\(\int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} (-10-2 x-20 x^2-6 x^3)+(-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3) \log (2 x)-x \log ^2(2 x)}{x} \, dx\) [1260]

Optimal result
Mathematica [B] (verified)
Rubi [B] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [B] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 99, antiderivative size = 23 \[ \int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} \left (-10-2 x-20 x^2-6 x^3\right )+\left (-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3\right ) \log (2 x)-x \log ^2(2 x)}{x} \, dx=(-5-x) \left (e^{2+e^3}+x^2+\log (2 x)\right )^2 \] Output:

(-x-5)*(x^2+ln(2*x)+exp(2+exp(3)))^2
                                                                                    
                                                                                    
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(76\) vs. \(2(23)=46\).

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.30 \[ \int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} \left (-10-2 x-20 x^2-6 x^3\right )+\left (-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3\right ) \log (2 x)-x \log ^2(2 x)}{x} \, dx=-x \left (e^{4+2 e^3}+2 e^{2+e^3} x (5+x)+x^3 (5+x)\right )-10 e^{2+e^3} \log (x)-2 x \left (e^{2+e^3}+x (5+x)\right ) \log (2 x)-(5+x) \log ^2(2 x) \] Input:

Integrate[(-(E^(4 + 2*E^3)*x) - 10*x^2 - 2*x^3 - 20*x^4 - 5*x^5 + E^(2 + E 
^3)*(-10 - 2*x - 20*x^2 - 6*x^3) + (-10 - 2*x - 2*E^(2 + E^3)*x - 20*x^2 - 
 6*x^3)*Log[2*x] - x*Log[2*x]^2)/x,x]
 

Output:

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(160\) vs. \(2(23)=46\).

Time = 0.53 (sec) , antiderivative size = 160, normalized size of antiderivative = 6.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {2010, 2009}

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.

\(\displaystyle \int \frac {-5 x^5-20 x^4-2 x^3-10 x^2+e^{2+e^3} \left (-6 x^3-20 x^2-2 x-10\right )+\left (-6 x^3-20 x^2-2 e^{2+e^3} x-2 x-10\right ) \log (2 x)-e^{4+2 e^3} x-x \log ^2(2 x)}{x} \, dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (\frac {2 \left (-3 x^3-10 x^2-\left (1+e^{2+e^3}\right ) x-5\right ) \log (2 x)}{x}+\frac {-5 x^5-20 x^4-2 \left (1+3 e^{2+e^3}\right ) x^3-10 \left (1+2 e^{2+e^3}\right ) x^2-2 e^{2+e^3} \left (1+\frac {e^{2+e^3}}{2}\right ) x-10 e^{2+e^3}}{x}-\log ^2(2 x)\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(-(E^(4 + 2*E^3)*x) - 10*x^2 - 2*x^3 - 20*x^4 - 5*x^5 + E^(2 + E^3)*(- 
10 - 2*x - 20*x^2 - 6*x^3) + (-10 - 2*x - 2*E^(2 + E^3)*x - 20*x^2 - 6*x^3 
)*Log[2*x] - x*Log[2*x]^2)/x,x]
 

Output:

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

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(86\) vs. \(2(21)=42\).

