3.43.57 \(\int \frac {-4 e^{4 e^x+4 x} \log ^3(\frac {3}{\log (4) \log (x)})+e^{4 e^x+4 x} (1+4 x+4 e^x x) \log (x) \log ^4(\frac {3}{\log (4) \log (x)})}{\log (x)} \, dx\)

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

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Rubi [F]  time = 2.28, 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 e^{4 e^x+4 x} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )+e^{4 e^x+4 x} \left (1+4 x+4 e^x x\right ) \log (x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*E^(4*E^x + 4*x)*Log[3/(Log[4]*Log[x])]^3 + E^(4*E^x + 4*x)*(1 + 4*x + 4*E^x*x)*Log[x]*Log[3/(Log[4]*Lo
g[x])]^4)/Log[x],x]

[Out]

-4*Defer[Int][(E^(4*(E^x + x))*Log[3/(Log[4]*Log[x])]^3)/Log[x], x] + Defer[Int][E^(4*(E^x + x))*Log[3/(Log[4]
*Log[x])]^4, x] + 4*Defer[Int][E^(4*(E^x + x))*x*Log[3/(Log[4]*Log[x])]^4, x] + 4*Defer[Int][E^(4*E^x + 5*x)*x
*Log[3/(Log[4]*Log[x])]^4, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (4 e^{4 e^x+5 x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right )+\frac {e^{4 \left (e^x+x\right )} \log ^3\left (\frac {3}{\log (4) \log (x)}\right ) \left (-4+\log (x) \log \left (\frac {3}{\log (4) \log (x)}\right )+4 x \log (x) \log \left (\frac {3}{\log (4) \log (x)}\right )\right )}{\log (x)}\right ) \, dx\\ &=4 \int e^{4 e^x+5 x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx+\int \frac {e^{4 \left (e^x+x\right )} \log ^3\left (\frac {3}{\log (4) \log (x)}\right ) \left (-4+\log (x) \log \left (\frac {3}{\log (4) \log (x)}\right )+4 x \log (x) \log \left (\frac {3}{\log (4) \log (x)}\right )\right )}{\log (x)} \, dx\\ &=4 \int e^{4 e^x+5 x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx+\int \frac {e^{4 \left (e^x+x\right )} \log ^3\left (\frac {3}{\log (4) \log (x)}\right ) \left (-4+(1+4 x) \log (x) \log \left (\frac {3}{\log (4) \log (x)}\right )\right )}{\log (x)} \, dx\\ &=4 \int e^{4 e^x+5 x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx+\int \left (-\frac {4 e^{4 \left (e^x+x\right )} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)}+e^{4 \left (e^x+x\right )} (1+4 x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right )\right ) \, dx\\ &=-\left (4 \int \frac {e^{4 \left (e^x+x\right )} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx\right )+4 \int e^{4 e^x+5 x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx+\int e^{4 \left (e^x+x\right )} (1+4 x) \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx\\ &=-\left (4 \int \frac {e^{4 \left (e^x+x\right )} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx\right )+4 \int e^{4 e^x+5 x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx+\int \left (e^{4 \left (e^x+x\right )} \log ^4\left (\frac {3}{\log (4) \log (x)}\right )+4 e^{4 \left (e^x+x\right )} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right )\right ) \, dx\\ &=-\left (4 \int \frac {e^{4 \left (e^x+x\right )} \log ^3\left (\frac {3}{\log (4) \log (x)}\right )}{\log (x)} \, dx\right )+4 \int e^{4 \left (e^x+x\right )} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx+4 \int e^{4 e^x+5 x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx+\int e^{4 \left (e^x+x\right )} \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.39, size = 26, normalized size = 1.00 \begin {gather*} e^{4 e^x+4 x} x \log ^4\left (\frac {3}{\log (4) \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^(4*E^x + 4*x)*Log[3/(Log[4]*Log[x])]^3 + E^(4*E^x + 4*x)*(1 + 4*x + 4*E^x*x)*Log[x]*Log[3/(Log
[4]*Log[x])]^4)/Log[x],x]

