\(\int \frac {e (-30 x^4-5 x^5)+(x^3+30 x^5+5 x^6) \log ^2(x)+(e (60 x^2+10 x^3)+(-x-60 x^3-10 x^4) \log ^2(x)) \log (2 x)+(e (-30-5 x)+(30 x+5 x^2) \log ^2(x)) \log ^2(2 x)+(6+x-12 x^2-2 x^3) \log ^2(x) \log (6+x)}{(30 x^5+5 x^6) \log ^2(x)+(-60 x^3-10 x^4) \log ^2(x) \log (2 x)+(30 x+5 x^2) \log ^2(x) \log ^2(2 x)} \, dx\) [2964]

Optimal result

Integrand size = 182, antiderivative size = 28 \[ \int \frac {e \left (-30 x^4-5 x^5\right )+\left (x^3+30 x^5+5 x^6\right ) \log ^2(x)+\left (e \left (60 x^2+10 x^3\right )+\left (-x-60 x^3-10 x^4\right ) \log ^2(x)\right ) \log (2 x)+\left (e (-30-5 x)+\left (30 x+5 x^2\right ) \log ^2(x)\right ) \log ^2(2 x)+\left (6+x-12 x^2-2 x^3\right ) \log ^2(x) \log (6+x)}{\left (30 x^5+5 x^6\right ) \log ^2(x)+\left (-60 x^3-10 x^4\right ) \log ^2(x) \log (2 x)+\left (30 x+5 x^2\right ) \log ^2(x) \log ^2(2 x)} \, dx=x+\frac {e}{\log (x)}+\frac {\log (6+x)}{5 \left (x^2-\log (2 x)\right )} \] Output:

exp(1)/ln(x)+1/5*ln(6+x)/(x^2-ln(2*x))+x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e \left (-30 x^4-5 x^5\right )+\left (x^3+30 x^5+5 x^6\right ) \log ^2(x)+\left (e \left (60 x^2+10 x^3\right )+\left (-x-60 x^3-10 x^4\right ) \log ^2(x)\right ) \log (2 x)+\left (e (-30-5 x)+\left (30 x+5 x^2\right ) \log ^2(x)\right ) \log ^2(2 x)+\left (6+x-12 x^2-2 x^3\right ) \log ^2(x) \log (6+x)}{\left (30 x^5+5 x^6\right ) \log ^2(x)+\left (-60 x^3-10 x^4\right ) \log ^2(x) \log (2 x)+\left (30 x+5 x^2\right ) \log ^2(x) \log ^2(2 x)} \, dx=\frac {1}{5} \left (5 x+\frac {5 e}{\log (x)}-\frac {\log (6+x)}{-x^2+\log (2 x)}\right ) \] Input:

Integrate[(E*(-30*x^4 - 5*x^5) + (x^3 + 30*x^5 + 5*x^6)*Log[x]^2 + (E*(60* 
x^2 + 10*x^3) + (-x - 60*x^3 - 10*x^4)*Log[x]^2)*Log[2*x] + (E*(-30 - 5*x) 
 + (30*x + 5*x^2)*Log[x]^2)*Log[2*x]^2 + (6 + x - 12*x^2 - 2*x^3)*Log[x]^2 
*Log[6 + x])/((30*x^5 + 5*x^6)*Log[x]^2 + (-60*x^3 - 10*x^4)*Log[x]^2*Log[ 
2*x] + (30*x + 5*x^2)*Log[x]^2*Log[2*x]^2),x]
 

Output:

(5*x + (5*E)/Log[x] - Log[6 + x]/(-x^2 + Log[2*x]))/5
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (\left (5 x^2+30 x\right ) \log ^2(x)+e (-5 x-30)\right ) \log ^2(2 x)+e \left (-5 x^5-30 x^4\right )+\left (-2 x^3-12 x^2+x+6\right ) \log (x+6) \log ^2(x)+\left (5 x^6+30 x^5+x^3\right ) \log ^2(x)+\left (\left (-10 x^4-60 x^3-x\right ) \log ^2(x)+e \left (10 x^3+60 x^2\right )\right ) \log (2 x)}{\left (5 x^2+30 x\right ) \log ^2(2 x) \log ^2(x)+\left (5 x^6+30 x^5\right ) \log ^2(x)+\left (-10 x^4-60 x^3\right ) \log (2 x) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {\frac {5 x^6+30 x^5+x^3-\left (10 x^4+60 x^3+x\right ) \log (2 x)+\left (-2 x^3-12 x^2+x+6\right ) \log (x+6)+5 (x+6) x \log ^2(2 x)}{(x+6) \left (x^2-\log (2 x)\right )^2}-\frac {5 e}{\log ^2(x)}}{5 x}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int -\frac {\frac {5 e}{\log ^2(x)}-\frac {5 x^6+30 x^5+x^3+5 (x+6) \log ^2(2 x) x-\left (10 x^4+60 x^3+x\right ) \log (2 x)+\left (-2 x^3-12 x^2+x+6\right ) \log (x+6)}{(x+6) \left (x^2-\log (2 x)\right )^2}}{x}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{5} \int \frac {\frac {5 e}{\log ^2(x)}-\frac {5 x^6+30 x^5+x^3+5 (x+6) \log ^2(2 x) x-\left (10 x^4+60 x^3+x\right ) \log (2 x)+\left (-2 x^3-12 x^2+x+6\right ) \log (x+6)}{(x+6) \left (x^2-\log (2 x)\right )^2}}{x}dx\)

\(\Big \downarrow \) 2010

\(\displaystyle -\frac {1}{5} \int \left (-\frac {5 x^5}{(x+6) \left (x^2-\log (2 x)\right )^2}-\frac {30 x^4}{(x+6) \left (x^2-\log (2 x)\right )^2}-\frac {x^2}{(x+6) \left (x^2-\log (2 x)\right )^2}-\frac {5 \log ^2(2 x)}{\left (x^2-\log (2 x)\right )^2}+\frac {\left (10 x^3+60 x^2+1\right ) \log (2 x)}{(x+6) \left (x^2-\log (2 x)\right )^2}+\frac {\left (2 x^2-1\right ) \log (x+6)}{\left (x^2-\log (2 x)\right )^2 x}+\frac {5 e}{\log ^2(x) x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} \left (\int \frac {1}{(x+6) \left (x^2-\log (2 x)\right )}dx+\int \frac {\log (x+6)}{x \left (x^2-\log (2 x)\right )^2}dx-2 \int \frac {x \log (x+6)}{\left (x^2-\log (2 x)\right )^2}dx+5 x+\frac {5 e}{\log (x)}\right )\)

Input:

Int[(E*(-30*x^4 - 5*x^5) + (x^3 + 30*x^5 + 5*x^6)*Log[x]^2 + (E*(60*x^2 + 
10*x^3) + (-x - 60*x^3 - 10*x^4)*Log[x]^2)*Log[2*x] + (E*(-30 - 5*x) + (30 
*x + 5*x^2)*Log[x]^2)*Log[2*x]^2 + (6 + x - 12*x^2 - 2*x^3)*Log[x]^2*Log[6 
 + x])/((30*x^5 + 5*x^6)*Log[x]^2 + (-60*x^3 - 10*x^4)*Log[x]^2*Log[2*x] + 
 (30*x + 5*x^2)*Log[x]^2*Log[2*x]^2),x]
 

Output:

$Aborted
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.41 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43

\[\frac {2 i \ln \left (6+x \right )}{5 \left (2 i x^{2}-2 i \ln \left (2\right )-2 i \ln \left (x \right )\right )}+\frac {x \ln \left (x \right )+{\mathrm e}}{\ln \left (x \right )}\]

Input:

int((((5*x^2+30*x)*ln(x)^2+(-5*x-30)*exp(1))*ln(2*x)^2+((-10*x^4-60*x^3-x) 
*ln(x)^2+(10*x^3+60*x^2)*exp(1))*ln(2*x)+(-2*x^3-12*x^2+x+6)*ln(x)^2*ln(6+ 
x)+(5*x^6+30*x^5+x^3)*ln(x)^2+(-5*x^5-30*x^4)*exp(1))/((5*x^2+30*x)*ln(x)^ 
2*ln(2*x)^2+(-10*x^4-60*x^3)*ln(x)^2*ln(2*x)+(5*x^6+30*x^5)*ln(x)^2),x)
 

Output:

2/5*I/(2*I*x^2-2*I*ln(2)-2*I*ln(x))*ln(6+x)+(x*ln(x)+exp(1))/ln(x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (27) = 54\).