Time = 1.50 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.78

method result size
risch \(\left (-x -5\right ) \ln \left (2 x \right )^{2}+\left (-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-2 x^{3}-10 x^{2}\right ) \ln \left (2 x \right )-x \,{\mathrm e}^{4+2 \,{\mathrm e}^{3}}-2 \,{\mathrm e}^{2+{\mathrm e}^{3}} x^{3}-10 \,{\mathrm e}^{2+{\mathrm e}^{3}} x^{2}-x^{5}-5 x^{4}-10 \ln \left (x \right ) {\mathrm e}^{2+{\mathrm e}^{3}}\) \(87\)
parallelrisch \(-x^{5}-2 \,{\mathrm e}^{2+{\mathrm e}^{3}} x^{3}-5 x^{4}-2 \ln \left (2 x \right ) x^{3}-x \,{\mathrm e}^{4+2 \,{\mathrm e}^{3}}-10 \,{\mathrm e}^{2+{\mathrm e}^{3}} x^{2}-2 \ln \left (2 x \right ) {\mathrm e}^{2+{\mathrm e}^{3}} x -10 x^{2} \ln \left (2 x \right )-x \ln \left (2 x \right )^{2}-10 \ln \left (2 x \right ) {\mathrm e}^{2+{\mathrm e}^{3}}-5 \ln \left (2 x \right )^{2}\) \(100\)
norman \(-10 \ln \left (2 x \right ) {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}-5 x^{4}-x^{5}-5 \ln \left (2 x \right )^{2}-x \ln \left (2 x \right )^{2}-10 x^{2} \ln \left (2 x \right )-2 \ln \left (2 x \right ) x^{3}-x \,{\mathrm e}^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}-10 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{2}-2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{3}-2 x \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} \ln \left (2 x \right )\) \(102\)
parts \(-x \ln \left (2 x \right )^{2}-x^{5}-2 \,{\mathrm e}^{2+{\mathrm e}^{3}} x^{3}-5 x^{4}-10 \,{\mathrm e}^{2+{\mathrm e}^{3}} x^{2}-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-x \,{\mathrm e}^{4+2 \,{\mathrm e}^{3}}-10 \ln \left (x \right ) {\mathrm e}^{2+{\mathrm e}^{3}}-2 \ln \left (2 x \right ) x^{3}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} \left (2 x \ln \left (2 x \right )-2 x \right )-10 x^{2} \ln \left (2 x \right )-5 \ln \left (2 x \right )^{2}\) \(112\)
derivativedivides \(-x^{5}-2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{3}-2 \ln \left (2 x \right ) x^{3}-5 x^{4}-x \,{\mathrm e}^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} \left (2 x \ln \left (2 x \right )-2 x \right )-10 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{2}-x \ln \left (2 x \right )^{2}-10 x^{2} \ln \left (2 x \right )-2 x \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}-10 \ln \left (2 x \right ) {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}-5 \ln \left (2 x \right )^{2}\) \(116\)
default \(-x^{5}-2 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{3}-2 \ln \left (2 x \right ) x^{3}-5 x^{4}-x \,{\mathrm e}^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}-{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} \left (2 x \ln \left (2 x \right )-2 x \right )-10 \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}} x^{2}-x \ln \left (2 x \right )^{2}-10 x^{2} \ln \left (2 x \right )-2 x \,{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}-10 \ln \left (2 x \right ) {\mathrm e}^{2} {\mathrm e}^{{\mathrm e}^{3}}-5 \ln \left (2 x \right )^{2}\) \(116\)
orering \(\frac {\left (5+x \right ) \left (-x \ln \left (2 x \right )^{2}+\left (-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-6 x^{3}-20 x^{2}-2 x -10\right ) \ln \left (2 x \right )-x \,{\mathrm e}^{4+2 \,{\mathrm e}^{3}}+\left (-6 x^{3}-20 x^{2}-2 x -10\right ) {\mathrm e}^{2+{\mathrm e}^{3}}-5 x^{5}-20 x^{4}-2 x^{3}-10 x^{2}\right )}{x}-\frac {x \left (48 x^{9}+600 x^{8}+2448 x^{7}+3540 x^{6}+1812 x^{5}-500 x^{2}-2800 x -375\right ) \left (\frac {-\ln \left (2 x \right )^{2}-2 \ln \left (2 x \right )+\left (-2 \,{\mathrm e}^{2+{\mathrm e}^{3}}-18 x^{2}-40 x -2\right ) \ln \left (2 x \right )+\frac {-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-6 x^{3}-20 x^{2}-2 x -10}{x}-{\mathrm e}^{4+2 \,{\mathrm e}^{3}}+\left (-18 x^{2}-40 x -2\right ) {\mathrm e}^{2+{\mathrm e}^{3}}-25 x^{4}-80 x^{3}-6 x^{2}-20 x}{x}-\frac {-x \ln \left (2 x \right )^{2}+\left (-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-6 x^{3}-20 x^{2}-2 x -10\right ) \ln \left (2 x \right )-x \,{\mathrm e}^{4+2 \,{\mathrm e}^{3}}+\left (-6 x^{3}-20 x^{2}-2 x -10\right ) {\mathrm e}^{2+{\mathrm e}^{3}}-5 x^{5}-20 x^{4}-2 x^{3}-10 x^{2}}{x^{2}}\right )}{120 x^{8}+480 x^{7}+676 x^{6}-80 x^{5}-490 x^{4}-200 x^{3}-349 x^{2}-20 x +25}+\frac {\left (8 x^{9}+120 x^{8}+612 x^{7}+1180 x^{6}+906 x^{5}-2199 x^{3}+500 x^{2}+1400 x +125\right ) x^{2} \left (\frac {-\frac {2 \ln \left (2 x \right )}{x}-\frac {2}{x}+\left (-36 x -40\right ) \ln \left (2 x \right )+\frac {-4 \,{\mathrm e}^{2+{\mathrm e}^{3}}-36 x^{2}-80 x -4}{x}-\frac {-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-6 x^{3}-20 x^{2}-2 x -10}{x^{2}}+\left (-36 x -40\right ) {\mathrm e}^{2+{\mathrm e}^{3}}-100 x^{3}-240 x^{2}-12 x -20}{x}-\frac {2 \left (-\ln \left (2 x \right )^{2}-2 \ln \left (2 x \right )+\left (-2 \,{\mathrm e}^{2+{\mathrm e}^{3}}-18 x^{2}-40 x -2\right ) \ln \left (2 x \right )+\frac {-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-6 x^{3}-20 x^{2}-2 x -10}{x}-{\mathrm e}^{4+2 \,{\mathrm e}^{3}}+\left (-18 x^{2}-40 x -2\right ) {\mathrm e}^{2+{\mathrm e}^{3}}-25 x^{4}-80 x^{3}-6 x^{2}-20 x \right )}{x^{2}}+\frac {-2 x \ln \left (2 x \right )^{2}+2 \left (-2 x \,{\mathrm e}^{2+{\mathrm e}^{3}}-6 x^{3}-20 x^{2}-2 x -10\right ) \ln \left (2 x \right )-2 x \,{\mathrm e}^{4+2 \,{\mathrm e}^{3}}+2 \left (-6 x^{3}-20 x^{2}-2 x -10\right ) {\mathrm e}^{2+{\mathrm e}^{3}}-10 x^{5}-40 x^{4}-4 x^{3}-20 x^{2}}{x^{3}}\right )}{\left (2 x^{2}+1\right ) \left (60 x^{6}+240 x^{5}+308 x^{4}-160 x^{3}-399 x^{2}-20 x +25\right )}\) \(780\)

