\(\int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+(-160 x^3-120 x^4) \log (4)+400 x^3 \log ^2(4)+(16 x+24 x^2+9 x^3+(-160 x-120 x^2) \log (4)+400 x \log ^2(4)) \log (x)+(-16 x^3-12 x^4+80 x^3 \log (4)+(-16 x-12 x^2+80 x \log (4)) \log (x)) \log (x^2+\log (x))+(4 x^3+4 x \log (x)) \log ^2(x^2+\log (x))} \, dx\) [2447]

Optimal result

Integrand size = 159, antiderivative size = 29 \[ \int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+400 x^3 \log ^2(4)+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx=\frac {10}{-x+2 \left (2+2 (x-5 \log (4))-\log \left (x^2+\log (x)\right )\right )} \] Output:

10/(3*x-40*ln(2)-2*ln(ln(x)+x^2)+4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+400 x^3 \log ^2(4)+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx=\frac {10}{4+3 x-20 \log (4)-2 \log \left (x^2+\log (x)\right )} \] Input:

Integrate[(20 + 40*x^2 - 30*x^3 - 30*x*Log[x])/(16*x^3 + 24*x^4 + 9*x^5 + 
(-160*x^3 - 120*x^4)*Log[4] + 400*x^3*Log[4]^2 + (16*x + 24*x^2 + 9*x^3 + 
(-160*x - 120*x^2)*Log[4] + 400*x*Log[4]^2)*Log[x] + (-16*x^3 - 12*x^4 + 8 
0*x^3*Log[4] + (-16*x - 12*x^2 + 80*x*Log[4])*Log[x])*Log[x^2 + Log[x]] + 
(4*x^3 + 4*x*Log[x])*Log[x^2 + Log[x]]^2),x]
 

Output:

10/(4 + 3*x - 20*Log[4] - 2*Log[x^2 + Log[x]])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6, 7239, 27, 7237}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {-30 x^3+40 x^2-30 x \log (x)+20}{9 x^5+24 x^4+16 x^3+400 x^3 \log ^2(4)+\left (-120 x^4-160 x^3\right ) \log (4)+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )+\left (9 x^3+24 x^2+\left (-120 x^2-160 x\right ) \log (4)+16 x+400 x \log ^2(4)\right ) \log (x)+\left (-12 x^4-16 x^3+80 x^3 \log (4)+\left (-12 x^2-16 x+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {-30 x^3+40 x^2-30 x \log (x)+20}{9 x^5+24 x^4+x^3 \left (16+400 \log ^2(4)\right )+\left (-120 x^4-160 x^3\right ) \log (4)+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )+\left (9 x^3+24 x^2+\left (-120 x^2-160 x\right ) \log (4)+16 x+400 x \log ^2(4)\right ) \log (x)+\left (-12 x^4-16 x^3+80 x^3 \log (4)+\left (-12 x^2-16 x+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )}dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {10 \left (-3 x^3+4 x^2-3 x \log (x)+2\right )}{x \left (x^2+\log (x)\right ) \left (-2 \log \left (x^2+\log (x)\right )+3 x+4 (1-5 \log (4))\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 10 \int \frac {-3 x^3+4 x^2-3 \log (x) x+2}{x \left (x^2+\log (x)\right ) \left (3 x-2 \log \left (x^2+\log (x)\right )+4 (1-5 \log (4))\right )^2}dx\)

\(\Big \downarrow \) 7237

\(\displaystyle \frac {10}{-2 \log \left (x^2+\log (x)\right )+3 x+4 (1-5 \log (4))}\)

Input:

Int[(20 + 40*x^2 - 30*x^3 - 30*x*Log[x])/(16*x^3 + 24*x^4 + 9*x^5 + (-160* 
x^3 - 120*x^4)*Log[4] + 400*x^3*Log[4]^2 + (16*x + 24*x^2 + 9*x^3 + (-160* 
x - 120*x^2)*Log[4] + 400*x*Log[4]^2)*Log[x] + (-16*x^3 - 12*x^4 + 80*x^3* 
Log[4] + (-16*x - 12*x^2 + 80*x*Log[4])*Log[x])*Log[x^2 + Log[x]] + (4*x^3 
 + 4*x*Log[x])*Log[x^2 + Log[x]]^2),x]
 

Output:

10/(3*x + 4*(1 - 5*Log[4]) - 2*Log[x^2 + Log[x]])
 

Defintions of rubi rules used

rule 6

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 7237

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si 
mp[q*(y^(m + 1)/(m + 1)), x] /;  !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
 

rule 7239

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

Maple [A] (verified)

