3.10.58 \(\int \frac {x (-3+x^2)}{(-1+x^2)^{2/3} (1-x^2+x^3)} \, dx\)

Optimal. Leaf size=73 \[ \log \left (\sqrt [3]{x^2-1}-x\right )-\frac {1}{2} \log \left (x^2+\sqrt [3]{x^2-1} x+\left (x^2-1\right )^{2/3}\right )+\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^2-1}+x}\right ) \]

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Rubi [F]  time = 0.91, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(3^(3/4)*Sqrt[2 + Sqrt[3]]*(1 + (-1 + x^2)^(1/3))*Sqrt[(1 - (-1 + x^2)^(1/3) + (-1 + x^2)^(2/3))/(1 + Sqrt[3]
+ (-1 + x^2)^(1/3))^2]*EllipticF[ArcSin[(1 - Sqrt[3] + (-1 + x^2)^(1/3))/(1 + Sqrt[3] + (-1 + x^2)^(1/3))], -7
- 4*Sqrt[3]])/(x*Sqrt[(1 + (-1 + x^2)^(1/3))/(1 + Sqrt[3] + (-1 + x^2)^(1/3))^2]) - Defer[Int][1/((-1 + x^2)^
(2/3)*(1 - x^2 + x^3)), x] - 3*Defer[Int][x/((-1 + x^2)^(2/3)*(1 - x^2 + x^3)), x] + Defer[Int][x^2/((-1 + x^2
)^(2/3)*(1 - x^2 + x^3)), x]

Rubi steps

\begin {align*} \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx &=\int \left (\frac {1}{\left (-1+x^2\right )^{2/3}}-\frac {1+3 x-x^2}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )}\right ) \, dx\\ &=\int \frac {1}{\left (-1+x^2\right )^{2/3}} \, dx-\int \frac {1+3 x-x^2}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx\\ &=\frac {\left (3 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^3}} \, dx,x,\sqrt [3]{-1+x^2}\right )}{2 x}-\int \left (\frac {1}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )}+\frac {3 x}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )}-\frac {x^2}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )}\right ) \, dx\\ &=\frac {3^{3/4} \sqrt {2+\sqrt {3}} \left (1+\sqrt [3]{-1+x^2}\right ) \sqrt {\frac {1-\sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}}{\left (1+\sqrt {3}+\sqrt [3]{-1+x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac {1-\sqrt {3}+\sqrt [3]{-1+x^2}}{1+\sqrt {3}+\sqrt [3]{-1+x^2}}\right )|-7-4 \sqrt {3}\right )}{x \sqrt {\frac {1+\sqrt [3]{-1+x^2}}{\left (1+\sqrt {3}+\sqrt [3]{-1+x^2}\right )^2}}}-3 \int \frac {x}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx-\int \frac {1}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx+\int \frac {x^2}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx\\ \end {align*}

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Mathematica [F]  time = 0.18, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (-3+x^2\right )}{\left (-1+x^2\right )^{2/3} \left (1-x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 2.08, size = 73, normalized size = 1.00 \begin {gather*} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*(-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

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fricas [A]  time = 0.47, size = 74, normalized size = 1.01 \begin {gather*} -\sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{2} - 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) + \log \left (-\frac {x - {\left (x^{2} - 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{2} - 1\right )}^{\frac {1}{3}} x + {\left (x^{2} - 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} - 3\right )} x}{{\left (x^{3} - x^{2} + 1\right )} {\left (x^{2} - 1\right )}^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.88, size = 288, normalized size = 3.95

method	result	size
trager	\(\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-2 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 x \left (x^{2}-1\right )^{\frac {2}{3}}+3 \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+2 x^{2}-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-2}{x^{3}-x^{2}+1}\right )-\ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {2}{3}} x +\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+2 x \left (x^{2}-1\right )^{\frac {2}{3}}+2 \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+x^{3}+x^{2}-1}{x^{3}-x^{2}+1}\right ) \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-\ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {2}{3}} x +\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+2 x \left (x^{2}-1\right )^{\frac {2}{3}}+2 \left (x^{2}-1\right )^{\frac {1}{3}} x^{2}+x^{3}+x^{2}-1}{x^{3}-x^{2}+1}\right )\)	\(288\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} - 3\right )} x}{{\left (x^{3} - x^{2} + 1\right )} {\left (x^{2} - 1\right )}^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\left (x^2-3\right )}{{\left (x^2-1\right )}^{2/3}\,\left (x^3-x^2+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (x^{2} - 3\right )}{\left (\left (x - 1\right ) \left (x + 1\right )\right )^{\frac {2}{3}} \left (x^{3} - x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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