3.18.15 \(\int \frac {e^{\frac {36+12 \log (\frac {4}{x})}{x \log (e^{-e^4 x} x)}} (-36+36 e^4 x+(-12+12 e^4 x) \log (\frac {4}{x})+(-48-12 \log (\frac {4}{x})) \log (e^{-e^4 x} x))}{x^2 \log ^2(e^{-e^4 x} x)} \, dx\)

Optimal. Leaf size=28 \[ e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \]

________________________________________________________________________________________

Rubi [F]  time = 8.11, 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 {36+12 \log \left (\frac {4}{x}\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((36 + 12*Log[4/x])/(x*Log[x/E^(E^4*x)]))*(-36 + 36*E^4*x + (-12 + 12*E^4*x)*Log[4/x] + (-48 - 12*Log[4
/x])*Log[x/E^(E^4*x)]))/(x^2*Log[x/E^(E^4*x)]^2),x]

[Out]

-36*Defer[Int][E^((12*(3 + Log[4/x]))/(x*Log[x/E^(E^4*x)]))/(x^2*Log[x/E^(E^4*x)]^2), x] + 36*Defer[Int][E^(4
+ (12*(3 + Log[4/x]))/(x*Log[x/E^(E^4*x)]))/(x*Log[x/E^(E^4*x)]^2), x] - 12*Defer[Int][(E^((12*(3 + Log[4/x]))
/(x*Log[x/E^(E^4*x)]))*Log[4/x])/(x^2*Log[x/E^(E^4*x)]^2), x] + 12*Defer[Int][(E^(4 + (12*(3 + Log[4/x]))/(x*L
og[x/E^(E^4*x)]))*Log[4/x])/(x*Log[x/E^(E^4*x)]^2), x] - 48*Defer[Int][E^((12*(3 + Log[4/x]))/(x*Log[x/E^(E^4*
x)]))/(x^2*Log[x/E^(E^4*x)]), x] - 12*Defer[Int][(E^((12*(3 + Log[4/x]))/(x*Log[x/E^(E^4*x)]))*Log[4/x])/(x^2*
Log[x/E^(E^4*x)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-36+36 e^4 x+\left (-12+12 e^4 x\right ) \log \left (\frac {4}{x}\right )+\left (-48-12 \log \left (\frac {4}{x}\right )\right ) \log \left (e^{-e^4 x} x\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx\\ &=\int \left (\frac {12 e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-1+e^4 x\right ) \left (3+\log \left (\frac {4}{x}\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )}-\frac {12 e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (4+\log \left (\frac {4}{x}\right )\right )}{x^2 \log \left (e^{-e^4 x} x\right )}\right ) \, dx\\ &=12 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-1+e^4 x\right ) \left (3+\log \left (\frac {4}{x}\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx-12 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (4+\log \left (\frac {4}{x}\right )\right )}{x^2 \log \left (e^{-e^4 x} x\right )} \, dx\\ &=12 \int \left (\frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-3-\log \left (\frac {4}{x}\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )}+\frac {e^{4+\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log ^2\left (e^{-e^4 x} x\right )}\right ) \, dx-12 \int \left (\frac {4 e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x^2 \log \left (e^{-e^4 x} x\right )}+\frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x^2 \log \left (e^{-e^4 x} x\right )}\right ) \, dx\\ &=12 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (-3-\log \left (\frac {4}{x}\right )\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx+12 \int \frac {e^{4+\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log ^2\left (e^{-e^4 x} x\right )} \, dx-12 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x^2 \log \left (e^{-e^4 x} x\right )} \, dx-48 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x^2 \log \left (e^{-e^4 x} x\right )} \, dx\\ &=12 \int \left (-\frac {3 e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x^2 \log ^2\left (e^{-e^4 x} x\right )}-\frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )}\right ) \, dx+12 \int \left (\frac {3 e^{4+\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x \log ^2\left (e^{-e^4 x} x\right )}+\frac {e^{4+\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x \log ^2\left (e^{-e^4 x} x\right )}\right ) \, dx-12 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x^2 \log \left (e^{-e^4 x} x\right )} \, dx-48 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x^2 \log \left (e^{-e^4 x} x\right )} \, dx\\ &=-\left (12 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx\right )+12 \int \frac {e^{4+\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x \log ^2\left (e^{-e^4 x} x\right )} \, dx-12 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \log \left (\frac {4}{x}\right )}{x^2 \log \left (e^{-e^4 x} x\right )} \, dx-36 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x^2 \log ^2\left (e^{-e^4 x} x\right )} \, dx+36 \int \frac {e^{4+\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x \log ^2\left (e^{-e^4 x} x\right )} \, dx-48 \int \frac {e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}}}{x^2 \log \left (e^{-e^4 x} x\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 28, normalized size = 1.00 \begin {gather*} e^{\frac {12 \left (3+\log \left (\frac {4}{x}\right )\right )}{x \log \left (e^{-e^4 x} x\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((36 + 12*Log[4/x])/(x*Log[x/E^(E^4*x)]))*(-36 + 36*E^4*x + (-12 + 12*E^4*x)*Log[4/x] + (-48 - 12
*Log[4/x])*Log[x/E^(E^4*x)]))/(x^2*Log[x/E^(E^4*x)]^2),x]

[Out]

E^((12*(3 + Log[4/x]))/(x*Log[x/E^(E^4*x)]))

