3.98.26 \(\int \frac {e^{4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log (x^2)+\log ^2(x^2)}}+\frac {e^{4 e^x+x}}{16-8 \log (x^2)+\log ^2(x^2)}} (16 e^{2 x} x+e^x (4+4 x)+(-e^x x-4 e^{2 x} x) \log (x^2))}{-64 x+48 x \log (x^2)-12 x \log ^2(x^2)+x \log ^3(x^2)} \, dx\)

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

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Rubi [F]  time = 20.18, 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 {\exp \left (4 e^x+e^{\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}}+\frac {e^{4 e^x+x}}{16-8 \log \left (x^2\right )+\log ^2\left (x^2\right )}\right ) \left (16 e^{2 x} x+e^x (4+4 x)+\left (-e^x x-4 e^{2 x} x\right ) \log \left (x^2\right )\right )}{-64 x+48 x \log \left (x^2\right )-12 x \log ^2\left (x^2\right )+x \log ^3\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(4*E^x + E^(E^(4*E^x + x)/(16 - 8*Log[x^2] + Log[x^2]^2)) + E^(4*E^x + x)/(16 - 8*Log[x^2] + Log[x^2]^2
))*(16*E^(2*x)*x + E^x*(4 + 4*x) + (-(E^x*x) - 4*E^(2*x)*x)*Log[x^2]))/(-64*x + 48*x*Log[x^2] - 12*x*Log[x^2]^
2 + x*Log[x^2]^3),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.33, size = 24, normalized size = 0.92 \begin {gather*} -e^{e^{\frac {e^{4 e^x+x}}{\left (-4+\log \left (x^2\right )\right )^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4*E^x + E^(E^(4*E^x + x)/(16 - 8*Log[x^2] + Log[x^2]^2)) + E^(4*E^x + x)/(16 - 8*Log[x^2] + Log[
x^2]^2))*(16*E^(2*x)*x + E^x*(4 + 4*x) + (-(E^x*x) - 4*E^(2*x)*x)*Log[x^2]))/(-64*x + 48*x*Log[x^2] - 12*x*Log
[x^2]^2 + x*Log[x^2]^3),x]

[Out]

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

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fricas [B]  time = 0.72, size = 120, normalized size = 4.62 \begin {gather*} -e^{\left (\frac {4 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16\right )} e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16}\right )} - 32 \, e^{x} \log \left (x^{2}\right ) + e^{\left (x + 4 \, e^{x}\right )} + 64 \, e^{x}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} - \frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} - 4 \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2-exp(x)*x)*log(x^2)+16*x*exp(x)^2+(4*x+4)*exp(x))*exp(4*exp(x))*exp(exp(x)*exp(4*exp(
x))/(log(x^2)^2-8*log(x^2)+16))*exp(exp(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16)))/(x*log(x^2)^3-12*x*l
og(x^2)^2+48*x*log(x^2)-64*x),x, algorithm="fricas")

[Out]

-e^((4*e^x*log(x^2)^2 + (log(x^2)^2 - 8*log(x^2) + 16)*e^(e^(x + 4*e^x)/(log(x^2)^2 - 8*log(x^2) + 16)) - 32*e
^x*log(x^2) + e^(x + 4*e^x) + 64*e^x)/(log(x^2)^2 - 8*log(x^2) + 16) - e^(x + 4*e^x)/(log(x^2)^2 - 8*log(x^2)
+ 16) - 4*e^x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (16 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x + 1\right )} e^{x} - {\left (4 \, x e^{\left (2 \, x\right )} + x e^{x}\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16} + 4 \, e^{x} + e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{\log \left (x^{2}\right )^{2} - 8 \, \log \left (x^{2}\right ) + 16}\right )}\right )}}{x \log \left (x^{2}\right )^{3} - 12 \, x \log \left (x^{2}\right )^{2} + 48 \, x \log \left (x^{2}\right ) - 64 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2-exp(x)*x)*log(x^2)+16*x*exp(x)^2+(4*x+4)*exp(x))*exp(4*exp(x))*exp(exp(x)*exp(4*exp(
x))/(log(x^2)^2-8*log(x^2)+16))*exp(exp(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16)))/(x*log(x^2)^3-12*x*l
og(x^2)^2+48*x*log(x^2)-64*x),x, algorithm="giac")

