∫ −1+e1+ex−x+e2(2x−x2) (−x+exx+e2(2x−2x2))____________________ e2+2ex−2x+2e2(2x−x2)x+x log2(2)−2x log(2) log(x)+x log2(x)+e1+ex−x+e2(2x−x2)(2x log(2)−2x log(x)) dx

3.11.82 $\int \frac {-1+e^{1+e^x-x+e^2 (2 x-x^2)} (-x+e^x x+e^2 (2 x-2 x^2))}{e^{2+2 e^x-2 x+2 e^2 (2 x-x^2)} x+x \log ^2(2)-2 x \log (2) \log (x)+x \log ^2(x)+e^{1+e^x-x+e^2 (2 x-x^2)} (2 x \log (2)-2 x \log (x))} \, dx$

Optimal. Leaf size=30 \[ \frac {1}{-e^{-1+e^x-(-2+x) \left (1+e^2 x\right )}-\log (2)+\log (x)} \]

________________________________________________________________________________________

Rubi [F] time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Veriﬁcation is not applicable to the result.

[In]

Int[(-1 + E^(1 + E^x - x + E^2*(2*x - x^2))*(-x + E^x*x + E^2*(2*x - 2*x^2)))/(E^(2 + 2*E^x - 2*x + 2*E^2*(2*x

- x^2))*x + x*Log[2]^2 - 2*x*Log[2]*Log[x] + x*Log[x]^2 + E^(1 + E^x - x + E^2*(2*x - x^2))*(2*x*Log[2] - 2*x

*Log[x])),x]

[Out]

$Aborted

Rubi steps

Aborted

________________________________________________________________________________________

Mathematica [A] time = 0.38, size = 55, normalized size = 1.83 \begin {gather*} -\frac {e^{e^2 x^2}}{e^{1+e^x-x+2 e^2 x}+e^{e^2 x^2} \log (2)-e^{e^2 x^2} \log (x)} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Integrate[(-1 + E^(1 + E^x - x + E^2*(2*x - x^2))*(-x + E^x*x + E^2*(2*x - 2*x^2)))/(E^(2 + 2*E^x - 2*x + 2*E^

2*(2*x - x^2))*x + x*Log[2]^2 - 2*x*Log[2]*Log[x] + x*Log[x]^2 + E^(1 + E^x - x + E^2*(2*x - x^2))*(2*x*Log[2]

- 2*x*Log[x])),x]

[Out]

-(E^(E^2*x^2)/(E^(1 + E^x - x + 2*E^2*x) + E^(E^2*x^2)*Log[2] - E^(E^2*x^2)*Log[x]))

________________________________________________________________________________________

fricas [A] time = 0.59, size = 30, normalized size = 1.00 \begin {gather*} -\frac {1}{e^{\left (-{\left (x^{2} - 2 \, x\right )} e^{2} - x + e^{x} + 1\right )} + \log \relax (2) - \log \relax (x)} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+(-2*x^2+2*x)*exp(2)-x)*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1)-1)/(x*exp(exp(x)+(-x^2+2*x)*exp(

2)-x+1)^2+(-2*x*log(x)+2*x*log(2))*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1)+x*log(x)^2-2*x*log(2)*log(x)+x*log(2)^2),

x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B] time = 2.49, size = 1135, normalized size = 37.83 result too large to display

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+(-2*x^2+2*x)*exp(2)-x)*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1)-1)/(x*exp(exp(x)+(-x^2+2*x)*exp(

2)-x+1)^2+(-2*x*log(x)+2*x*log(2))*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1)+x*log(x)^2-2*x*log(2)*log(x)+x*log(2)^2),

x, algorithm="giac")

[Out]

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

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

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

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

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

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

og(x) - e^x*log(2) + e^x*log(x) - e^(-x^2*e^2 + 2*x*e^2 + e^x + 1))/(2*x^2*e^(x + 2)*log(2)^3 - 6*x^2*e^(x + 2

)*log(2)^2*log(x) + 6*x^2*e^(x + 2)*log(2)*log(x)^2 - 2*x^2*e^(x + 2)*log(x)^3 + 4*x^2*e^(-x^2*e^2 + 2*x*e^2 +

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

2 + e^x + 3)*log(2)*log(x) + 3*x*e^(2*x)*log(2)^2*log(x) + 6*x*e^(x + 2)*log(2)^2*log(x) - 3*x*e^x*log(2)^2*lo

g(x) + 4*x^2*e^(-x^2*e^2 + 2*x*e^2 + e^x + 3)*log(x)^2 - 3*x*e^(2*x)*log(2)*log(x)^2 - 6*x*e^(x + 2)*log(2)*lo

g(x)^2 + 3*x*e^x*log(2)*log(x)^2 + x*e^(2*x)*log(x)^3 + 2*x*e^(x + 2)*log(x)^3 - x*e^x*log(x)^3 + 2*x^2*e^(-2*

x^2*e^2 + 4*x*e^2 - x + 2*e^x + 4)*log(2) - 2*x*e^(-x^2*e^2 + 2*x*e^2 + x + e^x + 1)*log(2)^2 - 4*x*e^(-x^2*e^

2 + 2*x*e^2 + e^x + 3)*log(2)^2 + 2*x*e^(-x^2*e^2 + 2*x*e^2 + e^x + 1)*log(2)^2 - 2*x^2*e^(-2*x^2*e^2 + 4*x*e^

