3.102.100 \(\int \frac {x+2 e^{e^{2 x}+2 x} x+(5-e^{e^{2 x}}-x) \log (5-e^{e^{2 x}}-x)}{(-5+e^{e^{2 x}}+x) \log ^2(5-e^{e^{2 x}}-x)} \, dx\)

Optimal. Leaf size=28 \[ -e^{9/2}-\frac {x}{\log \left (5-e^{e^{2 x}}-x\right )} \]

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Rubi [F]  time = 1.49, 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*} \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x + 2*E^(E^(2*x) + 2*x)*x + (5 - E^E^(2*x) - x)*Log[5 - E^E^(2*x) - x])/((-5 + E^E^(2*x) + x)*Log[5 - E^E
^(2*x) - x]^2),x]

[Out]

Defer[Int][x/((-5 + E^E^(2*x) + x)*Log[5 - E^E^(2*x) - x]^2), x] + 2*Defer[Int][(E^(E^(2*x) + 2*x)*x)/((-5 + E
^E^(2*x) + x)*Log[5 - E^E^(2*x) - x]^2), x] - Defer[Int][Log[5 - E^E^(2*x) - x]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 e^{e^{2 x}+2 x} x}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )}-\frac {-x-5 \log \left (5-e^{e^{2 x}}-x\right )+e^{e^{2 x}} \log \left (5-e^{e^{2 x}}-x\right )+x \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )}\right ) \, dx\\ &=2 \int \frac {e^{e^{2 x}+2 x} x}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx-\int \frac {-x-5 \log \left (5-e^{e^{2 x}}-x\right )+e^{e^{2 x}} \log \left (5-e^{e^{2 x}}-x\right )+x \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx\\ &=2 \int \frac {e^{e^{2 x}+2 x} x}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx-\int \frac {-\frac {x}{-5+e^{e^{2 x}}+x}+\log \left (5-e^{e^{2 x}}-x\right )}{\log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx\\ &=2 \int \frac {e^{e^{2 x}+2 x} x}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx-\int \left (-\frac {x}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )}+\frac {1}{\log \left (5-e^{e^{2 x}}-x\right )}\right ) \, dx\\ &=2 \int \frac {e^{e^{2 x}+2 x} x}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx+\int \frac {x}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx-\int \frac {1}{\log \left (5-e^{e^{2 x}}-x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.47, size = 20, normalized size = 0.71 \begin {gather*} -\frac {x}{\log \left (5-e^{e^{2 x}}-x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + 2*E^(E^(2*x) + 2*x)*x + (5 - E^E^(2*x) - x)*Log[5 - E^E^(2*x) - x])/((-5 + E^E^(2*x) + x)*Log[5
- E^E^(2*x) - x]^2),x]

[Out]

-(x/Log[5 - E^E^(2*x) - x])

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fricas [A]  time = 0.69, size = 30, normalized size = 1.07 \begin {gather*} -\frac {x}{\log \left (-{\left ({\left (x - 5\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(exp(x)^2)+5-x)*log(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(exp(x)^2)+x)/(exp(exp(x)^2)+x-5)/log(
-exp(exp(x)^2)+5-x)^2,x, algorithm="fricas")

[Out]

-x/log(-((x - 5)*e^(2*x) + e^(2*x + e^(2*x)))*e^(-2*x))

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giac [B]  time = 0.18, size = 281, normalized size = 10.04 \begin {gather*} -\frac {2 \, x e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )} \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right ) + 4 \, x e^{\left (4 \, x + 2 \, e^{\left (2 \, x\right )}\right )} \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right ) + 2 \, x e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )} \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right ) + x \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right )}{4 \, e^{\left (4 \, x + 2 \, e^{\left (2 \, x\right )}\right )} \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right ) \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right ) + 4 \, e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )} \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right ) \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right ) + \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right ) \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(exp(x)^2)+5-x)*log(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(exp(x)^2)+x)/(exp(exp(x)^2)+x-5)/log(
-exp(exp(x)^2)+5-x)^2,x, algorithm="giac")

