\(\int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx\) [9082]

Optimal result

Integrand size = 24, antiderivative size = 18 \[ \int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx=\log \left (5-x+\frac {x (9+x)}{10-x}\right ) \]

[Out]

ln(5-x+(x+9)*x/(10-x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2099, 642} \[ \int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx=\log \left (x^2-3 x+25\right )-\log (10-x) \]

[In]

Int[(5 - 20*x + x^2)/(-250 + 55*x - 13*x^2 + x^3),x]

[Out]

-Log[10 - x] + Log[25 - 3*x + x^2]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{10-x}+\frac {-3+2 x}{25-3 x+x^2}\right ) \, dx \\ & = -\log (10-x)+\int \frac {-3+2 x}{25-3 x+x^2} \, dx \\ & = -\log (10-x)+\log \left (25-3 x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx=-\log (10-x)+\log \left (25-3 x+x^2\right ) \]

[In]

Integrate[(5 - 20*x + x^2)/(-250 + 55*x - 13*x^2 + x^3),x]

[Out]

-Log[10 - x] + Log[25 - 3*x + x^2]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
default	\(-\ln \left (x -10\right )+\ln \left (x^{2}-3 x +25\right )\)	\(17\)
norman	\(-\ln \left (x -10\right )+\ln \left (x^{2}-3 x +25\right )\)	\(17\)
risch	\(-\ln \left (x -10\right )+\ln \left (x^{2}-3 x +25\right )\)	\(17\)
parallelrisch	\(-\ln \left (x -10\right )+\ln \left (x^{2}-3 x +25\right )\)	\(17\)

[In]

int((x^2-20*x+5)/(x^3-13*x^2+55*x-250),x,method=_RETURNVERBOSE)

[Out]

-ln(x-10)+ln(x^2-3*x+25)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx=\log \left (x^{2} - 3 \, x + 25\right ) - \log \left (x - 10\right ) \]

[In]

integrate((x^2-20*x+5)/(x^3-13*x^2+55*x-250),x, algorithm="fricas")

[Out]

log(x^2 - 3*x + 25) - log(x - 10)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx=- \log {\left (x - 10 \right )} + \log {\left (x^{2} - 3 x + 25 \right )} \]

[In]

integrate((x**2-20*x+5)/(x**3-13*x**2+55*x-250),x)

[Out]

-log(x - 10) + log(x**2 - 3*x + 25)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx=\log \left (x^{2} - 3 \, x + 25\right ) - \log \left (x - 10\right ) \]

[In]

integrate((x^2-20*x+5)/(x^3-13*x^2+55*x-250),x, algorithm="maxima")

[Out]

log(x^2 - 3*x + 25) - log(x - 10)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx=\log \left (x^{2} - 3 \, x + 25\right ) - \log \left ({\left | x - 10 \right |}\right ) \]

[In]

integrate((x^2-20*x+5)/(x^3-13*x^2+55*x-250),x, algorithm="giac")

[Out]

log(x^2 - 3*x + 25) - log(abs(x - 10))

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {5-20 x+x^2}{-250+55 x-13 x^2+x^3} \, dx=\ln \left (x^2-3\,x+25\right )-\ln \left (x-10\right ) \]

[In]

int((x^2 - 20*x + 5)/(55*x - 13*x^2 + x^3 - 250),x)

[Out]

log(x^2 - 3*x + 25) - log(x - 10)