[Out]

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

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fricas [A]  time = 0.92, size = 24, normalized size = 0.92 \begin {gather*} x e^{\left (4 \, x + 4 \, e^{x}\right )} \log \left (\frac {3}{2 \, \log \relax (2) \log \relax (x)}\right )^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x+4*x+1)*log(x)*exp(4*exp(x)+4*x)*log(3/2/log(2)/log(x))^4-4*exp(4*exp(x)+4*x)*log(3/2/lo
g(2)/log(x))^3)/log(x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, x e^{x} + 4 \, x + 1\right )} e^{\left (4 \, x + 4 \, e^{x}\right )} \log \relax (x) \log \left (\frac {3}{2 \, \log \relax (2) \log \relax (x)}\right )^{4} - 4 \, e^{\left (4 \, x + 4 \, e^{x}\right )} \log \left (\frac {3}{2 \, \log \relax (2) \log \relax (x)}\right )^{3}}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x+4*x+1)*log(x)*exp(4*exp(x)+4*x)*log(3/2/log(2)/log(x))^4-4*exp(4*exp(x)+4*x)*log(3/2/lo
g(2)/log(x))^3)/log(x),x, algorithm="giac")

[Out]

integrate(((4*x*e^x + 4*x + 1)*e^(4*x + 4*e^x)*log(x)*log(3/2/(log(2)*log(x)))^4 - 4*e^(4*x + 4*e^x)*log(3/2/(
log(2)*log(x)))^3)/log(x), x)

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maple [B]  time = 1.09, size = 381, normalized size = 14.65

method	result	size
risch	\({\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} x \ln \left (\ln \relax (x )\right )^{4}+2 \left (2 x \ln \relax (2)-2 x \ln \relax (3)+2 x \ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} \ln \left (\ln \relax (x )\right )^{3}+\frac {3 \left (4 x \ln \relax (2)^{2}+4 x \ln \relax (3)^{2}+4 x \ln \left (\ln \relax (2)\right )^{2}-8 x \ln \relax (2) \ln \relax (3)+8 \ln \relax (2) \ln \left (\ln \relax (2)\right ) x -8 \ln \relax (3) \ln \left (\ln \relax (2)\right ) x \right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} \ln \left (\ln \relax (x )\right )^{2}}{2}+\frac {\left (24 x \ln \relax (2) \ln \relax (3)^{2}-8 x \ln \relax (3)^{3}+8 x \ln \relax (2)^{3}+8 \ln \left (\ln \relax (2)\right )^{3} x -48 \ln \relax (2) \ln \relax (3) \ln \left (\ln \relax (2)\right ) x +24 \ln \relax (3)^{2} \ln \left (\ln \relax (2)\right ) x -24 \ln \relax (3) \ln \left (\ln \relax (2)\right )^{2} x -24 \ln \relax (2)^{2} \ln \relax (3) x +24 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right ) x +24 \ln \relax (2) \ln \left (\ln \relax (2)\right )^{2} x \right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+4 x} \ln \left (\ln \relax (x )\right )}{2}+\frac {\left (-64 x \ln \relax (3) \ln \relax (2)^{3}+16 x \ln \relax (2)^{4}+16 x \ln \relax (3)^{4}+16 \ln \left (\ln \relax (2)\right )^{4} x -192 \ln \relax (2)^{2} \ln \relax (3) \ln \left (\ln \relax (2)\right ) x +192 \ln \relax (2) \ln \relax (3)^{2} \ln \left (\ln \relax (2)\right ) x -192 \ln \relax (2) \ln \relax (3) \ln \left (\ln \relax (2)\right )^{2} x +64 \ln \relax (2)^{3} \ln \left (\ln \relax (2)\right ) x +96 \ln \relax (2)^{2} \ln \relax (3)^{2} x +96 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right )^{2} x -64 \ln \relax (2) \ln \relax (3)^{3} x +64 \ln \relax (2) \ln \left (\ln \relax (2)\right )^{3} x -64 \ln \relax (3)^{3} \ln \left (\ln \relax (2)\right ) x +96 \ln \relax (3)^{2} \ln \left (\ln \relax (2)\right )^{2} x -64 \ln \relax (3) \ln \left (\ln \relax (2)\right )^{3} x \right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+4 x}}{16}\)	\(381\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(x)*x+4*x+1)*ln(x)*exp(4*exp(x)+4*x)*ln(3/2/ln(2)/ln(x))^4-4*exp(4*exp(x)+4*x)*ln(3/2/ln(2)/ln(x))^
3)/ln(x),x,method=_RETURNVERBOSE)

