3.1354 \(\int \frac{-11+6 x}{(-1+2 x) (-1+x^2)} \, dx\)

Optimal. Leaf size=29 \[ \frac{16}{3} \log (1-2 x)-\frac{5}{2} \log (1-x)-\frac{17}{6} \log (x+1) \]

[Out]

(16*Log[1 - 2*x])/3 - (5*Log[1 - x])/2 - (17*Log[1 + x])/6

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Rubi [A]  time = 0.0225441, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {801} \[ \frac{16}{3} \log (1-2 x)-\frac{5}{2} \log (1-x)-\frac{17}{6} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Int[(-11 + 6*x)/((-1 + 2*x)*(-1 + x^2)),x]

[Out]

(16*Log[1 - 2*x])/3 - (5*Log[1 - x])/2 - (17*Log[1 + x])/6

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin{align*} \int \frac{-11+6 x}{(-1+2 x) \left (-1+x^2\right )} \, dx &=\int \left (-\frac{5}{2 (-1+x)}-\frac{17}{6 (1+x)}+\frac{32}{3 (-1+2 x)}\right ) \, dx\\ &=\frac{16}{3} \log (1-2 x)-\frac{5}{2} \log (1-x)-\frac{17}{6} \log (1+x)\\ \end{align*}

Mathematica [A]  time = 0.00981, size = 31, normalized size = 1.07 \[ -\frac{5}{2} \log (2-2 x)+\frac{16}{3} \log (2 x-1)-\frac{17}{6} \log (2 x+2) \]

Antiderivative was successfully verified.

[In]

Integrate[(-11 + 6*x)/((-1 + 2*x)*(-1 + x^2)),x]

[Out]

(-5*Log[2 - 2*x])/2 + (16*Log[-1 + 2*x])/3 - (17*Log[2 + 2*x])/6

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Maple [A]  time = 0.007, size = 22, normalized size = 0.8 \begin{align*} -{\frac{17\,\ln \left ( 1+x \right ) }{6}}-{\frac{5\,\ln \left ( -1+x \right ) }{2}}+{\frac{16\,\ln \left ( -1+2\,x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-11+6*x)/(-1+2*x)/(x^2-1),x)

[Out]

-17/6*ln(1+x)-5/2*ln(-1+x)+16/3*ln(-1+2*x)

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Maxima [A]  time = 0.996604, size = 28, normalized size = 0.97 \begin{align*} \frac{16}{3} \, \log \left (2 \, x - 1\right ) - \frac{17}{6} \, \log \left (x + 1\right ) - \frac{5}{2} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-11+6*x)/(-1+2*x)/(x^2-1),x, algorithm="maxima")

[Out]

16/3*log(2*x - 1) - 17/6*log(x + 1) - 5/2*log(x - 1)

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Fricas [A]  time = 2.36723, size = 73, normalized size = 2.52 \begin{align*} \frac{16}{3} \, \log \left (2 \, x - 1\right ) - \frac{17}{6} \, \log \left (x + 1\right ) - \frac{5}{2} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-11+6*x)/(-1+2*x)/(x^2-1),x, algorithm="fricas")

[Out]

16/3*log(2*x - 1) - 17/6*log(x + 1) - 5/2*log(x - 1)

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Sympy [A]  time = 0.217161, size = 26, normalized size = 0.9 \begin{align*} - \frac{5 \log{\left (x - 1 \right )}}{2} + \frac{16 \log{\left (x - \frac{1}{2} \right )}}{3} - \frac{17 \log{\left (x + 1 \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-11+6*x)/(-1+2*x)/(x**2-1),x)

[Out]

-5*log(x - 1)/2 + 16*log(x - 1/2)/3 - 17*log(x + 1)/6

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Giac [A]  time = 1.1632, size = 32, normalized size = 1.1 \begin{align*} \frac{16}{3} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) - \frac{17}{6} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{5}{2} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-11+6*x)/(-1+2*x)/(x^2-1),x, algorithm="giac")

[Out]

16/3*log(abs(2*x - 1)) - 17/6*log(abs(x + 1)) - 5/2*log(abs(x - 1))