3.77.24 \(\int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} (e^{4+2 e^x x} (-1+e^x x)+(8 x+8 x^2) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} (-8 x-8 x^2+2 \log (5)+e^x (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5))))}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx\)

Optimal. Leaf size=38 \[ \left (2-e^{-e^x+\frac {(2+2 x)^2}{e^{2+e^x x}-\log (5)}}\right ) x \]

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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[(2*E^(4 + 2*E^x*x) - 4*E^(2 + E^x*x)*Log[5] + 2*Log[5]^2 + E^((4 - E^(2 + x + E^x*x) + 8*x + 4*x^2 + E^x*L
og[5])/(E^(2 + E^x*x) - Log[5]))*(E^(4 + 2*E^x*x)*(-1 + E^x*x) + (8*x + 8*x^2)*Log[5] - Log[5]^2 + E^x*x*Log[5
]^2 + E^(2 + E^x*x)*(-8*x - 8*x^2 + 2*Log[5] + E^x*(4*x + 12*x^2 + 12*x^3 + 4*x^4 - 2*x*Log[5]))))/(E^(4 + 2*E
^x*x) - 2*E^(2 + E^x*x)*Log[5] + Log[5]^2),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [F]  time = 126.13, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 e^{4+2 e^x x}-4 e^{2+e^x x} \log (5)+2 \log ^2(5)+e^{\frac {4-e^{2+x+e^x x}+8 x+4 x^2+e^x \log (5)}{e^{2+e^x x}-\log (5)}} \left (e^{4+2 e^x x} \left (-1+e^x x\right )+\left (8 x+8 x^2\right ) \log (5)-\log ^2(5)+e^x x \log ^2(5)+e^{2+e^x x} \left (-8 x-8 x^2+2 \log (5)+e^x \left (4 x+12 x^2+12 x^3+4 x^4-2 x \log (5)\right )\right )\right )}{e^{4+2 e^x x}-2 e^{2+e^x x} \log (5)+\log ^2(5)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(2*E^(4 + 2*E^x*x) - 4*E^(2 + E^x*x)*Log[5] + 2*Log[5]^2 + E^((4 - E^(2 + x + E^x*x) + 8*x + 4*x^2 +
E^x*Log[5])/(E^(2 + E^x*x) - Log[5]))*(E^(4 + 2*E^x*x)*(-1 + E^x*x) + (8*x + 8*x^2)*Log[5] - Log[5]^2 + E^x*x
*Log[5]^2 + E^(2 + E^x*x)*(-8*x - 8*x^2 + 2*Log[5] + E^x*(4*x + 12*x^2 + 12*x^3 + 4*x^4 - 2*x*Log[5]))))/(E^(4
+ 2*E^x*x) - 2*E^(2 + E^x*x)*Log[5] + Log[5]^2),x]

[Out]

Integrate[(2*E^(4 + 2*E^x*x) - 4*E^(2 + E^x*x)*Log[5] + 2*Log[5]^2 + E^((4 - E^(2 + x + E^x*x) + 8*x + 4*x^2 +
E^x*Log[5])/(E^(2 + E^x*x) - Log[5]))*(E^(4 + 2*E^x*x)*(-1 + E^x*x) + (8*x + 8*x^2)*Log[5] - Log[5]^2 + E^x*x
*Log[5]^2 + E^(2 + E^x*x)*(-8*x - 8*x^2 + 2*Log[5] + E^x*(4*x + 12*x^2 + 12*x^3 + 4*x^4 - 2*x*Log[5]))))/(E^(4
+ 2*E^x*x) - 2*E^(2 + E^x*x)*Log[5] + Log[5]^2), x]

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fricas [A]  time = 1.98, size = 60, normalized size = 1.58 \begin {gather*} -x e^{\left (-\frac {4 \, {\left (x^{2} + 2 \, x + 1\right )} e^{x} + e^{\left (2 \, x\right )} \log \relax (5) - e^{\left (x e^{x} + 2 \, x + 2\right )}}{e^{x} \log \relax (5) - e^{\left (x e^{x} + x + 2\right )}}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x-1)*exp(exp(x)*x+2)^2+((-2*x*log(5)+4*x^4+12*x^3+12*x^2+4*x)*exp(x)+2*log(5)-8*x^2-8*x)*e
xp(exp(x)*x+2)+x*log(5)^2*exp(x)-log(5)^2+(8*x^2+8*x)*log(5))*exp((-exp(x)*exp(exp(x)*x+2)+exp(x)*log(5)+4*x^2
+8*x+4)/(exp(exp(x)*x+2)-log(5)))+2*exp(exp(x)*x+2)^2-4*log(5)*exp(exp(x)*x+2)+2*log(5)^2)/(exp(exp(x)*x+2)^2-
2*log(5)*exp(exp(x)*x+2)+log(5)^2),x, algorithm="fricas")

[Out]

