\(\int \frac {1}{(1+x) \sqrt [3]{-1+2 x+x^2}} \, dx\) [1834]

Optimal result

Integrand size = 18, antiderivative size = 125 \[ \int \frac {1}{(1+x) \sqrt [3]{-1+2 x+x^2}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \sqrt [3]{-1+2 x+x^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2+2^{2/3} \sqrt [3]{-1+2 x+x^2}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2+2^{2/3} \sqrt [3]{-1+2 x+x^2}-\sqrt [3]{2} \left (-1+2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \]

[Out]

1/4*3^(1/2)*arctan(-1/3*3^(1/2)+1/3*2^(2/3)*(x^2+2*x-1)^(1/3)*3^(1/2))*2^(2/3)-1/4*ln(2+2^(2/3)*(x^2+2*x-1)^(1
/3))*2^(2/3)+1/8*ln(-2+2^(2/3)*(x^2+2*x-1)^(1/3)-2^(1/3)*(x^2+2*x-1)^(2/3))*2^(2/3)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.66, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {708, 272, 58, 631, 210, 31} \[ \int \frac {1}{(1+x) \sqrt [3]{-1+2 x+x^2}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-2^{2/3} \sqrt [3]{(x+1)^2-2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}+\frac {\log (x+1)}{2 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{(x+1)^2-2}+\sqrt [3]{2}\right )}{4 \sqrt [3]{2}} \]

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[(1 - 2^(2/3)*(-2 + (1 + x)^2)^(1/3))/Sqrt[3]])/2^(1/3) + Log[1 + x]/(2*2^(1/3)) - (3*Log[
2^(1/3) + (-2 + (1 + x)^2)^(1/3)])/(4*2^(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 58

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] && NegQ
[(b*c - a*d)/b]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 631

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 708

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{x \sqrt [3]{-2+x^2}} \, dx,x,1+x\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt [3]{-2+x} x} \, dx,x,(1+x)^2\right ) \\ & = \frac {\log (1+x)}{2 \sqrt [3]{2}}+\frac {3}{4} \text {Subst}\left (\int \frac {1}{2^{2/3}-\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{-2+(1+x)^2}\right )-\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt [3]{2}+x} \, dx,x,\sqrt [3]{-2+(1+x)^2}\right )}{4 \sqrt [3]{2}} \\ & = \frac {\log (1+x)}{2 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{2}+\sqrt [3]{-2+(1+x)^2}\right )}{4 \sqrt [3]{2}}+\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2^{2/3} \sqrt [3]{-2+(1+x)^2}\right )}{2 \sqrt [3]{2}} \\ & = -\frac {\sqrt {3} \arctan \left (\frac {1-2^{2/3} \sqrt [3]{-2+(1+x)^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}+\frac {\log (1+x)}{2 \sqrt [3]{2}}-\frac {3 \log \left (\sqrt [3]{2}+\sqrt [3]{-2+(1+x)^2}\right )}{4 \sqrt [3]{2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(1+x) \sqrt [3]{-1+2 x+x^2}} \, dx=\frac {-2 \sqrt {3} \arctan \left (\frac {1-2^{2/3} \sqrt [3]{-1+2 x+x^2}}{\sqrt {3}}\right )-2 \log \left (2+2^{2/3} \sqrt [3]{-1+2 x+x^2}\right )+\log \left (-2+2^{2/3} \sqrt [3]{-1+2 x+x^2}-\sqrt [3]{2} \left (-1+2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 6.17 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(\frac {2^{\frac {2}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (x^{2}+2 x -1\right )^{\frac {1}{3}}-1\right )}{3}\right )-2 \ln \left (\left (x^{2}+2 x -1\right )^{\frac {1}{3}}+2^{\frac {1}{3}}\right )+\ln \left (\left (x^{2}+2 x -1\right )^{\frac {2}{3}}-2^{\frac {1}{3}} \left (x^{2}+2 x -1\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )\right )}{8}\)	\(81\)
trager	\(\text {Expression too large to display}\)	\(1198\)

[In]

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

[Out]

1/8*2^(2/3)*(2*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(x^2+2*x-1)^(1/3)-1))-2*ln((x^2+2*x-1)^(1/3)+2^(1/3))+ln((x
^2+2*x-1)^(2/3)-2^(1/3)*(x^2+2*x-1)^(1/3)+2^(2/3)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(1+x) \sqrt [3]{-1+2 x+x^2}} \, dx=\frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{6}} {\left (2 \, \sqrt {2} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} + 2 \, x - 1\right )}^{\frac {1}{3}} + 2^{\frac {5}{6}}\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{2} + 2 \, x - 1\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (x^{2} + 2 \, x - 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (x^{2} + 2 \, x - 1\right )}^{\frac {1}{3}}\right ) \]

[In]

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

[Out]

1/4*sqrt(3)*2^(2/3)*(-1)^(1/3)*arctan(1/6*sqrt(3)*2^(1/6)*(2*sqrt(2)*(-1)^(1/3)*(x^2 + 2*x - 1)^(1/3) + 2^(5/6
))) - 1/8*2^(2/3)*(-1)^(1/3)*log(-2^(1/3)*(-1)^(2/3)*(x^2 + 2*x - 1)^(1/3) - 2^(2/3)*(-1)^(1/3) + (x^2 + 2*x -
1)^(2/3)) + 1/4*2^(2/3)*(-1)^(1/3)*log(2^(1/3)*(-1)^(2/3) + (x^2 + 2*x - 1)^(1/3))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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