3.11.28 \(\int \frac {-2 e^{2 x} x^5+(-5 x^5+5 x^4 \log (4 e^{e^{2 x}+x})) \log (-x+\log (4 e^{e^{2 x}+x}))+(-x-x \log (x)+\log (4 e^{e^{2 x}+x}) (1+\log (x))) \log ^2(-x+\log (4 e^{e^{2 x}+x}))}{(-x+\log (4 e^{e^{2 x}+x})) \log ^2(-x+\log (4 e^{e^{2 x}+x}))} \, dx\) [1028]

Optimal. Leaf size=28 \[ x \left (\log (x)+\frac {x^4}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}\right ) \]

[Out]

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

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Rubi [F]
time = 0.59, 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 {-2 e^{2 x} x^5+\left (-5 x^5+5 x^4 \log \left (4 e^{e^{2 x}+x}\right )\right ) \log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )+\left (-x-x \log (x)+\log \left (4 e^{e^{2 x}+x}\right ) (1+\log (x))\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )}{\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right ) \log ^2\left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 28, normalized size = 1.00 \begin {gather*} x \log (x)+\frac {x^5}{\log \left (-x+\log \left (4 e^{e^{2 x}+x}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.48, size = 29, normalized size = 1.04

method	result	size
risch	\(x \ln \left (x \right )+\frac {x^{5}}{\ln \left (2 \ln \left (2\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{2 x}+x}\right )-x \right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((ln(x)+1)*ln(4*exp(exp(2*x)+x))-x*ln(x)-x)*ln(ln(4*exp(exp(2*x)+x))-x)^2+(5*x^4*ln(4*exp(exp(2*x)+x))-5*
x^5)*ln(ln(4*exp(exp(2*x)+x))-x)-2*x^5*exp(2*x))/(ln(4*exp(exp(2*x)+x))-x)/ln(ln(4*exp(exp(2*x)+x))-x)^2,x,met
hod=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.57, size = 31, normalized size = 1.11 \begin {gather*} \frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 31, normalized size = 1.11 \begin {gather*} \frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
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((((ln(x)+1)*ln(4*exp(exp(2*x)+x))-x*ln(x)-x)*ln(ln(4*exp(exp(2*x)+x))-x)**2+(5*x**4*ln(4*exp(exp(2*x
)+x))-5*x**5)*ln(ln(4*exp(exp(2*x)+x))-x)-2*x**5*exp(2*x))/(ln(4*exp(exp(2*x)+x))-x)/ln(ln(4*exp(exp(2*x)+x))-
x)**2,x)

[Out]

Timed out

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Giac [A]
time = 0.41, size = 31, normalized size = 1.11 \begin {gather*} \frac {x^{5} + x \log \left (x\right ) \log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )}{\log \left (e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.91, size = 19, normalized size = 0.68 \begin {gather*} \frac {x^5}{\ln \left ({\mathrm {e}}^{2\,x}+\ln \left (4\right )\right )}+x\,\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(4*exp(x + exp(2*x))) - x)^2*(x + x*log(x) - log(4*exp(x + exp(2*x)))*(log(x) + 1)) - log(log(4*ex
p(x + exp(2*x))) - x)*(5*x^4*log(4*exp(x + exp(2*x))) - 5*x^5) + 2*x^5*exp(2*x))/(log(log(4*exp(x + exp(2*x)))
- x)^2*(x - log(4*exp(x + exp(2*x))))),x)

[Out]

x^5/log(exp(2*x) + log(4)) + x*log(x)

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