3.22.13 \(\int \frac {e^{\frac {x}{5+\log (\frac {\log (3)}{x})}} (e^{2 x} (-224 e+200 x+224 x^2)+e^{2 x} (-84 e+80 x+84 x^2) \log (\frac {\log (3)}{x})+e^{2 x} (-8 e+8 x+8 x^2) \log ^2(\frac {\log (3)}{x}))}{25+10 \log (\frac {\log (3)}{x})+\log ^2(\frac {\log (3)}{x})} \, dx\)

Optimal. Leaf size=28 \[ 4 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (-e+x^2\right ) \]

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Rubi [F]  time = 6.17, 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 {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{25+10 \log \left (\frac {\log (3)}{x}\right )+\log ^2\left (\frac {\log (3)}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x/(5 + Log[Log[3]/x]))*(E^(2*x)*(-224*E + 200*x + 224*x^2) + E^(2*x)*(-84*E + 80*x + 84*x^2)*Log[Log[3
]/x] + E^(2*x)*(-8*E + 8*x + 8*x^2)*Log[Log[3]/x]^2))/(25 + 10*Log[Log[3]/x] + Log[Log[3]/x]^2),x]

[Out]

-8*Defer[Int][E^(1 + 2*x + x/(5 + Log[Log[3]/x])), x] + 8*Defer[Int][E^(2*x + x/(5 + Log[Log[3]/x]))*x, x] + 8
*Defer[Int][E^(2*x + x/(5 + Log[Log[3]/x]))*x^2, x] - 4*Defer[Int][E^(1 + 2*x + x/(5 + Log[Log[3]/x]))/(5 + Lo
g[Log[3]/x])^2, x] + 4*Defer[Int][(E^(2*x + x/(5 + Log[Log[3]/x]))*x^2)/(5 + Log[Log[3]/x])^2, x] - 4*Defer[In
t][E^(1 + 2*x + x/(5 + Log[Log[3]/x]))/(5 + Log[Log[3]/x]), x] + 4*Defer[Int][(E^(2*x + x/(5 + Log[Log[3]/x]))
*x^2)/(5 + Log[Log[3]/x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e^{2 x} \left (-224 e+200 x+224 x^2\right )+e^{2 x} \left (-84 e+80 x+84 x^2\right ) \log \left (\frac {\log (3)}{x}\right )+e^{2 x} \left (-8 e+8 x+8 x^2\right ) \log ^2\left (\frac {\log (3)}{x}\right )\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx\\ &=\int \left (-\frac {224 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {200 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {224 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}-\frac {84 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {80 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {84 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}-\frac {8 e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {8 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}+\frac {8 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2}\right ) \, dx\\ &=-\left (8 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx\right )+8 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+8 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \log ^2\left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+80 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx-84 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+84 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2 \log \left (\frac {\log (3)}{x}\right )}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+200 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx-224 \int \frac {e^{1+2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}}}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx+224 \int \frac {e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} x^2}{\left (5+\log \left (\frac {\log (3)}{x}\right )\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 28, normalized size = 1.00 \begin {gather*} -4 e^{2 x+\frac {x}{5+\log \left (\frac {\log (3)}{x}\right )}} \left (e-x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/(5 + Log[Log[3]/x]))*(E^(2*x)*(-224*E + 200*x + 224*x^2) + E^(2*x)*(-84*E + 80*x + 84*x^2)*Log
[Log[3]/x] + E^(2*x)*(-8*E + 8*x + 8*x^2)*Log[Log[3]/x]^2))/(25 + 10*Log[Log[3]/x] + Log[Log[3]/x]^2),x]

[Out]

-4*E^(2*x + x/(5 + Log[Log[3]/x]))*(E - x^2)

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fricas [A]  time = 1.09, size = 28, normalized size = 1.00 \begin {gather*} 4 \, {\left (x^{2} - e\right )} e^{\left (2 \, x + \frac {x}{\log \left (\frac {\log \relax (3)}{x}\right ) + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*exp(1)+8*x^2+8*x)*exp(x)^2*log(log(3)/x)^2+(-84*exp(1)+84*x^2+80*x)*exp(x)^2*log(log(3)/x)+(-22
4*exp(1)+224*x^2+200*x)*exp(x)^2)*exp(x/(log(log(3)/x)+5))/(log(log(3)/x)^2+10*log(log(3)/x)+25),x, algorithm=
"fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*exp(1)+8*x^2+8*x)*exp(x)^2*log(log(3)/x)^2+(-84*exp(1)+84*x^2+80*x)*exp(x)^2*log(log(3)/x)+(-22
4*exp(1)+224*x^2+200*x)*exp(x)^2)*exp(x/(log(log(3)/x)+5))/(log(log(3)/x)^2+10*log(log(3)/x)+25),x, algorithm=
"giac")

