\(\int \frac {2+3 x}{1+x} \, dx\) [6753]

Optimal result

Integrand size = 11, antiderivative size = 17 \[ \int \frac {2+3 x}{1+x} \, dx=-5+3 (5+x)-\log \left (\frac {1+x}{3}\right ) \]

[Out]

10+3*x-ln(1/3*x+1/3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {2+3 x}{1+x} \, dx=3 x-\log (x+1) \]

[In]

Int[(2 + 3*x)/(1 + x),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {2+3 x}{1+x} \, dx=3 (1+x)-\log (1+x) \]

[In]

Integrate[(2 + 3*x)/(1 + x),x]

[Out]

3*(1 + x) - Log[1 + x]

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65

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

[In]

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

[Out]

3*x-ln(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {2+3 x}{1+x} \, dx=3 \, x - \log \left (x + 1\right ) \]

[In]

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

[Out]

3*x - log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.41 \[ \int \frac {2+3 x}{1+x} \, dx=3 x - \log {\left (x + 1 \right )} \]

[In]

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

[Out]

3*x - log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {2+3 x}{1+x} \, dx=3 \, x - \log \left (x + 1\right ) \]

[In]

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

[Out]

3*x - log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {2+3 x}{1+x} \, dx=3 \, x - \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

3*x - log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 12.52 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {2+3 x}{1+x} \, dx=3\,x-\ln \left (x+1\right ) \]

[In]

int((3*x + 2)/(x + 1),x)

[Out]

3*x - log(x + 1)