3.102.78 \(\int \frac {e^{e^x+x} x+\frac {e^{2+2 \log ^2(5 x)} (-4 e^{6+2 e^4}+4 e^{6+2 e^4} \log (5 x))}{625 x^4}}{x} \, dx\)

Optimal. Leaf size=27 \[ e^{e^x}+e^{6+2 e^4+2 (1-\log (5 x))^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 28, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {14, 2282, 2194, 2288} \begin {gather*} \frac {e^{2 \left (\log ^2(5 x)+e^4+4\right )}}{625 x^4}+e^{e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^x + x)*x + (E^(2 + 2*Log[5*x]^2)*(-4*E^(6 + 2*E^4) + 4*E^(6 + 2*E^4)*Log[5*x]))/(625*x^4))/x,x]

[Out]

E^E^x + E^(2*(4 + E^4 + Log[5*x]^2))/(625*x^4)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{e^x+x}+\frac {4 e^{2 \left (4 \left (1+\frac {e^4}{4}\right )+\log ^2(5 x)\right )} (-1+\log (5 x))}{625 x^5}\right ) \, dx\\ &=\frac {4}{625} \int \frac {e^{2 \left (4 \left (1+\frac {e^4}{4}\right )+\log ^2(5 x)\right )} (-1+\log (5 x))}{x^5} \, dx+\int e^{e^x+x} \, dx\\ &=\frac {e^{2 \left (4+e^4+\log ^2(5 x)\right )}}{625 x^4}+\operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=e^{e^x}+\frac {e^{2 \left (4+e^4+\log ^2(5 x)\right )}}{625 x^4}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 31, normalized size = 1.15 \begin {gather*} \frac {1}{625} \left (625 e^{e^x}+\frac {e^{2 \left (4+e^4+\log ^2(5 x)\right )}}{x^4}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^x + x)*x + (E^(2 + 2*Log[5*x]^2)*(-4*E^(6 + 2*E^4) + 4*E^(6 + 2*E^4)*Log[5*x]))/(625*x^4))/x,x
]

[Out]

(625*E^E^x + E^(2*(4 + E^4 + Log[5*x]^2))/x^4)/625

________________________________________________________________________________________

fricas [A]  time = 0.88, size = 33, normalized size = 1.22 \begin {gather*} {\left (e^{\left (2 \, \log \left (5 \, x\right )^{2} + x + 2 \, e^{4} - 4 \, \log \left (5 \, x\right ) + 8\right )} + e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(2*log(5*x)^2 + x + 2*e^4 - 4*log(5*x) + 8) + e^(x + e^x))*e^(-x)

________________________________________________________________________________________

giac [A]  time = 0.28, size = 36, normalized size = 1.33 \begin {gather*} \frac {{\left (625 \, x^{4} e^{\left (x + e^{x}\right )} + e^{\left (2 \, \log \left (5 \, x\right )^{2} + x + 2 \, e^{4} + 8\right )}\right )} e^{\left (-x\right )}}{625 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/625*(625*x^4*e^(x + e^x) + e^(2*log(5*x)^2 + x + 2*e^4 + 8))*e^(-x)/x^4

________________________________________________________________________________________

maple [A]  time = 0.14, size = 25, normalized size = 0.93

method	result	size
risch	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{4}+8+2 \ln \left (5 x \right )^{2}}}{625 x^{4}}+{\mathrm e}^{{\mathrm e}^{x}}\)	\(25\)
default	\(\frac {{\mathrm e}^{6} {\mathrm e}^{2 \,{\mathrm e}^{4}} {\mathrm e}^{2 \ln \left (5 x \right )^{2}+2}}{625 x^{4}}+{\mathrm e}^{{\mathrm e}^{x}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/625/x^4*exp(2*exp(4)+8+2*ln(5*x)^2)+exp(exp(x))

________________________________________________________________________________________

maxima [C]  time = 0.52, size = 119, normalized size = 4.41 \begin {gather*} i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (i \, \sqrt {2} \log \left (5 \, x\right ) - i \, \sqrt {2}\right ) e^{\left (2 \, e^{4} + 6\right )} + \frac {1}{2} \, \sqrt {2} {\left (\frac {2 \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {2} \sqrt {-{\left (\log \left (5 \, x\right ) - 1\right )}^{2}}\right ) - 1\right )} {\left (\log \left (5 \, x\right ) - 1\right )}}{\sqrt {-{\left (\log \left (5 \, x\right ) - 1\right )}^{2}}} + \sqrt {2} e^{\left (2 \, {\left (\log \left (5 \, x\right ) - 1\right )}^{2}\right )}\right )} e^{\left (2 \, {\left (e^{2} + 2 \, e + 2\right )} {\left (e^{2} - 2 \, e + 2\right )} - 2\right )} + e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*sqrt(2)*sqrt(pi)*erf(I*sqrt(2)*log(5*x) - I*sqrt(2))*e^(2*e^4 + 6) + 1/2*sqrt(2)*(2*sqrt(pi)*(erf(sqrt(2)*sq
rt(-(log(5*x) - 1)^2)) - 1)*(log(5*x) - 1)/sqrt(-(log(5*x) - 1)^2) + sqrt(2)*e^(2*(log(5*x) - 1)^2))*e^(2*(e^2
+ 2*e + 2)*(e^2 - 2*e + 2) - 2) + e^(e^x)

________________________________________________________________________________________

mupad [B]  time = 7.66, size = 36, normalized size = 1.33 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^x}+\frac {x^{4\,\ln \relax (5)}\,{\mathrm {e}}^{2\,{\ln \relax (x)}^2}\,{\mathrm {e}}^{2\,{\mathrm {e}}^4}\,{\mathrm {e}}^8\,{\mathrm {e}}^{2\,{\ln \relax (5)}^2}}{625\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*log(5*x)^2 - 4*log(5*x) + 2)*(4*exp(2*exp(4))*exp(6) - 4*log(5*x)*exp(2*exp(4))*exp(6)) - x*exp(ex
p(x))*exp(x))/x,x)

[Out]

exp(exp(x)) + (x^(4*log(5))*exp(2*log(x)^2)*exp(2*exp(4))*exp(8)*exp(2*log(5)^2))/(625*x^4)

________________________________________________________________________________________

sympy [A]  time = 0.64, size = 31, normalized size = 1.15 \begin {gather*} e^{e^{x}} + \frac {e^{6} e^{2 \log {\left (5 x \right )}^{2} + 2} e^{2 e^{4}}}{625 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp(x)) + exp(6)*exp(2*log(5*x)**2 + 2)*exp(2*exp(4))/(625*x**4)

________________________________________________________________________________________