3.19.56 \(\int \frac {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} (2-x+2 x^2)} \, dx\)

Optimal. Leaf size=128 \[ \frac {\log \left (\sqrt [3]{x^2+x+2}+\sqrt [3]{2} x\right )}{\sqrt [3]{2}}-\frac {\log \left (2^{2/3} x^2-\sqrt [3]{2} \sqrt [3]{x^2+x+2} x+\left (x^2+x+2\right )^{2/3}\right )}{2 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {\sqrt [3]{x^2+x+2}}{\sqrt {3}}-\frac {2 \sqrt [3]{2} x}{\sqrt {3}}}{\sqrt [3]{x^2+x+2}}\right )}{\sqrt [3]{2}} \]

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Rubi [F]  time = 0.45, 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 {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(6 + 2*x + x^2)/((1 + x)*(2 + x + x^2)^(1/3)*(2 - x + 2*x^2)),x]

[Out]

(-3*((1 - I*Sqrt[7] + 2*x)/(1 + x))^(1/3)*((1 + I*Sqrt[7] + 2*x)/(1 + x))^(1/3)*AppellF1[2/3, 1/3, 1/3, 5/3, (
1 - I*Sqrt[7])/(2*(1 + x)), (1 + I*Sqrt[7])/(2*(1 + x))])/(2*2^(2/3)*(2 + x + x^2)^(1/3)) + Defer[Int][(4 - x)
/((2 + x + x^2)^(1/3)*(2 - x + 2*x^2)), x]

Rubi steps

\begin {align*} \int \frac {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx &=\int \frac {6+2 x+x^2}{\sqrt [3]{2+x+x^2} \left (2+x+x^2+2 x^3\right )} \, dx\\ &=\int \left (\frac {1}{(1+x) \sqrt [3]{2+x+x^2}}+\frac {4-x}{\sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )}\right ) \, dx\\ &=\int \frac {1}{(1+x) \sqrt [3]{2+x+x^2}} \, dx+\int \frac {4-x}{\sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx\\ &=-\frac {\left (\sqrt [3]{\frac {1-i \sqrt {7}+2 x}{1+x}} \sqrt [3]{\frac {1+i \sqrt {7}+2 x}{1+x}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{x} \sqrt [3]{1-\frac {1}{2} \left (1-i \sqrt {7}\right ) x} \sqrt [3]{1-\frac {1}{2} \left (1+i \sqrt {7}\right ) x}} \, dx,x,\frac {1}{1+x}\right )}{2^{2/3} \left (\frac {1}{1+x}\right )^{2/3} \sqrt [3]{2+x+x^2}}+\int \frac {4-x}{\sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx\\ &=-\frac {3 \sqrt [3]{\frac {1-i \sqrt {7}+2 x}{1+x}} \sqrt [3]{\frac {1+i \sqrt {7}+2 x}{1+x}} F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};\frac {1-i \sqrt {7}}{2 (1+x)},\frac {1+i \sqrt {7}}{2 (1+x)}\right )}{2\ 2^{2/3} \sqrt [3]{2+x+x^2}}+\int \frac {4-x}{\sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx\\ \end {align*}

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Mathematica [F]  time = 0.30, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(6 + 2*x + x^2)/((1 + x)*(2 + x + x^2)^(1/3)*(2 - x + 2*x^2)),x]

[Out]

Integrate[(6 + 2*x + x^2)/((1 + x)*(2 + x + x^2)^(1/3)*(2 - x + 2*x^2)), x]

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IntegrateAlgebraic [A]  time = 0.18, size = 128, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {-\frac {2 \sqrt [3]{2} x}{\sqrt {3}}+\frac {\sqrt [3]{2+x+x^2}}{\sqrt {3}}}{\sqrt [3]{2+x+x^2}}\right )}{\sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2} x+\sqrt [3]{2+x+x^2}\right )}{\sqrt [3]{2}}-\frac {\log \left (2^{2/3} x^2-\sqrt [3]{2} x \sqrt [3]{2+x+x^2}+\left (2+x+x^2\right )^{2/3}\right )}{2 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(6 + 2*x + x^2)/((1 + x)*(2 + x + x^2)^(1/3)*(2 - x + 2*x^2)),x]

[Out]

