3.55.100 \(\int \frac {e^x x+5 x \log (x^2)+e^{-e^{\log ^4(e^x+5 \log (x^2))}+x} (e^x x+5 x \log (x^2)+e^{\log ^4(e^x+5 \log (x^2))} (-40-4 e^x x) \log ^3(e^x+5 \log (x^2)))}{e^x x+5 x \log (x^2)} \, dx\)

Optimal. Leaf size=24 \[ 4+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x}+x \]

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Rubi [F]  time = 4.15, 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^x x+5 x \log \left (x^2\right )+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \left (e^x x+5 x \log \left (x^2\right )+e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )} \left (-40-4 e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )\right )}{e^x x+5 x \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*x + 5*x*Log[x^2] + E^(-E^Log[E^x + 5*Log[x^2]]^4 + x)*(E^x*x + 5*x*Log[x^2] + E^Log[E^x + 5*Log[x^2]]
^4*(-40 - 4*E^x*x)*Log[E^x + 5*Log[x^2]]^3))/(E^x*x + 5*x*Log[x^2]),x]

[Out]

x + Defer[Int][E^(-E^Log[E^x + 5*Log[x^2]]^4 + x), x] - 4*Defer[Int][E^(-E^Log[E^x + 5*Log[x^2]]^4 + x + Log[E
^x + 5*Log[x^2]]^4)*Log[E^x + 5*Log[x^2]]^3, x] - 40*Defer[Int][(E^(-E^Log[E^x + 5*Log[x^2]]^4 + x + Log[E^x +
5*Log[x^2]]^4)*Log[E^x + 5*Log[x^2]]^3)/(x*(E^x + 5*Log[x^2])), x] + 20*Defer[Int][(E^(-E^Log[E^x + 5*Log[x^2
]]^4 + x + Log[E^x + 5*Log[x^2]]^4)*Log[x^2]*Log[E^x + 5*Log[x^2]]^3)/(E^x + 5*Log[x^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}} \left (e^{e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}}+e^x\right )-\frac {4 \exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \left (10+e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )}{x \left (e^x+5 \log \left (x^2\right )\right )}\right ) \, dx\\ &=-\left (4 \int \frac {\exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \left (10+e^x x\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )}{x \left (e^x+5 \log \left (x^2\right )\right )} \, dx\right )+\int e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}} \left (e^{e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}}+e^x\right ) \, dx\\ &=-\left (4 \int \left (\exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )-\frac {5 \exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \left (-2+x \log \left (x^2\right )\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )}{x \left (e^x+5 \log \left (x^2\right )\right )}\right ) \, dx\right )+\int \left (1+e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x}\right ) \, dx\\ &=x-4 \int \exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right ) \, dx+20 \int \frac {\exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \left (-2+x \log \left (x^2\right )\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )}{x \left (e^x+5 \log \left (x^2\right )\right )} \, dx+\int e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \, dx\\ &=x-4 \int \exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right ) \, dx+20 \int \left (-\frac {2 \exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )}{x \left (e^x+5 \log \left (x^2\right )\right )}+\frac {\exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \log \left (x^2\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )}{e^x+5 \log \left (x^2\right )}\right ) \, dx+\int e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \, dx\\ &=x-4 \int \exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right ) \, dx+20 \int \frac {\exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \log \left (x^2\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )}{e^x+5 \log \left (x^2\right )} \, dx-40 \int \frac {\exp \left (-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x+\log ^4\left (e^x+5 \log \left (x^2\right )\right )\right ) \log ^3\left (e^x+5 \log \left (x^2\right )\right )}{x \left (e^x+5 \log \left (x^2\right )\right )} \, dx+\int e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 23, normalized size = 0.96 \begin {gather*} e^{-e^{\log ^4\left (e^x+5 \log \left (x^2\right )\right )}+x}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*x + 5*x*Log[x^2] + E^(-E^Log[E^x + 5*Log[x^2]]^4 + x)*(E^x*x + 5*x*Log[x^2] + E^Log[E^x + 5*Log
[x^2]]^4*(-40 - 4*E^x*x)*Log[E^x + 5*Log[x^2]]^3))/(E^x*x + 5*x*Log[x^2]),x]

