\(\int \frac {1}{2+2 x} \, dx\) [271]

Optimal result

Integrand size = 7, antiderivative size = 8 \[ \int \frac {1}{2+2 x} \, dx=\frac {1}{2} \log (1+x) \]

[Out]

1/2*ln(1+x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {31} \[ \int \frac {1}{2+2 x} \, dx=\frac {1}{2} \log (x+1) \]

[In]

Int[(2 + 2*x)^(-1),x]

[Out]

Log[1 + x]/2

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}& = \frac {1}{2} \log (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.25 \[ \int \frac {1}{2+2 x} \, dx=\frac {1}{2} \log (2+2 x) \]

[In]

Integrate[(2 + 2*x)^(-1),x]

[Out]

Log[2 + 2*x]/2

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {\ln \left (1+x \right )}{2}\)	\(7\)
meijerg	\(\frac {\ln \left (1+x \right )}{2}\)	\(7\)
risch	\(\frac {\ln \left (1+x \right )}{2}\)	\(7\)
parallelrisch	\(\frac {\ln \left (1+x \right )}{2}\)	\(7\)
norman	\(\frac {\ln \left (2+2 x \right )}{2}\)	\(9\)

[In]

int(1/(2+2*x),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{2+2 x} \, dx=\frac {1}{2} \, \log \left (x + 1\right ) \]

[In]

integrate(1/(2+2*x),x, algorithm="fricas")

[Out]

1/2*log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {1}{2+2 x} \, dx=\frac {\log {\left (2 x + 2 \right )}}{2} \]

[In]

integrate(1/(2+2*x),x)

[Out]

log(2*x + 2)/2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{2+2 x} \, dx=\frac {1}{2} \, \log \left (x + 1\right ) \]

[In]

integrate(1/(2+2*x),x, algorithm="maxima")

[Out]

1/2*log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {1}{2+2 x} \, dx=\frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

integrate(1/(2+2*x),x, algorithm="giac")

[Out]

1/2*log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int \frac {1}{2+2 x} \, dx=\frac {\ln \left (x+1\right )}{2} \]

[In]

int(1/(2*x + 2),x)

[Out]

log(x + 1)/2