3.11.25 \(\int \frac {e^x x^2+e^5 (-1-x^2)+(e^x x+e^5 (1-4 x+e^3 x-x^2)) \log (\frac {-e^x x+e^5 (-1+4 x-e^3 x+x^2)}{x}) \log (\log (\frac {-e^x x+e^5 (-1+4 x-e^3 x+x^2)}{x}))}{(e^x x+e^5 (1-4 x+e^3 x-x^2)) \log (\frac {-e^x x+e^5 (-1+4 x-e^3 x+x^2)}{x})} \, dx\) [1025]

3.11.25.1 Optimal result

Integrand size = 166, antiderivative size = 27 \begin {dmath*} \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log \left (\log \left (-e^x+e^5 \left (4-e^3-\frac {1}{x}+x\right )\right )\right ) \end {dmath*}

output

x*ln(ln(exp(5)*(x+4-exp(3)-1/x)-exp(x)))
 

3.11.25.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log \left (\log \left (-e^8-e^x+e^5 \left (4-\frac {1}{x}+x\right )\right )\right ) \end {dmath*}

input

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

output

x*Log[Log[-E^8 - E^x + E^5*(4 - x^(-1) + x)]]
 

3.11.25.3 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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^5 \left (x^3+\left (3-e^3\right ) x^2-x-1\right )}{\left (-e^5 x^2+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5\right ) \log \left (e^5 \left (x-\frac {1}{x}+4\right )-e^x-e^8\right )}+\frac {x+\log \left (e^5 \left (x-\frac {1}{x}+4\right )-e^x-e^8\right ) \log \left (\log \left (e^5 \left (x-\frac {1}{x}+4\right )-e^x-e^8\right )\right )}{\log \left (e^5 \left (x-\frac {1}{x}+4\right )-e^x-e^8\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle e^5 \left (3-e^3\right ) \int \frac {x^2}{\left (-e^5 x^2+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5\right ) \log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+e^5 \int \frac {1}{\left (e^5 x^2-e^x x+4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5\right ) \log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+e^5 \int \frac {x}{\left (e^5 x^2-e^x x+4 e^5 \left (1-\frac {e^3}{4}\right ) x-e^5\right ) \log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+e^5 \int \frac {x^3}{\left (-e^5 x^2+e^x x-4 e^5 \left (1-\frac {e^3}{4}\right ) x+e^5\right ) \log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+\int \frac {x}{\log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )}dx+\int \log \left (\log \left (e^5 \left (x+4-\frac {1}{x}\right )-e^x-e^8\right )\right )dx\)

input

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

output

$Aborted
 

3.11.25.3.1 Defintions of rubi rules used

rule 2009

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

rule 7292

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

rule 7293

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

3.11.25.4 Maple [A] (verified)

Time = 16.68 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15

method	result	size
parallelrisch	\(\ln \left (\ln \left (\frac {-{\mathrm e}^{x} x +\left (-x \,{\mathrm e}^{3}+x^{2}+4 x -1\right ) {\mathrm e}^{5}}{x}\right )\right ) x\)	\(31\)
risch	\(x \ln \left (i \pi -\ln \left (x \right )+\ln \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )\right )\right )}{2}+i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right )}^{2} \left (\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +\left (x \,{\mathrm e}^{3}-x^{2}-4 x +1\right ) {\mathrm e}^{5}\right )}{x}\right )-1\right )\right )\)	\(232\)

input

int(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*ln((-exp(x)*x+(-x*exp(3)+x^2+4 
*x-1)*exp(5))/x)*ln(ln((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x))+exp(x) 
*x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/ln((-exp(x)*x 
+(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x,method=_RETURNVERBOSE)
 

output

ln(ln((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x))*x
 

3.11.25.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \begin {dmath*} \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log \left (\log \left (-\frac {x e^{8} - {\left (x^{2} + 4 \, x - 1\right )} e^{5} + x e^{x}}{x}\right )\right ) \end {dmath*}

input

integrate(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*log((-exp(x)*x+(-x*exp(3 
)+x^2+4*x-1)*exp(5))/x)*log(log((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x 
))+exp(x)*x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/log( 
(-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x, algorithm=\
 

output

x*log(log(-(x*e^8 - (x^2 + 4*x - 1)*e^5 + x*e^x)/x))
 

