\(\int \frac {1+2 x}{2+3 x} \, dx\) [250]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {1+2 x}{2+3 x} \, dx=\frac {2 x}{3}-\frac {1}{9} \log (2+3 x) \]

[Out]

2/3*x-1/9*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, 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} \[ \int \frac {1+2 x}{2+3 x} \, dx=\frac {2 x}{3}-\frac {1}{9} \log (3 x+2) \]

[In]

Int[(1 + 2*x)/(2 + 3*x),x]

[Out]

(2*x)/3 - Log[2 + 3*x]/9

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 (\frac {2}{3}-\frac {1}{3 (2+3 x)}\right ) \, dx \\ & = \frac {2 x}{3}-\frac {1}{9} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {1+2 x}{2+3 x} \, dx=\frac {1}{9} (4+6 x-\log (2+3 x)) \]

[In]

Integrate[(1 + 2*x)/(2 + 3*x),x]

[Out]

(4 + 6*x - Log[2 + 3*x])/9

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

method result size
parallelrisch \(\frac {2 x}{3}-\frac {\ln \left (\frac {2}{3}+x \right )}{9}\) \(11\)
default \(\frac {2 x}{3}-\frac {\ln \left (2+3 x \right )}{9}\) \(13\)
norman \(\frac {2 x}{3}-\frac {\ln \left (2+3 x \right )}{9}\) \(13\)
meijerg \(-\frac {\ln \left (1+\frac {3 x}{2}\right )}{9}+\frac {2 x}{3}\) \(13\)
risch \(\frac {2 x}{3}-\frac {\ln \left (2+3 x \right )}{9}\) \(13\)

[In]

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

[Out]

2/3*x-1/9*ln(2/3+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1+2 x}{2+3 x} \, dx=\frac {2}{3} \, x - \frac {1}{9} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

2/3*x - 1/9*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1+2 x}{2+3 x} \, dx=\frac {2 x}{3} - \frac {\log {\left (3 x + 2 \right )}}{9} \]

[In]

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

[Out]

2*x/3 - log(3*x + 2)/9

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {1+2 x}{2+3 x} \, dx=\frac {2}{3} \, x - \frac {1}{9} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

2/3*x - 1/9*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {1+2 x}{2+3 x} \, dx=\frac {2}{3} \, x - \frac {1}{9} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

2/3*x - 1/9*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {1+2 x}{2+3 x} \, dx=\frac {2\,x}{3}-\frac {\ln \left (x+\frac {2}{3}\right )}{9} \]

[In]

int((2*x + 1)/(3*x + 2),x)

[Out]

(2*x)/3 - log(x + 2/3)/9