3.78.29 \(\int \frac {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} (\frac {1}{x})^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} (-10+10 \log (\frac {1}{4 x}))}{(i \pi +\log (\log (625)))^2} \, dx\)

Optimal. Leaf size=35 \[ 4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \]

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Rubi [F]  time = 0.69, 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 {4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right )}{(i \pi +\log (\log (625)))^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((x^(-1))^((10*x)/(I*Pi + Log[Log[625]])^2)*(-10 + 10*Log[1/(4*x)]))/(4^((10*x)/(I*Pi + Log[Log[625]])^2)*
(I*Pi + Log[Log[625]])^2),x]

[Out]

(-5*Defer[Int][2^(1 - (20*x)/(I*Pi + Log[Log[625]])^2)*(x^(-1))^((10*x)/(I*Pi + Log[Log[625]])^2), x])/(I*Pi +
Log[Log[625]])^2 + (5*Log[1/(4*x)]*Defer[Int][2^(1 - (20*x)/(I*Pi + Log[Log[625]])^2)*(x^(-1))^((10*x)/(I*Pi
+ Log[Log[625]])^2), x])/(I*Pi + Log[Log[625]])^2 + (5*Defer[Int][Defer[Int][2^(1 + (20*x)/(Pi - I*Log[Log[625
]])^2)*(x^(-1))^((10*x)/(I*Pi + Log[Log[625]])^2), x]/x, x])/(I*Pi + Log[Log[625]])^2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int 4^{-\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-10+10 \log \left (\frac {1}{4 x}\right )\right ) \, dx}{(i \pi +\log (\log (625)))^2}\\ &=\frac {\int 5\ 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-1+\log \left (\frac {1}{4 x}\right )\right ) \, dx}{(i \pi +\log (\log (625)))^2}\\ &=\frac {5 \int 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \left (-1+\log \left (\frac {1}{4 x}\right )\right ) \, dx}{(i \pi +\log (\log (625)))^2}\\ &=\frac {5 \int \left (-2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}}+2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \log \left (\frac {1}{4 x}\right )\right ) \, dx}{(i \pi +\log (\log (625)))^2}\\ &=-\frac {5 \int 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \, dx}{(i \pi +\log (\log (625)))^2}+\frac {5 \int 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \log \left (\frac {1}{4 x}\right ) \, dx}{(i \pi +\log (\log (625)))^2}\\ &=-\frac {5 \int 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \, dx}{(i \pi +\log (\log (625)))^2}+\frac {5 \int \frac {\int 2^{1+\frac {20 x}{(\pi -i \log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \, dx}{x} \, dx}{(i \pi +\log (\log (625)))^2}+\frac {\left (5 \log \left (\frac {1}{4 x}\right )\right ) \int 2^{1-\frac {20 x}{(i \pi +\log (\log (625)))^2}} \left (\frac {1}{x}\right )^{\frac {10 x}{(i \pi +\log (\log (625)))^2}} \, dx}{(i \pi +\log (\log (625)))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 58, normalized size = 1.66 \begin {gather*} -\frac {2^{\frac {20 x}{(\pi -i \log (\log (625)))^2}} \left (\frac {1}{x}\right )^{-\frac {10 x}{(\pi -i \log (\log (625)))^2}} (\pi -i \log (\log (625)))^2}{(i \pi +\log (\log (625)))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((x^(-1))^((10*x)/(I*Pi + Log[Log[625]])^2)*(-10 + 10*Log[1/(4*x)]))/(4^((10*x)/(I*Pi + Log[Log[625]
])^2)*(I*Pi + Log[Log[625]])^2),x]

[Out]

-((2^((20*x)/(Pi - I*Log[Log[625]])^2)*(Pi - I*Log[Log[625]])^2)/((x^(-1))^((10*x)/(Pi - I*Log[Log[625]])^2)*(
I*Pi + Log[Log[625]])^2))

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fricas [A]  time = 1.58, size = 16, normalized size = 0.46 \begin {gather*} \left (\frac {1}{4 \, x}\right )^{\frac {10 \, x}{\log \left (-4 \, \log \relax (5)\right )^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(10*(1/4/x)^(10*x/log(-4*log(5))^2)*(log(1/4/x) - 1)/log(-4*log(5))^2, x)

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maple [A]  time = 0.39, size = 20, normalized size = 0.57

