\(\int \frac {1}{(1+x) \sqrt [3]{1-x+x^2}} \, dx\) [2305]

Optimal result

Integrand size = 18, antiderivative size = 177 \[ \int \frac {1}{(1+x) \sqrt [3]{1-x+x^2}} \, dx=-\frac {\arctan \left (\frac {\frac {4}{3 \sqrt [6]{3}}-\frac {2 x}{3 \sqrt [6]{3}}+\frac {\sqrt [3]{1-x+x^2}}{\sqrt {3}}}{\sqrt [3]{1-x+x^2}}\right )}{3^{5/6}}+\frac {\log \left (-2 \sqrt [3]{3}+\sqrt [3]{3} x+3 \sqrt [3]{1-x+x^2}\right )}{3 \sqrt [3]{3}}-\frac {\log \left (4\ 3^{2/3}-4\ 3^{2/3} x+3^{2/3} x^2+\left (6 \sqrt [3]{3}-3 \sqrt [3]{3} x\right ) \sqrt [3]{1-x+x^2}+9 \left (1-x+x^2\right )^{2/3}\right )}{6 \sqrt [3]{3}} \]

[Out]

-1/3*arctan((4/9*3^(5/6)-2/9*x*3^(5/6)+1/3*(x^2-x+1)^(1/3)*3^(1/2))/(x^2-x+1)^(1/3))*3^(1/6)+1/9*ln(-2*3^(1/3)
+3^(1/3)*x+3*(x^2-x+1)^(1/3))*3^(2/3)-1/18*ln(4*3^(2/3)-4*3^(2/3)*x+3^(2/3)*x^2+(6*3^(1/3)-3*3^(1/3)*x)*(x^2-x
+1)^(1/3)+9*(x^2-x+1)^(2/3))*3^(2/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.50, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {764} \[ \int \frac {1}{(1+x) \sqrt [3]{1-x+x^2}} \, dx=-\frac {\arctan \left (\frac {2 (2-x)}{3 \sqrt [6]{3} \sqrt [3]{x^2-x+1}}+\frac {1}{\sqrt {3}}\right )}{3^{5/6}}+\frac {\log \left (-3^{2/3} \sqrt [3]{x^2-x+1}-x+2\right )}{2 \sqrt [3]{3}}-\frac {\log (x+1)}{2 \sqrt [3]{3}} \]

[In]

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

[Out]

-(ArcTan[1/Sqrt[3] + (2*(2 - x))/(3*3^(1/6)*(1 - x + x^2)^(1/3))]/3^(5/6)) - Log[1 + x]/(2*3^(1/3)) + Log[2 -
x - 3^(2/3)*(1 - x + x^2)^(1/3)]/(2*3^(1/3))

Rule 764

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 (2-x)}{3 \sqrt [6]{3} \sqrt [3]{1-x+x^2}}\right )}{3^{5/6}}-\frac {\log (1+x)}{2 \sqrt [3]{3}}+\frac {\log \left (2-x-3^{2/3} \sqrt [3]{1-x+x^2}\right )}{2 \sqrt [3]{3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(1+x) \sqrt [3]{1-x+x^2}} \, dx=\frac {-6 \arctan \left (\frac {4\ 3^{5/6}-2\ 3^{5/6} x+3 \sqrt {3} \sqrt [3]{1-x+x^2}}{9 \sqrt [3]{1-x+x^2}}\right )+\sqrt {3} \left (2 \log \left (-2 \sqrt [3]{3}+\sqrt [3]{3} x+3 \sqrt [3]{1-x+x^2}\right )-\log \left (4\ 3^{2/3}-4\ 3^{2/3} x+3^{2/3} x^2-3 \sqrt [3]{3} (-2+x) \sqrt [3]{1-x+x^2}+9 \left (1-x+x^2\right )^{2/3}\right )\right )}{6\ 3^{5/6}} \]

[In]

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

[Out]

(-6*ArcTan[(4*3^(5/6) - 2*3^(5/6)*x + 3*Sqrt[3]*(1 - x + x^2)^(1/3))/(9*(1 - x + x^2)^(1/3))] + Sqrt[3]*(2*Log
[-2*3^(1/3) + 3^(1/3)*x + 3*(1 - x + x^2)^(1/3)] - Log[4*3^(2/3) - 4*3^(2/3)*x + 3^(2/3)*x^2 - 3*3^(1/3)*(-2 +
x)*(1 - x + x^2)^(1/3) + 9*(1 - x + x^2)^(2/3)]))/(6*3^(5/6))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 11.02 (sec) , antiderivative size = 2133, normalized size of antiderivative = 12.05

method	result	size
trager	\(\text {Expression too large to display}\)	\(2133\)

[In]

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

[Out]

