3.59.20 \(\int \frac {5 x-x^3+e^2 (5-x^2)+e^{2 e^{2 e^{\log (x) \log (e^2+x)}}} (e^2+x+e^{2 e^{\log (x) \log (e^2+x)}+\log (x) \log (e^2+x)} (-4 x \log (x)+(-4 e^2-4 x) \log (e^2+x)))}{25 x-10 x^2+11 x^3-2 x^4+x^5+e^{4 e^{2 e^{\log (x) \log (e^2+x)}}} (e^2+x)+e^2 (25-10 x+11 x^2-2 x^3+x^4)+e^{2 e^{2 e^{\log (x) \log (e^2+x)}}} (10 x-2 x^2+2 x^3+e^2 (10-2 x+2 x^2))} \, dx\)

Optimal. Leaf size=31 \[ \frac {x}{5+e^{2 e^{2 e^{\log (x) \log \left (e^2+x\right )}}}-x+x^2} \]

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Rubi [F]  time = 13.87, 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 {5 x-x^3+e^2 \left (5-x^2\right )+e^{2 e^{2 e^{\log (x) \log \left (e^2+x\right )}}} \left (e^2+x+\exp \left (2 e^{\log (x) \log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \left (-4 x \log (x)+\left (-4 e^2-4 x\right ) \log \left (e^2+x\right )\right )\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+e^{4 e^{2 e^{\log (x) \log \left (e^2+x\right )}}} \left (e^2+x\right )+e^2 \left (25-10 x+11 x^2-2 x^3+x^4\right )+e^{2 e^{2 e^{\log (x) \log \left (e^2+x\right )}}} \left (10 x-2 x^2+2 x^3+e^2 \left (10-2 x+2 x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x - x^3 + E^2*(5 - x^2) + E^(2*E^(2*E^(Log[x]*Log[E^2 + x])))*(E^2 + x + E^(2*E^(Log[x]*Log[E^2 + x]) +
Log[x]*Log[E^2 + x])*(-4*x*Log[x] + (-4*E^2 - 4*x)*Log[E^2 + x])))/(25*x - 10*x^2 + 11*x^3 - 2*x^4 + x^5 + E^
(4*E^(2*E^(Log[x]*Log[E^2 + x])))*(E^2 + x) + E^2*(25 - 10*x + 11*x^2 - 2*x^3 + x^4) + E^(2*E^(2*E^(Log[x]*Log
[E^2 + x])))*(10*x - 2*x^2 + 2*x^3 + E^2*(10 - 2*x + 2*x^2))),x]

[Out]

