\(\int e^t \log (1+t) \, dt\) [165]

Optimal result

Integrand size = 8, antiderivative size = 18 \[ \int e^t \log (1+t) \, dt=-\frac {\operatorname {ExpIntegralEi}(1+t)}{e}+e^t \log (1+t) \]

[Out]

-Ei(1+t)/exp(1)+exp(t)*ln(1+t)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2225, 2634, 2209} \[ \int e^t \log (1+t) \, dt=e^t \log (t+1)-\frac {\operatorname {ExpIntegralEi}(t+1)}{e} \]

[In]

Int[E^t*Log[1 + t],t]

[Out]

-(ExpIntegralEi[1 + t]/E) + E^t*Log[1 + t]

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 2225

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 2634

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]

Rubi steps \begin{align*} \text {integral}& = e^t \log (1+t)-\int \frac {e^t}{1+t} \, dt \\ & = -\frac {\operatorname {ExpIntegralEi}(1+t)}{e}+e^t \log (1+t) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^t \log (1+t) \, dt=-\frac {\operatorname {ExpIntegralEi}(1+t)}{e}+e^t \log (1+t) \]

[In]

Integrate[E^t*Log[1 + t],t]

[Out]

-(ExpIntegralEi[1 + t]/E) + E^t*Log[1 + t]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
risch	\({\mathrm e}^{t} \ln \left (1+t \right )+{\mathrm e}^{-1} \operatorname {Ei}_{1}\left (-1-t \right )\)	\(19\)

[In]

int(exp(t)*ln(1+t),t,method=_RETURNVERBOSE)

[Out]

exp(t)*ln(1+t)+exp(-1)*Ei(1,-1-t)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int e^t \log (1+t) \, dt={\left (e^{\left (t + 1\right )} \log \left (t + 1\right ) - {\rm Ei}\left (t + 1\right )\right )} e^{\left (-1\right )} \]

[In]

integrate(exp(t)*log(1+t),t, algorithm="fricas")

[Out]

(e^(t + 1)*log(t + 1) - Ei(t + 1))*e^(-1)

Sympy [F]

\[ \int e^t \log (1+t) \, dt=\int e^{t} \log {\left (t + 1 \right )}\, dt \]

[In]

integrate(exp(t)*ln(1+t),t)

[Out]

Integral(exp(t)*log(t + 1), t)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^t \log (1+t) \, dt=e^{\left (-1\right )} E_{1}\left (-t - 1\right ) + e^{t} \log \left (t + 1\right ) \]

[In]

integrate(exp(t)*log(1+t),t, algorithm="maxima")

[Out]

e^(-1)*exp_integral_e(1, -t - 1) + e^t*log(t + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^t \log (1+t) \, dt=-{\rm Ei}\left (t + 1\right ) e^{\left (-1\right )} + e^{t} \log \left (t + 1\right ) \]

[In]

integrate(exp(t)*log(1+t),t, algorithm="giac")

[Out]

-Ei(t + 1)*e^(-1) + e^t*log(t + 1)

Mupad [F(-1)]

Timed out. \[ \int e^t \log (1+t) \, dt=\int \ln \left (t+1\right )\,{\mathrm {e}}^t \,d t \]

[In]

int(log(t + 1)*exp(t),t)

[Out]

int(log(t + 1)*exp(t), t)