3.32.69 \(\int \frac {e^{\frac {60-5 x \log (\frac {e}{4+4 \log (5)+\log ^2(5)})}{x \log (\frac {e}{4+4 \log (5)+\log ^2(5)})}} (-60 \log (x)+x \log (\frac {e}{4+4 \log (5)+\log ^2(5)}))}{x^2 \log (\frac {e}{4+4 \log (5)+\log ^2(5)})} \, dx\)

Optimal. Leaf size=23 \[ e^{-5+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \]

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Rubi [B]  time = 0.09, antiderivative size = 103, normalized size of antiderivative = 4.48, number of steps used = 2, number of rules used = 2, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {12, 2288} \begin {gather*} \frac {12 (2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \exp \left (\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\right )}{x^2 \log \left (\frac {e}{(2+\log (5))^2}\right ) \left (\frac {12-x \log \left (\frac {e}{(2+\log (5))^2}\right )}{x^2 \log \left (\frac {e}{(2+\log (5))^2}\right )}+\frac {1}{x}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(E^((60 - 5*x*Log[E/(4 + 4*Log[5] + Log[5]^2)])/(x*Log[E/(4 + 4*Log[5] + Log[5]^2)]))*(-60*Log[x] + x*Log[
E/(4 + 4*Log[5] + Log[5]^2)]))/(x^2*Log[E/(4 + 4*Log[5] + Log[5]^2)]),x]

[Out]

(12*E^(-5/Log[E/(2 + Log[5])^2] + 60/(x*Log[E/(2 + Log[5])^2]))*(2 + Log[5])^(10/Log[E/(2 + Log[5])^2])*Log[x]
)/(x^2*Log[E/(2 + Log[5])^2]*(x^(-1) + (12 - x*Log[E/(2 + Log[5])^2])/(x^2*Log[E/(2 + Log[5])^2])))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {\exp \left (\frac {60-5 x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}{x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )}\right ) \left (-60 \log (x)+x \log \left (\frac {e}{4+4 \log (5)+\log ^2(5)}\right )\right )}{x^2} \, dx}{\log \left (\frac {e}{(2+\log (5))^2}\right )}\\ &=\frac {12 \exp \left (-\frac {5}{\log \left (\frac {e}{(2+\log (5))^2}\right )}+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}\right ) (2+\log (5))^{\frac {10}{\log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x)}{x^2 \log \left (\frac {e}{(2+\log (5))^2}\right ) \left (\frac {1}{x}+\frac {12-x \log \left (\frac {e}{(2+\log (5))^2}\right )}{x^2 \log \left (\frac {e}{(2+\log (5))^2}\right )}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 1.00 \begin {gather*} e^{-5+\frac {60}{x \log \left (\frac {e}{(2+\log (5))^2}\right )}} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((60 - 5*x*Log[E/(4 + 4*Log[5] + Log[5]^2)])/(x*Log[E/(4 + 4*Log[5] + Log[5]^2)]))*(-60*Log[x] +
x*Log[E/(4 + 4*Log[5] + Log[5]^2)]))/(x^2*Log[E/(4 + 4*Log[5] + Log[5]^2)]),x]

[Out]

E^(-5 + 60/(x*Log[E/(2 + Log[5])^2]))*Log[x]

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fricas [B]  time = 0.56, size = 47, normalized size = 2.04 \begin {gather*} e^{\left (-\frac {5 \, {\left (x \log \left (\frac {e}{\log \relax (5)^{2} + 4 \, \log \relax (5) + 4}\right ) - 12\right )}}{x \log \left (\frac {e}{\log \relax (5)^{2} + 4 \, \log \relax (5) + 4}\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(exp(1)/(log(5)^2+4*log(5)+4))+60)/x/l
og(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="fricas")

[Out]

e^(-5*(x*log(e/(log(5)^2 + 4*log(5) + 4)) - 12)/(x*log(e/(log(5)^2 + 4*log(5) + 4))))*log(x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x \log \left (\frac {e}{\log \relax (5)^{2} + 4 \, \log \relax (5) + 4}\right ) - 60 \, \log \relax (x)\right )} e^{\left (-\frac {5 \, {\left (x \log \left (\frac {e}{\log \relax (5)^{2} + 4 \, \log \relax (5) + 4}\right ) - 12\right )}}{x \log \left (\frac {e}{\log \relax (5)^{2} + 4 \, \log \relax (5) + 4}\right )}\right )}}{x^{2} \log \left (\frac {e}{\log \relax (5)^{2} + 4 \, \log \relax (5) + 4}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(exp(1)/(log(5)^2+4*log(5)+4))+60)/x/l
og(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="giac")

[Out]

integrate((x*log(e/(log(5)^2 + 4*log(5) + 4)) - 60*log(x))*e^(-5*(x*log(e/(log(5)^2 + 4*log(5) + 4)) - 12)/(x*
log(e/(log(5)^2 + 4*log(5) + 4))))/(x^2*log(e/(log(5)^2 + 4*log(5) + 4))), x)

