\(\int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx\) [10155]

Optimal result

Integrand size = 33, antiderivative size = 27 \[ \int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx=\log \left (8 \left (5+\frac {x+\frac {1}{4} (-4+2 x+x \log (4))}{4 x^2}\right )\right ) \]

[Out]

ln(2*(3/2*x+1/2*x*ln(2)-1)/x^2+40)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {6, 1608, 814, 642} \[ \int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx=\log \left (-80 x^2-x (6+\log (4))+4\right )-2 \log (x) \]

[In]

Int[(8 - 6*x - x*Log[4])/(-4*x + 6*x^2 + 80*x^3 + x^2*Log[4]),x]

[Out]

-2*Log[x] + Log[4 - 80*x^2 - x*(6 + Log[4])]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

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 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {8+x (-6-\log (4))}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx \\ & = \int \frac {8+x (-6-\log (4))}{-4 x+80 x^3+x^2 (6+\log (4))} \, dx \\ & = \int \frac {8+x (-6-\log (4))}{x \left (-4+80 x^2+x (6+\log (4))\right )} \, dx \\ & = \int \left (-\frac {2}{x}+\frac {-6-160 x-\log (4)}{4-80 x^2-x (6+\log (4))}\right ) \, dx \\ & = -2 \log (x)+\int \frac {-6-160 x-\log (4)}{4-80 x^2+x (-6-\log (4))} \, dx \\ & = -2 \log (x)+\log \left (4-80 x^2-x (6+\log (4))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx=-2 \log (x)+\log \left (4-6 x-80 x^2-x \log (4)\right ) \]

[In]

Integrate[(8 - 6*x - x*Log[4])/(-4*x + 6*x^2 + 80*x^3 + x^2*Log[4]),x]

[Out]

-2*Log[x] + Log[4 - 6*x - 80*x^2 - x*Log[4]]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
risch	\(-2 \ln \left (x \right )+\ln \left (-2+40 x^{2}+x \left (3+\ln \left (2\right )\right )\right )\)	\(20\)
parallelrisch	\(-2 \ln \left (x \right )+\ln \left (\frac {x \ln \left (2\right )}{40}+x^{2}+\frac {3 x}{40}-\frac {1}{20}\right )\)	\(20\)
default	\(\ln \left (x \ln \left (2\right )+40 x^{2}+3 x -2\right )-2 \ln \left (x \right )\)	\(21\)
norman	\(\ln \left (x \ln \left (2\right )+40 x^{2}+3 x -2\right )-2 \ln \left (x \right )\)	\(21\)

[In]

int((-2*x*ln(2)-6*x+8)/(2*x^2*ln(2)+80*x^3+6*x^2-4*x),x,method=_RETURNVERBOSE)

[Out]

-2*ln(x)+ln(-2+40*x^2+x*(3+ln(2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx=\log \left (40 \, x^{2} + x \log \left (2\right ) + 3 \, x - 2\right ) - 2 \, \log \left (x\right ) \]

[In]

integrate((-2*x*log(2)-6*x+8)/(2*x^2*log(2)+80*x^3+6*x^2-4*x),x, algorithm="fricas")

[Out]

log(40*x^2 + x*log(2) + 3*x - 2) - 2*log(x)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx=- 2 \log {\left (x \right )} + \log {\left (x^{2} + x \left (\frac {\log {\left (2 \right )}}{40} + \frac {3}{40}\right ) - \frac {1}{20} \right )} \]

[In]

integrate((-2*x*ln(2)-6*x+8)/(2*x**2*ln(2)+80*x**3+6*x**2-4*x),x)

[Out]

-2*log(x) + log(x**2 + x*(log(2)/40 + 3/40) - 1/20)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx=\log \left (40 \, x^{2} + x {\left (\log \left (2\right ) + 3\right )} - 2\right ) - 2 \, \log \left (x\right ) \]

[In]

integrate((-2*x*log(2)-6*x+8)/(2*x^2*log(2)+80*x^3+6*x^2-4*x),x, algorithm="maxima")

[Out]

log(40*x^2 + x*(log(2) + 3) - 2) - 2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx=\log \left ({\left | 40 \, x^{2} + x \log \left (2\right ) + 3 \, x - 2 \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((-2*x*log(2)-6*x+8)/(2*x^2*log(2)+80*x^3+6*x^2-4*x),x, algorithm="giac")

[Out]

log(abs(40*x^2 + x*log(2) + 3*x - 2)) - 2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 14.88 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {8-6 x-x \log (4)}{-4 x+6 x^2+80 x^3+x^2 \log (4)} \, dx=\ln \left (6-3\,x\,\ln \left (2\right )-120\,x^2-9\,x\right )-2\,\ln \left (x\right ) \]

[In]

int(-(6*x + 2*x*log(2) - 8)/(2*x^2*log(2) - 4*x + 6*x^2 + 80*x^3),x)

[Out]

log(6 - 3*x*log(2) - 120*x^2 - 9*x) - 2*log(x)