2 - x + 2*e^x + 4)*log(x) + 4*x*e^(-x^2*e^2 + 2*x*e^2 + x + e^x + 1)*log(2)*log(x) + 8*x*e^(-x^2*e^2 + 2*x*e^2

+ e^x + 3)*log(2)*log(x) - 4*x*e^(-x^2*e^2 + 2*x*e^2 + e^x + 1)*log(2)*log(x) - 2*x*e^(-x^2*e^2 + 2*x*e^2 + x

+ e^x + 1)*log(x)^2 - 4*x*e^(-x^2*e^2 + 2*x*e^2 + e^x + 3)*log(x)^2 + 2*x*e^(-x^2*e^2 + 2*x*e^2 + e^x + 1)*lo

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

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

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

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

^x + 1)*log(x) - e^(-2*x^2*e^2 + 4*x*e^2 - x + 2*e^x + 2))

________________________________________________________________________________________

maple [A] time = 0.11, size = 32, normalized size = 1.07


method	result	size

risch	$-\frac {1}{\ln \relax (2)-\ln \relax (x )+{\mathrm e}^{-x^{2} {\mathrm e}^{2}+2 \,{\mathrm e}^{2} x +{\mathrm e}^{x}-x +1}}$	$32$

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)*x+(-2*x^2+2*x)*exp(2)-x)*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1)-1)/(x*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1

)^2+(-2*x*ln(x)+2*x*ln(2))*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1)+x*ln(x)^2-2*x*ln(2)*ln(x)+x*ln(2)^2),x,method=_RE

TURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A] time = 0.75, size = 41, normalized size = 1.37 \begin {gather*} -\frac {e^{\left (x^{2} e^{2} + x\right )}}{{\left (\log \relax (2) - \log \relax (x)\right )} e^{\left (x^{2} e^{2} + x\right )} + e^{\left (2 \, x e^{2} + e^{x} + 1\right )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+(-2*x^2+2*x)*exp(2)-x)*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1)-1)/(x*exp(exp(x)+(-x^2+2*x)*exp(

2)-x+1)^2+(-2*x*log(x)+2*x*log(2))*exp(exp(x)+(-x^2+2*x)*exp(2)-x+1)+x*log(x)^2-2*x*log(2)*log(x)+x*log(2)^2),

x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{{\mathrm {e}}^x-x+{\mathrm {e}}^2\,\left (2\,x-x^2\right )+1}\,\left ({\mathrm {e}}^2\,\left (2\,x-2\,x^2\right )-x+x\,{\mathrm {e}}^x\right )-1}{x\,{\ln \relax (x)}^2+{\mathrm {e}}^{{\mathrm {e}}^x-x+{\mathrm {e}}^2\,\left (2\,x-x^2\right )+1}\,\left (2\,x\,\ln \relax (2)-2\,x\,\ln \relax (x)\right )+x\,{\ln \relax (2)}^2+x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x-2\,x+2\,{\mathrm {e}}^2\,\left (2\,x-x^2\right )+2}-2\,x\,\ln \relax (2)\,\ln \relax (x)} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x) - x + exp(2)*(2*x - x^2) + 1)*(exp(2)*(2*x - 2*x^2) - x + x*exp(x)) - 1)/(x*log(x)^2 + exp(exp

(x) - x + exp(2)*(2*x - x^2) + 1)*(2*x*log(2) - 2*x*log(x)) + x*log(2)^2 + x*exp(2*exp(x) - 2*x + 2*exp(2)*(2*

x - x^2) + 2) - 2*x*log(2)*log(x)),x)

[Out]

int((exp(exp(x) - x + exp(2)*(2*x - x^2) + 1)*(exp(2)*(2*x - 2*x^2) - x + x*exp(x)) - 1)/(x*log(x)^2 + exp(exp

(x) - x + exp(2)*(2*x - x^2) + 1)*(2*x*log(2) - 2*x*log(x)) + x*log(2)^2 + x*exp(2*exp(x) - 2*x + 2*exp(2)*(2*

x - x^2) + 2) - 2*x*log(2)*log(x)), x)

________________________________________________________________________________________

sympy [A] time = 0.41, size = 27, normalized size = 0.90 \begin {gather*} - \frac {1}{e^{- x + \left (- x^{2} + 2 x\right ) e^{2} + e^{x} + 1} - \log {\relax (x )} + \log {\relax (2 )}} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)*x+(-2*x**2+2*x)*exp(2)-x)*exp(exp(x)+(-x**2+2*x)*exp(2)-x+1)-1)/(x*exp(exp(x)+(-x**2+2*x)*e

xp(2)-x+1)**2+(-2*x*ln(x)+2*x*ln(2))*exp(exp(x)+(-x**2+2*x)*exp(2)-x+1)+x*ln(x)**2-2*x*ln(2)*ln(x)+x*ln(2)**2)

,x)

[Out]

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

________________________________________________________________________________________

[next] [prev] [prev-tail] [front] [up]

3.11.82 \(\int \frac {-1+e^{1+e^x-x+e^2 (2 x-x^2)} (-x+e^x x+e^2 (2 x-2 x^2))}{e^{2+2 e^x-2 x+2 e^2 (2 x-x^2)} x+x \log ^2(2)-2 x \log (2) \log (x)+x \log ^2(x)+e^{1+e^x-x+e^2 (2 x-x^2)} (2 x \log (2)-2 x \log (x))} \, dx\)