\(\int \frac {1}{1+t} \, dt\) [60]

Optimal result

Integrand size = 5, antiderivative size = 4 \[ \int \frac {1}{1+t} \, dt=\log (1+t) \]

[Out]

ln(1+t)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {31} \[ \int \frac {1}{1+t} \, dt=\log (t+1) \]

[In]

Int[(1 + t)^(-1),t]

[Out]

Log[1 + t]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \log (1+t) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1+t} \, dt=\log (1+t) \]

[In]

Integrate[(1 + t)^(-1),t]

[Out]

Log[1 + t]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.25

method	result	size
default	\(\ln \left (1+t \right )\)	\(5\)
norman	\(\ln \left (1+t \right )\)	\(5\)
meijerg	\(\ln \left (1+t \right )\)	\(5\)
risch	\(\ln \left (1+t \right )\)	\(5\)
parallelrisch	\(\ln \left (1+t \right )\)	\(5\)

[In]

int(1/(1+t),t,method=_RETURNVERBOSE)

[Out]

ln(1+t)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1+t} \, dt=\log \left (t + 1\right ) \]

[In]

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

[Out]

log(t + 1)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.75 \[ \int \frac {1}{1+t} \, dt=\log {\left (t + 1 \right )} \]

[In]

integrate(1/(1+t),t)

[Out]

log(t + 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1+t} \, dt=\log \left (t + 1\right ) \]

[In]

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

[Out]

log(t + 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.25 \[ \int \frac {1}{1+t} \, dt=\log \left ({\left | t + 1 \right |}\right ) \]

[In]

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

[Out]

log(abs(t + 1))

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 4, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1+t} \, dt=\ln \left (t+1\right ) \]

[In]

int(1/(t + 1),t)

[Out]

log(t + 1)