\(\int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx\) [7734]

Optimal result

Integrand size = 29, antiderivative size = 15 \[ \int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx=-1209 \log (x)+\log \left (-5-x+\log ^2(4)\right ) \]

[Out]

ln(4*ln(2)^2-5-x)-1209*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6, 645} \[ \int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx=\log \left (x+5-\log ^2(4)\right )-1209 \log (x) \]

[In]

Int[(6045 + 1208*x - 1209*Log[4]^2)/(-5*x - x^2 + x*Log[4]^2),x]

[Out]

-1209*Log[x] + Log[5 + x - Log[4]^2]

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 645

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {6045+1208 x-1209 \log ^2(4)}{-x^2+x \left (-5+\log ^2(4)\right )} \, dx \\ & = \int \left (-\frac {1209}{x}+\frac {1}{5+x-\log ^2(4)}\right ) \, dx \\ & = -1209 \log (x)+\log \left (5+x-\log ^2(4)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx=-1209 \log (x)+\log \left (5+x-\log ^2(4)\right ) \]

[In]

Integrate[(6045 + 1208*x - 1209*Log[4]^2)/(-5*x - x^2 + x*Log[4]^2),x]

[Out]

-1209*Log[x] + Log[5 + x - Log[4]^2]

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07

method	result	size
default	\(-1209 \ln \left (x \right )+\ln \left (-4 \ln \left (2\right )^{2}+5+x \right )\)	\(16\)
risch	\(-1209 \ln \left (x \right )+\ln \left (-4 \ln \left (2\right )^{2}+5+x \right )\)	\(16\)
parallelrisch	\(-1209 \ln \left (x \right )+\ln \left (-4 \ln \left (2\right )^{2}+5+x \right )\)	\(16\)
norman	\(\ln \left (4 \ln \left (2\right )^{2}-5-x \right )-1209 \ln \left (x \right )\)	\(18\)
meijerg	\(\frac {4836 \ln \left (2\right )^{2} \left (-4 \ln \left (2\right )^{2}+5\right ) \left (\ln \left (x \right )+\ln \left (-\frac {1}{4 \ln \left (2\right )^{2}-5}\right )-\ln \left (1-\frac {x}{4 \ln \left (2\right )^{2}-5}\right )\right )}{\left (4 \ln \left (2\right )^{2}-5\right )^{2}}+\frac {1208 \left (-4 \ln \left (2\right )^{2}+5\right ) \ln \left (1-\frac {x}{4 \ln \left (2\right )^{2}-5}\right )}{4 \ln \left (2\right )^{2}-5}-\frac {6045 \left (-4 \ln \left (2\right )^{2}+5\right ) \left (\ln \left (x \right )+\ln \left (-\frac {1}{4 \ln \left (2\right )^{2}-5}\right )-\ln \left (1-\frac {x}{4 \ln \left (2\right )^{2}-5}\right )\right )}{\left (4 \ln \left (2\right )^{2}-5\right )^{2}}\)	\(150\)

[In]

int((-4836*ln(2)^2+1208*x+6045)/(4*x*ln(2)^2-x^2-5*x),x,method=_RETURNVERBOSE)

[Out]

-1209*ln(x)+ln(-4*ln(2)^2+5+x)

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx=\log \left (-4 \, \log \left (2\right )^{2} + x + 5\right ) - 1209 \, \log \left (x\right ) \]

[In]

integrate((-4836*log(2)^2+1208*x+6045)/(4*x*log(2)^2-x^2-5*x),x, algorithm="fricas")

[Out]

log(-4*log(2)^2 + x + 5) - 1209*log(x)

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx=- 1209 \log {\left (x \right )} + \log {\left (x - 4 \log {\left (2 \right )}^{2} + 5 \right )} \]

[In]

integrate((-4836*ln(2)**2+1208*x+6045)/(4*x*ln(2)**2-x**2-5*x),x)

[Out]

-1209*log(x) + log(x - 4*log(2)**2 + 5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx=\log \left (-4 \, \log \left (2\right )^{2} + x + 5\right ) - 1209 \, \log \left (x\right ) \]

[In]

integrate((-4836*log(2)^2+1208*x+6045)/(4*x*log(2)^2-x^2-5*x),x, algorithm="maxima")

[Out]

log(-4*log(2)^2 + x + 5) - 1209*log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx=\log \left ({\left | -4 \, \log \left (2\right )^{2} + x + 5 \right |}\right ) - 1209 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((-4836*log(2)^2+1208*x+6045)/(4*x*log(2)^2-x^2-5*x),x, algorithm="giac")

[Out]

log(abs(-4*log(2)^2 + x + 5)) - 1209*log(abs(x))

Mupad [B] (verification not implemented)

Time = 13.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {6045+1208 x-1209 \log ^2(4)}{-5 x-x^2+x \log ^2(4)} \, dx=\ln \left (x-4\,{\ln \left (2\right )}^2+5\right )-1209\,\ln \left (x\right ) \]

[In]

int(-(1208*x - 4836*log(2)^2 + 6045)/(5*x - 4*x*log(2)^2 + x^2),x)

[Out]

log(x - 4*log(2)^2 + 5) - 1209*log(x)