\(\int \frac {-3-9 x-6 x^2+12 x^3+(-1-3 x-2 x^2+4 x^3) \log (4)+(6 x+12 x^2+(2 x+4 x^2) \log (4)) \log (x)+(-6 x+6 x^2+(-2 x+2 x^2) \log (4)+(6 x+2 x \log (4)) \log (x)) \log (-1+x+\log (x))}{-2 x+2 x^3+(2 x+2 x^2) \log (x)+(-x+x^2+x \log (x)) \log (-1+x+\log (x))} \, dx\) [4131]

Optimal result

Integrand size = 139, antiderivative size = 26 \[ \int \frac {-3-9 x-6 x^2+12 x^3+\left (-1-3 x-2 x^2+4 x^3\right ) \log (4)+\left (6 x+12 x^2+\left (2 x+4 x^2\right ) \log (4)\right ) \log (x)+\left (-6 x+6 x^2+\left (-2 x+2 x^2\right ) \log (4)+(6 x+2 x \log (4)) \log (x)\right ) \log (-1+x+\log (x))}{-2 x+2 x^3+\left (2 x+2 x^2\right ) \log (x)+\left (-x+x^2+x \log (x)\right ) \log (-1+x+\log (x))} \, dx=(3+\log (4)) \left (2 x-\log \left (x+\frac {1}{2} (2+\log (-1+x+\log (x)))\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6820, 12, 6874, 6816} \[ \int \frac {-3-9 x-6 x^2+12 x^3+\left (-1-3 x-2 x^2+4 x^3\right ) \log (4)+\left (6 x+12 x^2+\left (2 x+4 x^2\right ) \log (4)\right ) \log (x)+\left (-6 x+6 x^2+\left (-2 x+2 x^2\right ) \log (4)+(6 x+2 x \log (4)) \log (x)\right ) \log (-1+x+\log (x))}{-2 x+2 x^3+\left (2 x+2 x^2\right ) \log (x)+\left (-x+x^2+x \log (x)\right ) \log (-1+x+\log (x))} \, dx=2 x (3+\log (4))-(3+\log (4)) \log (2 x+\log (x+\log (x)-1)+2) \]

[In]

Int[(-3 - 9*x - 6*x^2 + 12*x^3 + (-1 - 3*x - 2*x^2 + 4*x^3)*Log[4] + (6*x + 12*x^2 + (2*x + 4*x^2)*Log[4])*Log
[x] + (-6*x + 6*x^2 + (-2*x + 2*x^2)*Log[4] + (6*x + 2*x*Log[4])*Log[x])*Log[-1 + x + Log[x]])/(-2*x + 2*x^3 +
(2*x + 2*x^2)*Log[x] + (-x + x^2 + x*Log[x])*Log[-1 + x + Log[x]]),x]

[Out]

2*x*(3 + Log[4]) - (3 + Log[4])*Log[2 + 2*x + Log[-1 + x + Log[x]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 6820

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

Rule 6874

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

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-3-9 x-6 x^2+12 x^3+\left (-1-3 x-2 x^2+4 x^3\right ) \log (4)+\left (6 x+12 x^2+\left (2 x+4 x^2\right ) \log (4)\right ) \log (x)+\left (-6 x+6 x^2+\left (-2 x+2 x^2\right ) \log (4)+(6 x+2 x \log (4)) \log (x)\right ) \log (-1+x+\log (x))}{-2 x+2 x^3+\left (2 x+2 x^2\right ) \log (x)+\left (-x+x^2+x \log (x)\right ) \log (-1+x+\log (x))} \, dx=(3+\log (4)) (2 x-\log (2+2 x+\log (-1+x+\log (x)))) \]

[In]

Integrate[(-3 - 9*x - 6*x^2 + 12*x^3 + (-1 - 3*x - 2*x^2 + 4*x^3)*Log[4] + (6*x + 12*x^2 + (2*x + 4*x^2)*Log[4
])*Log[x] + (-6*x + 6*x^2 + (-2*x + 2*x^2)*Log[4] + (6*x + 2*x*Log[4])*Log[x])*Log[-1 + x + Log[x]])/(-2*x + 2
*x^3 + (2*x + 2*x^2)*Log[x] + (-x + x^2 + x*Log[x])*Log[-1 + x + Log[x]]),x]

[Out]

(3 + Log[4])*(2*x - Log[2 + 2*x + Log[-1 + x + Log[x]]])

Maple [A] (verified)

Time = 2.86 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54

method	result	size
risch	\(4 x \ln \left (2\right )+6 x -2 \ln \left (\ln \left (-1+\ln \left (x \right )+x \right )+2 x +2\right ) \ln \left (2\right )-3 \ln \left (\ln \left (-1+\ln \left (x \right )+x \right )+2 x +2\right )\)	\(40\)
parallelrisch	\(-2 \ln \left (2\right ) \ln \left (1+\frac {\ln \left (-1+\ln \left (x \right )+x \right )}{2}+x \right )+4 x \ln \left (2\right )-3 \ln \left (1+\frac {\ln \left (-1+\ln \left (x \right )+x \right )}{2}+x \right )+6 x\)	\(40\)
default	\(6 x -3 \ln \left (\ln \left (-1+\ln \left (x \right )+x \right )+2 x +2\right )+2 \ln \left (2\right ) \left (2 x -\ln \left (\ln \left (-1+\ln \left (x \right )+x \right )+2 x +2\right )\right )\)	\(41\)

