1+x}-\sqrt [3]{2} \sqrt [3]{x}\right )}{2 \sqrt [3]{2} \sqrt [3]{-x^2+x^3}}-\frac {\sqrt [3]{-1+x} x^{2/3} \log (1+x)}{2 \sqrt [3]{2} \sqrt [3]{-x^2+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.06 \[ \int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx=\frac {\sqrt [3]{-1+x} x^{2/3} \left (4 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{x}}\right )-2\ 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{2^{2/3} \sqrt [3]{-1+x}+\sqrt [3]{x}}\right )+2\ 2^{2/3} \log \left (2^{2/3} \sqrt [3]{-1+x}-2 \sqrt [3]{x}\right )-4 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{x}\right )+2 \log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{x}+x^{2/3}\right )-2^{2/3} \log \left (\sqrt [3]{2} (-1+x)^{2/3}+2^{2/3} \sqrt [3]{-1+x} \sqrt [3]{x}+2 x^{2/3}\right )\right )}{4 \sqrt [3]{(-1+x) x^2}} \]

[In]

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

[Out]

((-1 + x)^(1/3)*x^(2/3)*(4*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(2*(-1 + x)^(1/3) + x^(1/3))] - 2*2^(2/3)*Sqrt[3]*
ArcTan[(Sqrt[3]*x^(1/3))/(2^(2/3)*(-1 + x)^(1/3) + x^(1/3))] + 2*2^(2/3)*Log[2^(2/3)*(-1 + x)^(1/3) - 2*x^(1/3
)] - 4*Log[(-1 + x)^(1/3) - x^(1/3)] + 2*Log[(-1 + x)^(2/3) + (-1 + x)^(1/3)*x^(1/3) + x^(2/3)] - 2^(2/3)*Log[
2^(1/3)*(-1 + x)^(2/3) + 2^(2/3)*(-1 + x)^(1/3)*x^(1/3) + 2*x^(2/3)]))/(4*((-1 + x)*x^2)^(1/3))

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {2^{\frac {2}{3}} \ln \left (\frac {-2^{\frac {1}{3}} x +\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}{x}\right )}{2}-\frac {2^{\frac {2}{3}} \ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}}{x^{2}}\right )}{4}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x \right )}{3 x}\right )}{2}-\ln \left (\frac {-x +\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}{x}\right )+\frac {\ln \left (\frac {\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}+x \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x^{2}}{x^{2}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\left (2 \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )\)	\(179\)
trager	\(\text {Expression too large to display}\)	\(1152\)

[In]

int(x/(1+x)/(x^3-x^2)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/2*2^(2/3)*ln((-2^(1/3)*x+((-1+x)*x^2)^(1/3))/x)-1/4*2^(2/3)*ln((2^(2/3)*x^2+2^(1/3)*((-1+x)*x^2)^(1/3)*x+((-
1+x)*x^2)^(2/3))/x^2)+1/2*3^(1/2)*2^(2/3)*arctan(1/3*3^(1/2)*(2^(2/3)*((-1+x)*x^2)^(1/3)+x)/x)-ln((-x+((-1+x)*
x^2)^(1/3))/x)+1/2*ln((((-1+x)*x^2)^(2/3)+x*((-1+x)*x^2)^(1/3)+x^2)/x^2)-3^(1/2)*arctan(1/3*(2*((-1+x)*x^2)^(1
/3)+x)*3^(1/2)/x)

Fricas [A] (verification not implemented)

none

Time = 1.00 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.10 \[ \int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx=\frac {1}{4} \, {\left (2 \, \sqrt {\frac {3}{2}} \sqrt {-2^{\frac {1}{3}}} - 2^{\frac {2}{3}}\right )} \log \left (\frac {3 \, {\left (2^{\frac {2}{3}} \sqrt {\frac {3}{2}} x \sqrt {-2^{\frac {1}{3}}} + 2^{\frac {1}{3}} x + 2 \, {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}\right )}}{2 \, x}\right ) - \frac {1}{4} \, {\left (2 \, \sqrt {\frac {3}{2}} \sqrt {-2^{\frac {1}{3}}} + 2^{\frac {2}{3}}\right )} \log \left (-\frac {3 \, {\left (2^{\frac {2}{3}} \sqrt {\frac {3}{2}} x \sqrt {-2^{\frac {1}{3}}} - 2^{\frac {1}{3}} x - 2 \, {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}\right )}}{2 \, x}\right ) + \frac {1}{2} \cdot 2^{\frac {2}{3}} \log \left (-\frac {3 \, {\left (2^{\frac {1}{3}} x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}\right )}}{x}\right ) - \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \log \left (-\frac {3 \, {\left (x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}\right )}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[In]

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

[Out]

1/4*(2*sqrt(3/2)*sqrt(-2^(1/3)) - 2^(2/3))*log(3/2*(2^(2/3)*sqrt(3/2)*x*sqrt(-2^(1/3)) + 2^(1/3)*x + 2*(x^3 -
x^2)^(1/3))/x) - 1/4*(2*sqrt(3/2)*sqrt(-2^(1/3)) + 2^(2/3))*log(-3/2*(2^(2/3)*sqrt(3/2)*x*sqrt(-2^(1/3)) - 2^(
1/3)*x - 2*(x^3 - x^2)^(1/3))/x) + 1/2*2^(2/3)*log(-3*(2^(1/3)*x - (x^3 - x^2)^(1/3))/x) - sqrt(3)*arctan(1/3*
(sqrt(3)*x + 2*sqrt(3)*(x^3 - x^2)^(1/3))/x) - log(-3*(x - (x^3 - x^2)^(1/3))/x) + 1/2*log((x^2 + (x^3 - x^2)^
(1/3)*x + (x^3 - x^2)^(2/3))/x^2)

Sympy [F]

\[ \int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx=\int \frac {x}{\sqrt [3]{x^{2} \left (x - 1\right )} \left (x + 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx=\int { \frac {x}{{\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} {\left (x + 1\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.68 \[ \int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx=\frac {1}{2} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{4} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{2} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} \right |}\right ) - \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{2} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{(1+x) \sqrt [3]{-x^2+x^3}} \, dx=\int \frac {x}{{\left (x^3-x^2\right )}^{1/3}\,\left (x+1\right )} \,d x \]

[In]

int(x/((x^3 - x^2)^(1/3)*(x + 1)),x)

[Out]

int(x/((x^3 - x^2)^(1/3)*(x + 1)), x)