\(\int \frac {e^{-e^x+e^{-e^x+2 x} x (-2 x+x^2) \log ^4(2)} (-2 x+x^2) (e^{3 x} (2 x-x^2) \log ^4(2)+e^{2 x} (-4-x+2 x^2) \log ^4(2))}{-2+x} \, dx\) [96]

Optimal result

Integrand size = 85, antiderivative size = 24 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{e^{-e^x+2 x} (-2+x) x^2 \log ^4(2)} \]

[Out]

exp(ln(2)^4*exp(ln((-2+x)*x)-exp(x))*x*exp(x)^2)

Rubi [F]

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

[In]

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

[Out]

-4*Log[2]^4*Defer[Int][E^(-E^x + 2*x + E^(-E^x + 2*x)*x*(-2*x + x^2)*Log[2]^4)*x, x] - Log[2]^4*Defer[Int][E^(
-E^x + 2*x + E^(-E^x + 2*x)*x*(-2*x + x^2)*Log[2]^4)*x^2, x] + 2*Log[2]^4*Defer[Int][E^(-E^x + 3*x + E^(-E^x +
2*x)*x*(-2*x + x^2)*Log[2]^4)*x^2, x] + 2*Log[2]^4*Defer[Int][E^(-E^x + 2*x + E^(-E^x + 2*x)*x*(-2*x + x^2)*L
og[2]^4)*x^3, x] - Log[2]^4*Defer[Int][E^(-E^x + 3*x + E^(-E^x + 2*x)*x*(-2*x + x^2)*Log[2]^4)*x^3, x]

Rubi steps \begin{align*} \text {integral}& = \int \exp \left (-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right ) \, dx \\ & = \int \exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x \left (-4-x+2 e^x x+2 x^2-e^x x^2\right ) \log ^4(2) \, dx \\ & = \log ^4(2) \int \exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x \left (-4-x+2 e^x x+2 x^2-e^x x^2\right ) \, dx \\ & = \log ^4(2) \int \left (-\exp \left (-e^x+3 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) (-2+x) x^2+\exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x \left (-4-x+2 x^2\right )\right ) \, dx \\ & = -\left (\log ^4(2) \int \exp \left (-e^x+3 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) (-2+x) x^2 \, dx\right )+\log ^4(2) \int \exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x \left (-4-x+2 x^2\right ) \, dx \\ & = \log ^4(2) \int \left (-4 \exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x-\exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x^2+2 \exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x^3\right ) \, dx-\log ^4(2) \int \left (-2 \exp \left (-e^x+3 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x^2+\exp \left (-e^x+3 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x^3\right ) \, dx \\ & = -\left (\log ^4(2) \int \exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x^2 \, dx\right )-\log ^4(2) \int \exp \left (-e^x+3 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x^3 \, dx+\left (2 \log ^4(2)\right ) \int \exp \left (-e^x+3 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x^2 \, dx+\left (2 \log ^4(2)\right ) \int \exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x^3 \, dx-\left (4 \log ^4(2)\right ) \int \exp \left (-e^x+2 x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)\right ) x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{e^{-e^x+2 x} (-2+x) x^2 \log ^4(2)} \]

[In]

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

[Out]

E^(E^(-E^x + 2*x)*(-2 + x)*x^2*Log[2]^4)

Maple [A] (verified)

Time = 180.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08

method	result	size
parallelrisch	\({\mathrm e}^{x \ln \left (2\right )^{4} {\mathrm e}^{2 x} {\mathrm e}^{\ln \left (x^{2}-2 x \right )-{\mathrm e}^{x}}}\)	\(26\)
risch	\({\mathrm e}^{\ln \left (2\right )^{4} \left (-2+x \right ) x^{2} {\mathrm e}^{2 x -\frac {i \pi \operatorname {csgn}\left (i \left (-2+x \right ) x \right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i \left (-2+x \right ) x \right )^{2} \operatorname {csgn}\left (i x \right )}{2}+\frac {i \pi \operatorname {csgn}\left (i \left (-2+x \right ) x \right )^{2} \operatorname {csgn}\left (i \left (-2+x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (-2+x \right ) x \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-2+x \right )\right )}{2}-{\mathrm e}^{x}}}\)	\(100\)

[In]

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

[Out]

exp(x*ln(2)^4*exp(x)^2*exp(ln(x^2-2*x)-exp(x)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{\left (x e^{\left (2 \, x - e^{x} + \log \left (x^{2} - 2 \, x\right )\right )} \log \left (2\right )^{4}\right )} \]

[In]

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

[Out]

e^(x*e^(2*x - e^x + log(x^2 - 2*x))*log(2)^4)

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{x \left (x^{2} - 2 x\right ) e^{2 x} e^{- e^{x}} \log {\left (2 \right )}^{4}} \]

[In]

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

[Out]

exp(x*(x**2 - 2*x)*exp(2*x)*exp(-exp(x))*log(2)**4)

Maxima [A] (verification not implemented)

none

Time = 0.54 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx=e^{\left (x^{3} e^{\left (2 \, x - e^{x}\right )} \log \left (2\right )^{4} - 2 \, x^{2} e^{\left (2 \, x - e^{x}\right )} \log \left (2\right )^{4}\right )} \]

[In]

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

[Out]

e^(x^3*e^(2*x - e^x)*log(2)^4 - 2*x^2*e^(2*x - e^x)*log(2)^4)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.69 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {e^{-e^x+e^{-e^x+2 x} x \left (-2 x+x^2\right ) \log ^4(2)} \left (-2 x+x^2\right ) \left (e^{3 x} \left (2 x-x^2\right ) \log ^4(2)+e^{2 x} \left (-4-x+2 x^2\right ) \log ^4(2)\right )}{-2+x} \, dx={\mathrm {e}}^{x^3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\ln \left (2\right )}^4}\,{\mathrm {e}}^{-2\,x^2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\ln \left (2\right )}^4} \]

[In]

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

[Out]

exp(x^3*exp(2*x)*exp(-exp(x))*log(2)^4)*exp(-2*x^2*exp(2*x)*exp(-exp(x))*log(2)^4)