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

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fricas [B]  time = 17.58, size = 407, normalized size = 3.18 \begin {gather*} -\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (2^{\frac {5}{6}} {\left (8 \, x^{9} + 48 \, x^{8} + 18 \, x^{7} + 37 \, x^{6} - 147 \, x^{5} - 111 \, x^{4} - 107 \, x^{3} + 18 \, x^{2} + 12 \, x + 8\right )} + 12 \, \sqrt {2} {\left (4 \, x^{8} - 14 \, x^{7} - 13 \, x^{6} - 26 \, x^{5} + 5 \, x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} {\left (x^{2} + x + 2\right )}^{\frac {1}{3}} + 12 \cdot 2^{\frac {1}{6}} {\left (8 \, x^{7} + 2 \, x^{6} + x^{5} + 2 \, x^{4} - 5 \, x^{3} - 4 \, x^{2} - 4 \, x\right )} {\left (x^{2} + x + 2\right )}^{\frac {2}{3}}\right )}}{6 \, {\left (8 \, x^{9} - 96 \, x^{8} - 90 \, x^{7} - 179 \, x^{6} + 33 \, x^{5} + 33 \, x^{4} + 37 \, x^{3} + 18 \, x^{2} + 12 \, x + 8\right )}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (\frac {6 \cdot 2^{\frac {1}{3}} {\left (x^{2} + x + 2\right )}^{\frac {1}{3}} x^{2} + 2^{\frac {2}{3}} {\left (2 \, x^{3} + x^{2} + x + 2\right )} + 6 \, {\left (x^{2} + x + 2\right )}^{\frac {2}{3}} x}{2 \, x^{3} + x^{2} + x + 2}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (\frac {3 \cdot 2^{\frac {2}{3}} {\left (4 \, x^{4} - x^{3} - x^{2} - 2 \, x\right )} {\left (x^{2} + x + 2\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (4 \, x^{6} - 14 \, x^{5} - 13 \, x^{4} - 26 \, x^{3} + 5 \, x^{2} + 4 \, x + 4\right )} - 12 \, {\left (x^{5} - x^{4} - x^{3} - 2 \, x^{2}\right )} {\left (x^{2} + x + 2\right )}^{\frac {1}{3}}}{4 \, x^{6} + 4 \, x^{5} + 5 \, x^{4} + 10 \, x^{3} + 5 \, x^{2} + 4 \, x + 4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2*x+6)/(1+x)/(x^2+x+2)^(1/3)/(2*x^2-x+2),x, algorithm="fricas")

[Out]

-1/6*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(1/6)*(2^(5/6)*(8*x^9 + 48*x^8 + 18*x^7 + 37*x^6 - 147*x^5 - 111*x^4
- 107*x^3 + 18*x^2 + 12*x + 8) + 12*sqrt(2)*(4*x^8 - 14*x^7 - 13*x^6 - 26*x^5 + 5*x^4 + 4*x^3 + 4*x^2)*(x^2 +
x + 2)^(1/3) + 12*2^(1/6)*(8*x^7 + 2*x^6 + x^5 + 2*x^4 - 5*x^3 - 4*x^2 - 4*x)*(x^2 + x + 2)^(2/3))/(8*x^9 - 9
6*x^8 - 90*x^7 - 179*x^6 + 33*x^5 + 33*x^4 + 37*x^3 + 18*x^2 + 12*x + 8)) + 1/6*2^(2/3)*log((6*2^(1/3)*(x^2 +
x + 2)^(1/3)*x^2 + 2^(2/3)*(2*x^3 + x^2 + x + 2) + 6*(x^2 + x + 2)^(2/3)*x)/(2*x^3 + x^2 + x + 2)) - 1/12*2^(2
/3)*log((3*2^(2/3)*(4*x^4 - x^3 - x^2 - 2*x)*(x^2 + x + 2)^(2/3) + 2^(1/3)*(4*x^6 - 14*x^5 - 13*x^4 - 26*x^3 +
5*x^2 + 4*x + 4) - 12*(x^5 - x^4 - x^3 - 2*x^2)*(x^2 + x + 2)^(1/3))/(4*x^6 + 4*x^5 + 5*x^4 + 10*x^3 + 5*x^2
+ 4*x + 4))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} + 2 \, x + 6}{{\left (2 \, x^{2} - x + 2\right )} {\left (x^{2} + x + 2\right )}^{\frac {1}{3}} {\left (x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2*x+6)/(1+x)/(x^2+x+2)^(1/3)/(2*x^2-x+2),x, algorithm="giac")

[Out]

integrate((x^2 + 2*x + 6)/((2*x^2 - x + 2)*(x^2 + x + 2)^(1/3)*(x + 1)), x)

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maple [C]  time = 5.83, size = 943, normalized size = 7.37