-x*e^(-(4*(x^2 + 2*x + 1)*e^x + e^(2*x)*log(5) - e^(x*e^x + 2*x + 2))/(e^x*log(5) - e^(x*e^x + x + 2))) + 2*x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (x e^{x} \log \relax (5)^{2} + {\left (x e^{x} - 1\right )} e^{\left (2 \, x e^{x} + 4\right )} - 2 \, {\left (4 \, x^{2} - {\left (2 \, x^{4} + 6 \, x^{3} + 6 \, x^{2} - x \log \relax (5) + 2 \, x\right )} e^{x} + 4 \, x - \log \relax (5)\right )} e^{\left (x e^{x} + 2\right )} + 8 \, {\left (x^{2} + x\right )} \log \relax (5) - \log \relax (5)^{2}\right )} e^{\left (\frac {4 \, x^{2} + e^{x} \log \relax (5) + 8 \, x - e^{\left (x e^{x} + x + 2\right )} + 4}{e^{\left (x e^{x} + 2\right )} - \log \relax (5)}\right )} - 4 \, e^{\left (x e^{x} + 2\right )} \log \relax (5) + 2 \, \log \relax (5)^{2} + 2 \, e^{\left (2 \, x e^{x} + 4\right )}}{2 \, e^{\left (x e^{x} + 2\right )} \log \relax (5) - \log \relax (5)^{2} - e^{\left (2 \, x e^{x} + 4\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x-1)*exp(exp(x)*x+2)^2+((-2*x*log(5)+4*x^4+12*x^3+12*x^2+4*x)*exp(x)+2*log(5)-8*x^2-8*x)*e
xp(exp(x)*x+2)+x*log(5)^2*exp(x)-log(5)^2+(8*x^2+8*x)*log(5))*exp((-exp(x)*exp(exp(x)*x+2)+exp(x)*log(5)+4*x^2
+8*x+4)/(exp(exp(x)*x+2)-log(5)))+2*exp(exp(x)*x+2)^2-4*log(5)*exp(exp(x)*x+2)+2*log(5)^2)/(exp(exp(x)*x+2)^2-
2*log(5)*exp(exp(x)*x+2)+log(5)^2),x, algorithm="giac")

[Out]

integrate(-((x*e^x*log(5)^2 + (x*e^x - 1)*e^(2*x*e^x + 4) - 2*(4*x^2 - (2*x^4 + 6*x^3 + 6*x^2 - x*log(5) + 2*x
)*e^x + 4*x - log(5))*e^(x*e^x + 2) + 8*(x^2 + x)*log(5) - log(5)^2)*e^((4*x^2 + e^x*log(5) + 8*x - e^(x*e^x +
x + 2) + 4)/(e^(x*e^x + 2) - log(5))) - 4*e^(x*e^x + 2)*log(5) + 2*log(5)^2 + 2*e^(2*x*e^x + 4))/(2*e^(x*e^x
+ 2)*log(5) - log(5)^2 - e^(2*x*e^x + 4)), x)

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maple [A]  time = 0.22, size = 50, normalized size = 1.32

method	result	size
risch	\(-x \,{\mathrm e}^{-\frac {-{\mathrm e}^{x +{\mathrm e}^{x} x +2}+{\mathrm e}^{x} \ln \relax (5)+4 x^{2}+8 x +4}{-{\mathrm e}^{{\mathrm e}^{x} x +2}+\ln \relax (5)}}+2 x\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((exp(x)*x-1)*exp(exp(x)*x+2)^2+((-2*x*ln(5)+4*x^4+12*x^3+12*x^2+4*x)*exp(x)+2*ln(5)-8*x^2-8*x)*exp(exp(x
)*x+2)+x*ln(5)^2*exp(x)-ln(5)^2+(8*x^2+8*x)*ln(5))*exp((-exp(x)*exp(exp(x)*x+2)+exp(x)*ln(5)+4*x^2+8*x+4)/(exp
(exp(x)*x+2)-ln(5)))+2*exp(exp(x)*x+2)^2-4*ln(5)*exp(exp(x)*x+2)+2*ln(5)^2)/(exp(exp(x)*x+2)^2-2*ln(5)*exp(exp
(x)*x+2)+ln(5)^2),x,method=_RETURNVERBOSE)

[Out]

-x*exp(-(-exp(x+exp(x)*x+2)+exp(x)*ln(5)+4*x^2+8*x+4)/(-exp(exp(x)*x+2)+ln(5)))+2*x

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maxima [B]  time = 0.55, size = 104, normalized size = 2.74 \begin {gather*} -x e^{\left (\frac {4 \, x^{2}}{e^{\left (x e^{x} + 2\right )} - \log \relax (5)} + \frac {e^{x} \log \relax (5)}{e^{\left (x e^{x} + 2\right )} - \log \relax (5)} + \frac {8 \, x}{e^{\left (x e^{x} + 2\right )} - \log \relax (5)} - \frac {e^{\left (x e^{x} + x + 2\right )}}{e^{\left (x e^{x} + 2\right )} - \log \relax (5)} + \frac {4}{e^{\left (x e^{x} + 2\right )} - \log \relax (5)}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x-1)*exp(exp(x)*x+2)^2+((-2*x*log(5)+4*x^4+12*x^3+12*x^2+4*x)*exp(x)+2*log(5)-8*x^2-8*x)*e
xp(exp(x)*x+2)+x*log(5)^2*exp(x)-log(5)^2+(8*x^2+8*x)*log(5))*exp((-exp(x)*exp(exp(x)*x+2)+exp(x)*log(5)+4*x^2
+8*x+4)/(exp(exp(x)*x+2)-log(5)))+2*exp(exp(x)*x+2)^2-4*log(5)*exp(exp(x)*x+2)+2*log(5)^2)/(exp(exp(x)*x+2)^2-
2*log(5)*exp(exp(x)*x+2)+log(5)^2),x, algorithm="maxima")

