3.22.59 \(\int \frac {2 e^{-4+x+\frac {2 (2+5 x^2)}{x}} x^2 \log (1+e^{-4+x})+e^{\frac {2 (2+5 x^2)}{x}} (-4+10 x^2+e^{-4+x} (-4+10 x^2)) \log ^2(1+e^{-4+x})}{e^{\frac {2 (2+5 x^2)}{x}} (x^2+e^{-4+x} x^2) \log ^2(1+e^{-4+x})+(-2 x^2-2 e^{-4+x} x^2) \log (\log (4))} \, dx\)

Optimal. Leaf size=32 \[ \log \left (\frac {1}{2} e^{\frac {4}{x}+10 x} \log ^2\left (1+e^{-4+x}\right )-\log (\log (4))\right ) \]

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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 + x + (2*(2 + 5*x^2))/x)*x^2*Log[1 + E^(-4 + x)] + E^((2*(2 + 5*x^2))/x)*(-4 + 10*x^2 + E^(-4 + x
)*(-4 + 10*x^2))*Log[1 + E^(-4 + x)]^2)/(E^((2*(2 + 5*x^2))/x)*(x^2 + E^(-4 + x)*x^2)*Log[1 + E^(-4 + x)]^2 +
(-2*x^2 - 2*E^(-4 + x)*x^2)*Log[Log[4]]),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [F]  time = 3.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 e^{-4+x+\frac {2 \left (2+5 x^2\right )}{x}} x^2 \log \left (1+e^{-4+x}\right )+e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (-4+10 x^2+e^{-4+x} \left (-4+10 x^2\right )\right ) \log ^2\left (1+e^{-4+x}\right )}{e^{\frac {2 \left (2+5 x^2\right )}{x}} \left (x^2+e^{-4+x} x^2\right ) \log ^2\left (1+e^{-4+x}\right )+\left (-2 x^2-2 e^{-4+x} x^2\right ) \log (\log (4))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(2*E^(-4 + x + (2*(2 + 5*x^2))/x)*x^2*Log[1 + E^(-4 + x)] + E^((2*(2 + 5*x^2))/x)*(-4 + 10*x^2 + E^(
-4 + x)*(-4 + 10*x^2))*Log[1 + E^(-4 + x)]^2)/(E^((2*(2 + 5*x^2))/x)*(x^2 + E^(-4 + x)*x^2)*Log[1 + E^(-4 + x)
]^2 + (-2*x^2 - 2*E^(-4 + x)*x^2)*Log[Log[4]]),x]

[Out]

Integrate[(2*E^(-4 + x + (2*(2 + 5*x^2))/x)*x^2*Log[1 + E^(-4 + x)] + E^((2*(2 + 5*x^2))/x)*(-4 + 10*x^2 + E^(
-4 + x)*(-4 + 10*x^2))*Log[1 + E^(-4 + x)]^2)/(E^((2*(2 + 5*x^2))/x)*(x^2 + E^(-4 + x)*x^2)*Log[1 + E^(-4 + x)
]^2 + (-2*x^2 - 2*E^(-4 + x)*x^2)*Log[Log[4]]), x]

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fricas [B]  time = 0.64, size = 96, normalized size = 3.00 \begin {gather*} \frac {10 \, x^{2} + x \log \left ({\left (e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )} \log \left ({\left (e^{\left (\frac {11 \, x^{2} - 4 \, x + 4}{x}\right )} + e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )}\right )} e^{\left (-\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )}\right )^{2} - 2 \, \log \left (2 \, \log \relax (2)\right )\right )} e^{\left (-\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )}\right ) + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^2-4)*exp(x-4)+10*x^2-4)*exp((5*x^2+2)/x)^2*log(exp(x-4)+1)^2+2*x^2*exp(x-4)*exp((5*x^2+2)/x)
^2*log(exp(x-4)+1))/((x^2*exp(x-4)+x^2)*exp((5*x^2+2)/x)^2*log(exp(x-4)+1)^2+(-2*x^2*exp(x-4)-2*x^2)*log(2*log
(2))),x, algorithm="fricas")

[Out]

(10*x^2 + x*log((e^(2*(5*x^2 + 2)/x)*log((e^((11*x^2 - 4*x + 4)/x) + e^(2*(5*x^2 + 2)/x))*e^(-2*(5*x^2 + 2)/x)
)^2 - 2*log(2*log(2)))*e^(-2*(5*x^2 + 2)/x)) + 4)/x

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giac [B]  time = 0.38, size = 70, normalized size = 2.19 \begin {gather*} \log \left (-e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )} \log \left (e^{4} + e^{x}\right )^{2} + 8 \, e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )} \log \left (e^{4} + e^{x}\right ) - 16 \, e^{\left (\frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x}\right )} + 2 \, \log \relax (2) + 2 \, \log \left (\log \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^2-4)*exp(x-4)+10*x^2-4)*exp((5*x^2+2)/x)^2*log(exp(x-4)+1)^2+2*x^2*exp(x-4)*exp((5*x^2+2)/x)
^2*log(exp(x-4)+1))/((x^2*exp(x-4)+x^2)*exp((5*x^2+2)/x)^2*log(exp(x-4)+1)^2+(-2*x^2*exp(x-4)-2*x^2)*log(2*log
(2))),x, algorithm="giac")

