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

Optimal. Leaf size=14 \[ -\frac {1}{1+x}+2 \log (1+x) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \begin {gather*} 2 \log (x+1)-\frac {1}{x+1} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} -\frac {1}{1+x}+2 \log (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 2*x)/(1 + x)^2,x]

[Out]

-(1 + x)^(-1) + 2*Log[1 + x]

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Mathics [A]
time = 1.69, size = 17, normalized size = 1.21 \begin {gather*} \frac {-1+2 \text {Log}\left [1+x\right ] \left (1+x\right )}{1+x} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(3 + 2*x)/(1 + x)^2,x]')

[Out]

(-1 + 2 Log[1 + x] (1 + x)) / (1 + x)

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Maple [A]
time = 0.04, size = 15, normalized size = 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	\(-\frac {1}{1+x}+2 \ln \left (1+x \right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 14, normalized size = 1.00 \begin {gather*} -\frac {1}{x + 1} + 2 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(x + 1) + 2*log(x + 1)

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Fricas [A]
time = 0.34, size = 17, normalized size = 1.21 \begin {gather*} \frac {2 \, {\left (x + 1\right )} \log \left (x + 1\right ) - 1}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(x + 1)*log(x + 1) - 1)/(x + 1)

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Sympy [A]
time = 0.04, size = 10, normalized size = 0.71 \begin {gather*} 2 \log {\left (x + 1 \right )} - \frac {1}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x + 1) - 1/(x + 1)

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Giac [A]
time = 0.00, size = 13, normalized size = 0.93 \begin {gather*} -\frac 1{x+1}+2 \ln \left |x+1\right | \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(x + 1) + 2*log(abs(x + 1))

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Mupad [B]
time = 0.03, size = 14, normalized size = 1.00 \begin {gather*} 2\,\ln \left (x+1\right )-\frac {1}{x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 3)/(x + 1)^2,x)

[Out]

2*log(x + 1) - 1/(x + 1)

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