\(\int \frac {-1+2 x}{3+2 x} \, dx\) [26]

Optimal result

Integrand size = 13, antiderivative size = 10 \[ \int \frac {-1+2 x}{3+2 x} \, dx=x-2 \log (3+2 x) \]

[Out]

x-2*ln(3+2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int \frac {-1+2 x}{3+2 x} \, dx=x-2 \log (2 x+3) \]

[In]

Int[(-1 + 2*x)/(3 + 2*x),x]

[Out]

x - 2*Log[3 + 2*x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (1-\frac {4}{3+2 x}\right ) \, dx \\ & = x-2 \log (3+2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x}{3+2 x} \, dx=x-2 \log (3+2 x) \]

[In]

Integrate[(-1 + 2*x)/(3 + 2*x),x]

[Out]

x - 2*Log[3 + 2*x]

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90

method	result	size
parallelrisch	\(x -2 \ln \left (\frac {3}{2}+x \right )\)	\(9\)
default	\(x -2 \ln \left (3+2 x \right )\)	\(11\)
norman	\(x -2 \ln \left (3+2 x \right )\)	\(11\)
meijerg	\(-2 \ln \left (1+\frac {2 x}{3}\right )+x\)	\(11\)
risch	\(x -2 \ln \left (3+2 x \right )\)	\(11\)

[In]

int((2*x-1)/(3+2*x),x,method=_RETURNVERBOSE)

[Out]

x-2*ln(3/2+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x}{3+2 x} \, dx=x - 2 \, \log \left (2 \, x + 3\right ) \]

[In]

integrate((-1+2*x)/(3+2*x),x, algorithm="fricas")

[Out]

x - 2*log(2*x + 3)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {-1+2 x}{3+2 x} \, dx=x - 2 \log {\left (2 x + 3 \right )} \]

[In]

integrate((-1+2*x)/(3+2*x),x)

[Out]

x - 2*log(2*x + 3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x}{3+2 x} \, dx=x - 2 \, \log \left (2 \, x + 3\right ) \]

[In]

integrate((-1+2*x)/(3+2*x),x, algorithm="maxima")

[Out]

x - 2*log(2*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10 \[ \int \frac {-1+2 x}{3+2 x} \, dx=x - 2 \, \log \left ({\left | 2 \, x + 3 \right |}\right ) \]

[In]

integrate((-1+2*x)/(3+2*x),x, algorithm="giac")

[Out]

x - 2*log(abs(2*x + 3))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {-1+2 x}{3+2 x} \, dx=x-2\,\ln \left (x+\frac {3}{2}\right ) \]

[In]

int((2*x - 1)/(2*x + 3),x)

[Out]

x - 2*log(x + 3/2)