[Out]

log(-e^(2*(5*x^2 + 2)/x)*log(e^4 + e^x)^2 + 8*e^(2*(5*x^2 + 2)/x)*log(e^4 + e^x) - 16*e^(2*(5*x^2 + 2)/x) + 2*
log(2) + 2*log(log(2)))

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maple [A]  time = 0.14, size = 46, normalized size = 1.44

method	result	size
risch	\(\frac {10 x^{2}+4}{x}+\ln \left (\ln \left ({\mathrm e}^{x -4}+1\right )^{2}-2 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{-\frac {2 \left (5 x^{2}+2\right )}{x}}\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*x^2-4)*exp(x-4)+10*x^2-4)*exp((5*x^2+2)/x)^2*ln(exp(x-4)+1)^2+2*x^2*exp(x-4)*exp((5*x^2+2)/x)^2*ln(e
xp(x-4)+1))/((x^2*exp(x-4)+x^2)*exp((5*x^2+2)/x)^2*ln(exp(x-4)+1)^2+(-2*x^2*exp(x-4)-2*x^2)*ln(2*ln(2))),x,met
hod=_RETURNVERBOSE)

[Out]

2*(5*x^2+2)/x+ln(ln(exp(x-4)+1)^2-2*(ln(2)+ln(ln(2)))*exp(-2*(5*x^2+2)/x))

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maxima [B]  time = 0.70, size = 84, normalized size = 2.62 \begin {gather*} \frac {2 \, {\left (5 \, x^{2} + 2\right )}}{x} + \log \left ({\left (e^{\left (10 \, x + \frac {4}{x}\right )} \log \left (e^{4} + e^{x}\right )^{2} - 8 \, e^{\left (10 \, x + \frac {4}{x}\right )} \log \left (e^{4} + e^{x}\right ) + 16 \, e^{\left (10 \, x + \frac {4}{x}\right )} - 2 \, \log \relax (2) - 2 \, \log \left (\log \relax (2)\right )\right )} e^{\left (-10 \, x - \frac {4}{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^2-4)*exp(x-4)+10*x^2-4)*exp((5*x^2+2)/x)^2*log(exp(x-4)+1)^2+2*x^2*exp(x-4)*exp((5*x^2+2)/x)
^2*log(exp(x-4)+1))/((x^2*exp(x-4)+x^2)*exp((5*x^2+2)/x)^2*log(exp(x-4)+1)^2+(-2*x^2*exp(x-4)-2*x^2)*log(2*log
(2))),x, algorithm="maxima")

[Out]

2*(5*x^2 + 2)/x + log((e^(10*x + 4/x)*log(e^4 + e^x)^2 - 8*e^(10*x + 4/x)*log(e^4 + e^x) + 16*e^(10*x + 4/x) -
2*log(2) - 2*log(log(2)))*e^(-10*x - 4/x))

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mupad [B]  time = 1.53, size = 30, normalized size = 0.94 \begin {gather*} \ln \left ({\mathrm {e}}^{10\,x+\frac {4}{x}}\,{\ln \left ({\mathrm {e}}^{-4}\,{\mathrm {e}}^x+1\right )}^2+\ln \left (\frac {1}{4\,{\ln \relax (2)}^2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp(x - 4) + 1)^2*exp((2*(5*x^2 + 2))/x)*(exp(x - 4)*(10*x^2 - 4) + 10*x^2 - 4) + 2*x^2*log(exp(x -
4) + 1)*exp((2*(5*x^2 + 2))/x)*exp(x - 4))/(log(2*log(2))*(2*x^2*exp(x - 4) + 2*x^2) - log(exp(x - 4) + 1)^2*e
xp((2*(5*x^2 + 2))/x)*(x^2*exp(x - 4) + x^2)),x)

[Out]

log(log(1/(4*log(2)^2)) + exp(10*x + 4/x)*log(exp(-4)*exp(x) + 1)^2)

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sympy [A]  time = 2.11, size = 41, normalized size = 1.28 \begin {gather*} 10 x + \log {\left (\log {\left (e^{x - 4} + 1 \right )}^{2} + \left (- 2 \log {\relax (2 )} - 2 \log {\left (\log {\relax (2 )} \right )}\right ) e^{- \frac {2 \left (5 x^{2} + 2\right )}{x}} \right )} + \frac {4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x**2-4)*exp(x-4)+10*x**2-4)*exp((5*x**2+2)/x)**2*ln(exp(x-4)+1)**2+2*x**2*exp(x-4)*exp((5*x**2
+2)/x)**2*ln(exp(x-4)+1))/((x**2*exp(x-4)+x**2)*exp((5*x**2+2)/x)**2*ln(exp(x-4)+1)**2+(-2*x**2*exp(x-4)-2*x**
2)*ln(2*ln(2))),x)

[Out]

10*x + log(log(exp(x - 4) + 1)**2 + (-2*log(2) - 2*log(log(2)))*exp(-2*(5*x**2 + 2)/x)) + 4/x

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