[In]

int((((4*x*ln(2)+6*x)*ln(x)+2*(2*x^2-2*x)*ln(2)+6*x^2-6*x)*ln(-1+ln(x)+x)+(2*(4*x^2+2*x)*ln(2)+12*x^2+6*x)*ln(
x)+2*(4*x^3-2*x^2-3*x-1)*ln(2)+12*x^3-6*x^2-9*x-3)/((x*ln(x)+x^2-x)*ln(-1+ln(x)+x)+(2*x^2+2*x)*ln(x)+2*x^3-2*x
),x,method=_RETURNVERBOSE)

[Out]

4*x*ln(2)+6*x-2*ln(ln(-1+ln(x)+x)+2*x+2)*ln(2)-3*ln(ln(-1+ln(x)+x)+2*x+2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-3-9 x-6 x^2+12 x^3+\left (-1-3 x-2 x^2+4 x^3\right ) \log (4)+\left (6 x+12 x^2+\left (2 x+4 x^2\right ) \log (4)\right ) \log (x)+\left (-6 x+6 x^2+\left (-2 x+2 x^2\right ) \log (4)+(6 x+2 x \log (4)) \log (x)\right ) \log (-1+x+\log (x))}{-2 x+2 x^3+\left (2 x+2 x^2\right ) \log (x)+\left (-x+x^2+x \log (x)\right ) \log (-1+x+\log (x))} \, dx=4 \, x \log \left (2\right ) - {\left (2 \, \log \left (2\right ) + 3\right )} \log \left (2 \, x + \log \left (x + \log \left (x\right ) - 1\right ) + 2\right ) + 6 \, x \]

[In]

integrate((((4*x*log(2)+6*x)*log(x)+2*(2*x^2-2*x)*log(2)+6*x^2-6*x)*log(-1+log(x)+x)+(2*(4*x^2+2*x)*log(2)+12*
x^2+6*x)*log(x)+2*(4*x^3-2*x^2-3*x-1)*log(2)+12*x^3-6*x^2-9*x-3)/((x*log(x)+x^2-x)*log(-1+log(x)+x)+(2*x^2+2*x
)*log(x)+2*x^3-2*x),x, algorithm="fricas")

[Out]

4*x*log(2) - (2*log(2) + 3)*log(2*x + log(x + log(x) - 1) + 2) + 6*x

Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {-3-9 x-6 x^2+12 x^3+\left (-1-3 x-2 x^2+4 x^3\right ) \log (4)+\left (6 x+12 x^2+\left (2 x+4 x^2\right ) \log (4)\right ) \log (x)+\left (-6 x+6 x^2+\left (-2 x+2 x^2\right ) \log (4)+(6 x+2 x \log (4)) \log (x)\right ) \log (-1+x+\log (x))}{-2 x+2 x^3+\left (2 x+2 x^2\right ) \log (x)+\left (-x+x^2+x \log (x)\right ) \log (-1+x+\log (x))} \, dx=x \left (4 \log {\left (2 \right )} + 6\right ) + \left (-3 - 2 \log {\left (2 \right )}\right ) \log {\left (2 x + \log {\left (x + \log {\left (x \right )} - 1 \right )} + 2 \right )} \]

[In]

integrate((((4*x*ln(2)+6*x)*ln(x)+2*(2*x**2-2*x)*ln(2)+6*x**2-6*x)*ln(-1+ln(x)+x)+(2*(4*x**2+2*x)*ln(2)+12*x**
2+6*x)*ln(x)+2*(4*x**3-2*x**2-3*x-1)*ln(2)+12*x**3-6*x**2-9*x-3)/((x*ln(x)+x**2-x)*ln(-1+ln(x)+x)+(2*x**2+2*x)
*ln(x)+2*x**3-2*x),x)

[Out]

x*(4*log(2) + 6) + (-3 - 2*log(2))*log(2*x + log(x + log(x) - 1) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-3-9 x-6 x^2+12 x^3+\left (-1-3 x-2 x^2+4 x^3\right ) \log (4)+\left (6 x+12 x^2+\left (2 x+4 x^2\right ) \log (4)\right ) \log (x)+\left (-6 x+6 x^2+\left (-2 x+2 x^2\right ) \log (4)+(6 x+2 x \log (4)) \log (x)\right ) \log (-1+x+\log (x))}{-2 x+2 x^3+\left (2 x+2 x^2\right ) \log (x)+\left (-x+x^2+x \log (x)\right ) \log (-1+x+\log (x))} \, dx=2 \, x {\left (2 \, \log \left (2\right ) + 3\right )} - {\left (2 \, \log \left (2\right ) + 3\right )} \log \left (2 \, x + \log \left (x + \log \left (x\right ) - 1\right ) + 2\right ) \]

