3.56.93 \(\int \frac {2 e^x+e^{1+e^{-x} x} (x-x^2) \log (x) \log (125 e^2 \log ^2(x))}{e^{1+x+e^{-x} x} x \log (x) \log (125 e^2 \log ^2(x))+e^x x \log (x) \log (125 e^2 \log ^2(x)) \log (\log (125 e^2 \log ^2(x)))} \, dx\) [5593]

Optimal. Leaf size=24 \[ \log \left (e^{1+e^{-x} x}+\log \left (\log \left (125 e^2 \log ^2(x)\right )\right )\right ) \]

[Out]

ln(exp(x/exp(x))*exp(1)+ln(ln(125*exp(2)*ln(x)^2)))

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Rubi [A]
time = 1.05, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, integrand size = 95, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6873, 6816} \begin {gather*} \log \left (e^{e^{-x} x+1}+\log \left (\log \left (125 \log ^2(x)\right )+2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^x + E^(1 + x/E^x)*(x - x^2)*Log[x]*Log[125*E^2*Log[x]^2])/(E^(1 + x + x/E^x)*x*Log[x]*Log[125*E^2*Log
[x]^2] + E^x*x*Log[x]*Log[125*E^2*Log[x]^2]*Log[Log[125*E^2*Log[x]^2]]),x]

[Out]

Log[E^(1 + x/E^x) + Log[2 + Log[125*Log[x]^2]]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )\right )}{x \log (x) \left (2 \left (1+\frac {3 \log (5)}{2}\right )+\log \left (\log ^2(x)\right )\right ) \left (e^{1+e^{-x} x}+\log \left (2+\log \left (125 \log ^2(x)\right )\right )\right )} \, dx\\ &=\log \left (e^{1+e^{-x} x}+\log \left (2+\log \left (125 \log ^2(x)\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [F]
time = 0.41, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(2*E^x + E^(1 + x/E^x)*(x - x^2)*Log[x]*Log[125*E^2*Log[x]^2])/(E^(1 + x + x/E^x)*x*Log[x]*Log[125*E
^2*Log[x]^2] + E^x*x*Log[x]*Log[125*E^2*Log[x]^2]*Log[Log[125*E^2*Log[x]^2]]),x]

[Out]

Integrate[(2*E^x + E^(1 + x/E^x)*(x - x^2)*Log[x]*Log[125*E^2*Log[x]^2])/(E^(1 + x + x/E^x)*x*Log[x]*Log[125*E
^2*Log[x]^2] + E^x*x*Log[x]*Log[125*E^2*Log[x]^2]*Log[Log[125*E^2*Log[x]^2]]), x]

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 10.10, size = 55, normalized size = 2.29

method	result	size
risch	\(\ln \left ({\mathrm e}^{1+x \,{\mathrm e}^{-x}}+\ln \left (3 \ln \left (5\right )+2+2 \ln \left (\ln \left (x \right )\right )-\frac {i \pi \,\mathrm {csgn}\left (i \ln \left (x \right )^{2}\right ) \left (-\mathrm {csgn}\left (i \ln \left (x \right )^{2}\right )+\mathrm {csgn}\left (i \ln \left (x \right )\right )\right )^{2}}{2}\right )\right )\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+x)*exp(1)*ln(x)*exp(x/exp(x))*ln(125*exp(2)*ln(x)^2)+2*exp(x))/(x*exp(x)*ln(x)*ln(125*exp(2)*ln(x)^
2)*ln(ln(125*exp(2)*ln(x)^2))+x*exp(1)*exp(x)*ln(x)*exp(x/exp(x))*ln(125*exp(2)*ln(x)^2)),x,method=_RETURNVERB
OSE)

[Out]

ln(exp(1+x*exp(-x))+ln(3*ln(5)+2+2*ln(ln(x))-1/2*I*Pi*csgn(I*ln(x)^2)*(-csgn(I*ln(x)^2)+csgn(I*ln(x)))^2))

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Maxima [A]
time = 0.68, size = 23, normalized size = 0.96 \begin {gather*} \log \left (e^{\left (x e^{\left (-x\right )} + 1\right )} + \log \left (3 \, \log \left (5\right ) + 2 \, \log \left (\log \left (x\right )\right ) + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+x)*exp(1)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)+2*exp(x))/(x*exp(x)*log(x)*log(125*ex
p(2)*log(x)^2)*log(log(125*exp(2)*log(x)^2))+x*exp(1)*exp(x)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)),x,
algorithm="maxima")

[Out]

log(e^(x*e^(-x) + 1) + log(3*log(5) + 2*log(log(x)) + 2))

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Fricas [A]
time = 0.37, size = 34, normalized size = 1.42 \begin {gather*} \log \left ({\left (e^{x} \log \left (\log \left (125 \, e^{2} \log \left (x\right )^{2}\right )\right ) + e^{\left ({\left ({\left (x + 1\right )} e^{x} + x\right )} e^{\left (-x\right )}\right )}\right )} e^{\left (-x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+x)*exp(1)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)+2*exp(x))/(x*exp(x)*log(x)*log(125*ex
p(2)*log(x)^2)*log(log(125*exp(2)*log(x)^2))+x*exp(1)*exp(x)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)),x,
algorithm="fricas")

[Out]

log((e^x*log(log(125*e^2*log(x)^2)) + e^(((x + 1)*e^x + x)*e^(-x)))*e^(-x))

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Sympy [A]
time = 1.08, size = 24, normalized size = 1.00 \begin {gather*} \log {\left (e^{x e^{- x}} + \frac {\log {\left (\log {\left (125 e^{2} \log {\left (x \right )}^{2} \right )} \right )}}{e} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+x)*exp(1)*ln(x)*exp(x/exp(x))*ln(125*exp(2)*ln(x)**2)+2*exp(x))/(x*exp(x)*ln(x)*ln(125*exp(2
)*ln(x)**2)*ln(ln(125*exp(2)*ln(x)**2))+x*exp(1)*exp(x)*ln(x)*exp(x/exp(x))*ln(125*exp(2)*ln(x)**2)),x)

[Out]

log(exp(x*exp(-x)) + exp(-1)*log(log(125*exp(2)*log(x)**2)))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+x)*exp(1)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)+2*exp(x))/(x*exp(x)*log(x)*log(125*ex
p(2)*log(x)^2)*log(log(125*exp(2)*log(x)^2))+x*exp(1)*exp(x)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)),x,
algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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Mupad [B]
time = 5.04, size = 21, normalized size = 0.88 \begin {gather*} \ln \left ({\mathrm {e}}^{x\,{\mathrm {e}}^{-x}+1}+\ln \left (\ln \left (125\,{\mathrm {e}}^2\,{\ln \left (x\right )}^2\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(x) + exp(1)*exp(x*exp(-x))*log(125*exp(2)*log(x)^2)*log(x)*(x - x^2))/(x*exp(x)*log(125*exp(2)*log(
x)^2)*log(log(125*exp(2)*log(x)^2))*log(x) + x*exp(1)*exp(x*exp(-x))*exp(x)*log(125*exp(2)*log(x)^2)*log(x)),x
)

[Out]

log(exp(x*exp(-x) + 1) + log(log(125*exp(2)*log(x)^2)))

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