\(\int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} (2-x-16 x^2+x^3+4 x^4+(-2 x^2+x^3) \log (-2+x))}{-2 x^2+x^3} \, dx\) [7233]

Optimal result

Integrand size = 97, antiderivative size = 18 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{e^{x \left (8+\frac {1}{x^2}+2 x+\log (-2+x)\right )}} \]

[Out]

exp(exp((ln(-2+x)+8+1/x^2+2*x)*x))

Rubi [F]

\[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=\int \frac {\exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx \]

[In]

Int[(E^(E^((1 + 8*x^2 + 2*x^3 + x^2*Log[-2 + x])/x) + (1 + 8*x^2 + 2*x^3 + x^2*Log[-2 + x])/x)*(2 - x - 16*x^2
+ x^3 + 4*x^4 + (-2*x^2 + x^3)*Log[-2 + x]))/(-2*x^2 + x^3),x]

[Out]

-16*Defer[Int][E^(E^(x^(-1) + 8*x + 2*x^2)*(-2 + x)^x + x^(-1) + 8*x + 2*x^2)*(-2 + x)^(-1 + x), x] + 2*Defer[
Int][(E^(E^(x^(-1) + 8*x + 2*x^2)*(-2 + x)^x + x^(-1) + 8*x + 2*x^2)*(-2 + x)^(-1 + x))/x^2, x] - Defer[Int][(
E^(E^(x^(-1) + 8*x + 2*x^2)*(-2 + x)^x + x^(-1) + 8*x + 2*x^2)*(-2 + x)^(-1 + x))/x, x] + Defer[Int][E^(E^(x^(
-1) + 8*x + 2*x^2)*(-2 + x)^x + x^(-1) + 8*x + 2*x^2)*(-2 + x)^(-1 + x)*x, x] + 4*Defer[Int][E^(E^(x^(-1) + 8*
x + 2*x^2)*(-2 + x)^x + x^(-1) + 8*x + 2*x^2)*(-2 + x)^(-1 + x)*x^2, x] + Defer[Int][E^(E^((1 + 8*x^2 + 2*x^3
+ x^2*Log[-2 + x])/x) + (1 + 8*x^2 + 2*x^3 + x^2*Log[-2 + x])/x)*Log[-2 + x], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{(-2+x) x^2} \, dx \\ & = \int \left (\frac {\exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \left (2-x-16 x^2+x^3+4 x^4\right )}{(-2+x) x^2}+\exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \log (-2+x)\right ) \, dx \\ & = \int \frac {\exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \left (2-x-16 x^2+x^3+4 x^4\right )}{(-2+x) x^2} \, dx+\int \exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \log (-2+x) \, dx \\ & = \int \frac {\exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x} \left (2-x-16 x^2+x^3+4 x^4\right )}{x^2} \, dx+\int \exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \log (-2+x) \, dx \\ & = \int \left (-16 \exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x}+\frac {2 \exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x}}{x^2}-\frac {\exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x}}{x}+\exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x} x+4 \exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x} x^2\right ) \, dx+\int \exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \log (-2+x) \, dx \\ & = 2 \int \frac {\exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x}}{x^2} \, dx+4 \int \exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x} x^2 \, dx-16 \int \exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x} \, dx-\int \frac {\exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x}}{x} \, dx+\int \exp \left (e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x+\frac {1}{x}+8 x+2 x^2\right ) (-2+x)^{-1+x} x \, dx+\int \exp \left (e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}\right ) \log (-2+x) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{e^{\frac {1}{x}+8 x+2 x^2} (-2+x)^x} \]

[In]

Integrate[(E^(E^((1 + 8*x^2 + 2*x^3 + x^2*Log[-2 + x])/x) + (1 + 8*x^2 + 2*x^3 + x^2*Log[-2 + x])/x)*(2 - x -
16*x^2 + x^3 + 4*x^4 + (-2*x^2 + x^3)*Log[-2 + x]))/(-2*x^2 + x^3),x]

[Out]

E^(E^(x^(-1) + 8*x + 2*x^2)*(-2 + x)^x)

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39

method	result	size
risch	\({\mathrm e}^{\left (-2+x \right )^{x} {\mathrm e}^{\frac {2 x^{3}+8 x^{2}+1}{x}}}\)	\(25\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{\frac {x^{2} \ln \left (-2+x \right )+2 x^{3}+8 x^{2}+1}{x}}}\)	\(27\)

[In]

int(((x^3-2*x^2)*ln(-2+x)+4*x^4+x^3-16*x^2-x+2)*exp((x^2*ln(-2+x)+2*x^3+8*x^2+1)/x)*exp(exp((x^2*ln(-2+x)+2*x^
3+8*x^2+1)/x))/(x^3-2*x^2),x,method=_RETURNVERBOSE)

[Out]

exp((-2+x)^x*exp((2*x^3+8*x^2+1)/x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (16) = 32\).

Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 4.33 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + x e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )} + 1}{x} - \frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )} \]

[In]

integrate(((x^3-2*x^2)*log(-2+x)+4*x^4+x^3-16*x^2-x+2)*exp((x^2*log(-2+x)+2*x^3+8*x^2+1)/x)*exp(exp((x^2*log(-
2+x)+2*x^3+8*x^2+1)/x))/(x^3-2*x^2),x, algorithm="fricas")

[Out]

e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + x*e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x) + 1)/x - (2*x^3 + x^2*log(x
- 2) + 8*x^2 + 1)/x)

Sympy [A] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{e^{\frac {2 x^{3} + x^{2} \log {\left (x - 2 \right )} + 8 x^{2} + 1}{x}}} \]

[In]

integrate(((x**3-2*x**2)*ln(-2+x)+4*x**4+x**3-16*x**2-x+2)*exp((x**2*ln(-2+x)+2*x**3+8*x**2+1)/x)*exp(exp((x**
2*ln(-2+x)+2*x**3+8*x**2+1)/x))/(x**3-2*x**2),x)

[Out]

exp(exp((2*x**3 + x**2*log(x - 2) + 8*x**2 + 1)/x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=e^{\left (e^{\left (2 \, x^{2} + x \log \left (x - 2\right ) + 8 \, x + \frac {1}{x}\right )}\right )} \]

[In]

integrate(((x^3-2*x^2)*log(-2+x)+4*x^4+x^3-16*x^2-x+2)*exp((x^2*log(-2+x)+2*x^3+8*x^2+1)/x)*exp(exp((x^2*log(-
2+x)+2*x^3+8*x^2+1)/x))/(x^3-2*x^2),x, algorithm="maxima")

[Out]

e^(e^(2*x^2 + x*log(x - 2) + 8*x + 1/x))

Giac [F]

\[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx=\int { \frac {{\left (4 \, x^{4} + x^{3} - 16 \, x^{2} + {\left (x^{3} - 2 \, x^{2}\right )} \log \left (x - 2\right ) - x + 2\right )} e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x} + e^{\left (\frac {2 \, x^{3} + x^{2} \log \left (x - 2\right ) + 8 \, x^{2} + 1}{x}\right )}\right )}}{x^{3} - 2 \, x^{2}} \,d x } \]

[In]

integrate(((x^3-2*x^2)*log(-2+x)+4*x^4+x^3-16*x^2-x+2)*exp((x^2*log(-2+x)+2*x^3+8*x^2+1)/x)*exp(exp((x^2*log(-
2+x)+2*x^3+8*x^2+1)/x))/(x^3-2*x^2),x, algorithm="giac")

[Out]

integrate((4*x^4 + x^3 - 16*x^2 + (x^3 - 2*x^2)*log(x - 2) - x + 2)*e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x
+ e^((2*x^3 + x^2*log(x - 2) + 8*x^2 + 1)/x))/(x^3 - 2*x^2), x)

Mupad [B] (verification not implemented)

Time = 13.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {e^{e^{\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}}+\frac {1+8 x^2+2 x^3+x^2 \log (-2+x)}{x}} \left (2-x-16 x^2+x^3+4 x^4+\left (-2 x^2+x^3\right ) \log (-2+x)\right )}{-2 x^2+x^3} \, dx={\mathrm {e}}^{{\mathrm {e}}^{8\,x}\,{\mathrm {e}}^{1/x}\,{\mathrm {e}}^{2\,x^2}\,{\left (x-2\right )}^x} \]

[In]

int((exp(exp((x^2*log(x - 2) + 8*x^2 + 2*x^3 + 1)/x))*exp((x^2*log(x - 2) + 8*x^2 + 2*x^3 + 1)/x)*(x + log(x -
2)*(2*x^2 - x^3) + 16*x^2 - x^3 - 4*x^4 - 2))/(2*x^2 - x^3),x)

[Out]

exp(exp(8*x)*exp(1/x)*exp(2*x^2)*(x - 2)^x)