\(\int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx\) [1904]

Optimal result

Integrand size = 28, antiderivative size = 18 \[ \int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx=\log \left (30 \left (\frac {2}{5 (4-x)}+7 x\right )\right ) \]

[Out]

ln(210*x+12/(-x+4))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2099, 642} \[ \int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx=\log \left (-35 x^2+140 x+2\right )-\log (4-x) \]

[In]

Int[(562 - 280*x + 35*x^2)/(8 + 558*x - 280*x^2 + 35*x^3),x]

[Out]

-Log[4 - x] + Log[2 + 140*x - 35*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}{4-x}+\frac {70 (-2+x)}{-2-140 x+35 x^2}\right ) \, dx \\ & = -\log (4-x)+70 \int \frac {-2+x}{-2-140 x+35 x^2} \, dx \\ & = -\log (4-x)+\log \left (2+140 x-35 x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx=-\log (4-x)+\log \left (2+140 x-35 x^2\right ) \]

[In]

Integrate[(562 - 280*x + 35*x^2)/(8 + 558*x - 280*x^2 + 35*x^3),x]

[Out]

-Log[4 - x] + Log[2 + 140*x - 35*x^2]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(-\ln \left (x -4\right )+\ln \left (x^{2}-4 x -\frac {2}{35}\right )\)	\(17\)
default	\(-\ln \left (x -4\right )+\ln \left (35 x^{2}-140 x -2\right )\)	\(19\)
norman	\(-\ln \left (x -4\right )+\ln \left (35 x^{2}-140 x -2\right )\)	\(19\)
risch	\(-\ln \left (x -4\right )+\ln \left (35 x^{2}-140 x -2\right )\)	\(19\)

[In]

int((35*x^2-280*x+562)/(35*x^3-280*x^2+558*x+8),x,method=_RETURNVERBOSE)

[Out]

-ln(x-4)+ln(x^2-4*x-2/35)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx=\log \left (35 \, x^{2} - 140 \, x - 2\right ) - \log \left (x - 4\right ) \]

[In]

integrate((35*x^2-280*x+562)/(35*x^3-280*x^2+558*x+8),x, algorithm="fricas")

[Out]

log(35*x^2 - 140*x - 2) - log(x - 4)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx=- \log {\left (x - 4 \right )} + \log {\left (35 x^{2} - 140 x - 2 \right )} \]

[In]

integrate((35*x**2-280*x+562)/(35*x**3-280*x**2+558*x+8),x)

[Out]

-log(x - 4) + log(35*x**2 - 140*x - 2)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx=\log \left (35 \, x^{2} - 140 \, x - 2\right ) - \log \left (x - 4\right ) \]

[In]

integrate((35*x^2-280*x+562)/(35*x^3-280*x^2+558*x+8),x, algorithm="maxima")

[Out]

log(35*x^2 - 140*x - 2) - log(x - 4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx=\log \left ({\left | 35 \, x^{2} - 140 \, x - 2 \right |}\right ) - \log \left ({\left | x - 4 \right |}\right ) \]

[In]

integrate((35*x^2-280*x+562)/(35*x^3-280*x^2+558*x+8),x, algorithm="giac")

[Out]

log(abs(35*x^2 - 140*x - 2)) - log(abs(x - 4))

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {562-280 x+35 x^2}{8+558 x-280 x^2+35 x^3} \, dx=\ln \left (x^2-4\,x-\frac {2}{35}\right )-\ln \left (x-4\right ) \]

[In]

int((35*x^2 - 280*x + 562)/(558*x - 280*x^2 + 35*x^3 + 8),x)

[Out]

log(x^2 - 4*x - 2/35) - log(x - 4)