[Out]

exp(4*exp(x)+4*x)*x*ln(ln(x))^4+2*(2*x*ln(2)-2*x*ln(3)+2*x*ln(ln(2)))*exp(4*exp(x)+4*x)*ln(ln(x))^3+3/2*(4*x*l
n(2)^2+4*x*ln(3)^2+4*x*ln(ln(2))^2-8*x*ln(2)*ln(3)+8*ln(2)*ln(ln(2))*x-8*ln(3)*ln(ln(2))*x)*exp(4*exp(x)+4*x)*
ln(ln(x))^2+1/2*(24*x*ln(2)*ln(3)^2-8*x*ln(3)^3+8*x*ln(2)^3+8*ln(ln(2))^3*x-48*ln(2)*ln(3)*ln(ln(2))*x+24*ln(3
)^2*ln(ln(2))*x-24*ln(3)*ln(ln(2))^2*x-24*ln(2)^2*ln(3)*x+24*ln(2)^2*ln(ln(2))*x+24*ln(2)*ln(ln(2))^2*x)*exp(4
*exp(x)+4*x)*ln(ln(x))+1/16*(-64*x*ln(3)*ln(2)^3+16*x*ln(2)^4+16*x*ln(3)^4+16*ln(ln(2))^4*x-192*ln(2)^2*ln(3)*
ln(ln(2))*x+192*ln(2)*ln(3)^2*ln(ln(2))*x-192*ln(2)*ln(3)*ln(ln(2))^2*x+64*ln(2)^3*ln(ln(2))*x+96*ln(2)^2*ln(3
)^2*x+96*ln(2)^2*ln(ln(2))^2*x-64*ln(2)*ln(3)^3*x+64*ln(2)*ln(ln(2))^3*x-64*ln(3)^3*ln(ln(2))*x+96*ln(3)^2*ln(
ln(2))^2*x-64*ln(3)*ln(ln(2))^3*x)*exp(4*exp(x)+4*x)

