3.261 \(\int \frac {3+2 x^2}{(-1+x)^2 x} \, dx\)

Optimal. Leaf size=22 \[ \frac {5}{1-x}-\log (1-x)+3 \log (x) \]

[Out]

5/(1-x)-ln(1-x)+3*ln(x)

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {894} \[ \frac {5}{1-x}-\log (1-x)+3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 2*x^2)/((-1 + x)^2*x),x]

[Out]

5/(1 - x) - Log[1 - x] + 3*Log[x]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.91 \[ -\frac {5}{x-1}-\log (1-x)+3 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-5/(-1 + x) - Log[1 - x] + 3*Log[x]

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fricas [A]  time = 0.57, size = 24, normalized size = 1.09 \[ -\frac {{\left (x - 1\right )} \log \left (x - 1\right ) - 3 \, {\left (x - 1\right )} \log \relax (x) + 5}{x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((x - 1)*log(x - 1) - 3*(x - 1)*log(x) + 5)/(x - 1)

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giac [A]  time = 0.37, size = 28, normalized size = 1.27 \[ -\frac {5}{x - 1} + 2 \, \log \left ({\left | x - 1 \right |}\right ) + 3 \, \log \left ({\left | -\frac {1}{x - 1} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 19, normalized size = 0.86 \[ 3 \ln \relax (x )-\ln \left (x -1\right )-\frac {5}{x -1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/(x-1)-ln(x-1)+3*ln(x)

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maxima [A]  time = 1.09, size = 18, normalized size = 0.82 \[ -\frac {5}{x - 1} - \log \left (x - 1\right ) + 3 \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/(x - 1) - log(x - 1) + 3*log(x)

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mupad [B]  time = 0.04, size = 18, normalized size = 0.82 \[ 3\,\ln \relax (x)-\ln \left (x-1\right )-\frac {5}{x-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2 + 3)/(x*(x - 1)^2),x)

[Out]

3*log(x) - log(x - 1) - 5/(x - 1)

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sympy [A]  time = 0.11, size = 14, normalized size = 0.64 \[ 3 \log {\relax (x )} - \log {\left (x - 1 \right )} - \frac {5}{x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*log(x) - log(x - 1) - 5/(x - 1)

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