[Out]

integrate((16*x*e^(2*x) + 4*(x + 1)*e^x - (4*x*e^(2*x) + x*e^x)*log(x^2))*e^(e^(x + 4*e^x)/(log(x^2)^2 - 8*log
(x^2) + 16) + 4*e^x + e^(e^(x + 4*e^x)/(log(x^2)^2 - 8*log(x^2) + 16)))/(x*log(x^2)^3 - 12*x*log(x^2)^2 + 48*x
*log(x^2) - 64*x), x)

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maple [C]  time = 1.16, size = 71, normalized size = 2.73

method	result	size
risch	\(-{\mathrm e}^{{\mathrm e}^{\frac {4 \,{\mathrm e}^{4 \,{\mathrm e}^{x}+x}}{\left (-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+4 \ln \relax (x )-8\right )^{2}}}}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x*exp(x)^2-exp(x)*x)*ln(x^2)+16*x*exp(x)^2+(4*x+4)*exp(x))*exp(4*exp(x))*exp(exp(x)*exp(4*exp(x))/(ln
(x^2)^2-8*ln(x^2)+16))*exp(exp(exp(x)*exp(4*exp(x))/(ln(x^2)^2-8*ln(x^2)+16)))/(x*ln(x^2)^3-12*x*ln(x^2)^2+48*
x*ln(x^2)-64*x),x,method=_RETURNVERBOSE)

[Out]

-exp(exp(4*exp(4*exp(x)+x)/(-I*Pi*csgn(I*x^2)^3+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2-I*Pi*csgn(I*x)^2*csgn(I*x^2)+4*
ln(x)-8)^2))

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maxima [A]  time = 1.07, size = 25, normalized size = 0.96 \begin {gather*} -e^{\left (e^{\left (\frac {e^{\left (x + 4 \, e^{x}\right )}}{4 \, {\left (\log \relax (x)^{2} - 4 \, \log \relax (x) + 4\right )}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x*exp(x)^2-exp(x)*x)*log(x^2)+16*x*exp(x)^2+(4*x+4)*exp(x))*exp(4*exp(x))*exp(exp(x)*exp(4*exp(
x))/(log(x^2)^2-8*log(x^2)+16))*exp(exp(exp(x)*exp(4*exp(x))/(log(x^2)^2-8*log(x^2)+16)))/(x*log(x^2)^3-12*x*l
og(x^2)^2+48*x*log(x^2)-64*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.10, size = 28, normalized size = 1.08 \begin {gather*} -{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,{\mathrm {e}}^x}\,{\mathrm {e}}^x}{{\ln \left (x^2\right )}^2-8\,\ln \left (x^2\right )+16}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(4*exp(x))*exp(x))/(log(x^2)^2 - 8*log(x^2) + 16))*exp(4*exp(x))*exp(exp((exp(4*exp(x))*exp(x))/
(log(x^2)^2 - 8*log(x^2) + 16)))*(16*x*exp(2*x) + exp(x)*(4*x + 4) - log(x^2)*(4*x*exp(2*x) + x*exp(x))))/(64*
x - 48*x*log(x^2) + 12*x*log(x^2)^2 - x*log(x^2)^3),x)

[Out]

-exp(exp((exp(4*exp(x))*exp(x))/(log(x^2)^2 - 8*log(x^2) + 16)))

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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*x*exp(x)**2-exp(x)*x)*ln(x**2)+16*x*exp(x)**2+(4*x+4)*exp(x))*exp(4*exp(x))*exp(exp(x)*exp(4*ex
p(x))/(ln(x**2)**2-8*ln(x**2)+16))*exp(exp(exp(x)*exp(4*exp(x))/(ln(x**2)**2-8*ln(x**2)+16)))/(x*ln(x**2)**3-1
2*x*ln(x**2)**2+48*x*ln(x**2)-64*x),x)

[Out]

Timed out

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