Time = 3.72 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {10}{40 \ln \left (2\right )+2 \ln \left (\ln \left (x \right )+x^{2}\right )-3 x -4}\)	\(23\)
risch	\(-\frac {10}{40 \ln \left (2\right )+2 \ln \left (\ln \left (x \right )+x^{2}\right )-3 x -4}\)	\(23\)
parallelrisch	\(-\frac {10}{40 \ln \left (2\right )+2 \ln \left (\ln \left (x \right )+x^{2}\right )-3 x -4}\)	\(23\)

Input:

int((-30*x*ln(x)-30*x^3+40*x^2+20)/((4*x*ln(x)+4*x^3)*ln(ln(x)+x^2)^2+((16 
0*x*ln(2)-12*x^2-16*x)*ln(x)+160*x^3*ln(2)-12*x^4-16*x^3)*ln(ln(x)+x^2)+(1 
600*x*ln(2)^2+2*(-120*x^2-160*x)*ln(2)+9*x^3+24*x^2+16*x)*ln(x)+1600*x^3*l 
n(2)^2+2*(-120*x^4-160*x^3)*ln(2)+9*x^5+24*x^4+16*x^3),x,method=_RETURNVER 
BOSE)
 

Output:

-10/(40*ln(2)+2*ln(ln(x)+x^2)-3*x-4)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+400 x^3 \log ^2(4)+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx=\frac {10}{3 \, x - 40 \, \log \left (2\right ) - 2 \, \log \left (x^{2} + \log \left (x\right )\right ) + 4} \] Input:

integrate((-30*x*log(x)-30*x^3+40*x^2+20)/((4*x*log(x)+4*x^3)*log(log(x)+x 
^2)^2+((160*x*log(2)-12*x^2-16*x)*log(x)+160*x^3*log(2)-12*x^4-16*x^3)*log 
(log(x)+x^2)+(1600*x*log(2)^2+2*(-120*x^2-160*x)*log(2)+9*x^3+24*x^2+16*x) 
*log(x)+1600*x^3*log(2)^2+2*(-120*x^4-160*x^3)*log(2)+9*x^5+24*x^4+16*x^3) 
,x, algorithm="fricas")
 

Output:

10/(3*x - 40*log(2) - 2*log(x^2 + log(x)) + 4)
 

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+400 x^3 \log ^2(4)+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx=- \frac {5}{- \frac {3 x}{2} + \log {\left (x^{2} + \log {\left (x \right )} \right )} - 2 + 20 \log {\left (2 \right )}} \] Input:

integrate((-30*x*ln(x)-30*x**3+40*x**2+20)/((4*x*ln(x)+4*x**3)*ln(ln(x)+x* 
*2)**2+((160*x*ln(2)-12*x**2-16*x)*ln(x)+160*x**3*ln(2)-12*x**4-16*x**3)*l 
n(ln(x)+x**2)+(1600*x*ln(2)**2+2*(-120*x**2-160*x)*ln(2)+9*x**3+24*x**2+16 
*x)*ln(x)+1600*x**3*ln(2)**2+2*(-120*x**4-160*x**3)*ln(2)+9*x**5+24*x**4+1 
6*x**3),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+400 x^3 \log ^2(4)+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx=\frac {10}{3 \, x - 40 \, \log \left (2\right ) - 2 \, \log \left (x^{2} + \log \left (x\right )\right ) + 4} \] Input:

integrate((-30*x*log(x)-30*x^3+40*x^2+20)/((4*x*log(x)+4*x^3)*log(log(x)+x 
^2)^2+((160*x*log(2)-12*x^2-16*x)*log(x)+160*x^3*log(2)-12*x^4-16*x^3)*log 
(log(x)+x^2)+(1600*x*log(2)^2+2*(-120*x^2-160*x)*log(2)+9*x^3+24*x^2+16*x) 
*log(x)+1600*x^3*log(2)^2+2*(-120*x^4-160*x^3)*log(2)+9*x^5+24*x^4+16*x^3) 
,x, algorithm="maxima")
 

Output:

10/(3*x - 40*log(2) - 2*log(x^2 + log(x)) + 4)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+400 x^3 \log ^2(4)+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx=\frac {10}{3 \, x - 40 \, \log \left (2\right ) - 2 \, \log \left (x^{2} + \log \left (x\right )\right ) + 4} \] Input:

integrate((-30*x*log(x)-30*x^3+40*x^2+20)/((4*x*log(x)+4*x^3)*log(log(x)+x 
^2)^2+((160*x*log(2)-12*x^2-16*x)*log(x)+160*x^3*log(2)-12*x^4-16*x^3)*log 
(log(x)+x^2)+(1600*x*log(2)^2+2*(-120*x^2-160*x)*log(2)+9*x^3+24*x^2+16*x) 
*log(x)+1600*x^3*log(2)^2+2*(-120*x^4-160*x^3)*log(2)+9*x^5+24*x^4+16*x^3) 
,x, algorithm="giac")
 