method	result	size
default	\({\mathrm e}^{\frac {10 x \ln \left (\frac {1}{4 x}\right )}{\ln \left (-4 \ln \relax (5)\right )^{2}}}\)	\(20\)
norman	\(-\frac {\left (-4 i \ln \relax (2) \pi -2 i \ln \left (\ln \relax (5)\right ) \pi +\pi ^{2}-4 \ln \relax (2)^{2}-4 \ln \relax (2) \ln \left (\ln \relax (5)\right )-\ln \left (\ln \relax (5)\right )^{2}\right ) {\mathrm e}^{\frac {10 x \ln \left (\frac {1}{4 x}\right )}{\ln \left (-4 \ln \relax (5)\right )^{2}}}}{\left (2 \ln \relax (2)+\ln \left (\ln \relax (5)\right )+i \pi \right ) \ln \left (-4 \ln \relax (5)\right )}\)	\(80\)
risch	\(\frac {4 \left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \relax (5)}\right )\right )^{2}}} \ln \relax (2)^{2}}{\left (2 \ln \relax (2)+\ln \left (\ln \relax (5)\right )+i \pi \right )^{2}}+\frac {4 \left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \relax (5)}\right )\right )^{2}}} \ln \relax (2) \ln \left (\ln \relax (5)\right )}{\left (2 \ln \relax (2)+\ln \left (\ln \relax (5)\right )+i \pi \right )^{2}}+\frac {\left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \relax (5)}\right )\right )^{2}}} \ln \left (\ln \relax (5)\right )^{2}}{\left (2 \ln \relax (2)+\ln \left (\ln \relax (5)\right )+i \pi \right )^{2}}+\frac {4 i \left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \relax (5)}\right )\right )^{2}}} \ln \relax (2) \pi }{\left (2 \ln \relax (2)+\ln \left (\ln \relax (5)\right )+i \pi \right )^{2}}+\frac {2 i \left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \relax (5)}\right )\right )^{2}}} \ln \left (\ln \relax (5)\right ) \pi }{\left (2 \ln \relax (2)+\ln \left (\ln \relax (5)\right )+i \pi \right )^{2}}-\frac {\left (\frac {1}{4 x}\right )^{\frac {10 x}{\left (i \pi +2 \ln \left (2 \sqrt {\ln \relax (5)}\right )\right )^{2}}} \pi ^{2}}{\left (2 \ln \relax (2)+\ln \left (\ln \relax (5)\right )+i \pi \right )^{2}}\)	\(273\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*ln(1/4/x)-10)*exp(5*x*ln(1/4/x)/ln(-4*ln(5))^2)^2/ln(-4*ln(5))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.50, size = 93, normalized size = 2.66 \begin {gather*} \frac {{\left (4 \, \log \relax (2)^{2} + 4 \, \log \relax (2) \log \left (-\log \relax (5)\right ) + \log \left (-\log \relax (5)\right )^{2}\right )} e^{\left (-\frac {20 \, x \log \relax (2)}{4 \, \log \relax (2)^{2} + 4 \, \log \relax (2) \log \left (-\log \relax (5)\right ) + \log \left (-\log \relax (5)\right )^{2}} - \frac {10 \, x \log \relax (x)}{4 \, \log \relax (2)^{2} + 4 \, \log \relax (2) \log \left (-\log \relax (5)\right ) + \log \left (-\log \relax (5)\right )^{2}}\right )}}{\log \left (-4 \, \log \relax (5)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*log(2)^2 + 4*log(2)*log(-log(5)) + log(-log(5))^2)*e^(-20*x*log(2)/(4*log(2)^2 + 4*log(2)*log(-log(5)) + lo
g(-log(5))^2) - 10*x*log(x)/(4*log(2)^2 + 4*log(2)*log(-log(5)) + log(-log(5))^2))/log(-4*log(5))^2

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mupad [B]  time = 5.56, size = 93, normalized size = 2.66 \begin {gather*} {\mathrm {e}}^{-\frac {20\,x\,\ln \relax (2)}{{\ln \left (\ln \relax (5)\right )}^2+4\,\ln \relax (2)\,\ln \left (\ln \relax (5)\right )-\pi ^2+4\,{\ln \relax (2)}^2+\pi \,\ln \relax (2)\,4{}\mathrm {i}+\pi \,\ln \left (\ln \relax (5)\right )\,2{}\mathrm {i}}}\,{\mathrm {e}}^{\frac {10\,x\,\ln \left (\frac {1}{x}\right )}{{\ln \left (\ln \relax (5)\right )}^2+4\,\ln \relax (2)\,\ln \left (\ln \relax (5)\right )-\pi ^2+4\,{\ln \relax (2)}^2+\pi \,\ln \relax (2)\,4{}\mathrm {i}+\pi \,\ln \left (\ln \relax (5)\right )\,2{}\mathrm {i}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(-(20*x*log(2))/(pi*log(2)*4i + log(log(5))^2 + pi*log(log(5))*2i + 4*log(2)*log(log(5)) - pi^2 + 4*log(2)^
2))*exp((10*x*log(1/x))/(pi*log(2)*4i + log(log(5))^2 + pi*log(log(5))*2i + 4*log(2)*log(log(5)) - pi^2 + 4*lo
g(2)^2))

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sympy [B]  time = 19.53, size = 109, normalized size = 3.11 \begin {gather*} e^{- \frac {20 x \log {\relax (2 )}}{- \pi ^{2} + \log {\left (\log {\relax (5 )} \right )}^{2} + 4 \log {\relax (2 )} \log {\left (\log {\relax (5 )} \right )} + 4 \log {\relax (2 )}^{2} + 2 i \pi \log {\left (\log {\relax (5 )} \right )} + 4 i \pi \log {\relax (2 )}}} e^{\frac {10 x \log {\left (\frac {1}{x} \right )}}{- \pi ^{2} + \log {\left (\log {\relax (5 )} \right )}^{2} + 4 \log {\relax (2 )} \log {\left (\log {\relax (5 )} \right )} + 4 \log {\relax (2 )}^{2} + 2 i \pi \log {\left (\log {\relax (5 )} \right )} + 4 i \pi \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(-20*x*log(2)/(-pi**2 + log(log(5))**2 + 4*log(2)*log(log(5)) + 4*log(2)**2 + 2*I*pi*log(log(5)) + 4*I*pi*l
og(2)))*exp(10*x*log(1/x)/(-pi**2 + log(log(5))**2 + 4*log(2)*log(log(5)) + 4*log(2)**2 + 2*I*pi*log(log(5)) +
4*I*pi*log(2)))

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