5*Defer[Int][(5 + E^(2*E^(2*x^Log[E^2 + x])) - x + x^2)^(-2), x] + Defer[Int][E^(2*E^(2*x^Log[E^2 + x]))/(5 +
E^(2*E^(2*x^Log[E^2 + x])) - x + x^2)^2, x] - Defer[Int][x^2/(5 + E^(2*E^(2*x^Log[E^2 + x])) - x + x^2)^2, x]
- 4*Defer[Int][(E^(2*E^(2*x^Log[E^2 + x]) + 2*x^Log[E^2 + x] + Log[x]*Log[E^2 + x])*Log[x])/(5 + E^(2*E^(2*x^L
og[E^2 + x])) - x + x^2)^2, x] + 4*Defer[Int][(E^(2 + 2*E^(2*x^Log[E^2 + x]) + 2*x^Log[E^2 + x] + Log[x]*Log[E
^2 + x])*Log[x])/((E^2 + x)*(5 + E^(2*E^(2*x^Log[E^2 + x])) - x + x^2)^2), x] - 4*Defer[Int][(E^(2*E^(2*x^Log[
E^2 + x]) + 2*x^Log[E^2 + x] + Log[x]*Log[E^2 + x])*Log[E^2 + x])/(5 + E^(2*E^(2*x^Log[E^2 + x])) - x + x^2)^2
, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4 \exp \left (2 \left (e^{2 x^{\log \left (e^2+x\right )}}+x^{\log \left (e^2+x\right )}\right )+\log (x) \log \left (e^2+x\right )\right ) x \log (x)-\left (e^2+x\right ) \left (-5-e^{2 e^{2 x^{\log \left (e^2+x\right )}}}+x^2+4 \exp \left (2 \left (e^{2 x^{\log \left (e^2+x\right )}}+x^{\log \left (e^2+x\right )}\right )+\log (x) \log \left (e^2+x\right )\right ) \log \left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\\ &=\int \left (\frac {5}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}+\frac {e^{2 e^{2 x^{\log \left (e^2+x\right )}}}}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}-\frac {x^2}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}-\frac {4 \exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \left (x \log (x)+e^2 \log \left (e^2+x\right )+x \log \left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}\right ) \, dx\\ &=-\left (4 \int \frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \left (x \log (x)+e^2 \log \left (e^2+x\right )+x \log \left (e^2+x\right )\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\right )+5 \int \frac {1}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx+\int \frac {e^{2 e^{2 x^{\log \left (e^2+x\right )}}}}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx-\int \frac {x^2}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\\ &=-\left (4 \int \left (\frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) x \log (x)}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}+\frac {\exp \left (2+2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log \left (e^2+x\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}+\frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) x \log \left (e^2+x\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}\right ) \, dx\right )+5 \int \frac {1}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx+\int \frac {e^{2 e^{2 x^{\log \left (e^2+x\right )}}}}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx-\int \frac {x^2}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\\ &=-\left (4 \int \frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) x \log (x)}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\right )-4 \int \frac {\exp \left (2+2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log \left (e^2+x\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx-4 \int \frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) x \log \left (e^2+x\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx+5 \int \frac {1}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx+\int \frac {e^{2 e^{2 x^{\log \left (e^2+x\right )}}}}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx-\int \frac {x^2}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\\ &=-\left (4 \int \left (\frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log (x)}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}-\frac {\exp \left (2+2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log (x)}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}\right ) \, dx\right )-4 \int \frac {\exp \left (2+2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log \left (e^2+x\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx-4 \int \left (\frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log \left (e^2+x\right )}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}-\frac {\exp \left (2+2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log \left (e^2+x\right )}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2}\right ) \, dx+5 \int \frac {1}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx+\int \frac {e^{2 e^{2 x^{\log \left (e^2+x\right )}}}}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx-\int \frac {x^2}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\\ &=-\left (4 \int \frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log (x)}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\right )+4 \int \frac {\exp \left (2+2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log (x)}{\left (e^2+x\right ) \left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx-4 \int \frac {\exp \left (2 e^{2 x^{\log \left (e^2+x\right )}}+2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )\right ) \log \left (e^2+x\right )}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx+5 \int \frac {1}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx+\int \frac {e^{2 e^{2 x^{\log \left (e^2+x\right )}}}}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx-\int \frac {x^2}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.81, size = 201, normalized size = 6.48 \begin {gather*} \frac {x \left (4 e^{2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )} x \left (5-x+x^2\right ) \log (x)+\left (e^2+x\right ) \left (x-2 x^2+4 e^{2 x^{\log \left (e^2+x\right )}+\log (x) \log \left (e^2+x\right )} \left (5-x+x^2\right ) \log \left (e^2+x\right )\right )\right )}{\left (5+e^{2 e^{2 x^{\log \left (e^2+x\right )}}}-x+x^2\right ) \left (4 e^{2 x^{\log \left (e^2+x\right )}} x^{1+\log \left (e^2+x\right )} \left (5-x+x^2\right ) \log (x)+\left (e^2+x\right ) \left (x-2 x^2+4 e^{2 x^{\log \left (e^2+x\right )}} x^{\log \left (e^2+x\right )} \left (5-x+x^2\right ) \log \left (e^2+x\right )\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x - x^3 + E^2*(5 - x^2) + E^(2*E^(2*E^(Log[x]*Log[E^2 + x])))*(E^2 + x + E^(2*E^(Log[x]*Log[E^2 +
x]) + Log[x]*Log[E^2 + x])*(-4*x*Log[x] + (-4*E^2 - 4*x)*Log[E^2 + x])))/(25*x - 10*x^2 + 11*x^3 - 2*x^4 + x^
5 + E^(4*E^(2*E^(Log[x]*Log[E^2 + x])))*(E^2 + x) + E^2*(25 - 10*x + 11*x^2 - 2*x^3 + x^4) + E^(2*E^(2*E^(Log[
x]*Log[E^2 + x])))*(10*x - 2*x^2 + 2*x^3 + E^2*(10 - 2*x + 2*x^2))),x]

[Out]

(x*(4*E^(2*x^Log[E^2 + x] + Log[x]*Log[E^2 + x])*x*(5 - x + x^2)*Log[x] + (E^2 + x)*(x - 2*x^2 + 4*E^(2*x^Log[
E^2 + x] + Log[x]*Log[E^2 + x])*(5 - x + x^2)*Log[E^2 + x])))/((5 + E^(2*E^(2*x^Log[E^2 + x])) - x + x^2)*(4*E
^(2*x^Log[E^2 + x])*x^(1 + Log[E^2 + x])*(5 - x + x^2)*Log[x] + (E^2 + x)*(x - 2*x^2 + 4*E^(2*x^Log[E^2 + x])*
x^Log[E^2 + x]*(5 - x + x^2)*Log[E^2 + x])))

