\(\int \frac {1}{5-6 x+x^2} \, dx\) [44]

Optimal result

Integrand size = 10, antiderivative size = 21 \[ \int \frac {1}{5-6 x+x^2} \, dx=-\frac {1}{4} \log (1-x)+\frac {1}{4} \log (5-x) \]

[Out]

-1/4*ln(1-x)+1/4*ln(5-x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {630, 31} \[ \int \frac {1}{5-6 x+x^2} \, dx=\frac {1}{4} \log (5-x)-\frac {1}{4} \log (1-x) \]

[In]

Int[(5 - 6*x + x^2)^(-1),x]

[Out]

-1/4*Log[1 - x] + Log[5 - x]/4

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 630

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {1}{-5+x} \, dx-\frac {1}{4} \int \frac {1}{-1+x} \, dx \\ & = -\frac {1}{4} \log (1-x)+\frac {1}{4} \log (5-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5-6 x+x^2} \, dx=-\frac {1}{4} \log (1-x)+\frac {1}{4} \log (5-x) \]

[In]

Integrate[(5 - 6*x + x^2)^(-1),x]

[Out]

-1/4*Log[1 - x] + Log[5 - x]/4

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
default	\(-\frac {\ln \left (-1+x \right )}{4}+\frac {\ln \left (x -5\right )}{4}\)	\(14\)
norman	\(-\frac {\ln \left (-1+x \right )}{4}+\frac {\ln \left (x -5\right )}{4}\)	\(14\)
risch	\(-\frac {\ln \left (-1+x \right )}{4}+\frac {\ln \left (x -5\right )}{4}\)	\(14\)
parallelrisch	\(-\frac {\ln \left (-1+x \right )}{4}+\frac {\ln \left (x -5\right )}{4}\)	\(14\)

[In]

int(1/(x^2-6*x+5),x,method=_RETURNVERBOSE)

[Out]

-1/4*ln(-1+x)+1/4*ln(x-5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5-6 x+x^2} \, dx=-\frac {1}{4} \, \log \left (x - 1\right ) + \frac {1}{4} \, \log \left (x - 5\right ) \]

[In]

integrate(1/(x^2-6*x+5),x, algorithm="fricas")

[Out]

-1/4*log(x - 1) + 1/4*log(x - 5)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {1}{5-6 x+x^2} \, dx=\frac {\log {\left (x - 5 \right )}}{4} - \frac {\log {\left (x - 1 \right )}}{4} \]

[In]

integrate(1/(x**2-6*x+5),x)

[Out]

log(x - 5)/4 - log(x - 1)/4

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5-6 x+x^2} \, dx=-\frac {1}{4} \, \log \left (x - 1\right ) + \frac {1}{4} \, \log \left (x - 5\right ) \]

[In]

integrate(1/(x^2-6*x+5),x, algorithm="maxima")

[Out]

-1/4*log(x - 1) + 1/4*log(x - 5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {1}{5-6 x+x^2} \, dx=-\frac {1}{4} \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x - 5 \right |}\right ) \]

[In]

integrate(1/(x^2-6*x+5),x, algorithm="giac")

[Out]

-1/4*log(abs(x - 1)) + 1/4*log(abs(x - 5))

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.38 \[ \int \frac {1}{5-6 x+x^2} \, dx=-\frac {\mathrm {atanh}\left (\frac {x}{2}-\frac {3}{2}\right )}{2} \]

[In]

int(1/(x^2 - 6*x + 5),x)

[Out]

-atanh(x/2 - 3/2)/2