\(\int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx\) [4512]

Optimal result

Integrand size = 31, antiderivative size = 20 \[ \int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx=\log \left (x-4 \left (5+\frac {5}{16 x}+\frac {3 \log (2)}{8}\right )\right ) \]

[Out]

ln(x-3/2*ln(2)-20-5/4/x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {6, 1608, 1642, 642} \[ \int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx=\log \left (-4 x^2+x (80+\log (64))+5\right )-\log (x) \]

[In]

Int[(-5 - 4*x^2)/(5*x + 80*x^2 - 4*x^3 + 6*x^2*Log[2]),x]

[Out]

-Log[x] + Log[5 - 4*x^2 + x*(80 + Log[64])]

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 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]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-5-4 x^2}{5 x-4 x^3+x^2 (80+6 \log (2))} \, dx \\ & = \int \frac {-5-4 x^2}{x \left (5-4 x^2+x (80+6 \log (2))\right )} \, dx \\ & = \int \left (-\frac {1}{x}+\frac {80-8 x+\log (64)}{5-4 x^2+x (80+\log (64))}\right ) \, dx \\ & = -\log (x)+\int \frac {80-8 x+\log (64)}{5-4 x^2+x (80+\log (64))} \, dx \\ & = -\log (x)+\log \left (5-4 x^2+x (80+\log (64))\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx=-\log (x)+\log \left (5+80 x-4 x^2+2 x \log (8)\right ) \]

[In]

Integrate[(-5 - 4*x^2)/(5*x + 80*x^2 - 4*x^3 + 6*x^2*Log[2]),x]

[Out]

-Log[x] + Log[5 + 80*x - 4*x^2 + 2*x*Log[8]]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(-\ln \left (x \right )+\ln \left (-\frac {3 x \ln \left (2\right )}{2}+x^{2}-20 x -\frac {5}{4}\right )\)	\(20\)
default	\(-\ln \left (x \right )+\ln \left (-6 x \ln \left (2\right )+4 x^{2}-80 x -5\right )\)	\(22\)
norman	\(-\ln \left (x \right )+\ln \left (6 x \ln \left (2\right )-4 x^{2}+80 x +5\right )\)	\(22\)
risch	\(-\ln \left (-x \right )+\ln \left (-5+4 x^{2}+\left (-6 \ln \left (2\right )-80\right ) x \right )\)	\(24\)

[In]

int((-4*x^2-5)/(6*x^2*ln(2)-4*x^3+80*x^2+5*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(-3/2*x*ln(2)+x^2-20*x-5/4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx=\log \left (4 \, x^{2} - 6 \, x \log \left (2\right ) - 80 \, x - 5\right ) - \log \left (x\right ) \]

[In]

integrate((-4*x^2-5)/(6*x^2*log(2)-4*x^3+80*x^2+5*x),x, algorithm="fricas")

[Out]

log(4*x^2 - 6*x*log(2) - 80*x - 5) - log(x)

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx=- \log {\left (x \right )} + \log {\left (x^{2} + x \left (-20 - \frac {3 \log {\left (2 \right )}}{2}\right ) - \frac {5}{4} \right )} \]

[In]

integrate((-4*x**2-5)/(6*x**2*ln(2)-4*x**3+80*x**2+5*x),x)

[Out]

-log(x) + log(x**2 + x*(-20 - 3*log(2)/2) - 5/4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx=\log \left (4 \, x^{2} - 2 \, x {\left (3 \, \log \left (2\right ) + 40\right )} - 5\right ) - \log \left (x\right ) \]

[In]

integrate((-4*x^2-5)/(6*x^2*log(2)-4*x^3+80*x^2+5*x),x, algorithm="maxima")

[Out]

log(4*x^2 - 2*x*(3*log(2) + 40) - 5) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx=\log \left ({\left | 4 \, x^{2} - 6 \, x \log \left (2\right ) - 80 \, x - 5 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]

[In]

integrate((-4*x^2-5)/(6*x^2*log(2)-4*x^3+80*x^2+5*x),x, algorithm="giac")

[Out]

log(abs(4*x^2 - 6*x*log(2) - 80*x - 5)) - log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-5-4 x^2}{5 x+80 x^2-4 x^3+6 x^2 \log (2)} \, dx=\ln \left (x^2-\frac {3\,x\,\ln \left (2\right )}{2}-20\,x-\frac {5}{4}\right )-\ln \left (x\right ) \]

[In]

int(-(4*x^2 + 5)/(5*x + 6*x^2*log(2) + 80*x^2 - 4*x^3),x)

[Out]

log(x^2 - (3*x*log(2))/2 - 20*x - 5/4) - log(x)