[In]

integrate((((4*x*log(2)+6*x)*log(x)+2*(2*x^2-2*x)*log(2)+6*x^2-6*x)*log(-1+log(x)+x)+(2*(4*x^2+2*x)*log(2)+12*
x^2+6*x)*log(x)+2*(4*x^3-2*x^2-3*x-1)*log(2)+12*x^3-6*x^2-9*x-3)/((x*log(x)+x^2-x)*log(-1+log(x)+x)+(2*x^2+2*x
)*log(x)+2*x^3-2*x),x, algorithm="maxima")

[Out]

2*x*(2*log(2) + 3) - (2*log(2) + 3)*log(2*x + log(x + log(x) - 1) + 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-3-9 x-6 x^2+12 x^3+\left (-1-3 x-2 x^2+4 x^3\right ) \log (4)+\left (6 x+12 x^2+\left (2 x+4 x^2\right ) \log (4)\right ) \log (x)+\left (-6 x+6 x^2+\left (-2 x+2 x^2\right ) \log (4)+(6 x+2 x \log (4)) \log (x)\right ) \log (-1+x+\log (x))}{-2 x+2 x^3+\left (2 x+2 x^2\right ) \log (x)+\left (-x+x^2+x \log (x)\right ) \log (-1+x+\log (x))} \, dx=2 \, x {\left (2 \, \log \left (2\right ) + 3\right )} - {\left (2 \, \log \left (2\right ) + 3\right )} \log \left (2 \, x + \log \left (x + \log \left (x\right ) - 1\right ) + 2\right ) \]

[In]

integrate((((4*x*log(2)+6*x)*log(x)+2*(2*x^2-2*x)*log(2)+6*x^2-6*x)*log(-1+log(x)+x)+(2*(4*x^2+2*x)*log(2)+12*
x^2+6*x)*log(x)+2*(4*x^3-2*x^2-3*x-1)*log(2)+12*x^3-6*x^2-9*x-3)/((x*log(x)+x^2-x)*log(-1+log(x)+x)+(2*x^2+2*x
)*log(x)+2*x^3-2*x),x, algorithm="giac")

[Out]

2*x*(2*log(2) + 3) - (2*log(2) + 3)*log(2*x + log(x + log(x) - 1) + 2)

Mupad [F(-1)]

Timed out. \[ \int \frac {-3-9 x-6 x^2+12 x^3+\left (-1-3 x-2 x^2+4 x^3\right ) \log (4)+\left (6 x+12 x^2+\left (2 x+4 x^2\right ) \log (4)\right ) \log (x)+\left (-6 x+6 x^2+\left (-2 x+2 x^2\right ) \log (4)+(6 x+2 x \log (4)) \log (x)\right ) \log (-1+x+\log (x))}{-2 x+2 x^3+\left (2 x+2 x^2\right ) \log (x)+\left (-x+x^2+x \log (x)\right ) \log (-1+x+\log (x))} \, dx=-\int \frac {9\,x+2\,\ln \left (2\right )\,\left (-4\,x^3+2\,x^2+3\,x+1\right )+6\,x^2-12\,x^3-\ln \left (x\right )\,\left (6\,x+2\,\ln \left (2\right )\,\left (4\,x^2+2\,x\right )+12\,x^2\right )+\ln \left (x+\ln \left (x\right )-1\right )\,\left (6\,x+2\,\ln \left (2\right )\,\left (2\,x-2\,x^2\right )-\ln \left (x\right )\,\left (6\,x+4\,x\,\ln \left (2\right )\right )-6\,x^2\right )+3}{\ln \left (x+\ln \left (x\right )-1\right )\,\left (x\,\ln \left (x\right )-x+x^2\right )-2\,x+\ln \left (x\right )\,\left (2\,x^2+2\,x\right )+2\,x^3} \,d x \]

[In]

int(-(9*x + 2*log(2)*(3*x + 2*x^2 - 4*x^3 + 1) + 6*x^2 - 12*x^3 - log(x)*(6*x + 2*log(2)*(2*x + 4*x^2) + 12*x^
2) + log(x + log(x) - 1)*(6*x + 2*log(2)*(2*x - 2*x^2) - log(x)*(6*x + 4*x*log(2)) - 6*x^2) + 3)/(log(x + log(
x) - 1)*(x*log(x) - x + x^2) - 2*x + log(x)*(2*x + 2*x^2) + 2*x^3),x)

[Out]

-int((9*x + 2*log(2)*(3*x + 2*x^2 - 4*x^3 + 1) + 6*x^2 - 12*x^3 - log(x)*(6*x + 2*log(2)*(2*x + 4*x^2) + 12*x^
2) + log(x + log(x) - 1)*(6*x + 2*log(2)*(2*x - 2*x^2) - log(x)*(6*x + 4*x*log(2)) - 6*x^2) + 3)/(log(x + log(
x) - 1)*(x*log(x) - x + x^2) - 2*x + log(x)*(2*x + 2*x^2) + 2*x^3), x)