3.16.1 \(\int \frac {(2+x)^2}{x (4-2 x+x^2) \sqrt [3]{1+x+x^2}} \, dx\)

Optimal. Leaf size=105 \[ \log \left (\sqrt [3]{x^2+x+1}+x-1\right )-\frac {1}{2} \log \left (x^2+\left (x^2+x+1\right )^{2/3}+(1-x) \sqrt [3]{x^2+x+1}-2 x+1\right )-\sqrt {3} \tan ^{-1}\left (\frac {\frac {\sqrt [3]{x^2+x+1}}{\sqrt {3}}-\frac {2 x}{\sqrt {3}}+\frac {2}{\sqrt {3}}}{\sqrt [3]{x^2+x+1}}\right ) \]

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Rubi [F]  time = 0.52, 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 {(2+x)^2}{x \left (4-2 x+x^2\right ) \sqrt [3]{1+x+x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 + x)^2/(x*(4 - 2*x + x^2)*(1 + x + x^2)^(1/3)),x]

[Out]

(-3*((1 - I*Sqrt[3] + 2*x)/x)^(1/3)*((1 + I*Sqrt[3] + 2*x)/x)^(1/3)*AppellF1[2/3, 1/3, 1/3, 5/3, -1/2*(1 - I*S
qrt[3])/x, -1/2*(1 + I*Sqrt[3])/x])/(2*2^(2/3)*(1 + x + x^2)^(1/3)) + 6*Defer[Int][1/((4 - 2*x + x^2)*(1 + x +
x^2)^(1/3)), x]

Rubi steps

\begin {align*} \int \frac {(2+x)^2}{x \left (4-2 x+x^2\right ) \sqrt [3]{1+x+x^2}} \, dx &=\int \left (\frac {1}{x \sqrt [3]{1+x+x^2}}+\frac {6}{\left (4-2 x+x^2\right ) \sqrt [3]{1+x+x^2}}\right ) \, dx\\ &=6 \int \frac {1}{\left (4-2 x+x^2\right ) \sqrt [3]{1+x+x^2}} \, dx+\int \frac {1}{x \sqrt [3]{1+x+x^2}} \, dx\\ &=6 \int \frac {1}{\left (4-2 x+x^2\right ) \sqrt [3]{1+x+x^2}} \, dx-\frac {\left (\sqrt [3]{\frac {1-i \sqrt {3}+2 x}{x}} \sqrt [3]{\frac {1+i \sqrt {3}+2 x}{x}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{x} \sqrt [3]{1+\frac {1}{2} \left (1-i \sqrt {3}\right ) x} \sqrt [3]{1+\frac {1}{2} \left (1+i \sqrt {3}\right ) x}} \, dx,x,\frac {1}{x}\right )}{2^{2/3} \left (\frac {1}{x}\right )^{2/3} \sqrt [3]{1+x+x^2}}\\ &=-\frac {3 \sqrt [3]{\frac {1-i \sqrt {3}+2 x}{x}} \sqrt [3]{\frac {1+i \sqrt {3}+2 x}{x}} F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};-\frac {1-i \sqrt {3}}{2 x},-\frac {1+i \sqrt {3}}{2 x}\right )}{2\ 2^{2/3} \sqrt [3]{1+x+x^2}}+6 \int \frac {1}{\left (4-2 x+x^2\right ) \sqrt [3]{1+x+x^2}} \, dx\\ \end {align*}

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Mathematica [F]  time = 0.26, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(2+x)^2}{x \left (4-2 x+x^2\right ) \sqrt [3]{1+x+x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(2 + x)^2/(x*(4 - 2*x + x^2)*(1 + x + x^2)^(1/3)),x]

[Out]

Integrate[(2 + x)^2/(x*(4 - 2*x + x^2)*(1 + x + x^2)^(1/3)), x]

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IntegrateAlgebraic [A]  time = 0.11, size = 105, normalized size = 1.00 \begin {gather*} -\sqrt {3} \tan ^{-1}\left (\frac {\frac {2}{\sqrt {3}}-\frac {2 x}{\sqrt {3}}+\frac {\sqrt [3]{1+x+x^2}}{\sqrt {3}}}{\sqrt [3]{1+x+x^2}}\right )+\log \left (-1+x+\sqrt [3]{1+x+x^2}\right )-\frac {1}{2} \log \left (1-2 x+x^2+(1-x) \sqrt [3]{1+x+x^2}+\left (1+x+x^2\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2 + x)^2/(x*(4 - 2*x + x^2)*(1 + x + x^2)^(1/3)),x]

[Out]

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

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fricas [A]  time = 1.37, size = 139, normalized size = 1.32 \begin {gather*} -\sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} {\left (x^{2} + x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )} + 2 \, \sqrt {3} {\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} + \sqrt {3} {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}}{x^{3} - 11 \, x^{2} - 5 \, x - 9}\right ) + \frac {1}{2} \, \log \left (\frac {x^{3} - 2 \, x^{2} + 3 \, {\left (x^{2} + x + 1\right )}^{\frac {2}{3}} {\left (x - 1\right )} + 3 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 1\right )} + 4 \, x}{x^{3} - 2 \, x^{2} + 4 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)^2/x/(x^2-2*x+4)/(x^2+x+1)^(1/3),x, algorithm="fricas")

