\(\int \frac {e^{1+x} x-\log (3 e^{4+e^{1+x}})}{x^2} \, dx\) [7517]

Optimal result

Integrand size = 26, antiderivative size = 16 \[ \int \frac {e^{1+x} x-\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx=\frac {\log \left (3 e^{4+e^{1+x}}\right )}{x} \]

[Out]

ln(3/exp(-exp(1+x)-4))/x

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {14, 2209, 2631} \[ \int \frac {e^{1+x} x-\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx=\frac {\log \left (3 e^{e^{x+1}+4}\right )}{x} \]

[In]

Int[(E^(1 + x)*x - Log[3*E^(4 + E^(1 + x))])/x^2,x]

[Out]

Log[3*E^(4 + E^(1 + x))]/x

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 2209

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

Rule 2631

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1)*(Log[u]/(b*(m + 1))), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[(a + b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^{1+x}}{x}-\frac {\log \left (3 e^{4+e^{1+x}}\right )}{x^2}\right ) \, dx \\ & = \int \frac {e^{1+x}}{x} \, dx-\int \frac {\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx \\ & = e \operatorname {ExpIntegralEi}(x)+\frac {\log \left (3 e^{4+e^{1+x}}\right )}{x}-\int \frac {e^{1+x}}{x} \, dx \\ & = \frac {\log \left (3 e^{4+e^{1+x}}\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1+x} x-\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx=\frac {\log \left (3 e^{4+e^{1+x}}\right )}{x} \]

[In]

Integrate[(E^(1 + x)*x - Log[3*E^(4 + E^(1 + x))])/x^2,x]

[Out]

Log[3*E^(4 + E^(1 + x))]/x

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {\ln \left (3 \,{\mathrm e}^{{\mathrm e}^{1+x}+4}\right )}{x}\)	\(19\)
parallelrisch	\(\frac {\ln \left (3 \,{\mathrm e}^{{\mathrm e}^{1+x}+4}\right )}{x}\)	\(19\)
parts	\(\frac {\ln \left (3 \,{\mathrm e}^{{\mathrm e}^{1+x}+4}\right )}{x}\)	\(19\)
risch	\(-\frac {\ln \left ({\mathrm e}^{-{\mathrm e}^{1+x}-4}\right )}{x}+\frac {\ln \left (3\right )}{x}\)	\(23\)

[In]

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

[Out]

ln(3/exp(-exp(1+x)-4))/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{1+x} x-\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx=\frac {e^{\left (x + 1\right )} + \log \left (3\right ) + 4}{x} \]

[In]

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

[Out]

(e^(x + 1) + log(3) + 4)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {e^{1+x} x-\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx=\frac {e^{x + 1}}{x} - \frac {-4 - \log {\left (3 \right )}}{x} \]

[In]

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

[Out]

exp(x + 1)/x - (-4 - log(3))/x

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {e^{1+x} x-\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx=\frac {\log \left (3 \, e^{\left (e^{\left (x + 1\right )} + 4\right )}\right )}{x} \]

[In]

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

[Out]

log(3*e^(e^(x + 1) + 4))/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{1+x} x-\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx=\frac {e^{\left (x + 1\right )} + \log \left (3\right ) + 4}{x} \]

[In]

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

[Out]

(e^(x + 1) + log(3) + 4)/x

Mupad [B] (verification not implemented)

Time = 11.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{1+x} x-\log \left (3 e^{4+e^{1+x}}\right )}{x^2} \, dx=\frac {{\mathrm {e}}^{x+1}+\ln \left (3\right )+4}{x} \]

[In]

int(-(log(3*exp(exp(x + 1) + 4)) - x*exp(x + 1))/x^2,x)

[Out]

(exp(x + 1) + log(3) + 4)/x