[Out]

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

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fricas [A]  time = 1.91, size = 20, normalized size = 0.83 \begin {gather*} x + e^{\left (x - e^{\left (\log \left (e^{x} + 5 \, \log \left (x^{2}\right )\right )^{4}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(x)*x-40)*log(5*log(x^2)+exp(x))^3*exp(log(5*log(x^2)+exp(x))^4)+5*x*log(x^2)+exp(x)*x)*exp
(-exp(log(5*log(x^2)+exp(x))^4)+x)+5*x*log(x^2)+exp(x)*x)/(5*x*log(x^2)+exp(x)*x),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(x)*x-40)*log(5*log(x^2)+exp(x))^3*exp(log(5*log(x^2)+exp(x))^4)+5*x*log(x^2)+exp(x)*x)*exp
(-exp(log(5*log(x^2)+exp(x))^4)+x)+5*x*log(x^2)+exp(x)*x)/(5*x*log(x^2)+exp(x)*x),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.21, size = 48, normalized size = 2.00

method	result	size
risch	\(x +{\mathrm e}^{x} {\mathrm e}^{-{\mathrm e}^{\ln \left (10 \ln \relax (x )-\frac {5 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}+{\mathrm e}^{x}\right )^{4}}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*exp(x)*x-40)*ln(5*ln(x^2)+exp(x))^3*exp(ln(5*ln(x^2)+exp(x))^4)+5*x*ln(x^2)+exp(x)*x)*exp(-exp(ln(5*
ln(x^2)+exp(x))^4)+x)+5*x*ln(x^2)+exp(x)*x)/(5*x*ln(x^2)+exp(x)*x),x,method=_RETURNVERBOSE)

[Out]

x+exp(x)*exp(-exp(ln(10*ln(x)-5/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2+exp(x))^4))

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maxima [A]  time = 0.50, size = 18, normalized size = 0.75 \begin {gather*} x + e^{\left (x - e^{\left (\log \left (e^{x} + 10 \, \log \relax (x)\right )^{4}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*exp(x)*x-40)*log(5*log(x^2)+exp(x))^3*exp(log(5*log(x^2)+exp(x))^4)+5*x*log(x^2)+exp(x)*x)*exp
(-exp(log(5*log(x^2)+exp(x))^4)+x)+5*x*log(x^2)+exp(x)*x)/(5*x*log(x^2)+exp(x)*x),x, algorithm="maxima")

[Out]

x + e^(x - e^(log(e^x + 10*log(x))^4))

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mupad [B]  time = 3.74, size = 19, normalized size = 0.79 \begin {gather*} x+{\mathrm {e}}^{-{\mathrm {e}}^{{\ln \left (\ln \left (x^{10}\right )+{\mathrm {e}}^x\right )}^4}}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*log(x^2) + exp(x - exp(log(5*log(x^2) + exp(x))^4))*(5*x*log(x^2) + x*exp(x) - log(5*log(x^2) + exp(x
))^3*exp(log(5*log(x^2) + exp(x))^4)*(4*x*exp(x) + 40)) + x*exp(x))/(5*x*log(x^2) + x*exp(x)),x)

[Out]

x + exp(-exp(log(log(x^10) + exp(x))^4))*exp(x)

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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-40)*ln(5*ln(x**2)+exp(x))**3*exp(ln(5*ln(x**2)+exp(x))**4)+5*x*ln(x**2)+exp(x)*x)*exp
(-exp(ln(5*ln(x**2)+exp(x))**4)+x)+5*x*ln(x**2)+exp(x)*x)/(5*x*ln(x**2)+exp(x)*x),x)

[Out]

Timed out

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