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

Optimal. Leaf size=30 \[ x+\frac{7}{2} \log (1-x)-25 \log (2-x)+\frac{61}{2} \log (3-x) \]

[Out]

x + (7*Log[1 - x])/2 - 25*Log[2 - x] + (61*Log[3 - x])/2

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Rubi [A]  time = 0.0576631, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.034, Rules used = {1612} \[ x+\frac{7}{2} \log (1-x)-25 \log (2-x)+\frac{61}{2} \log (3-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x + (7*Log[1 - x])/2 - 25*Log[2 - x] + (61*Log[3 - x])/2

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{1+2 x+3 x^2+x^3}{(-3+x) (-2+x) (-1+x)} \, dx &=\int \left (1+\frac{61}{2 (-3+x)}-\frac{25}{-2+x}+\frac{7}{2 (-1+x)}\right ) \, dx\\ &=x+\frac{7}{2} \log (1-x)-25 \log (2-x)+\frac{61}{2} \log (3-x)\\ \end{align*}

Mathematica [A]  time = 0.011634, size = 24, normalized size = 0.8 \[ x+\frac{61}{2} \log (x-3)-25 \log (x-2)+\frac{7}{2} \log (x-1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x + (61*Log[-3 + x])/2 - 25*Log[-2 + x] + (7*Log[-1 + x])/2

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Maple [A]  time = 0.007, size = 21, normalized size = 0.7 \begin{align*} x+{\frac{7\,\ln \left ( x-1 \right ) }{2}}+{\frac{61\,\ln \left ( -3+x \right ) }{2}}-25\,\ln \left ( -2+x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+7/2*ln(x-1)+61/2*ln(-3+x)-25*ln(-2+x)

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Maxima [A]  time = 0.967553, size = 27, normalized size = 0.9 \begin{align*} x + \frac{7}{2} \, \log \left (x - 1\right ) - 25 \, \log \left (x - 2\right ) + \frac{61}{2} \, \log \left (x - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 7/2*log(x - 1) - 25*log(x - 2) + 61/2*log(x - 3)

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Fricas [A]  time = 1.5394, size = 73, normalized size = 2.43 \begin{align*} x + \frac{7}{2} \, \log \left (x - 1\right ) - 25 \, \log \left (x - 2\right ) + \frac{61}{2} \, \log \left (x - 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 7/2*log(x - 1) - 25*log(x - 2) + 61/2*log(x - 3)

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Sympy [A]  time = 0.13527, size = 24, normalized size = 0.8 \begin{align*} x + \frac{61 \log{\left (x - 3 \right )}}{2} - 25 \log{\left (x - 2 \right )} + \frac{7 \log{\left (x - 1 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 61*log(x - 3)/2 - 25*log(x - 2) + 7*log(x - 1)/2

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Giac [A]  time = 1.12603, size = 31, normalized size = 1.03 \begin{align*} x + \frac{7}{2} \, \log \left ({\left | x - 1 \right |}\right ) - 25 \, \log \left ({\left | x - 2 \right |}\right ) + \frac{61}{2} \, \log \left ({\left | x - 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 7/2*log(abs(x - 1)) - 25*log(abs(x - 2)) + 61/2*log(abs(x - 3))