\(\int \frac {-3+x^2}{\sqrt [3]{-1+x^2} (-1+x^2+x^3)} \, dx\) [1060]

Optimal result

Integrand size = 25, antiderivative size = 80 \[ \int \frac {-3+x^2}{\sqrt [3]{-1+x^2} \left (-1+x^2+x^3\right )} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x^2}}{-2 x+\sqrt [3]{-1+x^2}}\right )+\log \left (x+\sqrt [3]{-1+x^2}\right )-\frac {1}{2} \log \left (x^2-x \sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right ) \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {-3+x^2}{\sqrt [3]{-1+x^2} \left (-1+x^2+x^3\right )} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{-1+x^2}}{-2 x+\sqrt [3]{-1+x^2}}\right )+\log \left (x+\sqrt [3]{-1+x^2}\right )-\frac {1}{2} \log \left (x^2-x \sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right ) \]

[In]

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

[Out]

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

Maple [C] (verified)

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

Time = 1.46 (sec) , antiderivative size = 284, normalized size of antiderivative = 3.55

method	result	size
trager	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {2}{3}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+2 x \left (x^{2}-1\right )^{\frac {2}{3}}-2 \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+x^{3}-x^{2}+1}{x^{3}+x^{2}-1}\right )-\ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-3 x \left (x^{2}-1\right )^{\frac {2}{3}}+3 \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+2 x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2}{x^{3}+x^{2}-1}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-\ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-3 x \left (x^{2}-1\right )^{\frac {2}{3}}+3 \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+2 x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2}{x^{3}+x^{2}-1}\right )\)	\(284\)

[In]

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

[Out]

RootOf(_Z^2+_Z+1)*ln(-(RootOf(_Z^2+_Z+1)*(x^2-1)^(2/3)*x-RootOf(_Z^2+_Z+1)*(x^2-1)^(1/3)*x^2+RootOf(_Z^2+_Z+1)
*x^3+2*x*(x^2-1)^(2/3)-2*(x^2-1)^(1/3)*x^2+x^3-x^2+1)/(x^3+x^2-1))-ln((RootOf(_Z^2+_Z+1)^2*x^3+2*RootOf(_Z^2+_
Z+1)*x^3+RootOf(_Z^2+_Z+1)*x^2-3*x*(x^2-1)^(2/3)+3*(x^2-1)^(1/3)*x^2+2*x^2-RootOf(_Z^2+_Z+1)-2)/(x^3+x^2-1))*R
ootOf(_Z^2+_Z+1)-ln((RootOf(_Z^2+_Z+1)^2*x^3+2*RootOf(_Z^2+_Z+1)*x^3+RootOf(_Z^2+_Z+1)*x^2-3*x*(x^2-1)^(2/3)+3
*(x^2-1)^(1/3)*x^2+2*x^2-RootOf(_Z^2+_Z+1)-2)/(x^3+x^2-1))

Fricas [A] (verification not implemented)

none

Time = 0.70 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int \frac {-3+x^2}{\sqrt [3]{-1+x^2} \left (-1+x^2+x^3\right )} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} x^{3} + 2 \, \sqrt {3} {\left (x^{2} - 1\right )}^{\frac {1}{3}} x^{2} + 4 \, \sqrt {3} {\left (x^{2} - 1\right )}^{\frac {2}{3}} x}{x^{3} - 8 \, x^{2} + 8}\right ) + \frac {1}{2} \, \log \left (\frac {x^{3} + 3 \, {\left (x^{2} - 1\right )}^{\frac {1}{3}} x^{2} + x^{2} + 3 \, {\left (x^{2} - 1\right )}^{\frac {2}{3}} x - 1}{x^{3} + x^{2} - 1}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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