\(\int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx\) [100]

Optimal result

Integrand size = 21, antiderivative size = 41 \[ \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx=-\frac {1}{6} \log (1-x)+\frac {1}{2} \log (2-x)-\frac {1}{2} \log (3-x)+\frac {1}{6} \log (4-x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {186} \[ \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx=-\frac {1}{6} \log (1-x)+\frac {1}{2} \log (2-x)-\frac {1}{2} \log (3-x)+\frac {1}{6} \log (4-x) \]

[In]

Int[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

Rule 186

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx=-\frac {1}{6} \log (1-x)+\frac {1}{2} \log (2-x)-\frac {1}{2} \log (3-x)+\frac {1}{6} \log (4-x) \]

[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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx=-\frac {1}{6} \, \log \left (x - 1\right ) + \frac {1}{2} \, \log \left (x - 2\right ) - \frac {1}{2} \, \log \left (x - 3\right ) + \frac {1}{6} \, \log \left (x - 4\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx=\frac {\log {\left (x - 4 \right )}}{6} - \frac {\log {\left (x - 3 \right )}}{2} + \frac {\log {\left (x - 2 \right )}}{2} - \frac {\log {\left (x - 1 \right )}}{6} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx=-\frac {1}{6} \, \log \left (x - 1\right ) + \frac {1}{2} \, \log \left (x - 2\right ) - \frac {1}{2} \, \log \left (x - 3\right ) + \frac {1}{6} \, \log \left (x - 4\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx=-\frac {1}{6} \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x - 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | x - 3 \right |}\right ) + \frac {1}{6} \, \log \left ({\left | x - 4 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.37 \[ \int \frac {1}{(-4+x) (-3+x) (-2+x) (-1+x)} \, dx=\mathrm {atanh}\left (2\,x-5\right )-\frac {\mathrm {atanh}\left (\frac {2\,x}{3}-\frac {5}{3}\right )}{3} \]

[In]

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

[Out]

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