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maple [B]  time = 0.13, size = 48, normalized size = 2.09

method	result	size
default	\({\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \relax (5)^{2}+4 \ln \relax (5)+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \relax (5)^{2}+4 \ln \relax (5)+4}\right )}} \ln \relax (x )\)	\(48\)
norman	\({\mathrm e}^{\frac {-5 x \ln \left (\frac {{\mathrm e}}{\ln \relax (5)^{2}+4 \ln \relax (5)+4}\right )+60}{x \ln \left (\frac {{\mathrm e}}{\ln \relax (5)^{2}+4 \ln \relax (5)+4}\right )}} \ln \relax (x )\)	\(48\)
risch	\(-\frac {\left (2 \ln \left (2+\ln \relax (5)\right )-1\right ) {\mathrm e}^{-\frac {5 \left (2 x \ln \left (2+\ln \relax (5)\right )-x +12\right )}{x \left (2 \ln \left (2+\ln \relax (5)\right )-1\right )}} \ln \relax (x )}{-2 \ln \left (2+\ln \relax (5)\right )+1}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-60*ln(x)+x*ln(exp(1)/(ln(5)^2+4*ln(5)+4)))*exp((-5*x*ln(exp(1)/(ln(5)^2+4*ln(5)+4))+60)/x/ln(exp(1)/(ln(
5)^2+4*ln(5)+4)))/x^2/ln(exp(1)/(ln(5)^2+4*ln(5)+4)),x,method=_RETURNVERBOSE)

[Out]

exp((-5*x*ln(exp(1)/(ln(5)^2+4*ln(5)+4))+60)/x/ln(exp(1)/(ln(5)^2+4*ln(5)+4)))*ln(x)

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maxima [A]  time = 0.81, size = 29, normalized size = 1.26 \begin {gather*} e^{\left (\frac {60}{x \log \left (\frac {e}{\log \relax (5)^{2} + 4 \, \log \relax (5) + 4}\right )} - 5\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*log(x)+x*log(exp(1)/(log(5)^2+4*log(5)+4)))*exp((-5*x*log(exp(1)/(log(5)^2+4*log(5)+4))+60)/x/l
og(exp(1)/(log(5)^2+4*log(5)+4)))/x^2/log(exp(1)/(log(5)^2+4*log(5)+4)),x, algorithm="maxima")

[Out]

e^(60/(x*log(e/(log(5)^2 + 4*log(5) + 4))) - 5)*log(x)

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mupad [B]  time = 2.28, size = 72, normalized size = 3.13 \begin {gather*} \frac {{\mathrm {e}}^{\frac {60}{x-x\,\ln \left (4\,\ln \relax (5)+{\ln \relax (5)}^2+4\right )}}\,{\mathrm {e}}^{\frac {5}{\ln \left (4\,\ln \relax (5)+{\ln \relax (5)}^2+4\right )-1}}\,\ln \relax (x)}{{\left (4\,\ln \relax (5)+{\ln \relax (5)}^2+4\right )}^{\frac {5}{\ln \left (4\,\ln \relax (5)+{\ln \relax (5)}^2+4\right )-1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(5*x*log(exp(1)/(4*log(5) + log(5)^2 + 4)) - 60)/(x*log(exp(1)/(4*log(5) + log(5)^2 + 4))))*(60*log
(x) - x*log(exp(1)/(4*log(5) + log(5)^2 + 4))))/(x^2*log(exp(1)/(4*log(5) + log(5)^2 + 4))),x)

[Out]

(exp(60/(x - x*log(4*log(5) + log(5)^2 + 4)))*exp(5/(log(4*log(5) + log(5)^2 + 4) - 1))*log(x))/(4*log(5) + lo
g(5)^2 + 4)^(5/(log(4*log(5) + log(5)^2 + 4) - 1))

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sympy [B]  time = 0.53, size = 44, normalized size = 1.91 \begin {gather*} e^{\frac {- 5 x \log {\left (\frac {e}{\log {\relax (5 )}^{2} + 4 + 4 \log {\relax (5 )}} \right )} + 60}{x \log {\left (\frac {e}{\log {\relax (5 )}^{2} + 4 + 4 \log {\relax (5 )}} \right )}}} \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*ln(x)+x*ln(exp(1)/(ln(5)**2+4*ln(5)+4)))*exp((-5*x*ln(exp(1)/(ln(5)**2+4*ln(5)+4))+60)/x/ln(exp
(1)/(ln(5)**2+4*ln(5)+4)))/x**2/ln(exp(1)/(ln(5)**2+4*ln(5)+4)),x)

[Out]

exp((-5*x*log(E/(log(5)**2 + 4 + 4*log(5))) + 60)/(x*log(E/(log(5)**2 + 4 + 4*log(5)))))*log(x)

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