3.88.89 \(\int \frac {e^{2 e^3 x+x \log (x)+e^{2 e^{2 e^5}} (2 e^3+\log (x))} (e^{2 e^{2 e^5}}+x+2 e^3 x+x \log (x))}{x} \, dx\)

Optimal. Leaf size=24 \[ e^{\left (e^{2 e^{2 e^5}}+x\right ) \left (2 e^3+\log (x)\right )} \]

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Rubi [A]  time = 1.23, antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 2, number of rules used = 2, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6, 6706} \begin {gather*} x^x e^{2 e^3 x+e^{2 e^{2 e^5}} \left (\log (x)+2 e^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*E^3*x + E^(2*E^(2*E^5))*(2*E^3 + Log[x]))*x^x

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.70, size = 36, normalized size = 1.50 \begin {gather*} e^{2 e^3 \left (e^{2 e^{2 e^5}}+x\right )} x^{e^{2 e^{2 e^5}}+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2*E^3*(E^(2*E^(2*E^5)) + x))*x^(E^(2*E^(2*E^5)) + x)

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fricas [A]  time = 0.51, size = 27, normalized size = 1.12 \begin {gather*} e^{\left (2 \, x e^{3} + {\left (2 \, e^{3} + \log \relax (x)\right )} e^{\left (2 \, e^{\left (2 \, e^{5}\right )}\right )} + x \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x*e^3 + (2*e^3 + log(x))*e^(2*e^(2*e^5)) + x*log(x))

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giac [A]  time = 0.19, size = 34, normalized size = 1.42 \begin {gather*} e^{\left (2 \, x e^{3} + x \log \relax (x) + e^{\left (2 \, e^{\left (2 \, e^{5}\right )}\right )} \log \relax (x) + 2 \, e^{\left (2 \, e^{\left (2 \, e^{5}\right )} + 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x*e^3 + x*log(x) + e^(2*e^(2*e^5))*log(x) + 2*e^(2*e^(2*e^5) + 3))

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maple [A]  time = 0.06, size = 28, normalized size = 1.17

method	result	size
norman	\({\mathrm e}^{\left (\ln \relax (x )+2 \,{\mathrm e}^{3}\right ) {\mathrm e}^{2 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}}+x \ln \relax (x )+2 x \,{\mathrm e}^{3}}\)	\(28\)
risch	\(x^{x} x^{{\mathrm e}^{2 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}}} {\mathrm e}^{2 \,{\mathrm e}^{2 \,{\mathrm e}^{2 \,{\mathrm e}^{5}}+3}+2 x \,{\mathrm e}^{3}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((ln(x)+2*exp(3))*exp(exp(2*exp(5)))^2+x*ln(x)+2*x*exp(3))

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maxima [A]  time = 0.51, size = 34, normalized size = 1.42 \begin {gather*} e^{\left (2 \, x e^{3} + x \log \relax (x) + e^{\left (2 \, e^{\left (2 \, e^{5}\right )}\right )} \log \relax (x) + 2 \, e^{\left (2 \, e^{\left (2 \, e^{5}\right )} + 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x*e^3 + x*log(x) + e^(2*e^(2*e^5))*log(x) + 2*e^(2*e^(2*e^5) + 3))

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mupad [B]  time = 5.62, size = 32, normalized size = 1.33 \begin {gather*} x^{x+{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,{\mathrm {e}}^5}}}\,{\mathrm {e}}^{2\,{\mathrm {e}}^3\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,{\mathrm {e}}^5}}}\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x*exp(3) + exp(2*exp(2*exp(5)))*(2*exp(3) + log(x)) + x*log(x))*(x + exp(2*exp(2*exp(5))) + 2*x*exp
(3) + x*log(x)))/x,x)

[Out]

x^(x + exp(2*exp(2*exp(5))))*exp(2*exp(3)*exp(2*exp(2*exp(5))))*exp(2*x*exp(3))

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sympy [A]  time = 0.35, size = 31, normalized size = 1.29 \begin {gather*} e^{x \log {\relax (x )} + 2 x e^{3} + \left (\log {\relax (x )} + 2 e^{3}\right ) e^{2 e^{2 e^{5}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(2*exp(5)))**2+x*ln(x)+2*x*exp(3)+x)*exp((ln(x)+2*exp(3))*exp(exp(2*exp(5)))**2+x*ln(x)+2*x*
exp(3))/x,x)

[Out]

exp(x*log(x) + 2*x*exp(3) + (log(x) + 2*exp(3))*exp(2*exp(2*exp(5))))

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