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int \frac {e \left (-30 x^4-5 x^5\right )+\left (x^3+30 x^5+5 x^6\right ) \log ^2(x)+\left (e \left (60 x^2+10 x^3\right )+\left (-x-60 x^3-10 x^4\right ) \log ^2(x)\right ) \log (2 x)+\left (e (-30-5 x)+\left (30 x+5 x^2\right ) \log ^2(x)\right ) \log ^2(2 x)+\left (6+x-12 x^2-2 x^3\right ) \log ^2(x) \log (6+x)}{\left (30 x^5+5 x^6\right ) \log ^2(x)+\left (-60 x^3-10 x^4\right ) \log ^2(x) \log (2 x)+\left (30 x+5 x^2\right ) \log ^2(x) \log ^2(2 x)} \, dx=\frac {5 \, x^{2} e - 5 \, x \log \left (x\right )^{2} - 5 \, e \log \left (2\right ) + 5 \, {\left (x^{3} - x \log \left (2\right ) - e\right )} \log \left (x\right ) + \log \left (x + 6\right ) \log \left (x\right )}{5 \, {\left ({\left (x^{2} - \log \left (2\right )\right )} \log \left (x\right ) - \log \left (x\right )^{2}\right )}} \] Input:

integrate((((5*x^2+30*x)*log(x)^2+(-5*x-30)*exp(1))*log(2*x)^2+((-10*x^4-6 
0*x^3-x)*log(x)^2+(10*x^3+60*x^2)*exp(1))*log(2*x)+(-2*x^3-12*x^2+x+6)*log 
(x)^2*log(6+x)+(5*x^6+30*x^5+x^3)*log(x)^2+(-5*x^5-30*x^4)*exp(1))/((5*x^2 
+30*x)*log(x)^2*log(2*x)^2+(-10*x^4-60*x^3)*log(x)^2*log(2*x)+(5*x^6+30*x^ 
5)*log(x)^2),x, algorithm="fricas")
 

Output:

1/5*(5*x^2*e - 5*x*log(x)^2 - 5*e*log(2) + 5*(x^3 - x*log(2) - e)*log(x) + 
 log(x + 6)*log(x))/((x^2 - log(2))*log(x) - log(x)^2)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {e \left (-30 x^4-5 x^5\right )+\left (x^3+30 x^5+5 x^6\right ) \log ^2(x)+\left (e \left (60 x^2+10 x^3\right )+\left (-x-60 x^3-10 x^4\right ) \log ^2(x)\right ) \log (2 x)+\left (e (-30-5 x)+\left (30 x+5 x^2\right ) \log ^2(x)\right ) \log ^2(2 x)+\left (6+x-12 x^2-2 x^3\right ) \log ^2(x) \log (6+x)}{\left (30 x^5+5 x^6\right ) \log ^2(x)+\left (-60 x^3-10 x^4\right ) \log ^2(x) \log (2 x)+\left (30 x+5 x^2\right ) \log ^2(x) \log ^2(2 x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((5*x**2+30*x)*ln(x)**2+(-5*x-30)*exp(1))*ln(2*x)**2+((-10*x**4 
-60*x**3-x)*ln(x)**2+(10*x**3+60*x**2)*exp(1))*ln(2*x)+(-2*x**3-12*x**2+x+ 
6)*ln(x)**2*ln(6+x)+(5*x**6+30*x**5+x**3)*ln(x)**2+(-5*x**5-30*x**4)*exp(1 
))/((5*x**2+30*x)*ln(x)**2*ln(2*x)**2+(-10*x**4-60*x**3)*ln(x)**2*ln(2*x)+ 
(5*x**6+30*x**5)*ln(x)**2),x)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (27) = 54\).

Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int \frac {e \left (-30 x^4-5 x^5\right )+\left (x^3+30 x^5+5 x^6\right ) \log ^2(x)+\left (e \left (60 x^2+10 x^3\right )+\left (-x-60 x^3-10 x^4\right ) \log ^2(x)\right ) \log (2 x)+\left (e (-30-5 x)+\left (30 x+5 x^2\right ) \log ^2(x)\right ) \log ^2(2 x)+\left (6+x-12 x^2-2 x^3\right ) \log ^2(x) \log (6+x)}{\left (30 x^5+5 x^6\right ) \log ^2(x)+\left (-60 x^3-10 x^4\right ) \log ^2(x) \log (2 x)+\left (30 x+5 x^2\right ) \log ^2(x) \log ^2(2 x)} \, dx=\frac {5 \, x^{2} e - 5 \, x \log \left (x\right )^{2} - 5 \, e \log \left (2\right ) + 5 \, {\left (x^{3} - x \log \left (2\right ) - e\right )} \log \left (x\right ) + \log \left (x + 6\right ) \log \left (x\right )}{5 \, {\left ({\left (x^{2} - \log \left (2\right )\right )} \log \left (x\right ) - \log \left (x\right )^{2}\right )}} \] Input:

integrate((((5*x^2+30*x)*log(x)^2+(-5*x-30)*exp(1))*log(2*x)^2+((-10*x^4-6 
0*x^3-x)*log(x)^2+(10*x^3+60*x^2)*exp(1))*log(2*x)+(-2*x^3-12*x^2+x+6)*log 
(x)^2*log(6+x)+(5*x^6+30*x^5+x^3)*log(x)^2+(-5*x^5-30*x^4)*exp(1))/((5*x^2 
+30*x)*log(x)^2*log(2*x)^2+(-10*x^4-60*x^3)*log(x)^2*log(2*x)+(5*x^6+30*x^ 
5)*log(x)^2),x, algorithm="maxima")
 

Output:

1/5*(5*x^2*e - 5*x*log(x)^2 - 5*e*log(2) + 5*(x^3 - x*log(2) - e)*log(x) + 
 log(x + 6)*log(x))/((x^2 - log(2))*log(x) - log(x)^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (27) = 54\).

Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.54 \[ \int \frac {e \left (-30 x^4-5 x^5\right )+\left (x^3+30 x^5+5 x^6\right ) \log ^2(x)+\left (e \left (60 x^2+10 x^3\right )+\left (-x-60 x^3-10 x^4\right ) \log ^2(x)\right ) \log (2 x)+\left (e (-30-5 x)+\left (30 x+5 x^2\right ) \log ^2(x)\right ) \log ^2(2 x)+\left (6+x-12 x^2-2 x^3\right ) \log ^2(x) \log (6+x)}{\left (30 x^5+5 x^6\right ) \log ^2(x)+\left (-60 x^3-10 x^4\right ) \log ^2(x) \log (2 x)+\left (30 x+5 x^2\right ) \log ^2(x) \log ^2(2 x)} \, dx=\frac {5 \, x^{3} \log \left (x\right ) + 5 \, x^{2} e - 5 \, x \log \left (2\right ) \log \left (x\right ) - 5 \, x \log \left (x\right )^{2} - 5 \, e \log \left (2\right ) - 5 \, e \log \left (x\right ) + \log \left (x + 6\right ) \log \left (x\right )}{5 \, {\left (x^{2} \log \left (x\right ) - \log \left (2\right ) \log \left (x\right ) - \log \left (x\right )^{2}\right )}} \] Input:

integrate((((5*x^2+30*x)*log(x)^2+(-5*x-30)*exp(1))*log(2*x)^2+((-10*x^4-6 
0*x^3-x)*log(x)^2+(10*x^3+60*x^2)*exp(1))*log(2*x)+(-2*x^3-12*x^2+x+6)*log 
(x)^2*log(6+x)+(5*x^6+30*x^5+x^3)*log(x)^2+(-5*x^5-30*x^4)*exp(1))/((5*x^2 
+30*x)*log(x)^2*log(2*x)^2+(-10*x^4-60*x^3)*log(x)^2*log(2*x)+(5*x^6+30*x^ 
5)*log(x)^2),x, algorithm="giac")
 

Output:

1/5*(5*x^3*log(x) + 5*x^2*e - 5*x*log(2)*log(x) - 5*x*log(x)^2 - 5*e*log(2 
) - 5*e*log(x) + log(x + 6)*log(x))/(x^2*log(x) - log(2)*log(x) - log(x)^2 
)
 

