3.90.79 \(\int \frac {(-1920 x-576 x^2+e^{4/x} (640 x+192 x^2)) \log (-3+e^{4/x}) \log (\log (-3+e^{4/x}))+512 e^{4/x} \log (\frac {5}{\log (\log (-3+e^{4/x}))})+(-384 x+128 e^{4/x} x) \log (-3+e^{4/x}) \log (\log (-3+e^{4/x})) \log ^2(\frac {5}{\log (\log (-3+e^{4/x}))})}{(-3+e^{4/x}) \log (-3+e^{4/x}) \log (\log (-3+e^{4/x}))} \, dx\)

Optimal. Leaf size=26 \[ 64 x^2 \left (5+x+\log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 1.32, 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 (-1920 x-576 x^2+e^{4/x} \left (640 x+192 x^2\right )\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )+512 e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )+\left (-384 x+128 e^{4/x} x\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right ) \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-1920*x - 576*x^2 + E^(4/x)*(640*x + 192*x^2))*Log[-3 + E^(4/x)]*Log[Log[-3 + E^(4/x)]] + 512*E^(4/x)*Lo
g[5/Log[Log[-3 + E^(4/x)]]] + (-384*x + 128*E^(4/x)*x)*Log[-3 + E^(4/x)]*Log[Log[-3 + E^(4/x)]]*Log[5/Log[Log[
-3 + E^(4/x)]]]^2)/((-3 + E^(4/x))*Log[-3 + E^(4/x)]*Log[Log[-3 + E^(4/x)]]),x]

[Out]

320*x^2 + 64*x^3 + 512*Defer[Int][(E^(4/x)*Log[5/Log[Log[-3 + E^(4/x)]]])/((-3 + E^(4/x))*Log[-3 + E^(4/x)]*Lo
g[Log[-3 + E^(4/x)]]), x] + 128*Defer[Int][x*Log[5/Log[Log[-3 + E^(4/x)]]]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int 64 \left (\frac {8 e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )}+x \left (10+3 x+2 \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right )\right ) \, dx\\ &=64 \int \left (\frac {8 e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )}+x \left (10+3 x+2 \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right )\right ) \, dx\\ &=64 \int x \left (10+3 x+2 \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ &=64 \int \left (x (10+3 x)+2 x \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ &=64 \int x (10+3 x) \, dx+128 \int x \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ &=64 \int \left (10 x+3 x^2\right ) \, dx+128 \int x \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ &=320 x^2+64 x^3+128 \int x \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right ) \, dx+512 \int \frac {e^{4/x} \log \left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )}{\left (-3+e^{4/x}\right ) \log \left (-3+e^{4/x}\right ) \log \left (\log \left (-3+e^{4/x}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 33, normalized size = 1.27 \begin {gather*} 64 \left (5 x^2+x^3+x^2 \log ^2\left (\frac {5}{\log \left (\log \left (-3+e^{4/x}\right )\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-1920*x - 576*x^2 + E^(4/x)*(640*x + 192*x^2))*Log[-3 + E^(4/x)]*Log[Log[-3 + E^(4/x)]] + 512*E^(4
/x)*Log[5/Log[Log[-3 + E^(4/x)]]] + (-384*x + 128*E^(4/x)*x)*Log[-3 + E^(4/x)]*Log[Log[-3 + E^(4/x)]]*Log[5/Lo
g[Log[-3 + E^(4/x)]]]^2)/((-3 + E^(4/x))*Log[-3 + E^(4/x)]*Log[Log[-3 + E^(4/x)]]),x]

[Out]

64*(5*x^2 + x^3 + x^2*Log[5/Log[Log[-3 + E^(4/x)]]]^2)

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 33, normalized size = 1.27 \begin {gather*} 64 \, x^{2} \log \left (\frac {5}{\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )}\right )^{2} + 64 \, x^{3} + 320 \, x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((128*x*exp(4/x)-384*x)*log(exp(4/x)-3)*log(log(exp(4/x)-3))*log(5/log(log(exp(4/x)-3)))^2+512*exp(4
/x)*log(5/log(log(exp(4/x)-3)))+((192*x^2+640*x)*exp(4/x)-576*x^2-1920*x)*log(exp(4/x)-3)*log(log(exp(4/x)-3))
)/(exp(4/x)-3)/log(exp(4/x)-3)/log(log(exp(4/x)-3)),x, algorithm="fricas")

[Out]

64*x^2*log(5/log(log(e^(4/x) - 3)))^2 + 64*x^3 + 320*x^2

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {64 \, {\left (2 \, {\left (x e^{\frac {4}{x}} - 3 \, x\right )} \log \left (\frac {5}{\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )}\right )^{2} \log \left (e^{\frac {4}{x}} - 3\right ) \log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right ) - {\left (9 \, x^{2} - {\left (3 \, x^{2} + 10 \, x\right )} e^{\frac {4}{x}} + 30 \, x\right )} \log \left (e^{\frac {4}{x}} - 3\right ) \log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right ) + 8 \, e^{\frac {4}{x}} \log \left (\frac {5}{\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )}\right )\right )}}{{\left (e^{\frac {4}{x}} - 3\right )} \log \left (e^{\frac {4}{x}} - 3\right ) \log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((128*x*exp(4/x)-384*x)*log(exp(4/x)-3)*log(log(exp(4/x)-3))*log(5/log(log(exp(4/x)-3)))^2+512*exp(4
/x)*log(5/log(log(exp(4/x)-3)))+((192*x^2+640*x)*exp(4/x)-576*x^2-1920*x)*log(exp(4/x)-3)*log(log(exp(4/x)-3))
)/(exp(4/x)-3)/log(exp(4/x)-3)/log(log(exp(4/x)-3)),x, algorithm="giac")