[Out]

-(2*x*e^(2*x + e^(2*x))*log(-(x*e^(2*x) - 5*e^(2*x) + e^(2*x + e^(2*x)))*e^(-2*x)) + 4*x*e^(4*x + 2*e^(2*x))*l
og(-x - e^(e^(2*x)) + 5) + 2*x*e^(2*x + e^(2*x))*log(-x - e^(e^(2*x)) + 5) + x*log(-(x*e^(2*x) - 5*e^(2*x) + e
^(2*x + e^(2*x)))*e^(-2*x)))/(4*e^(4*x + 2*e^(2*x))*log(-(x*e^(2*x) - 5*e^(2*x) + e^(2*x + e^(2*x)))*e^(-2*x))
*log(-x - e^(e^(2*x)) + 5) + 4*e^(2*x + e^(2*x))*log(-(x*e^(2*x) - 5*e^(2*x) + e^(2*x + e^(2*x)))*e^(-2*x))*lo
g(-x - e^(e^(2*x)) + 5) + log(-(x*e^(2*x) - 5*e^(2*x) + e^(2*x + e^(2*x)))*e^(-2*x))*log(-x - e^(e^(2*x)) + 5)
)

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maple [A]  time = 0.06, size = 19, normalized size = 0.68

method	result	size
risch	\(-\frac {x}{\ln \left (-{\mathrm e}^{{\mathrm e}^{2 x}}+5-x \right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(exp(x)^2)+5-x)*ln(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(exp(x)^2)+x)/(exp(exp(x)^2)+x-5)/ln(-exp(exp
(x)^2)+5-x)^2,x,method=_RETURNVERBOSE)

[Out]

-x/ln(-exp(exp(2*x))+5-x)

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maxima [A]  time = 0.40, size = 18, normalized size = 0.64 \begin {gather*} -\frac {x}{\log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(exp(x)^2)+5-x)*log(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(exp(x)^2)+x)/(exp(exp(x)^2)+x-5)/log(
-exp(exp(x)^2)+5-x)^2,x, algorithm="maxima")

[Out]

-x/log(-x - e^(e^(2*x)) + 5)

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mupad [B]  time = 7.08, size = 106, normalized size = 3.79 \begin {gather*} -\frac {{\mathrm {e}}^{-2\,x}\,\left (x\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{2\,x+{\mathrm {e}}^{2\,x}}\,\ln \left (5-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x\right )-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\ln \left (5-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x\right )+2\,x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^{2\,x}}\right )}{\ln \left (5-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x\right )\,\left (2\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^{2\,x}}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log(5 - exp(exp(2*x)) - x)*(x + exp(exp(2*x)) - 5) + 2*x*exp(2*x)*exp(exp(2*x)))/(log(5 - exp(exp(2*x
)) - x)^2*(x + exp(exp(2*x)) - 5)),x)

[Out]

-(exp(-2*x)*(x*exp(2*x) + exp(2*x + exp(2*x))*log(5 - exp(exp(2*x)) - x) - exp(2*x)*exp(exp(2*x))*log(5 - exp(
exp(2*x)) - x) + 2*x*exp(2*x)*exp(2*x + exp(2*x))))/(log(5 - exp(exp(2*x)) - x)*(2*exp(2*x + exp(2*x)) + 1))

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sympy [A]  time = 0.33, size = 14, normalized size = 0.50 \begin {gather*} - \frac {x}{\log {\left (- x - e^{e^{2 x}} + 5 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(exp(x)**2)+5-x)*ln(-exp(exp(x)**2)+5-x)+2*x*exp(x)**2*exp(exp(x)**2)+x)/(exp(exp(x)**2)+x-5)/
ln(-exp(exp(x)**2)+5-x)**2,x)

[Out]

-x/log(-x - exp(exp(2*x)) + 5)

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