Input:

int((-x*ln(2*x)^2+(-2*x*exp(2+exp(3))-6*x^3-20*x^2-2*x-10)*ln(2*x)-x*exp(2 
+exp(3))^2+(-6*x^3-20*x^2-2*x-10)*exp(2+exp(3))-5*x^5-20*x^4-2*x^3-10*x^2) 
/x,x,method=_RETURNVERBOSE)
 

Output:

(-x-5)*ln(2*x)^2+(-2*x*exp(2+exp(3))-2*x^3-10*x^2)*ln(2*x)-x*exp(4+2*exp(3 
))-2*exp(2+exp(3))*x^3-10*exp(2+exp(3))*x^2-x^5-5*x^4-10*ln(x)*exp(2+exp(3 
))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (20) = 40\).

Time = 0.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.13 \[ \int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} \left (-10-2 x-20 x^2-6 x^3\right )+\left (-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3\right ) \log (2 x)-x \log ^2(2 x)}{x} \, dx=-x^{5} - 5 \, x^{4} - {\left (x + 5\right )} \log \left (2 \, x\right )^{2} - x e^{\left (2 \, e^{3} + 4\right )} - 2 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (e^{3} + 2\right )} - 2 \, {\left (x^{3} + 5 \, x^{2} + {\left (x + 5\right )} e^{\left (e^{3} + 2\right )}\right )} \log \left (2 \, x\right ) \] Input:

integrate((-x*log(2*x)^2+(-2*x*exp(2+exp(3))-6*x^3-20*x^2-2*x-10)*log(2*x) 
-x*exp(2+exp(3))^2+(-6*x^3-20*x^2-2*x-10)*exp(2+exp(3))-5*x^5-20*x^4-2*x^3 
-10*x^2)/x,x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (20) = 40\).