Mupad [B] (verification not implemented)

Time = 3.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e \left (-30 x^4-5 x^5\right )+\left (x^3+30 x^5+5 x^6\right ) \log ^2(x)+\left (e \left (60 x^2+10 x^3\right )+\left (-x-60 x^3-10 x^4\right ) \log ^2(x)\right ) \log (2 x)+\left (e (-30-5 x)+\left (30 x+5 x^2\right ) \log ^2(x)\right ) \log ^2(2 x)+\left (6+x-12 x^2-2 x^3\right ) \log ^2(x) \log (6+x)}{\left (30 x^5+5 x^6\right ) \log ^2(x)+\left (-60 x^3-10 x^4\right ) \log ^2(x) \log (2 x)+\left (30 x+5 x^2\right ) \log ^2(x) \log ^2(2 x)} \, dx=x+\frac {\mathrm {e}}{\ln \left (x\right )}-\frac {\ln \left (x+6\right )}{5\,\left (\ln \left (2\,x\right )-x^2\right )} \] Input:

int((log(x)^2*(x^3 + 30*x^5 + 5*x^6) + log(2*x)^2*(log(x)^2*(30*x + 5*x^2) 
 - exp(1)*(5*x + 30)) + log(2*x)*(exp(1)*(60*x^2 + 10*x^3) - log(x)^2*(x + 
 60*x^3 + 10*x^4)) - exp(1)*(30*x^4 + 5*x^5) + log(x + 6)*log(x)^2*(x - 12 
*x^2 - 2*x^3 + 6))/(log(x)^2*(30*x^5 + 5*x^6) + log(2*x)^2*log(x)^2*(30*x 
+ 5*x^2) - log(2*x)*log(x)^2*(60*x^3 + 10*x^4)),x)
 

Output:

x + exp(1)/log(x) - log(x + 6)/(5*(log(2*x) - x^2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.00 \[ \int \frac {e \left (-30 x^4-5 x^5\right )+\left (x^3+30 x^5+5 x^6\right ) \log ^2(x)+\left (e \left (60 x^2+10 x^3\right )+\left (-x-60 x^3-10 x^4\right ) \log ^2(x)\right ) \log (2 x)+\left (e (-30-5 x)+\left (30 x+5 x^2\right ) \log ^2(x)\right ) \log ^2(2 x)+\left (6+x-12 x^2-2 x^3\right ) \log ^2(x) \log (6+x)}{\left (30 x^5+5 x^6\right ) \log ^2(x)+\left (-60 x^3-10 x^4\right ) \log ^2(x) \log (2 x)+\left (30 x+5 x^2\right ) \log ^2(x) \log ^2(2 x)} \, dx=\frac {-\mathrm {log}\left (x +6\right ) \mathrm {log}\left (x \right )+5 \,\mathrm {log}\left (2 x \right ) \mathrm {log}\left (x \right ) x +5 \,\mathrm {log}\left (2 x \right ) e -5 \,\mathrm {log}\left (x \right ) x^{3}-5 e \,x^{2}}{5 \,\mathrm {log}\left (x \right ) \left (\mathrm {log}\left (2 x \right )-x^{2}\right )} \] Input:

int((((5*x^2+30*x)*log(x)^2+(-5*x-30)*exp(1))*log(2*x)^2+((-10*x^4-60*x^3- 
x)*log(x)^2+(10*x^3+60*x^2)*exp(1))*log(2*x)+(-2*x^3-12*x^2+x+6)*log(x)^2* 
log(6+x)+(5*x^6+30*x^5+x^3)*log(x)^2+(-5*x^5-30*x^4)*exp(1))/((5*x^2+30*x) 
*log(x)^2*log(2*x)^2+(-10*x^4-60*x^3)*log(x)^2*log(2*x)+(5*x^6+30*x^5)*log 
(x)^2),x)
 

Output:

( - log(x + 6)*log(x) + 5*log(2*x)*log(x)*x + 5*log(2*x)*e - 5*log(x)*x**3 
 - 5*e*x**2)/(5*log(x)*(log(2*x) - x**2))