________________________________________________________________________________________

fricas [A]  time = 0.69, size = 33, normalized size = 1.18 \begin {gather*} e^{\left (-\frac {12 \, {\left (\log \left (\frac {4}{x}\right ) + 3\right )}}{x^{2} e^{4} - 2 \, x \log \relax (2) + x \log \left (\frac {4}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(4/x)-48)*log(x/exp(x*exp(4)))+(12*x*exp(4)-12)*log(4/x)+36*x*exp(4)-36)*exp((12*log(4/x)+3
6)/x/log(x/exp(x*exp(4))))/x^2/log(x/exp(x*exp(4)))^2,x, algorithm="fricas")

[Out]

e^(-12*(log(4/x) + 3)/(x^2*e^4 - 2*x*log(2) + x*log(4/x)))

________________________________________________________________________________________

giac [A]  time = 1.29, size = 40, normalized size = 1.43 \begin {gather*} e^{\left (\frac {12 \, \log \left (\frac {4}{x}\right )}{x \log \left (x e^{\left (-x e^{4}\right )}\right )} + \frac {36}{x \log \left (x e^{\left (-x e^{4}\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(4/x)-48)*log(x/exp(x*exp(4)))+(12*x*exp(4)-12)*log(4/x)+36*x*exp(4)-36)*exp((12*log(4/x)+3
6)/x/log(x/exp(x*exp(4))))/x^2/log(x/exp(x*exp(4)))^2,x, algorithm="giac")

[Out]

e^(12*log(4/x)/(x*log(x*e^(-x*e^4))) + 36/(x*log(x*e^(-x*e^4))))

________________________________________________________________________________________

maple [C]  time = 0.77, size = 128, normalized size = 4.57

method	result	size
risch	\({\mathrm e}^{\frac {48 \ln \relax (2)-24 \ln \relax (x )+72}{x \left (-i \mathrm {csgn}\left (i x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right )^{3} \pi +i \mathrm {csgn}\left (i x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right )^{2} \mathrm {csgn}\left (i x \right ) \pi +i \mathrm {csgn}\left (i x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x \,{\mathrm e}^{4}}\right ) \pi -i \mathrm {csgn}\left (i x \,{\mathrm e}^{-x \,{\mathrm e}^{4}}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x \,{\mathrm e}^{4}}\right ) \pi +2 \ln \relax (x )-2 \ln \left ({\mathrm e}^{x \,{\mathrm e}^{4}}\right )\right )}}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*ln(4/x)-48)*ln(x/exp(x*exp(4)))+(12*x*exp(4)-12)*ln(4/x)+36*x*exp(4)-36)*exp((12*ln(4/x)+36)/x/ln(x/
exp(x*exp(4))))/x^2/ln(x/exp(x*exp(4)))^2,x,method=_RETURNVERBOSE)

[Out]

exp(24*(2*ln(2)-ln(x)+3)/x/(-I*csgn(I*x*exp(-x*exp(4)))^3*Pi+I*csgn(I*x*exp(-x*exp(4)))^2*csgn(I*x)*Pi+I*csgn(
I*x*exp(-x*exp(4)))^2*csgn(I*exp(-x*exp(4)))*Pi-I*csgn(I*x*exp(-x*exp(4)))*csgn(I*x)*csgn(I*exp(-x*exp(4)))*Pi
+2*ln(x)-2*ln(exp(x*exp(4)))))

________________________________________________________________________________________

maxima [B]  time = 0.86, size = 82, normalized size = 2.93 \begin {gather*} e^{\left (-\frac {24 \, e^{4} \log \relax (2)}{x e^{4} \log \relax (x) - \log \relax (x)^{2}} - \frac {36 \, e^{4}}{x e^{4} \log \relax (x) - \log \relax (x)^{2}} + \frac {12 \, e^{4}}{x e^{4} - \log \relax (x)} - \frac {12}{x} + \frac {24 \, \log \relax (2)}{x \log \relax (x)} + \frac {36}{x \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*log(4/x)-48)*log(x/exp(x*exp(4)))+(12*x*exp(4)-12)*log(4/x)+36*x*exp(4)-36)*exp((12*log(4/x)+3
6)/x/log(x/exp(x*exp(4))))/x^2/log(x/exp(x*exp(4)))^2,x, algorithm="maxima")

[Out]

e^(-24*e^4*log(2)/(x*e^4*log(x) - log(x)^2) - 36*e^4/(x*e^4*log(x) - log(x)^2) + 12*e^4/(x*e^4 - log(x)) - 12/
x + 24*log(2)/(x*log(x)) + 36/(x*log(x)))

________________________________________________________________________________________

mupad [B]  time = 1.66, size = 60, normalized size = 2.14 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {36}{x^2\,{\mathrm {e}}^4-x\,\ln \relax (x)}}}{2^{\frac {24}{x^2\,{\mathrm {e}}^4-x\,\ln \relax (x)}}\,{\left (\frac {1}{x}\right )}^{\frac {12}{x^2\,{\mathrm {e}}^4-x\,\ln \relax (x)}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((12*log(4/x) + 36)/(x*log(x*exp(-x*exp(4)))))*(36*x*exp(4) + log(4/x)*(12*x*exp(4) - 12) - log(x*exp(
-x*exp(4)))*(12*log(4/x) + 48) - 36))/(x^2*log(x*exp(-x*exp(4)))^2),x)

[Out]

exp(-36/(x^2*exp(4) - x*log(x)))/(2^(24/(x^2*exp(4) - x*log(x)))*(1/x)^(12/(x^2*exp(4) - x*log(x))))

________________________________________________________________________________________

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(((-12*ln(4/x)-48)*ln(x/exp(x*exp(4)))+(12*x*exp(4)-12)*ln(4/x)+36*x*exp(4)-36)*exp((12*ln(4/x)+36)/x
/ln(x/exp(x*exp(4))))/x**2/ln(x/exp(x*exp(4)))**2,x)

[Out]

Timed out

________________________________________________________________________________________