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fricas [A]  time = 0.52, size = 27, normalized size = 0.87 \begin {gather*} \frac {x}{x^{2} - x + e^{\left (2 \, e^{\left (2 \, e^{\left (\log \left (x + e^{2}\right ) \log \relax (x)\right )}\right )}\right )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-4*exp(2)-4*x)*log(x+exp(2))-4*x*log(x))*exp(log(x)*log(x+exp(2)))*exp(exp(log(x)*log(x+exp(2)))
)^2+x+exp(2))*exp(exp(exp(log(x)*log(x+exp(2))))^2)^2+(-x^2+5)*exp(2)-x^3+5*x)/((x+exp(2))*exp(exp(exp(log(x)*
log(x+exp(2))))^2)^4+((2*x^2-2*x+10)*exp(2)+2*x^3-2*x^2+10*x)*exp(exp(exp(log(x)*log(x+exp(2))))^2)^2+(x^4-2*x
^3+11*x^2-10*x+25)*exp(2)+x^5-2*x^4+11*x^3-10*x^2+25*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 28.45, size = 27, normalized size = 0.87 \begin {gather*} \frac {x}{x^{2} - x + e^{\left (2 \, e^{\left (2 \, e^{\left (\log \left (x + e^{2}\right ) \log \relax (x)\right )}\right )}\right )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-4*exp(2)-4*x)*log(x+exp(2))-4*x*log(x))*exp(log(x)*log(x+exp(2)))*exp(exp(log(x)*log(x+exp(2)))
)^2+x+exp(2))*exp(exp(exp(log(x)*log(x+exp(2))))^2)^2+(-x^2+5)*exp(2)-x^3+5*x)/((x+exp(2))*exp(exp(exp(log(x)*
log(x+exp(2))))^2)^4+((2*x^2-2*x+10)*exp(2)+2*x^3-2*x^2+10*x)*exp(exp(exp(log(x)*log(x+exp(2))))^2)^2+(x^4-2*x
^3+11*x^2-10*x+25)*exp(2)+x^5-2*x^4+11*x^3-10*x^2+25*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.13, size = 26, normalized size = 0.84

method	result	size
risch	\(\frac {x}{x^{2}-x +{\mathrm e}^{2 \,{\mathrm e}^{2 \left (x +{\mathrm e}^{2}\right )^{\ln \relax (x )}}}+5}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-4*exp(2)-4*x)*ln(x+exp(2))-4*x*ln(x))*exp(ln(x)*ln(x+exp(2)))*exp(exp(ln(x)*ln(x+exp(2))))^2+x+exp(2)
)*exp(exp(exp(ln(x)*ln(x+exp(2))))^2)^2+(-x^2+5)*exp(2)-x^3+5*x)/((x+exp(2))*exp(exp(exp(ln(x)*ln(x+exp(2))))^
2)^4+((2*x^2-2*x+10)*exp(2)+2*x^3-2*x^2+10*x)*exp(exp(exp(ln(x)*ln(x+exp(2))))^2)^2+(x^4-2*x^3+11*x^2-10*x+25)
*exp(2)+x^5-2*x^4+11*x^3-10*x^2+25*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.72, size = 27, normalized size = 0.87 \begin {gather*} \frac {x}{x^{2} - x + e^{\left (2 \, e^{\left (2 \, e^{\left (\log \left (x + e^{2}\right ) \log \relax (x)\right )}\right )}\right )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-4*exp(2)-4*x)*log(x+exp(2))-4*x*log(x))*exp(log(x)*log(x+exp(2)))*exp(exp(log(x)*log(x+exp(2)))
)^2+x+exp(2))*exp(exp(exp(log(x)*log(x+exp(2))))^2)^2+(-x^2+5)*exp(2)-x^3+5*x)/((x+exp(2))*exp(exp(exp(log(x)*
log(x+exp(2))))^2)^4+((2*x^2-2*x+10)*exp(2)+2*x^3-2*x^2+10*x)*exp(exp(exp(log(x)*log(x+exp(2))))^2)^2+(x^4-2*x
^3+11*x^2-10*x+25)*exp(2)+x^5-2*x^4+11*x^3-10*x^2+25*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.93, size = 732, normalized size = 23.61 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + exp(2*exp(2*exp(log(x + exp(2))*log(x))))*(x + exp(2) - exp(2*exp(log(x + exp(2))*log(x)))*exp(log(
x + exp(2))*log(x))*(log(x + exp(2))*(4*x + 4*exp(2)) + 4*x*log(x))) - x^3 - exp(2)*(x^2 - 5))/(25*x + exp(2*e
xp(2*exp(log(x + exp(2))*log(x))))*(10*x + exp(2)*(2*x^2 - 2*x + 10) - 2*x^2 + 2*x^3) + exp(2)*(11*x^2 - 10*x
- 2*x^3 + x^4 + 25) + exp(4*exp(2*exp(log(x + exp(2))*log(x))))*(x + exp(2)) - 10*x^2 + 11*x^3 - 2*x^4 + x^5),
x)

