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

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1593, 894} \[ x+2 \log (1-x)-\log (x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)/(-x + x^2),x]

[Out]

x + 2*Log[1 - x] - 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]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \[ x+2 \log (1-x)-\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^2)/(-x + x^2),x]

[Out]

x + 2*Log[1 - x] - Log[x]

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fricas [A]  time = 0.75, size = 12, normalized size = 0.86 \[ x + 2 \, \log \left (x - 1\right ) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*log(x - 1) - log(x)

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giac [A]  time = 0.26, size = 14, normalized size = 1.00 \[ x + 2 \, \log \left ({\left | x - 1 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*log(abs(x - 1)) - log(abs(x))

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maple [A]  time = 0.00, size = 13, normalized size = 0.93 \[ x -\ln \relax (x )+2 \ln \left (x -1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.91, size = 12, normalized size = 0.86 \[ x + 2 \, \log \left (x - 1\right ) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*log(x - 1) - log(x)

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mupad [B]  time = 0.04, size = 12, normalized size = 0.86 \[ x+2\,\ln \left (x-1\right )-\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2 + 1)/(x - x^2),x)

[Out]

x + 2*log(x - 1) - log(x)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.71 \[ x - \log {\relax (x )} + 2 \log {\left (x - 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - log(x) + 2*log(x - 1)

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