[Out]

undef

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maple [A]  time = 0.57, size = 36, normalized size = 1.29

method	result	size
risch	\(-4 \left ({\mathrm e}-x^{2}\right ) {\mathrm e}^{\frac {x \left (-2 \ln \left (\ln \relax (3)\right )+2 \ln \relax (x )-11\right )}{-\ln \left (\ln \relax (3)\right )+\ln \relax (x )-5}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*exp(1)+8*x^2+8*x)*exp(x)^2*ln(ln(3)/x)^2+(-84*exp(1)+84*x^2+80*x)*exp(x)^2*ln(ln(3)/x)+(-224*exp(1)+2
24*x^2+200*x)*exp(x)^2)*exp(x/(ln(ln(3)/x)+5))/(ln(ln(3)/x)^2+10*ln(ln(3)/x)+25),x,method=_RETURNVERBOSE)

[Out]

-4*(exp(1)-x^2)*exp(x*(-2*ln(ln(3))+2*ln(x)-11)/(-ln(ln(3))+ln(x)-5))

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maxima [A]  time = 0.66, size = 29, normalized size = 1.04 \begin {gather*} 4 \, {\left (x^{2} - e\right )} e^{\left (2 \, x - \frac {x}{\log \relax (x) - \log \left (\log \relax (3)\right ) - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*exp(1)+8*x^2+8*x)*exp(x)^2*log(log(3)/x)^2+(-84*exp(1)+84*x^2+80*x)*exp(x)^2*log(log(3)/x)+(-22
4*exp(1)+224*x^2+200*x)*exp(x)^2)*exp(x/(log(log(3)/x)+5))/(log(log(3)/x)^2+10*log(log(3)/x)+25),x, algorithm=
"maxima")

[Out]

4*(x^2 - e)*e^(2*x - x/(log(x) - log(log(3)) - 5))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {x}{\ln \left (\frac {\ln \relax (3)}{x}\right )+5}}\,\left ({\mathrm {e}}^{2\,x}\,\left (8\,x^2+8\,x-8\,\mathrm {e}\right )\,{\ln \left (\frac {\ln \relax (3)}{x}\right )}^2+{\mathrm {e}}^{2\,x}\,\left (84\,x^2+80\,x-84\,\mathrm {e}\right )\,\ln \left (\frac {\ln \relax (3)}{x}\right )+{\mathrm {e}}^{2\,x}\,\left (224\,x^2+200\,x-224\,\mathrm {e}\right )\right )}{{\ln \left (\frac {\ln \relax (3)}{x}\right )}^2+10\,\ln \left (\frac {\ln \relax (3)}{x}\right )+25} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/(log(log(3)/x) + 5))*(exp(2*x)*(200*x - 224*exp(1) + 224*x^2) + log(log(3)/x)^2*exp(2*x)*(8*x - 8*e
xp(1) + 8*x^2) + log(log(3)/x)*exp(2*x)*(80*x - 84*exp(1) + 84*x^2)))/(10*log(log(3)/x) + log(log(3)/x)^2 + 25
),x)

[Out]

int((exp(x/(log(log(3)/x) + 5))*(exp(2*x)*(200*x - 224*exp(1) + 224*x^2) + log(log(3)/x)^2*exp(2*x)*(8*x - 8*e
xp(1) + 8*x^2) + log(log(3)/x)*exp(2*x)*(80*x - 84*exp(1) + 84*x^2)))/(10*log(log(3)/x) + log(log(3)/x)^2 + 25
), x)

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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(((-8*exp(1)+8*x**2+8*x)*exp(x)**2*ln(ln(3)/x)**2+(-84*exp(1)+84*x**2+80*x)*exp(x)**2*ln(ln(3)/x)+(-2
24*exp(1)+224*x**2+200*x)*exp(x)**2)*exp(x/(ln(ln(3)/x)+5))/(ln(ln(3)/x)**2+10*ln(ln(3)/x)+25),x)

[Out]

Timed out

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