3.1.41 \(\int \frac {e^{-e^2-e^3-\frac {1}{4} e^{-e^2-e^3} x} (4 e^{e^2+e^3}+(-4 e^{e^2+e^3}-x) \log (5 x))}{4 x^2} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^{-\frac {1}{4} e^{-e^2-e^3} x} \log (5 x)}{x} \]

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Rubi [A]  time = 1.27, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 6, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 6742, 2177, 2178, 2197, 2554} \begin {gather*} \frac {e^{-\frac {1}{4} e^{-e^2 (1+e)} x} \log (5 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[5*x]/(E^(x/(4*E^(E^2*(1 + E))))*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.37, size = 25, normalized size = 0.89 \begin {gather*} \frac {e^{-\frac {1}{4} e^{-e^2 (1+e)} x} \log (5 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[5*x]/(E^(x/(4*E^(E^2*(1 + E))))*x)

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fricas [A]  time = 0.81, size = 41, normalized size = 1.46 \begin {gather*} \frac {e^{\left (-\frac {1}{4} \, {\left (4 \, {\left (e^{3} + e^{2}\right )} e^{\left (e^{3} + e^{2}\right )} + x\right )} e^{\left (-e^{3} - e^{2}\right )} + e^{3} + e^{2}\right )} \log \left (5 \, x\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-4*exp(exp(3)+exp(2))-x)*log(5*x)+4*exp(exp(3)+exp(2)))/x^2/exp(exp(3)+exp(2))/exp(1/4*x/exp(e
xp(3)+exp(2))),x, algorithm="fricas")

[Out]

e^(-1/4*(4*(e^3 + e^2)*e^(e^3 + e^2) + x)*e^(-e^3 - e^2) + e^3 + e^2)*log(5*x)/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (x + 4 \, e^{\left (e^{3} + e^{2}\right )}\right )} \log \left (5 \, x\right ) - 4 \, e^{\left (e^{3} + e^{2}\right )}\right )} e^{\left (-\frac {1}{4} \, x e^{\left (-e^{3} - e^{2}\right )} - e^{3} - e^{2}\right )}}{4 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-4*exp(exp(3)+exp(2))-x)*log(5*x)+4*exp(exp(3)+exp(2)))/x^2/exp(exp(3)+exp(2))/exp(1/4*x/exp(e
xp(3)+exp(2))),x, algorithm="giac")

[Out]

integrate(-1/4*((x + 4*e^(e^3 + e^2))*log(5*x) - 4*e^(e^3 + e^2))*e^(-1/4*x*e^(-e^3 - e^2) - e^3 - e^2)/x^2, x
)

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maple [A]  time = 0.11, size = 23, normalized size = 0.82

method	result	size
norman	\(\frac {{\mathrm e}^{-\frac {x \,{\mathrm e}^{-{\mathrm e}^{3}-{\mathrm e}^{2}}}{4}} \ln \left (5 x \right )}{x}\)	\(23\)
risch	\(\frac {{\mathrm e}^{-\frac {x \,{\mathrm e}^{-{\mathrm e}^{3}-{\mathrm e}^{2}}}{4}} \ln \left (5 x \right )}{x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-4*exp(exp(3)+exp(2))-x)*ln(5*x)+4*exp(exp(3)+exp(2)))/x^2/exp(exp(3)+exp(2))/exp(1/4*x/exp(exp(3)+e
xp(2))),x,method=_RETURNVERBOSE)

[Out]

ln(5*x)/exp(1/4*x/exp(exp(3)+exp(2)))/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{4} \, e^{\left (-e^{3} - e^{2}\right )} \Gamma \left (-1, \frac {1}{4} \, x e^{\left (-e^{3} - e^{2}\right )}\right ) + \frac {e^{\left (-\frac {1}{4} \, x e^{\left (-e^{3} - e^{2}\right )}\right )} \log \relax (x)}{x} - \frac {1}{4} \, \int \frac {{\left (4 \, {\left (\log \relax (5) + 1\right )} e^{\left (e^{3} + e^{2}\right )} + x \log \relax (5)\right )} e^{\left (-\frac {1}{4} \, x e^{\left (-e^{3} - e^{2}\right )} - e^{3} - e^{2}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-4*exp(exp(3)+exp(2))-x)*log(5*x)+4*exp(exp(3)+exp(2)))/x^2/exp(exp(3)+exp(2))/exp(1/4*x/exp(e
xp(3)+exp(2))),x, algorithm="maxima")

[Out]

-1/4*e^(-e^3 - e^2)*gamma(-1, 1/4*x*e^(-e^3 - e^2)) + e^(-1/4*x*e^(-e^3 - e^2))*log(x)/x - 1/4*integrate((4*(l
og(5) + 1)*e^(e^3 + e^2) + x*log(5))*e^(-1/4*x*e^(-e^3 - e^2) - e^3 - e^2)/x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^{-{\mathrm {e}}^2-{\mathrm {e}}^3}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{-{\mathrm {e}}^2-{\mathrm {e}}^3}}{4}}\,\left ({\mathrm {e}}^{{\mathrm {e}}^2+{\mathrm {e}}^3}-\frac {\ln \left (5\,x\right )\,\left (x+4\,{\mathrm {e}}^{{\mathrm {e}}^2+{\mathrm {e}}^3}\right )}{4}\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.29, size = 19, normalized size = 0.68 \begin {gather*} \frac {e^{- \frac {x}{4 e^{e^{2} + e^{3}}}} \log {\left (5 x \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-4*exp(exp(3)+exp(2))-x)*ln(5*x)+4*exp(exp(3)+exp(2)))/x**2/exp(exp(3)+exp(2))/exp(1/4*x/exp(e
xp(3)+exp(2))),x)

[Out]

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

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