\(\int \frac {(1+e^5-2 x) \log ^2(1+e^5-2 x)+e^{\frac {x}{\log (1+e^5-2 x)}} (-2 x^2 \log (\frac {3 x}{2})+(-x-e^5 x+2 x^2) \log (1+e^5-2 x) \log (\frac {3 x}{2}))}{(x+e^5 x-2 x^2) \log ^2(1+e^5-2 x) \log (\frac {3 x}{2})} \, dx\) [922]

Optimal result

Integrand size = 114, antiderivative size = 27 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=\log \left (e^{-e^{\frac {x}{\log \left (1+e^5-2 x\right )}}} \log \left (\frac {3 x}{2}\right )\right ) \] Output:

ln(ln(3/2*x)/exp(exp(x/ln(exp(5)+1-2*x))))
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=-e^{\frac {x}{\log \left (1+e^5-2 x\right )}}+\log \left (\log \left (\frac {3 x}{2}\right )\right ) \] Input:

Integrate[((1 + E^5 - 2*x)*Log[1 + E^5 - 2*x]^2 + E^(x/Log[1 + E^5 - 2*x]) 
*(-2*x^2*Log[(3*x)/2] + (-x - E^5*x + 2*x^2)*Log[1 + E^5 - 2*x]*Log[(3*x)/ 
2]))/((x + E^5*x - 2*x^2)*Log[1 + E^5 - 2*x]^2*Log[(3*x)/2]),x]
 

Output:

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

Rubi [A] (verified)

Time = 2.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6, 2026, 7293, 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.

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 2026

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[((1 + E^5 - 2*x)*Log[1 + E^5 - 2*x]^2 + E^(x/Log[1 + E^5 - 2*x])*(-2*x 
^2*Log[(3*x)/2] + (-x - E^5*x + 2*x^2)*Log[1 + E^5 - 2*x]*Log[(3*x)/2]))/( 
(x + E^5*x - 2*x^2)*Log[1 + E^5 - 2*x]^2*Log[(3*x)/2]),x]
 

Output:

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

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 2009

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

rule 2026

Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 7293

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

Maple [A] (verified)

Time = 17.87 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
default	\(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\)	\(22\)
risch	\(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\)	\(22\)
parallelrisch	\(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\)	\(22\)
parts	\(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\)	\(22\)

Input:

int((((-x*exp(5)+2*x^2-x)*ln(3/2*x)*ln(exp(5)+1-2*x)-2*x^2*ln(3/2*x))*exp( 
x/ln(exp(5)+1-2*x))+(exp(5)+1-2*x)*ln(exp(5)+1-2*x)^2)/(x*exp(5)-2*x^2+x)/ 
ln(3/2*x)/ln(exp(5)+1-2*x)^2,x,method=_RETURNVERBOSE)
 

Output:

ln(ln(3/2*x))-exp(x/ln(exp(5)+1-2*x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=-e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )} + \log \left (\log \left (\frac {3}{2} \, x\right )\right ) \] Input:

integrate((((-x*exp(5)+2*x^2-x)*log(3/2*x)*log(exp(5)+1-2*x)-2*x^2*log(3/2 
*x))*exp(x/log(exp(5)+1-2*x))+(exp(5)+1-2*x)*log(exp(5)+1-2*x)^2)/(x*exp(5 
)-2*x^2+x)/log(3/2*x)/log(exp(5)+1-2*x)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((((-x*exp(5)+2*x**2-x)*ln(3/2*x)*ln(exp(5)+1-2*x)-2*x**2*ln(3/2* 
x))*exp(x/ln(exp(5)+1-2*x))+(exp(5)+1-2*x)*ln(exp(5)+1-2*x)**2)/(x*exp(5)- 
2*x**2+x)/ln(3/2*x)/ln(exp(5)+1-2*x)**2,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=-e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )} + \log \left (\log \left (3\right ) - \log \left (2\right ) + \log \left (x\right )\right ) \] Input:

integrate((((-x*exp(5)+2*x^2-x)*log(3/2*x)*log(exp(5)+1-2*x)-2*x^2*log(3/2 
*x))*exp(x/log(exp(5)+1-2*x))+(exp(5)+1-2*x)*log(exp(5)+1-2*x)^2)/(x*exp(5 
)-2*x^2+x)/log(3/2*x)/log(exp(5)+1-2*x)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=\int { \frac {{\left (2 \, x - e^{5} - 1\right )} \log \left (-2 \, x + e^{5} + 1\right )^{2} + {\left (2 \, x^{2} \log \left (\frac {3}{2} \, x\right ) - {\left (2 \, x^{2} - x e^{5} - x\right )} \log \left (\frac {3}{2} \, x\right ) \log \left (-2 \, x + e^{5} + 1\right )\right )} e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )}}{{\left (2 \, x^{2} - x e^{5} - x\right )} \log \left (\frac {3}{2} \, x\right ) \log \left (-2 \, x + e^{5} + 1\right )^{2}} \,d x } \] Input:

integrate((((-x*exp(5)+2*x^2-x)*log(3/2*x)*log(exp(5)+1-2*x)-2*x^2*log(3/2 
*x))*exp(x/log(exp(5)+1-2*x))+(exp(5)+1-2*x)*log(exp(5)+1-2*x)^2)/(x*exp(5 
)-2*x^2+x)/log(3/2*x)/log(exp(5)+1-2*x)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=\ln \left (\ln \left (\frac {3\,x}{2}\right )\right )-{\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^5-2\,x+1\right )}} \] Input:

int((log(exp(5) - 2*x + 1)^2*(exp(5) - 2*x + 1) - exp(x/log(exp(5) - 2*x + 
 1))*(2*x^2*log((3*x)/2) + log(exp(5) - 2*x + 1)*log((3*x)/2)*(x + x*exp(5 
) - 2*x^2)))/(log(exp(5) - 2*x + 1)^2*log((3*x)/2)*(x + x*exp(5) - 2*x^2)) 
,x)
 

Output:

log(log((3*x)/2)) - exp(x/log(exp(5) - 2*x + 1))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=-e^{\frac {x}{\mathrm {log}\left (e^{5}-2 x +1\right )}}+\mathrm {log}\left (\mathrm {log}\left (\frac {3 x}{2}\right )\right ) \] Input:

int((((-x*exp(5)+2*x^2-x)*log(3/2*x)*log(exp(5)+1-2*x)-2*x^2*log(3/2*x))*e 
xp(x/log(exp(5)+1-2*x))+(exp(5)+1-2*x)*log(exp(5)+1-2*x)^2)/(x*exp(5)-2*x^ 
2+x)/log(3/2*x)/log(exp(5)+1-2*x)^2,x)
 

Output:

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