\(\int \frac {2-e^x-x-x^2+(64 x^2+64 e^x x^2+128 x^3) \log ^3(-2+e^x+x+x^2)}{-2 x^2+e^x x^2+x^3+x^4} \, dx\) [7768]

Optimal result

Integrand size = 69, antiderivative size = 18 \[ \int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx=\frac {1}{x}+16 \log ^4\left (-2+e^x+x+x^2\right ) \]

[Out]

16*ln(exp(x)+x^2+x-2)^4+1/x

Rubi [F]

\[ \int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx=\int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx \]

[In]

Int[(2 - E^x - x - x^2 + (64*x^2 + 64*E^x*x^2 + 128*x^3)*Log[-2 + E^x + x + x^2]^3)/(-2*x^2 + E^x*x^2 + x^3 +
x^4),x]

[Out]

x^(-1) + 64*Defer[Int][Log[-2 + E^x + x + x^2]^3, x] + 192*Defer[Int][Log[-2 + E^x + x + x^2]^3/(-2 + E^x + x
+ x^2), x] + 64*Defer[Int][(x*Log[-2 + E^x + x + x^2]^3)/(-2 + E^x + x + x^2), x] - 64*Defer[Int][(x^2*Log[-2
+ E^x + x + x^2]^3)/(-2 + E^x + x + x^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {64 \left (-3-x+x^2\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2+e^x+x+x^2}+\frac {-1+64 x^2 \log ^3\left (-2+e^x+x+x^2\right )}{x^2}\right ) \, dx \\ & = -\left (64 \int \frac {\left (-3-x+x^2\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2+e^x+x+x^2} \, dx\right )+\int \frac {-1+64 x^2 \log ^3\left (-2+e^x+x+x^2\right )}{x^2} \, dx \\ & = -\left (64 \int \left (-\frac {3 \log ^3\left (-2+e^x+x+x^2\right )}{-2+e^x+x+x^2}-\frac {x \log ^3\left (-2+e^x+x+x^2\right )}{-2+e^x+x+x^2}+\frac {x^2 \log ^3\left (-2+e^x+x+x^2\right )}{-2+e^x+x+x^2}\right ) \, dx\right )+\int \left (-\frac {1}{x^2}+64 \log ^3\left (-2+e^x+x+x^2\right )\right ) \, dx \\ & = \frac {1}{x}+64 \int \log ^3\left (-2+e^x+x+x^2\right ) \, dx+64 \int \frac {x \log ^3\left (-2+e^x+x+x^2\right )}{-2+e^x+x+x^2} \, dx-64 \int \frac {x^2 \log ^3\left (-2+e^x+x+x^2\right )}{-2+e^x+x+x^2} \, dx+192 \int \frac {\log ^3\left (-2+e^x+x+x^2\right )}{-2+e^x+x+x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx=\frac {1}{x}+16 \log ^4\left (-2+e^x+x+x^2\right ) \]

[In]

Integrate[(2 - E^x - x - x^2 + (64*x^2 + 64*E^x*x^2 + 128*x^3)*Log[-2 + E^x + x + x^2]^3)/(-2*x^2 + E^x*x^2 +
x^3 + x^4),x]

[Out]

x^(-1) + 16*Log[-2 + E^x + x + x^2]^4

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00

method	result	size
risch	\(16 \ln \left ({\mathrm e}^{x}+x^{2}+x -2\right )^{4}+\frac {1}{x}\)	\(18\)
parallelrisch	\(-\frac {-32 \ln \left ({\mathrm e}^{x}+x^{2}+x -2\right )^{4} x -2+2 x}{2 x}\)	\(25\)

[In]

int(((64*exp(x)*x^2+128*x^3+64*x^2)*ln(exp(x)+x^2+x-2)^3-exp(x)-x^2-x+2)/(exp(x)*x^2+x^4+x^3-2*x^2),x,method=_
RETURNVERBOSE)

[Out]

16*ln(exp(x)+x^2+x-2)^4+1/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx=\frac {16 \, x \log \left (x^{2} + x + e^{x} - 2\right )^{4} + 1}{x} \]

[In]

integrate(((64*exp(x)*x^2+128*x^3+64*x^2)*log(exp(x)+x^2+x-2)^3-exp(x)-x^2-x+2)/(exp(x)*x^2+x^4+x^3-2*x^2),x,
algorithm="fricas")

[Out]

(16*x*log(x^2 + x + e^x - 2)^4 + 1)/x

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx=16 \log {\left (x^{2} + x + e^{x} - 2 \right )}^{4} + \frac {1}{x} \]

[In]

integrate(((64*exp(x)*x**2+128*x**3+64*x**2)*ln(exp(x)+x**2+x-2)**3-exp(x)-x**2-x+2)/(exp(x)*x**2+x**4+x**3-2*
x**2),x)

[Out]

16*log(x**2 + x + exp(x) - 2)**4 + 1/x

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx=\frac {16 \, x \log \left (x^{2} + x + e^{x} - 2\right )^{4} + 1}{x} \]

[In]

integrate(((64*exp(x)*x^2+128*x^3+64*x^2)*log(exp(x)+x^2+x-2)^3-exp(x)-x^2-x+2)/(exp(x)*x^2+x^4+x^3-2*x^2),x,
algorithm="maxima")

[Out]

(16*x*log(x^2 + x + e^x - 2)^4 + 1)/x

Giac [F]

\[ \int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx=\int { \frac {64 \, {\left (2 \, x^{3} + x^{2} e^{x} + x^{2}\right )} \log \left (x^{2} + x + e^{x} - 2\right )^{3} - x^{2} - x - e^{x} + 2}{x^{4} + x^{3} + x^{2} e^{x} - 2 \, x^{2}} \,d x } \]

[In]

integrate(((64*exp(x)*x^2+128*x^3+64*x^2)*log(exp(x)+x^2+x-2)^3-exp(x)-x^2-x+2)/(exp(x)*x^2+x^4+x^3-2*x^2),x,
algorithm="giac")

[Out]

integrate((64*(2*x^3 + x^2*e^x + x^2)*log(x^2 + x + e^x - 2)^3 - x^2 - x - e^x + 2)/(x^4 + x^3 + x^2*e^x - 2*x
^2), x)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {2-e^x-x-x^2+\left (64 x^2+64 e^x x^2+128 x^3\right ) \log ^3\left (-2+e^x+x+x^2\right )}{-2 x^2+e^x x^2+x^3+x^4} \, dx=16\,{\ln \left (x+{\mathrm {e}}^x+x^2-2\right )}^4+\frac {1}{x} \]

[In]

int(-(x + exp(x) - log(x + exp(x) + x^2 - 2)^3*(64*x^2*exp(x) + 64*x^2 + 128*x^3) + x^2 - 2)/(x^2*exp(x) - 2*x
^2 + x^3 + x^4),x)

[Out]

16*log(x + exp(x) + x^2 - 2)^4 + 1/x