3.361 \(\int \frac {4+3 x+x^2}{(-3+x) (-2+x) (-1+x)} \, dx\)

Optimal. Leaf size=25 \[ 4 \log (1-x)-14 \log (2-x)+11 \log (3-x) \]

[Out]

4*ln(1-x)-14*ln(2-x)+11*ln(3-x)

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Rubi [A]  time = 0.05, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1612} \[ 4 \log (1-x)-14 \log (2-x)+11 \log (3-x) \]

Antiderivative was successfully verified.

[In]

Int[(4 + 3*x + x^2)/((-3 + x)*(-2 + x)*(-1 + x)),x]

[Out]

4*Log[1 - x] - 14*Log[2 - x] + 11*Log[3 - x]

Rule 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

Rubi steps

\begin {align*} \int \frac {4+3 x+x^2}{(-3+x) (-2+x) (-1+x)} \, dx &=\int \left (\frac {11}{-3+x}-\frac {14}{-2+x}+\frac {4}{-1+x}\right ) \, dx\\ &=4 \log (1-x)-14 \log (2-x)+11 \log (3-x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.76 \[ 11 \log (x-3)-14 \log (x-2)+4 \log (x-1) \]

Antiderivative was successfully verified.

[In]

Integrate[(4 + 3*x + x^2)/((-3 + x)*(-2 + x)*(-1 + x)),x]

[Out]

11*Log[-3 + x] - 14*Log[-2 + x] + 4*Log[-1 + x]

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fricas [A]  time = 0.52, size = 19, normalized size = 0.76 \[ 4 \, \log \left (x - 1\right ) - 14 \, \log \left (x - 2\right ) + 11 \, \log \left (x - 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+3*x+4)/(-3+x)/(-2+x)/(-1+x),x, algorithm="fricas")

[Out]

4*log(x - 1) - 14*log(x - 2) + 11*log(x - 3)

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giac [A]  time = 0.39, size = 22, normalized size = 0.88 \[ 4 \, \log \left ({\left | x - 1 \right |}\right ) - 14 \, \log \left ({\left | x - 2 \right |}\right ) + 11 \, \log \left ({\left | x - 3 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+3*x+4)/(-3+x)/(-2+x)/(-1+x),x, algorithm="giac")

[Out]

4*log(abs(x - 1)) - 14*log(abs(x - 2)) + 11*log(abs(x - 3))

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maple [A]  time = 0.01, size = 20, normalized size = 0.80 \[ 11 \ln \left (x -3\right )-14 \ln \left (x -2\right )+4 \ln \left (x -1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+3*x+4)/(x-3)/(x-2)/(x-1),x)

[Out]

4*ln(x-1)-14*ln(x-2)+11*ln(x-3)

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maxima [A]  time = 1.07, size = 19, normalized size = 0.76 \[ 4 \, \log \left (x - 1\right ) - 14 \, \log \left (x - 2\right ) + 11 \, \log \left (x - 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+3*x+4)/(-3+x)/(-2+x)/(-1+x),x, algorithm="maxima")

[Out]

4*log(x - 1) - 14*log(x - 2) + 11*log(x - 3)

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mupad [B]  time = 2.14, size = 19, normalized size = 0.76 \[ 4\,\ln \left (x-1\right )-14\,\ln \left (x-2\right )+11\,\ln \left (x-3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + x^2 + 4)/((x - 1)*(x - 2)*(x - 3)),x)

[Out]

4*log(x - 1) - 14*log(x - 2) + 11*log(x - 3)

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sympy [A]  time = 0.14, size = 19, normalized size = 0.76 \[ 11 \log {\left (x - 3 \right )} - 14 \log {\left (x - 2 \right )} + 4 \log {\left (x - 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+3*x+4)/(-3+x)/(-2+x)/(-1+x),x)

[Out]

11*log(x - 3) - 14*log(x - 2) + 4*log(x - 1)

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