method	result	size
trager	\(-\frac {\ln \left (\frac {\RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-4\right )^{3} x^{3}+4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} x^{3}+3 \left (x^{2}+x +2\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} x^{2}+6 \left (x^{2}+x +2\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}-4\right ) \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x^{2}+2 \RootOf \left (\textit {\_Z}^{3}-4\right ) x^{3}+8 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x^{3}+6 \left (x^{2}+x +2\right )^{\frac {2}{3}} x -\RootOf \left (\textit {\_Z}^{3}-4\right ) x^{2}-4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x^{2}-\RootOf \left (\textit {\_Z}^{3}-4\right ) x -4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x -2 \RootOf \left (\textit {\_Z}^{3}-4\right )-8 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right )}{\left (2 x^{2}-x +2\right ) \left (1+x \right )}\right ) \RootOf \left (\textit {\_Z}^{3}-4\right )}{2}-\ln \left (\frac {\RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-4\right )^{3} x^{3}+4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} x^{3}+3 \left (x^{2}+x +2\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} x^{2}+6 \left (x^{2}+x +2\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}-4\right ) \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x^{2}+2 \RootOf \left (\textit {\_Z}^{3}-4\right ) x^{3}+8 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x^{3}+6 \left (x^{2}+x +2\right )^{\frac {2}{3}} x -\RootOf \left (\textit {\_Z}^{3}-4\right ) x^{2}-4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x^{2}-\RootOf \left (\textit {\_Z}^{3}-4\right ) x -4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x -2 \RootOf \left (\textit {\_Z}^{3}-4\right )-8 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right )}{\left (2 x^{2}-x +2\right ) \left (1+x \right )}\right ) \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right )+\frac {\RootOf \left (\textit {\_Z}^{3}-4\right ) \ln \left (-\frac {2 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} x^{3}+\left (x^{2}+x +2\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x -2 \left (x^{2}+x +2\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} x^{2}-2 \left (x^{2}+x +2\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}-4\right ) \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x^{2}-2 \left (x^{2}+x +2\right )^{\frac {2}{3}} x +2 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x^{2}+2 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right ) x +4 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+4 \textit {\_Z}^{2}\right )}{\left (2 x^{2}-x +2\right ) \left (1+x \right )}\right )}{2}\)	\(943\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+2*x+6)/(1+x)/(x^2+x+2)^(1/3)/(2*x^2-x+2),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln((RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^3*x^3+4*RootOf(RootOf(_Z^3-4)^2+2*
_Z*RootOf(_Z^3-4)+4*_Z^2)^2*RootOf(_Z^3-4)^2*x^3+3*(x^2+x+2)^(1/3)*RootOf(_Z^3-4)^2*x^2+6*(x^2+x+2)^(1/3)*Root
Of(_Z^3-4)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x^2+2*RootOf(_Z^3-4)*x^3+8*RootOf(RootOf(_Z^3-4
)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x^3+6*(x^2+x+2)^(2/3)*x-RootOf(_Z^3-4)*x^2-4*RootOf(RootOf(_Z^3-4)^2+2*_Z*Root
Of(_Z^3-4)+4*_Z^2)*x^2-RootOf(_Z^3-4)*x-4*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x-2*RootOf(_Z^3-
4)-8*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2))/(2*x^2-x+2)/(1+x))*RootOf(_Z^3-4)-ln((RootOf(RootOf(
_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^3*x^3+4*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^
2)^2*RootOf(_Z^3-4)^2*x^3+3*(x^2+x+2)^(1/3)*RootOf(_Z^3-4)^2*x^2+6*(x^2+x+2)^(1/3)*RootOf(_Z^3-4)*RootOf(RootO
f(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x^2+2*RootOf(_Z^3-4)*x^3+8*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)
+4*_Z^2)*x^3+6*(x^2+x+2)^(2/3)*x-RootOf(_Z^3-4)*x^2-4*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x^2-
RootOf(_Z^3-4)*x-4*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x-2*RootOf(_Z^3-4)-8*RootOf(RootOf(_Z^3
-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2))/(2*x^2-x+2)/(1+x))*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)+1/2*
RootOf(_Z^3-4)*ln(-(2*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)^2*RootOf(_Z^3-4)^2*x^3+(x^2+x+2)^(2/
3)*RootOf(_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x-2*(x^2+x+2)^(1/3)*RootOf(_Z^3-4)^2*x
^2-2*(x^2+x+2)^(1/3)*RootOf(_Z^3-4)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x^2-2*(x^2+x+2)^(2/3)*
x+2*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x^2+2*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z
^2)*x+4*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2))/(2*x^2-x+2)/(1+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} + 2 \, x + 6}{{\left (2 \, x^{2} - x + 2\right )} {\left (x^{2} + x + 2\right )}^{\frac {1}{3}} {\left (x + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2*x+6)/(1+x)/(x^2+x+2)^(1/3)/(2*x^2-x+2),x, algorithm="maxima")

[Out]

integrate((x^2 + 2*x + 6)/((2*x^2 - x + 2)*(x^2 + x + 2)^(1/3)*(x + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2+2\,x+6}{\left (x+1\right )\,\left (2\,x^2-x+2\right )\,{\left (x^2+x+2\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + x^2 + 6)/((x + 1)*(2*x^2 - x + 2)*(x + x^2 + 2)^(1/3)),x)

[Out]

int((2*x + x^2 + 6)/((x + 1)*(2*x^2 - x + 2)*(x + x^2 + 2)^(1/3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} + 2 x + 6}{\left (x + 1\right ) \sqrt [3]{x^{2} + x + 2} \left (2 x^{2} - x + 2\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+2*x+6)/(1+x)/(x**2+x+2)**(1/3)/(2*x**2-x+2),x)

[Out]

Integral((x**2 + 2*x + 6)/((x + 1)*(x**2 + x + 2)**(1/3)*(2*x**2 - x + 2)), x)

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