[Out]

((x*exp(2) + x^2)^2*(x*exp(2) - 2*x^2*exp(2) + x^2 - 2*x^3 + 20*x^log(x + exp(2))*exp(2*x^log(x + exp(2)) + 2)
*log(x + exp(2)) + 4*x^log(x + exp(2))*x^2*exp(2*x^log(x + exp(2)) + 2)*log(x + exp(2)) + 20*x*x^log(x + exp(2
))*exp(2*x^log(x + exp(2)))*log(x + exp(2)) + 20*x*x^log(x + exp(2))*exp(2*x^log(x + exp(2)))*log(x) - 4*x*x^l
og(x + exp(2))*exp(2*x^log(x + exp(2)) + 2)*log(x + exp(2)) - 4*x^log(x + exp(2))*x^2*exp(2*x^log(x + exp(2)))
*log(x + exp(2)) + 4*x^log(x + exp(2))*x^3*exp(2*x^log(x + exp(2)))*log(x + exp(2)) - 4*x^log(x + exp(2))*x^2*
exp(2*x^log(x + exp(2)))*log(x) + 4*x^log(x + exp(2))*x^3*exp(2*x^log(x + exp(2)))*log(x)))/((x + exp(2))*(exp
(2*exp(2*x^log(x + exp(2)))) - x + x^2 + 5)*(2*x^3*exp(2) + x^2*exp(4) - 4*x^4*exp(2) - 2*x^3*exp(4) + x^4 - 2
*x^5 + 40*x^log(x + exp(2))*x^2*exp(2*x^log(x + exp(2)) + 2)*log(x + exp(2)) - 8*x^log(x + exp(2))*x^3*exp(2*x
^log(x + exp(2)) + 2)*log(x + exp(2)) - 4*x^log(x + exp(2))*x^2*exp(2*x^log(x + exp(2)) + 4)*log(x + exp(2)) +
8*x^log(x + exp(2))*x^4*exp(2*x^log(x + exp(2)) + 2)*log(x + exp(2)) + 4*x^log(x + exp(2))*x^3*exp(2*x^log(x
+ exp(2)) + 4)*log(x + exp(2)) + 20*x^log(x + exp(2))*x^2*exp(2*x^log(x + exp(2)) + 2)*log(x) - 4*x^log(x + ex
p(2))*x^3*exp(2*x^log(x + exp(2)) + 2)*log(x) + 4*x^log(x + exp(2))*x^4*exp(2*x^log(x + exp(2)) + 2)*log(x) +
20*x*x^log(x + exp(2))*exp(2*x^log(x + exp(2)) + 4)*log(x + exp(2)) + 20*x^log(x + exp(2))*x^3*exp(2*x^log(x +
exp(2)))*log(x + exp(2)) - 4*x^log(x + exp(2))*x^4*exp(2*x^log(x + exp(2)))*log(x + exp(2)) + 4*x^log(x + exp
(2))*x^5*exp(2*x^log(x + exp(2)))*log(x + exp(2)) + 20*x^log(x + exp(2))*x^3*exp(2*x^log(x + exp(2)))*log(x) -
4*x^log(x + exp(2))*x^4*exp(2*x^log(x + exp(2)))*log(x) + 4*x^log(x + exp(2))*x^5*exp(2*x^log(x + exp(2)))*lo
g(x)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-4*exp(2)-4*x)*ln(x+exp(2))-4*x*ln(x))*exp(ln(x)*ln(x+exp(2)))*exp(exp(ln(x)*ln(x+exp(2))))**2+x
+exp(2))*exp(exp(exp(ln(x)*ln(x+exp(2))))**2)**2+(-x**2+5)*exp(2)-x**3+5*x)/((x+exp(2))*exp(exp(exp(ln(x)*ln(x
+exp(2))))**2)**4+((2*x**2-2*x+10)*exp(2)+2*x**3-2*x**2+10*x)*exp(exp(exp(ln(x)*ln(x+exp(2))))**2)**2+(x**4-2*
x**3+11*x**2-10*x+25)*exp(2)+x**5-2*x**4+11*x**3-10*x**2+25*x),x)

[Out]

Timed out

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