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

Optimal. Leaf size=16 \[ \frac {\log \left (3 e^{4+e^{1+x}}\right )}{x} \]

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Rubi [A]  time = 0.06, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {14, 2178, 2551} \begin {gather*} \frac {\log \left (3 e^{e^{x+1}+4}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[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 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 2551

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 {gather*} \begin {aligned} \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 \text {Ei}(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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 1.00 \begin {gather*} \frac {\log \left (3 e^{4+e^{1+x}}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.83, size = 12, normalized size = 0.75 \begin {gather*} \frac {e^{\left (x + 1\right )} + \log \relax (3) + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(x + 1) + log(3) + 4)/x

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giac [A]  time = 0.22, size = 12, normalized size = 0.75 \begin {gather*} \frac {e^{\left (x + 1\right )} + \log \relax (3) + 4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(x + 1) + log(3) + 4)/x

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maple [A]  time = 0.06, size = 19, normalized size = 1.19

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 14, normalized size = 0.88 \begin {gather*} \frac {\log \left (3 \, e^{\left (e^{\left (x + 1\right )} + 4\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3*e^(e^(x + 1) + 4))/x

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mupad [B]  time = 0.12, size = 12, normalized size = 0.75 \begin {gather*} \frac {{\mathrm {e}}^{x+1}+\ln \relax (3)+4}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(x + 1) + log(3) + 4)/x

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sympy [A]  time = 0.13, size = 14, normalized size = 0.88 \begin {gather*} \frac {e^{x + 1}}{x} - \frac {-4 - \log {\relax (3 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x + 1)/x - (-4 - log(3))/x

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