3.165 \(\int e^t \log (1+t) \, dt\)

Optimal. Leaf size=18 \[ e^t \log (t+1)-\frac {\text {Ei}(t+1)}{e} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2194, 2554, 2178} \[ e^t \log (t+1)-\frac {\text {ExpIntegralEi}(t+1)}{e} \]

Antiderivative was successfully verified.

[In]

Int[E^t*Log[1 + t],t]

[Out]

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

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 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 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]

Rubi steps

\begin {align*} \int e^t \log (1+t) \, dt &=e^t \log (1+t)-\int \frac {e^t}{1+t} \, dt\\ &=-\frac {\text {Ei}(1+t)}{e}+e^t \log (1+t)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.00 \[ e^t \log (t+1)-\frac {\text {Ei}(t+1)}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[E^t*Log[1 + t],t]

[Out]

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

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fricas [A]  time = 0.41, size = 19, normalized size = 1.06 \[ {\left (e^{\left (t + 1\right )} \log \left (t + 1\right ) - {\rm Ei}\left (t + 1\right )\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.01, size = 16, normalized size = 0.89 \[ -{\rm Ei}\left (t + 1\right ) e^{\left (-1\right )} + e^{t} \log \left (t + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Ei(t + 1)*e^(-1) + e^t*log(t + 1)

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maple [A]  time = 0.04, size = 19, normalized size = 1.06 \[ {\mathrm e}^{t} \ln \left (t +1\right )+{\mathrm e}^{-1} \Ei \left (1, -t -1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.77, size = 18, normalized size = 1.00 \[ e^{\left (-1\right )} E_{1}\left (-t - 1\right ) + e^{t} \log \left (t + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \[ \int \ln \left (t+1\right )\,{\mathrm {e}}^t \,d t \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(t + 1)*exp(t),t)

[Out]

int(log(t + 1)*exp(t), t)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{t} \log {\left (t + 1 \right )}\, dt \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(t)*log(t + 1), t)

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