[Out]

integrate(64*(2*(x*e^(4/x) - 3*x)*log(5/log(log(e^(4/x) - 3)))^2*log(e^(4/x) - 3)*log(log(e^(4/x) - 3)) - (9*x
^2 - (3*x^2 + 10*x)*e^(4/x) + 30*x)*log(e^(4/x) - 3)*log(log(e^(4/x) - 3)) + 8*e^(4/x)*log(5/log(log(e^(4/x) -
3))))/((e^(4/x) - 3)*log(e^(4/x) - 3)*log(log(e^(4/x) - 3))), x)

________________________________________________________________________________________

maple [B]  time = 0.54, size = 57, normalized size = 2.19

method	result	size
risch	\(64 x^{2} \ln \left (\ln \left (\ln \left ({\mathrm e}^{\frac {4}{x}}-3\right )\right )\right )^{2}-128 \ln \relax (5) x^{2} \ln \left (\ln \left (\ln \left ({\mathrm e}^{\frac {4}{x}}-3\right )\right )\right )+64 x^{2} \ln \relax (5)^{2}+64 x^{3}+320 x^{2}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((128*x*exp(4/x)-384*x)*ln(exp(4/x)-3)*ln(ln(exp(4/x)-3))*ln(5/ln(ln(exp(4/x)-3)))^2+512*exp(4/x)*ln(5/ln(
ln(exp(4/x)-3)))+((192*x^2+640*x)*exp(4/x)-576*x^2-1920*x)*ln(exp(4/x)-3)*ln(ln(exp(4/x)-3)))/(exp(4/x)-3)/ln(
exp(4/x)-3)/ln(ln(exp(4/x)-3)),x,method=_RETURNVERBOSE)

[Out]

64*x^2*ln(ln(ln(exp(4/x)-3)))^2-128*ln(5)*x^2*ln(ln(ln(exp(4/x)-3)))+64*x^2*ln(5)^2+64*x^3+320*x^2

________________________________________________________________________________________

maxima [B]  time = 0.55, size = 53, normalized size = 2.04 \begin {gather*} -128 \, x^{2} \log \relax (5) \log \left (\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )\right ) + 64 \, x^{2} \log \left (\log \left (\log \left (e^{\frac {4}{x}} - 3\right )\right )\right )^{2} + 64 \, {\left (\log \relax (5)^{2} + 5\right )} x^{2} + 64 \, x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((128*x*exp(4/x)-384*x)*log(exp(4/x)-3)*log(log(exp(4/x)-3))*log(5/log(log(exp(4/x)-3)))^2+512*exp(4
/x)*log(5/log(log(exp(4/x)-3)))+((192*x^2+640*x)*exp(4/x)-576*x^2-1920*x)*log(exp(4/x)-3)*log(log(exp(4/x)-3))
)/(exp(4/x)-3)/log(exp(4/x)-3)/log(log(exp(4/x)-3)),x, algorithm="maxima")

[Out]

-128*x^2*log(5)*log(log(log(e^(4/x) - 3))) + 64*x^2*log(log(log(e^(4/x) - 3)))^2 + 64*(log(5)^2 + 5)*x^2 + 64*
x^3

________________________________________________________________________________________

mupad [B]  time = 8.55, size = 25, normalized size = 0.96 \begin {gather*} 64\,x^2\,\left ({\ln \left (\frac {5}{\ln \left (\ln \left ({\mathrm {e}}^{4/x}-3\right )\right )}\right )}^2+x+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp(4/x) - 3)*log(log(exp(4/x) - 3))*(1920*x - exp(4/x)*(640*x + 192*x^2) + 576*x^2) - 512*exp(4/x)*
log(5/log(log(exp(4/x) - 3))) + log(exp(4/x) - 3)*log(log(exp(4/x) - 3))*log(5/log(log(exp(4/x) - 3)))^2*(384*
x - 128*x*exp(4/x)))/(log(exp(4/x) - 3)*log(log(exp(4/x) - 3))*(exp(4/x) - 3)),x)

[Out]

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

________________________________________________________________________________________

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(((128*x*exp(4/x)-384*x)*ln(exp(4/x)-3)*ln(ln(exp(4/x)-3))*ln(5/ln(ln(exp(4/x)-3)))**2+512*exp(4/x)*l
n(5/ln(ln(exp(4/x)-3)))+((192*x**2+640*x)*exp(4/x)-576*x**2-1920*x)*ln(exp(4/x)-3)*ln(ln(exp(4/x)-3)))/(exp(4/
x)-3)/ln(exp(4/x)-3)/ln(ln(exp(4/x)-3)),x)

[Out]

Timed out

________________________________________________________________________________________