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

Optimal. Leaf size=25 \[ e^{10}-2 x-\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \]

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Rubi [F]  time = 1.05, 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 {\left (-10-2 e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )+\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \left (-\frac {1}{2} e^{\frac {e^x}{2}+x} x^2+\left (-10 x-2 e^{\frac {e^x}{2}} x\right ) \log \left (5+e^{\frac {e^x}{2}}\right ) \log \left (\log \left (5+e^{\frac {e^x}{2}}\right )\right )\right )}{\left (5+e^{\frac {e^x}{2}}\right ) \log \left (5+e^{\frac {e^x}{2}}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 22, normalized size = 0.88 \begin {gather*} -2 x-\log ^{x^2}\left (5+e^{\frac {e^x}{2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*x - Log[5 + E^(E^x/2)]^x^2

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fricas [A]  time = 0.83, size = 21, normalized size = 0.84 \begin {gather*} -2 \, x - \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (x^{2} e^{\left (x + e^{\left (x - \log \relax (2)\right )} - \log \relax (2)\right )} + 2 \, {\left (x e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5 \, x\right )} \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right ) \log \left (\log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )\right )\right )} \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )^{\left (x^{2}\right )} + 2 \, {\left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )} \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )}{{\left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )} \log \left (e^{\left (e^{\left (x - \log \relax (2)\right )}\right )} + 5\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 19, normalized size = 0.76

method	result	size
risch	\(-2 x -\ln \left ({\mathrm e}^{\frac {{\mathrm e}^{x}}{2}}+5\right )^{x^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x*exp(exp(x-ln(2)))-10*x)*ln(exp(exp(x-ln(2)))+5)*ln(ln(exp(exp(x-ln(2)))+5))-x^2*exp(x-ln(2))*exp(e
xp(x-ln(2))))*exp(x^2*ln(ln(exp(exp(x-ln(2)))+5)))+(-2*exp(exp(x-ln(2)))-10)*ln(exp(exp(x-ln(2)))+5))/(exp(exp
(x-ln(2)))+5)/ln(exp(exp(x-ln(2)))+5),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.73, size = 18, normalized size = 0.72 \begin {gather*} -2 \, x - \log \left (e^{\left (\frac {1}{2} \, e^{x}\right )} + 5\right )^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x - log(e^(1/2*e^x) + 5)^(x^2)

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mupad [B]  time = 1.15, size = 18, normalized size = 0.72 \begin {gather*} -2\,x-{\ln \left ({\mathrm {e}}^{\frac {{\mathrm {e}}^x}{2}}+5\right )}^{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x^2*log(log(exp(exp(x - log(2))) + 5)))*(x^2*exp(x - log(2))*exp(exp(x - log(2))) + log(exp(exp(x -
log(2))) + 5)*log(log(exp(exp(x - log(2))) + 5))*(10*x + 2*x*exp(exp(x - log(2))))) + log(exp(exp(x - log(2)))
+ 5)*(2*exp(exp(x - log(2))) + 10))/(log(exp(exp(x - log(2))) + 5)*(exp(exp(x - log(2))) + 5)),x)

[Out]

- 2*x - log(exp(exp(x)/2) + 5)^(x^2)

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

[Out]

Timed out

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