\(\int \frac {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} (2-x+2 x^2)} \, dx\) [1856]

Optimal result

Integrand size = 36, antiderivative size = 128 \[ \int \frac {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {-\frac {2 \sqrt [3]{2} x}{\sqrt {3}}+\frac {\sqrt [3]{2+x+x^2}}{\sqrt {3}}}{\sqrt [3]{2+x+x^2}}\right )}{\sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2} x+\sqrt [3]{2+x+x^2}\right )}{\sqrt [3]{2}}-\frac {\log \left (2^{2/3} x^2-\sqrt [3]{2} x \sqrt [3]{2+x+x^2}+\left (2+x+x^2\right )^{2/3}\right )}{2 \sqrt [3]{2}} \]

[Out]

-1/2*3^(1/2)*arctan((-2/3*2^(1/3)*x*3^(1/2)+1/3*(x^2+x+2)^(1/3)*3^(1/2))/(x^2+x+2)^(1/3))*2^(2/3)+1/2*ln(2^(1/
3)*x+(x^2+x+2)^(1/3))*2^(2/3)-1/4*ln(2^(2/3)*x^2-2^(1/3)*x*(x^2+x+2)^(1/3)+(x^2+x+2)^(2/3))*2^(2/3)

Rubi [F]

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

[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*} \text {integral}& = \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 ) \text {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}} \operatorname {AppellF1}\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*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.81 \[ \int \frac {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{2+x+x^2}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{2} x+\sqrt [3]{2+x+x^2}\right )+\log \left (2^{2/3} x^2-\sqrt [3]{2} x \sqrt [3]{2+x+x^2}+\left (2+x+x^2\right )^{2/3}\right )}{2 \sqrt [3]{2}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 5.21 (sec) , antiderivative size = 671, normalized size of antiderivative = 5.24

method	result	size
trager	\(\text {Expression too large to display}\)	\(671\)

[In]

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

[Out]

RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*ln((RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*Ro
otOf(_Z^3-4)^3*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(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootO
f(_Z^3-4)*x^2+2*RootOf(_Z^3-4)*x^3-RootOf(_Z^3-4)*x^2+4*(x^2+x+2)^(2/3)*x-RootOf(_Z^3-4)*x-2*RootOf(_Z^3-4))/(
2*x^2-x+2)/(1+x))-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+(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+(x^2+x+2)^(1/3)*RootOf(_Z^3-4)^2*
x^2-2*(x^2+x+2)^(1/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)*x^2+RootOf(_Z^3-4)*x^
2-2*(x^2+x+2)^(2/3)*x+RootOf(_Z^3-4)*x+2*RootOf(_Z^3-4))/(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+(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+(x^2+x+2)^(1/3)*RootOf(_Z^3-4)^2*x^2-2*(x^2+x+2)^(1/3)*RootOf(RootOf(_Z^3-
4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)*x^2+RootOf(_Z^3-4)*x^2-2*(x^2+x+2)^(2/3)*x+RootOf(_Z^3-4)*x+2*
RootOf(_Z^3-4))/(2*x^2-x+2)/(1+x))*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 407 vs. \(2 (99) = 198\).

Time = 10.35 (sec) , antiderivative size = 407, normalized size of antiderivative = 3.18 \[ \int \frac {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx=-\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (2^{\frac {5}{6}} {\left (8 \, x^{9} + 48 \, x^{8} + 18 \, x^{7} + 37 \, x^{6} - 147 \, x^{5} - 111 \, x^{4} - 107 \, x^{3} + 18 \, x^{2} + 12 \, x + 8\right )} + 12 \, \sqrt {2} {\left (4 \, x^{8} - 14 \, x^{7} - 13 \, x^{6} - 26 \, x^{5} + 5 \, x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} {\left (x^{2} + x + 2\right )}^{\frac {1}{3}} + 12 \cdot 2^{\frac {1}{6}} {\left (8 \, x^{7} + 2 \, x^{6} + x^{5} + 2 \, x^{4} - 5 \, x^{3} - 4 \, x^{2} - 4 \, x\right )} {\left (x^{2} + x + 2\right )}^{\frac {2}{3}}\right )}}{6 \, {\left (8 \, x^{9} - 96 \, x^{8} - 90 \, x^{7} - 179 \, x^{6} + 33 \, x^{5} + 33 \, x^{4} + 37 \, x^{3} + 18 \, x^{2} + 12 \, x + 8\right )}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (\frac {6 \cdot 2^{\frac {1}{3}} {\left (x^{2} + x + 2\right )}^{\frac {1}{3}} x^{2} + 2^{\frac {2}{3}} {\left (2 \, x^{3} + x^{2} + x + 2\right )} + 6 \, {\left (x^{2} + x + 2\right )}^{\frac {2}{3}} x}{2 \, x^{3} + x^{2} + x + 2}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (\frac {3 \cdot 2^{\frac {2}{3}} {\left (4 \, x^{4} - x^{3} - x^{2} - 2 \, x\right )} {\left (x^{2} + x + 2\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (4 \, x^{6} - 14 \, x^{5} - 13 \, x^{4} - 26 \, x^{3} + 5 \, x^{2} + 4 \, x + 4\right )} - 12 \, {\left (x^{5} - x^{4} - x^{3} - 2 \, x^{2}\right )} {\left (x^{2} + x + 2\right )}^{\frac {1}{3}}}{4 \, x^{6} + 4 \, x^{5} + 5 \, x^{4} + 10 \, x^{3} + 5 \, x^{2} + 4 \, x + 4}\right ) \]

[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))

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {6+2 x+x^2}{(1+x) \sqrt [3]{2+x+x^2} \left (2-x+2 x^2\right )} \, dx=\int \frac {x^2+2\,x+6}{\left (x+1\right )\,\left (2\,x^2-x+2\right )\,{\left (x^2+x+2\right )}^{1/3}} \,d x \]

[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)