\(\int \frac {e^t}{(1+t)^2} \, dt\) [164]

Optimal result

Integrand size = 9, antiderivative size = 19 \[ \int \frac {e^t}{(1+t)^2} \, dt=-\frac {e^t}{1+t}+\frac {\operatorname {ExpIntegralEi}(1+t)}{e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2208, 2209} \[ \int \frac {e^t}{(1+t)^2} \, dt=\frac {\operatorname {ExpIntegralEi}(t+1)}{e}-\frac {e^t}{t+1} \]

[In]

Int[E^t/(1 + t)^2,t]

[Out]

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

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

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]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^t}{(1+t)^2} \, dt=-\frac {e^t}{1+t}+\frac {\operatorname {ExpIntegralEi}(1+t)}{e} \]

[In]

Integrate[E^t/(1 + t)^2,t]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16

method	result	size
default	\(-\frac {{\mathrm e}^{t}}{1+t}-{\mathrm e}^{-1} \operatorname {Ei}_{1}\left (-1-t \right )\)	\(22\)
risch	\(-\frac {{\mathrm e}^{t}}{1+t}-{\mathrm e}^{-1} \operatorname {Ei}_{1}\left (-1-t \right )\)	\(22\)

[In]

int(exp(t)/(1+t)^2,t,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {e^t}{(1+t)^2} \, dt=\frac {{\left ({\left (t + 1\right )} {\rm Ei}\left (t + 1\right ) - e^{\left (t + 1\right )}\right )} e^{\left (-1\right )}}{t + 1} \]

[In]

integrate(exp(t)/(1+t)^2,t, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {e^t}{(1+t)^2} \, dt=\int \frac {e^{t}}{\left (t + 1\right )^{2}}\, dt \]

[In]

integrate(exp(t)/(1+t)**2,t)

[Out]

Integral(exp(t)/(t + 1)**2, t)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^t}{(1+t)^2} \, dt=-\frac {e^{\left (-1\right )} E_{2}\left (-t - 1\right )}{t + 1} \]

[In]

integrate(exp(t)/(1+t)^2,t, algorithm="maxima")

[Out]

-e^(-1)*exp_integral_e(2, -t - 1)/(t + 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (17) = 34\).

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 4.21 \[ \int \frac {e^t}{(1+t)^2} \, dt=\frac {{\left (t + 1\right )} {\left (\frac {1}{t + 1} - 1\right )} {\rm Ei}\left (-{\left (t + 1\right )} {\left (\frac {1}{t + 1} - 1\right )} + 1\right ) - {\rm Ei}\left (-{\left (t + 1\right )} {\left (\frac {1}{t + 1} - 1\right )} + 1\right ) + e^{\left (-{\left (t + 1\right )} {\left (\frac {1}{t + 1} - 1\right )} + 1\right )}}{{\left (t + 1\right )} {\left (\frac {1}{t + 1} - 1\right )} e - e} \]

[In]

integrate(exp(t)/(1+t)^2,t, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^t}{(1+t)^2} \, dt=\mathrm {ei}\left (t+1\right )\,{\mathrm {e}}^{-1}-\frac {{\mathrm {e}}^t}{t+1} \]

[In]

int(exp(t)/(t + 1)^2,t)

[Out]

ei(t + 1)*exp(-1) - exp(t)/(t + 1)