[Out]

-x*e^(4*x^2/(e^(x*e^x + 2) - log(5)) + e^x*log(5)/(e^(x*e^x + 2) - log(5)) + 8*x/(e^(x*e^x + 2) - log(5)) - e^
(x*e^x + x + 2)/(e^(x*e^x + 2) - log(5)) + 4/(e^(x*e^x + 2) - log(5))) + 2*x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {2\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x+4}-4\,{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}\,\ln \relax (5)+{\mathrm {e}}^{-\frac {8\,x+{\mathrm {e}}^x\,\ln \relax (5)-{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}\,{\mathrm {e}}^x+4\,x^2+4}{\ln \relax (5)-{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}}}\,\left (\ln \relax (5)\,\left (8\,x^2+8\,x\right )-{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}\,\left (8\,x-2\,\ln \relax (5)-{\mathrm {e}}^x\,\left (4\,x-2\,x\,\ln \relax (5)+12\,x^2+12\,x^3+4\,x^4\right )+8\,x^2\right )-{\ln \relax (5)}^2+{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x+4}\,\left (x\,{\mathrm {e}}^x-1\right )+x\,{\mathrm {e}}^x\,{\ln \relax (5)}^2\right )+2\,{\ln \relax (5)}^2}{{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x+4}-2\,{\mathrm {e}}^{x\,{\mathrm {e}}^x+2}\,\ln \relax (5)+{\ln \relax (5)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(2*x*exp(x) + 4) - 4*exp(x*exp(x) + 2)*log(5) + exp(-(8*x + exp(x)*log(5) - exp(x*exp(x) + 2)*exp(x)
+ 4*x^2 + 4)/(log(5) - exp(x*exp(x) + 2)))*(log(5)*(8*x + 8*x^2) - exp(x*exp(x) + 2)*(8*x - 2*log(5) - exp(x)
*(4*x - 2*x*log(5) + 12*x^2 + 12*x^3 + 4*x^4) + 8*x^2) - log(5)^2 + exp(2*x*exp(x) + 4)*(x*exp(x) - 1) + x*exp
(x)*log(5)^2) + 2*log(5)^2)/(exp(2*x*exp(x) + 4) - 2*exp(x*exp(x) + 2)*log(5) + log(5)^2),x)

[Out]

int((2*exp(2*x*exp(x) + 4) - 4*exp(x*exp(x) + 2)*log(5) + exp(-(8*x + exp(x)*log(5) - exp(x*exp(x) + 2)*exp(x)
+ 4*x^2 + 4)/(log(5) - exp(x*exp(x) + 2)))*(log(5)*(8*x + 8*x^2) - exp(x*exp(x) + 2)*(8*x - 2*log(5) - exp(x)
*(4*x - 2*x*log(5) + 12*x^2 + 12*x^3 + 4*x^4) + 8*x^2) - log(5)^2 + exp(2*x*exp(x) + 4)*(x*exp(x) - 1) + x*exp
(x)*log(5)^2) + 2*log(5)^2)/(exp(2*x*exp(x) + 4) - 2*exp(x*exp(x) + 2)*log(5) + log(5)^2), x)

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sympy [A]  time = 8.04, size = 46, normalized size = 1.21 \begin {gather*} - x e^{\frac {4 x^{2} + 8 x - e^{x} e^{x e^{x} + 2} + e^{x} \log {\relax (5 )} + 4}{e^{x e^{x} + 2} - \log {\relax (5 )}}} + 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(x)*x-1)*exp(exp(x)*x+2)**2+((-2*x*ln(5)+4*x**4+12*x**3+12*x**2+4*x)*exp(x)+2*ln(5)-8*x**2-8*x
)*exp(exp(x)*x+2)+x*ln(5)**2*exp(x)-ln(5)**2+(8*x**2+8*x)*ln(5))*exp((-exp(x)*exp(exp(x)*x+2)+exp(x)*ln(5)+4*x
**2+8*x+4)/(exp(exp(x)*x+2)-ln(5)))+2*exp(exp(x)*x+2)**2-4*ln(5)*exp(exp(x)*x+2)+2*ln(5)**2)/(exp(exp(x)*x+2)*
*2-2*ln(5)*exp(exp(x)*x+2)+ln(5)**2),x)

[Out]

-x*exp((4*x**2 + 8*x - exp(x)*exp(x*exp(x) + 2) + exp(x)*log(5) + 4)/(exp(x*exp(x) + 2) - log(5))) + 2*x

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