Time = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.35 \[ \int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} \left (-10-2 x-20 x^2-6 x^3\right )+\left (-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3\right ) \log (2 x)-x \log ^2(2 x)}{x} \, dx=- x^{5} - 5 x^{4} - 2 x^{3} e^{2} e^{e^{3}} - 10 x^{2} e^{2} e^{e^{3}} - x e^{4} e^{2 e^{3}} + \left (- x - 5\right ) \log {\left (2 x \right )}^{2} + \left (- 2 x^{3} - 10 x^{2} - 2 x e^{2} e^{e^{3}}\right ) \log {\left (2 x \right )} - 10 e^{2} e^{e^{3}} \log {\left (x \right )} \] Input:

integrate((-x*ln(2*x)**2+(-2*x*exp(2+exp(3))-6*x**3-20*x**2-2*x-10)*ln(2*x 
)-x*exp(2+exp(3))**2+(-6*x**3-20*x**2-2*x-10)*exp(2+exp(3))-5*x**5-20*x**4 
-2*x**3-10*x**2)/x,x)
 

Output:

-x**5 - 5*x**4 - 2*x**3*exp(2)*exp(exp(3)) - 10*x**2*exp(2)*exp(exp(3)) - 
x*exp(4)*exp(2*exp(3)) + (-x - 5)*log(2*x)**2 + (-2*x**3 - 10*x**2 - 2*x*e 
xp(2)*exp(exp(3)))*log(2*x) - 10*exp(2)*exp(exp(3))*log(x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (20) = 40\).

Time = 0.03 (sec) , antiderivative size = 128, normalized size of antiderivative = 5.57 \[ \int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} \left (-10-2 x-20 x^2-6 x^3\right )+\left (-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3\right ) \log (2 x)-x \log ^2(2 x)}{x} \, dx=-x^{5} - 5 \, x^{4} - 2 \, x^{3} e^{\left (e^{3} + 2\right )} - 2 \, x^{3} \log \left (2 \, x\right ) - 10 \, x^{2} e^{\left (e^{3} + 2\right )} - 10 \, x^{2} \log \left (2 \, x\right ) - {\left (\log \left (2 \, x\right )^{2} - 2 \, \log \left (2 \, x\right ) + 2\right )} x - x e^{\left (2 \, e^{3} + 4\right )} - 2 \, {\left (x \log \left (2 \, x\right ) - x\right )} e^{\left (e^{3} + 2\right )} - 2 \, x e^{\left (e^{3} + 2\right )} - 2 \, x \log \left (2 \, x\right ) - 5 \, \log \left (2 \, x\right )^{2} - 10 \, e^{\left (e^{3} + 2\right )} \log \left (x\right ) + 2 \, x \] Input:

integrate((-x*log(2*x)^2+(-2*x*exp(2+exp(3))-6*x^3-20*x^2-2*x-10)*log(2*x) 
-x*exp(2+exp(3))^2+(-6*x^3-20*x^2-2*x-10)*exp(2+exp(3))-5*x^5-20*x^4-2*x^3 
-10*x^2)/x,x, algorithm="maxima")
 

Output:

-x^5 - 5*x^4 - 2*x^3*e^(e^3 + 2) - 2*x^3*log(2*x) - 10*x^2*e^(e^3 + 2) - 1 
0*x^2*log(2*x) - (log(2*x)^2 - 2*log(2*x) + 2)*x - x*e^(2*e^3 + 4) - 2*(x* 
log(2*x) - x)*e^(e^3 + 2) - 2*x*e^(e^3 + 2) - 2*x*log(2*x) - 5*log(2*x)^2 
- 10*e^(e^3 + 2)*log(x) + 2*x
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (20) = 40\).

Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.22 \[ \int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} \left (-10-2 x-20 x^2-6 x^3\right )+\left (-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3\right ) \log (2 x)-x \log ^2(2 x)}{x} \, dx=-x^{5} - 5 \, x^{4} - 2 \, x^{3} e^{\left (e^{3} + 2\right )} - 2 \, x^{3} \log \left (2 \, x\right ) - 10 \, x^{2} e^{\left (e^{3} + 2\right )} - 10 \, x^{2} \log \left (2 \, x\right ) - 2 \, x e^{\left (e^{3} + 2\right )} \log \left (2 \, x\right ) - x \log \left (2 \, x\right )^{2} - x e^{\left (2 \, e^{3} + 4\right )} - 5 \, \log \left (2 \, x\right )^{2} - 10 \, e^{\left (e^{3} + 2\right )} \log \left (x\right ) \] Input:

integrate((-x*log(2*x)^2+(-2*x*exp(2+exp(3))-6*x^3-20*x^2-2*x-10)*log(2*x) 
-x*exp(2+exp(3))^2+(-6*x^3-20*x^2-2*x-10)*exp(2+exp(3))-5*x^5-20*x^4-2*x^3 
-10*x^2)/x,x, algorithm="giac")
 