1/3*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*ln(-(21999*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)
+9*_Z^2)*RootOf(_Z^3-9)^3*x^3-118773*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^
3+43998*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^2-237546*RootOf(RootOf(_Z^3-9)^
2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^2-967776*(x^2-x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootO
f(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2*x+87996*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^
3*x-475092*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x+1935552*(x^2-x+1)^(2/3)*Ro
otOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2-203855*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2*x^2
+967776*(x^2-x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x^2+815420*(x^2-x+1
)^(1/3)*RootOf(_Z^3-9)^2*x-3871104*(x^2-x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(
_Z^3-9)*x+124661*RootOf(_Z^3-9)*x^3-673047*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^3-815420*(x^2
-x+1)^(1/3)*RootOf(_Z^3-9)^2+3871104*(x^2-x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootO
f(_Z^3-9)-1341939*RootOf(_Z^3-9)*x^2+7245153*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^2+1834695*x
*(x^2-x+1)^(2/3)+2089905*RootOf(_Z^3-9)*x-11283435*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x-36693
90*(x^2-x+1)^(2/3)-1591261*RootOf(_Z^3-9)+8591247*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2))/(1+x)^3
)-1/9*ln((61590*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^3+118773*RootOf(RootOf(
_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^3+123180*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9
)+9*_Z^2)*RootOf(_Z^3-9)^3*x^2+237546*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x
^2-967776*(x^2-x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2*x+246360*RootOf
(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x+475092*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z
^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x+1935552*(x^2-x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)
*RootOf(_Z^3-9)^2+526447*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2*x^2+967776*(x^2-x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3
*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x^2-2105788*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2*x-3871104*(x^2-x+1)^(1/
3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x-164240*RootOf(_Z^3-9)*x^3-316728*RootO
f(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^3+2105788*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2+3871104*(x^2-x+1)^
(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)+4126530*RootOf(_Z^3-9)*x^2+7957791*Ro
otOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x^2-4738023*x*(x^2-x+1)^(2/3)-5111970*RootOf(_Z^3-9)*x-98581
59*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*x+9476046*(x^2-x+1)^(2/3)+4455010*RootOf(_Z^3-9)+859124
7*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2))/(1+x)^3)*RootOf(_Z^3-9)-1/3*ln((61590*RootOf(RootOf(_Z^
3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^3+118773*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_
Z^2)^2*RootOf(_Z^3-9)^2*x^3+123180*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x^2+23
7546*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*x^2-967776*(x^2-x+1)^(2/3)*RootOf(
RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2*x+246360*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^
3-9)+9*_Z^2)*RootOf(_Z^3-9)^3*x+475092*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)^2*RootOf(_Z^3-9)^2*
x+1935552*(x^2-x+1)^(2/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)^2+526447*(x^2-x+1
)^(1/3)*RootOf(_Z^3-9)^2*x^2+967776*(x^2-x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf
(_Z^3-9)*x^2-2105788*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2*x-3871104*(x^2-x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*R
ootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)*x-164240*RootOf(_Z^3-9)*x^3-316728*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z
^3-9)+9*_Z^2)*x^3+2105788*(x^2-x+1)^(1/3)*RootOf(_Z^3-9)^2+3871104*(x^2-x+1)^(1/3)*RootOf(RootOf(_Z^3-9)^2+3*_
Z*RootOf(_Z^3-9)+9*_Z^2)*RootOf(_Z^3-9)+4126530*RootOf(_Z^3-9)*x^2+7957791*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf
(_Z^3-9)+9*_Z^2)*x^2-4738023*x*(x^2-x+1)^(2/3)-5111970*RootOf(_Z^3-9)*x-9858159*RootOf(RootOf(_Z^3-9)^2+3*_Z*R
ootOf(_Z^3-9)+9*_Z^2)*x+9476046*(x^2-x+1)^(2/3)+4455010*RootOf(_Z^3-9)+8591247*RootOf(RootOf(_Z^3-9)^2+3*_Z*Ro
otOf(_Z^3-9)+9*_Z^2))/(1+x)^3)*RootOf(RootOf(_Z^3-9)^2+3*_Z*RootOf(_Z^3-9)+9*_Z^2)

Fricas [A] (verification not implemented)

none

Time = 1.39 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(1+x) \sqrt [3]{1-x+x^2}} \, dx=-\frac {1}{18} \cdot 3^{\frac {2}{3}} \log \left (\frac {3 \cdot 3^{\frac {2}{3}} {\left (x^{2} - x + 1\right )}^{\frac {2}{3}} + 3^{\frac {1}{3}} {\left (x^{2} - 4 \, x + 4\right )} - 3 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x - 2\right )}}{x^{2} + 2 \, x + 1}\right ) + \frac {1}{9} \cdot 3^{\frac {2}{3}} \log \left (\frac {3^{\frac {1}{3}} {\left (x - 2\right )} + 3 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{3}}}{x + 1}\right ) - \frac {1}{3} \cdot 3^{\frac {1}{6}} \arctan \left (\frac {3^{\frac {1}{6}} {\left (6 \cdot 3^{\frac {2}{3}} {\left (x^{2} - x + 1\right )}^{\frac {2}{3}} {\left (x - 2\right )} + 3^{\frac {1}{3}} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 1\right )} + 6 \, {\left (x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 4 \, x + 4\right )}\right )}}{3 \, {\left (x^{3} - 15 \, x^{2} + 21 \, x - 17\right )}}\right ) \]

[In]

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

[Out]

-1/18*3^(2/3)*log((3*3^(2/3)*(x^2 - x + 1)^(2/3) + 3^(1/3)*(x^2 - 4*x + 4) - 3*(x^2 - x + 1)^(1/3)*(x - 2))/(x
^2 + 2*x + 1)) + 1/9*3^(2/3)*log((3^(1/3)*(x - 2) + 3*(x^2 - x + 1)^(1/3))/(x + 1)) - 1/3*3^(1/6)*arctan(1/3*3
^(1/6)*(6*3^(2/3)*(x^2 - x + 1)^(2/3)*(x - 2) + 3^(1/3)*(x^3 + 3*x^2 + 3*x + 1) + 6*(x^2 - x + 1)^(1/3)*(x^2 -
4*x + 4))/(x^3 - 15*x^2 + 21*x - 17))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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