3.91.62 \(\int \frac {-x+e^x x+(e^x-x-\log (3)) \log (e^x-x-\log (3)) \log (\log (e^x-x-\log (3)))}{(e^x-x-\log (3)) \log (e^x-x-\log (3))} \, dx\)

Optimal. Leaf size=15 \[ x \log \left (\log \left (e^x-x-\log (3)\right )\right ) \]

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Rubi [F]  time = 1.50, 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+e^x x+\left (e^x-x-\log (3)\right ) \log \left (e^x-x-\log (3)\right ) \log \left (\log \left (e^x-x-\log (3)\right )\right )}{\left (e^x-x-\log (3)\right ) \log \left (e^x-x-\log (3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-x + E^x*x + (E^x - x - Log[3])*Log[E^x - x - Log[3]]*Log[Log[E^x - x - Log[3]]])/((E^x - x - Log[3])*Log
[E^x - x - Log[3]]),x]

[Out]

Defer[Int][x/Log[E^x - x - Log[3]], x] + (1 - Log[3])*Defer[Int][x/((-E^x + x + Log[3])*Log[E^x - x - Log[3]])
, x] - Defer[Int][x^2/((-E^x + x + Log[3])*Log[E^x - x - Log[3]]), x] + Defer[Int][Log[Log[E^x - x - Log[3]]],
x]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 15, normalized size = 1.00 \begin {gather*} x \log \left (\log \left (e^x-x-\log (3)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + E^x*x + (E^x - x - Log[3])*Log[E^x - x - Log[3]]*Log[Log[E^x - x - Log[3]]])/((E^x - x - Log[3
])*Log[E^x - x - Log[3]]),x]

[Out]

x*Log[Log[E^x - x - Log[3]]]

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fricas [A]  time = 0.54, size = 14, normalized size = 0.93 \begin {gather*} x \log \left (\log \left (-x + e^{x} - \log \relax (3)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-log(3)-x)*log(exp(x)-log(3)-x)*log(log(exp(x)-log(3)-x))+exp(x)*x-x)/(exp(x)-log(3)-x)/log(
exp(x)-log(3)-x),x, algorithm="fricas")

[Out]

x*log(log(-x + e^x - log(3)))

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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 (3)\right )} \log \left (-x + e^{x} - \log \relax (3)\right ) \log \left (\log \left (-x + e^{x} - \log \relax (3)\right )\right ) - x e^{x} + x}{{\left (x - e^{x} + \log \relax (3)\right )} \log \left (-x + e^{x} - \log \relax (3)\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-log(3)-x)*log(exp(x)-log(3)-x)*log(log(exp(x)-log(3)-x))+exp(x)*x-x)/(exp(x)-log(3)-x)/log(
exp(x)-log(3)-x),x, algorithm="giac")

[Out]

integrate(((x - e^x + log(3))*log(-x + e^x - log(3))*log(log(-x + e^x - log(3))) - x*e^x + x)/((x - e^x + log(
3))*log(-x + e^x - log(3))), x)

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maple [A]  time = 0.04, size = 15, normalized size = 1.00

method	result	size
risch	\(\ln \left (\ln \left ({\mathrm e}^{x}-\ln \relax (3)-x \right )\right ) x\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)-ln(3)-x)*ln(exp(x)-ln(3)-x)*ln(ln(exp(x)-ln(3)-x))+exp(x)*x-x)/(exp(x)-ln(3)-x)/ln(exp(x)-ln(3)-x
),x,method=_RETURNVERBOSE)

[Out]

ln(ln(exp(x)-ln(3)-x))*x

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maxima [A]  time = 0.49, size = 14, normalized size = 0.93 \begin {gather*} x \log \left (\log \left (-x + e^{x} - \log \relax (3)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-log(3)-x)*log(exp(x)-log(3)-x)*log(log(exp(x)-log(3)-x))+exp(x)*x-x)/(exp(x)-log(3)-x)/log(
exp(x)-log(3)-x),x, algorithm="maxima")

[Out]

x*log(log(-x + e^x - log(3)))

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mupad [B]  time = 7.33, size = 14, normalized size = 0.93 \begin {gather*} x\,\ln \left (\ln \left ({\mathrm {e}}^x-\ln \relax (3)-x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - x*exp(x) + log(exp(x) - log(3) - x)*log(log(exp(x) - log(3) - x))*(x + log(3) - exp(x)))/(log(exp(x)
- log(3) - x)*(x + log(3) - exp(x))),x)

[Out]

x*log(log(exp(x) - log(3) - x))

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sympy [A]  time = 1.40, size = 12, normalized size = 0.80 \begin {gather*} x \log {\left (\log {\left (- x + e^{x} - \log {\relax (3 )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)-ln(3)-x)*ln(exp(x)-ln(3)-x)*ln(ln(exp(x)-ln(3)-x))+exp(x)*x-x)/(exp(x)-ln(3)-x)/ln(exp(x)-l
n(3)-x),x)

[Out]

x*log(log(-x + exp(x) - log(3)))

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