[Out]

-sqrt(3)*arctan((4*sqrt(3)*(x^2 + x + 1)^(2/3)*(x - 1) + 2*sqrt(3)*(x^2 + x + 1)^(1/3)*(x^2 - 2*x + 1) + sqrt(
3)*(x^3 - 3*x^2 + 3*x - 1))/(x^3 - 11*x^2 - 5*x - 9)) + 1/2*log((x^3 - 2*x^2 + 3*(x^2 + x + 1)^(2/3)*(x - 1) +
3*(x^2 + x + 1)^(1/3)*(x^2 - 2*x + 1) + 4*x)/(x^3 - 2*x^2 + 4*x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x + 2\right )}^{2}}{{\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 4\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)^2/x/(x^2-2*x+4)/(x^2+x+1)^(1/3),x, algorithm="giac")

[Out]

integrate((x + 2)^2/((x^2 + x + 1)^(1/3)*(x^2 - 2*x + 4)*x), x)

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maple [C]  time = 1.91, size = 398, normalized size = 3.79

method	result	size
trager	\(\ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+x +1\right )^{\frac {1}{3}} x^{2}+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x -6 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+x +1\right )^{\frac {1}{3}} x -10 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-3 x \left (x^{2}+x +1\right )^{\frac {2}{3}}+2 x^{3}-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+3 \left (x^{2}+x +1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+8 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x +3 \left (x^{2}+x +1\right )^{\frac {2}{3}}-8 x^{2}-4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+4 x -4}{\left (x^{2}-2 x +4\right ) x}\right )+\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}-6 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+x +1\right )^{\frac {1}{3}} x^{2}+\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+6 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +6 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+x +1\right )^{\frac {1}{3}} x -\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+3 x \left (x^{2}+x +1\right )^{\frac {2}{3}}-2 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-3 \left (x^{2}+x +1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+5 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x -3 \left (x^{2}+x +1\right )^{\frac {2}{3}}+x^{2}+\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+x +1}{\left (x^{2}-2 x +4\right ) x}\right )\)	\(398\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+x)^2/x/(x^2-2*x+4)/(x^2+x+1)^(1/3),x,method=_RETURNVERBOSE)

[Out]

ln(-(RootOf(_Z^2+_Z+1)^2*x^3-3*RootOf(_Z^2+_Z+1)^2*x^2+3*RootOf(_Z^2+_Z+1)*(x^2+x+1)^(1/3)*x^2+3*RootOf(_Z^2+_
Z+1)*x^3+3*RootOf(_Z^2+_Z+1)^2*x-6*RootOf(_Z^2+_Z+1)*(x^2+x+1)^(1/3)*x-10*RootOf(_Z^2+_Z+1)*x^2-3*x*(x^2+x+1)^
(2/3)+2*x^3-RootOf(_Z^2+_Z+1)^2+3*(x^2+x+1)^(1/3)*RootOf(_Z^2+_Z+1)+8*RootOf(_Z^2+_Z+1)*x+3*(x^2+x+1)^(2/3)-8*
x^2-4*RootOf(_Z^2+_Z+1)+4*x-4)/(x^2-2*x+4)/x)+RootOf(_Z^2+_Z+1)*ln((2*RootOf(_Z^2+_Z+1)^2*x^3-6*RootOf(_Z^2+_Z
+1)^2*x^2-3*RootOf(_Z^2+_Z+1)*(x^2+x+1)^(1/3)*x^2+RootOf(_Z^2+_Z+1)*x^3+6*RootOf(_Z^2+_Z+1)^2*x+6*RootOf(_Z^2+
_Z+1)*(x^2+x+1)^(1/3)*x-RootOf(_Z^2+_Z+1)*x^2+3*x*(x^2+x+1)^(2/3)-2*RootOf(_Z^2+_Z+1)^2-3*(x^2+x+1)^(1/3)*Root
Of(_Z^2+_Z+1)+5*RootOf(_Z^2+_Z+1)*x-3*(x^2+x+1)^(2/3)+x^2+RootOf(_Z^2+_Z+1)+x+1)/(x^2-2*x+4)/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x + 2\right )}^{2}}{{\left (x^{2} + x + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x + 4\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)^2/x/(x^2-2*x+4)/(x^2+x+1)^(1/3),x, algorithm="maxima")

[Out]

integrate((x + 2)^2/((x^2 + x + 1)^(1/3)*(x^2 - 2*x + 4)*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x+2\right )}^2}{x\,\left (x^2-2\,x+4\right )\,{\left (x^2+x+1\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 2)^2/(x*(x^2 - 2*x + 4)*(x + x^2 + 1)^(1/3)),x)

[Out]

int((x + 2)^2/(x*(x^2 - 2*x + 4)*(x + x^2 + 1)^(1/3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x + 2\right )^{2}}{x \left (x^{2} - 2 x + 4\right ) \sqrt [3]{x^{2} + x + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)**2/x/(x**2-2*x+4)/(x**2+x+1)**(1/3),x)

[Out]

Integral((x + 2)**2/(x*(x**2 - 2*x + 4)*(x**2 + x + 1)**(1/3)), x)

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