Output:

10/(3*x - 40*log(2) - 2*log(x^2 + log(x)) + 4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+400 x^3 \log ^2(4)+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx=-\int \frac {30\,x\,\ln \left (x\right )-40\,x^2+30\,x^3-20}{1600\,x^3\,{\ln \left (2\right )}^2-\ln \left (\ln \left (x\right )+x^2\right )\,\left (16\,x^3-160\,x^3\,\ln \left (2\right )+12\,x^4+\ln \left (x\right )\,\left (16\,x-160\,x\,\ln \left (2\right )+12\,x^2\right )\right )+{\ln \left (\ln \left (x\right )+x^2\right )}^2\,\left (4\,x\,\ln \left (x\right )+4\,x^3\right )-2\,\ln \left (2\right )\,\left (120\,x^4+160\,x^3\right )+\ln \left (x\right )\,\left (16\,x-2\,\ln \left (2\right )\,\left (120\,x^2+160\,x\right )+1600\,x\,{\ln \left (2\right )}^2+24\,x^2+9\,x^3\right )+16\,x^3+24\,x^4+9\,x^5} \,d x \] Input:

int(-(30*x*log(x) - 40*x^2 + 30*x^3 - 20)/(1600*x^3*log(2)^2 - log(log(x) 
+ x^2)*(16*x^3 - 160*x^3*log(2) + 12*x^4 + log(x)*(16*x - 160*x*log(2) + 1 
2*x^2)) + log(log(x) + x^2)^2*(4*x*log(x) + 4*x^3) - 2*log(2)*(160*x^3 + 1 
20*x^4) + log(x)*(16*x - 2*log(2)*(160*x + 120*x^2) + 1600*x*log(2)^2 + 24 
*x^2 + 9*x^3) + 16*x^3 + 24*x^4 + 9*x^5),x)
 

Output:

-int((30*x*log(x) - 40*x^2 + 30*x^3 - 20)/(1600*x^3*log(2)^2 - log(log(x) 
+ x^2)*(16*x^3 - 160*x^3*log(2) + 12*x^4 + log(x)*(16*x - 160*x*log(2) + 1 
2*x^2)) + log(log(x) + x^2)^2*(4*x*log(x) + 4*x^3) - 2*log(2)*(160*x^3 + 1 
20*x^4) + log(x)*(16*x - 2*log(2)*(160*x + 120*x^2) + 1600*x*log(2)^2 + 24 
*x^2 + 9*x^3) + 16*x^3 + 24*x^4 + 9*x^5), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {20+40 x^2-30 x^3-30 x \log (x)}{16 x^3+24 x^4+9 x^5+\left (-160 x^3-120 x^4\right ) \log (4)+400 x^3 \log ^2(4)+\left (16 x+24 x^2+9 x^3+\left (-160 x-120 x^2\right ) \log (4)+400 x \log ^2(4)\right ) \log (x)+\left (-16 x^3-12 x^4+80 x^3 \log (4)+\left (-16 x-12 x^2+80 x \log (4)\right ) \log (x)\right ) \log \left (x^2+\log (x)\right )+\left (4 x^3+4 x \log (x)\right ) \log ^2\left (x^2+\log (x)\right )} \, dx=-\frac {10}{2 \,\mathrm {log}\left (\mathrm {log}\left (x \right )+x^{2}\right )+40 \,\mathrm {log}\left (2\right )-3 x -4} \] Input:

int((-30*x*log(x)-30*x^3+40*x^2+20)/((4*x*log(x)+4*x^3)*log(log(x)+x^2)^2+ 
((160*x*log(2)-12*x^2-16*x)*log(x)+160*x^3*log(2)-12*x^4-16*x^3)*log(log(x 
)+x^2)+(1600*x*log(2)^2+2*(-120*x^2-160*x)*log(2)+9*x^3+24*x^2+16*x)*log(x 
)+1600*x^3*log(2)^2+2*(-120*x^4-160*x^3)*log(2)+9*x^5+24*x^4+16*x^3),x)
 

Output:

( - 10)/(2*log(log(x) + x**2) + 40*log(2) - 3*x - 4)