\(\int \frac {-10-10 x+10 \log (2 x)+(2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)) \log (4 x)}{(-x-x^2+x \log (2 x)) \log (4 x)} \, dx\) [301]

Optimal result

Integrand size = 64, antiderivative size = 25 \[ \int \frac {-10-10 x+10 \log (2 x)+\left (2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)\right ) \log (4 x)}{\left (-x-x^2+x \log (2 x)\right ) \log (4 x)} \, dx=x^2+\log \left ((1+x-\log (2 x))^2\right )+5 \log \left (\log ^2(4 x)\right ) \]

[Out]

ln((x-ln(2*x)+1)^2)+5*ln(ln(4*x)^2)+x^2

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6874, 6816, 2339, 29} \[ \int \frac {-10-10 x+10 \log (2 x)+\left (2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)\right ) \log (4 x)}{\left (-x-x^2+x \log (2 x)\right ) \log (4 x)} \, dx=x^2+2 \log (x-\log (2 x)+1)+10 \log (\log (4 x)) \]

[In]

Int[(-10 - 10*x + 10*Log[2*x] + (2 - 2*x - 2*x^2 - 2*x^3 + 2*x^2*Log[2*x])*Log[4*x])/((-x - x^2 + x*Log[2*x])*
Log[4*x]),x]

[Out]

x^2 + 2*Log[1 + x - Log[2*x]] + 10*Log[Log[4*x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 \left (-1+x+x^2+x^3-x^2 \log (2 x)\right )}{x (1+x-\log (2 x))}+\frac {10}{x \log (4 x)}\right ) \, dx \\ & = 2 \int \frac {-1+x+x^2+x^3-x^2 \log (2 x)}{x (1+x-\log (2 x))} \, dx+10 \int \frac {1}{x \log (4 x)} \, dx \\ & = 2 \int \left (x+\frac {-1+x}{x (1+x-\log (2 x))}\right ) \, dx+10 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (4 x)\right ) \\ & = x^2+10 \log (\log (4 x))+2 \int \frac {-1+x}{x (1+x-\log (2 x))} \, dx \\ & = x^2+2 \log (1+x-\log (2 x))+10 \log (\log (4 x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-10-10 x+10 \log (2 x)+\left (2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)\right ) \log (4 x)}{\left (-x-x^2+x \log (2 x)\right ) \log (4 x)} \, dx=2 \left (\frac {x^2}{2}+\log (1+x-\log (2 x))+5 \log (\log (4 x))\right ) \]

[In]

Integrate[(-10 - 10*x + 10*Log[2*x] + (2 - 2*x - 2*x^2 - 2*x^3 + 2*x^2*Log[2*x])*Log[4*x])/((-x - x^2 + x*Log[
2*x])*Log[4*x]),x]

[Out]

2*(x^2/2 + Log[1 + x - Log[2*x]] + 5*Log[Log[4*x]])

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(x^{2}+10 \ln \left (\ln \left (4 x \right )\right )+2 \ln \left (x -\ln \left (2 x \right )+1\right )\)	\(24\)
default	\(x^{2}+10 \ln \left (\ln \left (x \right )+2 \ln \left (2\right )\right )+2 \ln \left (-x +\ln \left (2\right )+\ln \left (x \right )-1\right )\)	\(27\)
risch	\(x^{2}+2 \ln \left (\ln \left (x \right )-\frac {i \left (2 i \ln \left (2\right )-2 i x -2 i\right )}{2}\right )+10 \ln \left (\ln \left (x \right )+2 \ln \left (2\right )\right )\)	\(36\)

[In]

int(((2*x^2*ln(2*x)-2*x^3-2*x^2-2*x+2)*ln(4*x)+10*ln(2*x)-10*x-10)/(x*ln(2*x)-x^2-x)/ln(4*x),x,method=_RETURNV
ERBOSE)

