\(\int \frac {3+2 x}{(1+x)^2} \, dx\) [184]

Optimal result

Integrand size = 11, antiderivative size = 14 \[ \int \frac {3+2 x}{(1+x)^2} \, dx=-\frac {1}{1+x}+2 \log (1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int \frac {3+2 x}{(1+x)^2} \, dx=2 \log (x+1)-\frac {1}{x+1} \]

[In]

Int[(3 + 2*x)/(1 + x)^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {3+2 x}{(1+x)^2} \, dx=-\frac {1}{1+x}+2 \log (1+x) \]

[In]

Integrate[(3 + 2*x)/(1 + x)^2,x]

[Out]

-(1 + x)^(-1) + 2*Log[1 + x]

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {1}{1+x}+2 \ln \left (1+x \right )\)	\(15\)
norman	\(-\frac {1}{1+x}+2 \ln \left (1+x \right )\)	\(15\)
meijerg	\(\frac {x}{1+x}+2 \ln \left (1+x \right )\)	\(15\)
risch	\(-\frac {1}{1+x}+2 \ln \left (1+x \right )\)	\(15\)
parallelrisch	\(\frac {2 \ln \left (1+x \right ) x -1+2 \ln \left (1+x \right )}{1+x}\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {3+2 x}{(1+x)^2} \, dx=\frac {2 \, {\left (x + 1\right )} \log \left (x + 1\right ) - 1}{x + 1} \]

[In]

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

[Out]

(2*(x + 1)*log(x + 1) - 1)/(x + 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {3+2 x}{(1+x)^2} \, dx=2 \log {\left (x + 1 \right )} - \frac {1}{x + 1} \]

[In]

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

[Out]

2*log(x + 1) - 1/(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {3+2 x}{(1+x)^2} \, dx=-\frac {1}{x + 1} + 2 \, \log \left (x + 1\right ) \]

[In]

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

[Out]

-1/(x + 1) + 2*log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {3+2 x}{(1+x)^2} \, dx=-\frac {1}{x + 1} + 2 \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

-1/(x + 1) + 2*log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {3+2 x}{(1+x)^2} \, dx=2\,\ln \left (x+1\right )-\frac {1}{x+1} \]

[In]

int((2*x + 3)/(x + 1)^2,x)

[Out]

2*log(x + 1) - 1/(x + 1)