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maxima [B]  time = 0.55, size = 291, normalized size = 11.19 \begin {gather*} -{\left (4 \, x {\left (\log \relax (3) - \log \relax (2) - \log \left (\log \relax (2)\right )\right )} e^{\left (4 \, x\right )} \log \left (\log \relax (x)\right )^{3} - x e^{\left (4 \, x\right )} \log \left (\log \relax (x)\right )^{4} - 6 \, {\left (\log \relax (3)^{2} - 2 \, {\left (\log \relax (3) - \log \left (\log \relax (2)\right )\right )} \log \relax (2) + \log \relax (2)^{2} - 2 \, \log \relax (3) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )} x e^{\left (4 \, x\right )} \log \left (\log \relax (x)\right )^{2} + 4 \, {\left (\log \relax (3)^{3} + 3 \, {\left (\log \relax (3) - \log \left (\log \relax (2)\right )\right )} \log \relax (2)^{2} - \log \relax (2)^{3} - 3 \, \log \relax (3)^{2} \log \left (\log \relax (2)\right ) + 3 \, \log \relax (3) \log \left (\log \relax (2)\right )^{2} - \log \left (\log \relax (2)\right )^{3} - 3 \, {\left (\log \relax (3)^{2} - 2 \, \log \relax (3) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )} \log \relax (2)\right )} x e^{\left (4 \, x\right )} \log \left (\log \relax (x)\right ) - {\left (\log \relax (3)^{4} - 4 \, {\left (\log \relax (3) - \log \left (\log \relax (2)\right )\right )} \log \relax (2)^{3} + \log \relax (2)^{4} - 4 \, \log \relax (3)^{3} \log \left (\log \relax (2)\right ) + 6 \, \log \relax (3)^{2} \log \left (\log \relax (2)\right )^{2} - 4 \, \log \relax (3) \log \left (\log \relax (2)\right )^{3} + \log \left (\log \relax (2)\right )^{4} + 6 \, {\left (\log \relax (3)^{2} - 2 \, \log \relax (3) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )} \log \relax (2)^{2} - 4 \, {\left (\log \relax (3)^{3} - 3 \, \log \relax (3)^{2} \log \left (\log \relax (2)\right ) + 3 \, \log \relax (3) \log \left (\log \relax (2)\right )^{2} - \log \left (\log \relax (2)\right )^{3}\right )} \log \relax (2)\right )} x e^{\left (4 \, x\right )}\right )} e^{\left (4 \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x+4*x+1)*log(x)*exp(4*exp(x)+4*x)*log(3/2/log(2)/log(x))^4-4*exp(4*exp(x)+4*x)*log(3/2/lo
g(2)/log(x))^3)/log(x),x, algorithm="maxima")

[Out]

-(4*x*(log(3) - log(2) - log(log(2)))*e^(4*x)*log(log(x))^3 - x*e^(4*x)*log(log(x))^4 - 6*(log(3)^2 - 2*(log(3
) - log(log(2)))*log(2) + log(2)^2 - 2*log(3)*log(log(2)) + log(log(2))^2)*x*e^(4*x)*log(log(x))^2 + 4*(log(3)
^3 + 3*(log(3) - log(log(2)))*log(2)^2 - log(2)^3 - 3*log(3)^2*log(log(2)) + 3*log(3)*log(log(2))^2 - log(log(
2))^3 - 3*(log(3)^2 - 2*log(3)*log(log(2)) + log(log(2))^2)*log(2))*x*e^(4*x)*log(log(x)) - (log(3)^4 - 4*(log
(3) - log(log(2)))*log(2)^3 + log(2)^4 - 4*log(3)^3*log(log(2)) + 6*log(3)^2*log(log(2))^2 - 4*log(3)*log(log(
2))^3 + log(log(2))^4 + 6*(log(3)^2 - 2*log(3)*log(log(2)) + log(log(2))^2)*log(2)^2 - 4*(log(3)^3 - 3*log(3)^
2*log(log(2)) + 3*log(3)*log(log(2))^2 - log(log(2))^3)*log(2))*x*e^(4*x))*e^(4*e^x)

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mupad [B]  time = 3.27, size = 24, normalized size = 0.92 \begin {gather*} x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,{\ln \left (\frac {3}{2\,\ln \relax (2)\,\ln \relax (x)}\right )}^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*exp(4*x + 4*exp(x))*log(3/(2*log(2)*log(x)))^3 - exp(4*x + 4*exp(x))*log(3/(2*log(2)*log(x)))^4*log(x)
*(4*x + 4*x*exp(x) + 1))/log(x),x)

[Out]

x*exp(4*x)*exp(4*exp(x))*log(3/(2*log(2)*log(x)))^4

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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(((4*exp(x)*x+4*x+1)*ln(x)*exp(4*exp(x)+4*x)*ln(3/2/ln(2)/ln(x))**4-4*exp(4*exp(x)+4*x)*ln(3/2/ln(2)/
ln(x))**3)/ln(x),x)

[Out]

Timed out

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