3.3.13 \(\int \frac {1}{-3+2 x+x^2} \, dx\) [213]

Optimal. Leaf size=19 \[ \frac {1}{4} \log (1-x)-\frac {1}{4} \log (3+x) \]

[Out]

1/4*ln(1-x)-1/4*ln(3+x)

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Rubi [A]
time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {630, 31} \begin {gather*} \frac {1}{4} \log (1-x)-\frac {1}{4} \log (x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + 2*x + x^2)^(-1),x]

[Out]

Log[1 - x]/4 - Log[3 + x]/4

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {1}{-3+2 x+x^2} \, dx &=\frac {1}{4} \int \frac {1}{-1+x} \, dx-\frac {1}{4} \int \frac {1}{3+x} \, dx\\ &=\frac {1}{4} \log (1-x)-\frac {1}{4} \log (3+x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 19, normalized size = 1.00 \begin {gather*} \frac {1}{4} \log (1-x)-\frac {1}{4} \log (3+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + 2*x + x^2)^(-1),x]

[Out]

Log[1 - x]/4 - Log[3 + x]/4

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Mathics [A]
time = 1.84, size = 13, normalized size = 0.68 \begin {gather*} -\frac {\text {Log}\left [3+x\right ]}{4}+\frac {\text {Log}\left [-1+x\right ]}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[3 + x] / 4 + Log[-1 + x] / 4

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Maple [A]
time = 0.06, size = 14, normalized size = 0.74

method	result	size
default	\(\frac {\ln \left (-1+x \right )}{4}-\frac {\ln \left (3+x \right )}{4}\)	\(14\)
norman	\(\frac {\ln \left (-1+x \right )}{4}-\frac {\ln \left (3+x \right )}{4}\)	\(14\)
risch	\(\frac {\ln \left (-1+x \right )}{4}-\frac {\ln \left (3+x \right )}{4}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*ln(-1+x)-1/4*ln(3+x)

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Maxima [A]
time = 0.26, size = 13, normalized size = 0.68 \begin {gather*} -\frac {1}{4} \, \log \left (x + 3\right ) + \frac {1}{4} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(x + 3) + 1/4*log(x - 1)

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Fricas [A]
time = 0.32, size = 13, normalized size = 0.68 \begin {gather*} -\frac {1}{4} \, \log \left (x + 3\right ) + \frac {1}{4} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(x + 3) + 1/4*log(x - 1)

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Sympy [A]
time = 0.05, size = 12, normalized size = 0.63 \begin {gather*} \frac {\log {\left (x - 1 \right )}}{4} - \frac {\log {\left (x + 3 \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 1)/4 - log(x + 3)/4

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Giac [A]
time = 0.00, size = 18, normalized size = 0.95 \begin {gather*} \frac {\ln \left |x-1\right |}{4}-\frac {\ln \left |x+3\right |}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(abs(x + 3)) + 1/4*log(abs(x - 1))

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Mupad [B]
time = 0.08, size = 8, normalized size = 0.42 \begin {gather*} -\frac {\mathrm {atanh}\left (\frac {x}{2}+\frac {1}{2}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x + x^2 - 3),x)

[Out]

-atanh(x/2 + 1/2)/2

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