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

Optimal. Leaf size=41 \[ -\frac {1}{6} \log (1-x)+\frac {1}{2} \log (2-x)-\frac {1}{2} \log (3-x)+\frac {1}{6} \log (4-x) \]

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {180} \[ -\frac {1}{6} \log (1-x)+\frac {1}{2} \log (2-x)-\frac {1}{2} \log (3-x)+\frac {1}{6} \log (4-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[1 - x]/6 + Log[2 - x]/2 - Log[3 - x]/2 + Log[4 - x]/6

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rubi steps

\begin {align*} \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx &=\int \left (\frac {1}{6 (-4+x)}-\frac {1}{2 (-3+x)}+\frac {1}{2 (-2+x)}-\frac {1}{6 (-1+x)}\right ) \, dx\\ &=-\frac {1}{6} \log (1-x)+\frac {1}{2} \log (2-x)-\frac {1}{2} \log (3-x)+\frac {1}{6} \log (4-x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 1.00 \[ -\frac {1}{6} \log (1-x)+\frac {1}{2} \log (2-x)-\frac {1}{2} \log (3-x)+\frac {1}{6} \log (4-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*Log[1 - x] + Log[2 - x]/2 - Log[3 - x]/2 + Log[4 - x]/6

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IntegrateAlgebraic [A]  time = 0.01, size = 25, normalized size = 0.61 \[ \frac {1}{6} \log (x-4)-\frac {1}{6} \log (x-1)-\tanh ^{-1}(5-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[5 - 2*x] + Log[-4 + x]/6 - Log[-1 + x]/6

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fricas [A]  time = 0.90, size = 25, normalized size = 0.61 \[ -\frac {1}{6} \, \log \left (x - 1\right ) + \frac {1}{2} \, \log \left (x - 2\right ) - \frac {1}{2} \, \log \left (x - 3\right ) + \frac {1}{6} \, \log \left (x - 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*log(x - 1) + 1/2*log(x - 2) - 1/2*log(x - 3) + 1/6*log(x - 4)

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giac [A]  time = 0.89, size = 29, normalized size = 0.71 \[ -\frac {1}{6} \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x - 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x - 3 \right |}\right ) + \frac {1}{6} \, \log \left ({\left | x - 4 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*log(abs(x - 1)) + 1/2*log(abs(x - 2)) - 1/2*log(abs(x - 3)) + 1/6*log(abs(x - 4))

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maple [A]  time = 0.33, size = 26, normalized size = 0.63

method	result	size
default	\(\frac {\ln \left (-2+x \right )}{2}+\frac {\ln \left (-4+x \right )}{6}-\frac {\ln \left (-3+x \right )}{2}-\frac {\ln \left (-1+x \right )}{6}\)	\(26\)
norman	\(\frac {\ln \left (-2+x \right )}{2}+\frac {\ln \left (-4+x \right )}{6}-\frac {\ln \left (-3+x \right )}{2}-\frac {\ln \left (-1+x \right )}{6}\)	\(26\)
risch	\(\frac {\ln \left (-2+x \right )}{2}+\frac {\ln \left (-4+x \right )}{6}-\frac {\ln \left (-3+x \right )}{2}-\frac {\ln \left (-1+x \right )}{6}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-4+x)/(-3+x)/(-2+x)/(-1+x),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(-2+x)+1/6*ln(-4+x)-1/2*ln(-3+x)-1/6*ln(-1+x)

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maxima [A]  time = 0.43, size = 25, normalized size = 0.61 \[ -\frac {1}{6} \, \log \left (x - 1\right ) + \frac {1}{2} \, \log \left (x - 2\right ) - \frac {1}{2} \, \log \left (x - 3\right ) + \frac {1}{6} \, \log \left (x - 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*log(x - 1) + 1/2*log(x - 2) - 1/2*log(x - 3) + 1/6*log(x - 4)

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mupad [B]  time = 0.25, size = 15, normalized size = 0.37 \[ \mathrm {atanh}\left (2\,x-5\right )-\frac {\mathrm {atanh}\left (\frac {2\,x}{3}-\frac {5}{3}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(2*x - 5) - atanh((2*x)/3 - 5/3)/3

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sympy [A]  time = 0.19, size = 26, normalized size = 0.63 \[ \frac {\log {\left (x - 4 \right )}}{6} - \frac {\log {\left (x - 3 \right )}}{2} + \frac {\log {\left (x - 2 \right )}}{2} - \frac {\log {\left (x - 1 \right )}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 4)/6 - log(x - 3)/2 + log(x - 2)/2 - log(x - 1)/6

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