[Out]

x^2+10*ln(ln(4*x))+2*ln(x-ln(2*x)+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-10-10 x+10 \log (2 x)+\left (2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)\right ) \log (4 x)}{\left (-x-x^2+x \log (2 x)\right ) \log (4 x)} \, dx=x^{2} + 2 \, \log \left (-x + \log \left (2 \, x\right ) - 1\right ) + 10 \, \log \left (\log \left (2\right ) + \log \left (2 \, x\right )\right ) \]

[In]

integrate(((2*x^2*log(2*x)-2*x^3-2*x^2-2*x+2)*log(4*x)+10*log(2*x)-10*x-10)/(x*log(2*x)-x^2-x)/log(4*x),x, alg
orithm="fricas")

[Out]

x^2 + 2*log(-x + log(2*x) - 1) + 10*log(log(2) + log(2*x))

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-10-10 x+10 \log (2 x)+\left (2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)\right ) \log (4 x)}{\left (-x-x^2+x \log (2 x)\right ) \log (4 x)} \, dx=x^{2} + 10 \log {\left (\log {\left (2 x \right )} + \log {\left (2 \right )} \right )} + 2 \log {\left (- x + \log {\left (2 x \right )} - 1 \right )} \]

[In]

integrate(((2*x**2*ln(2*x)-2*x**3-2*x**2-2*x+2)*ln(4*x)+10*ln(2*x)-10*x-10)/(x*ln(2*x)-x**2-x)/ln(4*x),x)

[Out]

x**2 + 10*log(log(2*x) + log(2)) + 2*log(-x + log(2*x) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-10-10 x+10 \log (2 x)+\left (2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)\right ) \log (4 x)}{\left (-x-x^2+x \log (2 x)\right ) \log (4 x)} \, dx=x^{2} + 2 \, \log \left (-x + \log \left (2\right ) + \log \left (x\right ) - 1\right ) + 10 \, \log \left (2 \, \log \left (2\right ) + \log \left (x\right )\right ) \]

[In]

integrate(((2*x^2*log(2*x)-2*x^3-2*x^2-2*x+2)*log(4*x)+10*log(2*x)-10*x-10)/(x*log(2*x)-x^2-x)/log(4*x),x, alg
orithm="maxima")

[Out]

x^2 + 2*log(-x + log(2) + log(x) - 1) + 10*log(2*log(2) + log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {-10-10 x+10 \log (2 x)+\left (2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)\right ) \log (4 x)}{\left (-x-x^2+x \log (2 x)\right ) \log (4 x)} \, dx=x^{2} + 2 \, \log \left (2 \, x - 2 \, \log \left (2 \, x\right ) + 2\right ) + 10 \, \log \left (\log \left (2\right ) + \log \left (2 \, x\right )\right ) \]

[In]

integrate(((2*x^2*log(2*x)-2*x^3-2*x^2-2*x+2)*log(4*x)+10*log(2*x)-10*x-10)/(x*log(2*x)-x^2-x)/log(4*x),x, alg
orithm="giac")

[Out]

x^2 + 2*log(2*x - 2*log(2*x) + 2) + 10*log(log(2) + log(2*x))

Mupad [B] (verification not implemented)

Time = 8.99 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {-10-10 x+10 \log (2 x)+\left (2-2 x-2 x^2-2 x^3+2 x^2 \log (2 x)\right ) \log (4 x)}{\left (-x-x^2+x \log (2 x)\right ) \log (4 x)} \, dx=12\,\ln \left (x\right )-10\,\ln \left (x-1\right )+2\,\ln \left (\frac {8\,x-\ln \left (256\right )-8\,\ln \left (x\right )+8}{x}\right )+10\,\ln \left (\frac {\left (16\,\ln \left (2\right )+8\,\ln \left (x\right )\right )\,\left (x-1\right )}{x}\right )+x^2 \]

[In]

int((10*x - 10*log(2*x) + log(4*x)*(2*x - 2*x^2*log(2*x) + 2*x^2 + 2*x^3 - 2) + 10)/(log(4*x)*(x - x*log(2*x)
+ x^2)),x)

[Out]

12*log(x) - 10*log(x - 1) + 2*log((8*x - log(256) - 8*log(x) + 8)/x) + 10*log(((16*log(2) + 8*log(x))*(x - 1))
/x) + x^2