Output:

-x^5 - 5*x^4 - 2*x^3*e^(e^3 + 2) - 2*x^3*log(2*x) - 10*x^2*e^(e^3 + 2) - 1 
0*x^2*log(2*x) - 2*x*e^(e^3 + 2)*log(2*x) - x*log(2*x)^2 - x*e^(2*e^3 + 4) 
 - 5*log(2*x)^2 - 10*e^(e^3 + 2)*log(x)
 

Mupad [B] (verification not implemented)

Time = 3.01 (sec) , antiderivative size = 94, normalized size of antiderivative = 4.09 \[ \int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} \left (-10-2 x-20 x^2-6 x^3\right )+\left (-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3\right ) \log (2 x)-x \log ^2(2 x)}{x} \, dx=-x\,\left ({\ln \left (2\,x\right )}^2+2\,{\mathrm {e}}^{{\mathrm {e}}^3+2}\,\ln \left (2\,x\right )+{\mathrm {e}}^{2\,{\mathrm {e}}^3+4}\right )-10\,{\mathrm {e}}^{{\mathrm {e}}^3+2}\,\ln \left (x\right )-5\,{\ln \left (2\,x\right )}^2-x^3\,\left (2\,{\mathrm {e}}^{{\mathrm {e}}^3+2}+2\,\ln \left (2\,x\right )\right )-x^2\,\left (10\,{\mathrm {e}}^{{\mathrm {e}}^3+2}+10\,\ln \left (2\,x\right )\right )-5\,x^4-x^5 \] Input:

int(-(exp(exp(3) + 2)*(2*x + 20*x^2 + 6*x^3 + 10) + log(2*x)*(2*x + 2*x*ex 
p(exp(3) + 2) + 20*x^2 + 6*x^3 + 10) + x*exp(2*exp(3) + 4) + x*log(2*x)^2 
+ 10*x^2 + 2*x^3 + 20*x^4 + 5*x^5)/x,x)
 

Output:

- x*(exp(2*exp(3) + 4) + log(2*x)^2 + 2*log(2*x)*exp(exp(3) + 2)) - 10*exp 
(exp(3) + 2)*log(x) - 5*log(2*x)^2 - x^3*(2*exp(exp(3) + 2) + 2*log(2*x)) 
- x^2*(10*exp(exp(3) + 2) + 10*log(2*x)) - 5*x^4 - x^5
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.96 \[ \int \frac {-e^{4+2 e^3} x-10 x^2-2 x^3-20 x^4-5 x^5+e^{2+e^3} \left (-10-2 x-20 x^2-6 x^3\right )+\left (-10-2 x-2 e^{2+e^3} x-20 x^2-6 x^3\right ) \log (2 x)-x \log ^2(2 x)}{x} \, dx=-e^{2 e^{3}} e^{4} x -2 e^{e^{3}} \mathrm {log}\left (2 x \right ) e^{2} x -10 e^{e^{3}} \mathrm {log}\left (2 x \right ) e^{2}-2 e^{e^{3}} e^{2} x^{3}-10 e^{e^{3}} e^{2} x^{2}-\mathrm {log}\left (2 x \right )^{2} x -5 \mathrm {log}\left (2 x \right )^{2}-2 \,\mathrm {log}\left (2 x \right ) x^{3}-10 \,\mathrm {log}\left (2 x \right ) x^{2}-x^{5}-5 x^{4} \] Input:

int((-x*log(2*x)^2+(-2*x*exp(2+exp(3))-6*x^3-20*x^2-2*x-10)*log(2*x)-x*exp 
(2+exp(3))^2+(-6*x^3-20*x^2-2*x-10)*exp(2+exp(3))-5*x^5-20*x^4-2*x^3-10*x^ 
2)/x,x)
 

Output:

 - e**(2*e**3)*e**4*x - 2*e**(e**3)*log(2*x)*e**2*x - 10*e**(e**3)*log(2*x 
)*e**2 - 2*e**(e**3)*e**2*x**3 - 10*e**(e**3)*e**2*x**2 - log(2*x)**2*x - 
5*log(2*x)**2 - 2*log(2*x)*x**3 - 10*log(2*x)*x**2 - x**5 - 5*x**4