3.11.25.6 Sympy [A] (verification not implemented)

Time = 4.97 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log {\left (\log {\left (\frac {- x e^{x} + \left (x^{2} - x e^{3} + 4 x - 1\right ) e^{5}}{x} \right )} \right )} \end {dmath*}

input

integrate(((exp(x)*x+(x*exp(3)-x**2-4*x+1)*exp(5))*ln((-exp(x)*x+(-x*exp(3 
)+x**2+4*x-1)*exp(5))/x)*ln(ln((-exp(x)*x+(-x*exp(3)+x**2+4*x-1)*exp(5))/x 
))+exp(x)*x**2+(-x**2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x**2-4*x+1)*exp(5))/l 
n((-exp(x)*x+(-x*exp(3)+x**2+4*x-1)*exp(5))/x),x)
 

output

x*log(log((-x*exp(x) + (x**2 - x*exp(3) + 4*x - 1)*exp(5))/x))
 

3.11.25.7 Maxima [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \begin {dmath*} \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x \log \left (\log \left (x^{2} e^{5} - x {\left (e^{8} - 4 \, e^{5}\right )} - x e^{x} - e^{5}\right ) - \log \left (x\right )\right ) \end {dmath*}

input

integrate(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*log((-exp(x)*x+(-x*exp(3 
)+x^2+4*x-1)*exp(5))/x)*log(log((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x 
))+exp(x)*x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/log( 
(-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x, algorithm=\
 

output

x*log(log(x^2*e^5 - x*(e^8 - 4*e^5) - x*e^x - e^5) - log(x))
 

3.11.25.8 Giac [F]

\begin {dmath*} \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=\int { -\frac {x^{2} e^{x} - {\left ({\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}\right )} \log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right ) \log \left (\log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right )\right ) - {\left (x^{2} + 1\right )} e^{5}}{{\left ({\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}\right )} \log \left (\frac {{\left (x^{2} - x e^{3} + 4 \, x - 1\right )} e^{5} - x e^{x}}{x}\right )} \,d x } \end {dmath*}

input

integrate(((exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))*log((-exp(x)*x+(-x*exp(3 
)+x^2+4*x-1)*exp(5))/x)*log(log((-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x 
))+exp(x)*x^2+(-x^2-1)*exp(5))/(exp(x)*x+(x*exp(3)-x^2-4*x+1)*exp(5))/log( 
(-exp(x)*x+(-x*exp(3)+x^2+4*x-1)*exp(5))/x),x, algorithm=\
 

output

integrate(-(x^2*e^x - ((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)*log(((x^2 - x* 
e^3 + 4*x - 1)*e^5 - x*e^x)/x)*log(log(((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^ 
x)/x)) - (x^2 + 1)*e^5)/(((x^2 - x*e^3 + 4*x - 1)*e^5 - x*e^x)*log(((x^2 - 
 x*e^3 + 4*x - 1)*e^5 - x*e^x)/x)), x)
 

3.11.25.9 Mupad [B] (verification not implemented)

Time = 19.90 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \begin {dmath*} \int \frac {e^x x^2+e^5 \left (-1-x^2\right )+\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right ) \log \left (\log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )\right )}{\left (e^x x+e^5 \left (1-4 x+e^3 x-x^2\right )\right ) \log \left (\frac {-e^x x+e^5 \left (-1+4 x-e^3 x+x^2\right )}{x}\right )} \, dx=x\,\ln \left (\ln \left (\frac {{\mathrm {e}}^5\,\left (4\,x-x\,{\mathrm {e}}^3+x^2-1\right )-x\,{\mathrm {e}}^x}{x}\right )\right ) \end {dmath*}

input

int((exp(5)*(x^2 + 1) - x^2*exp(x) + log(log((exp(5)*(4*x - x*exp(3) + x^2 
 - 1) - x*exp(x))/x))*log((exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp(x))/x 
)*(exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp(x)))/(log((exp(5)*(4*x - x*ex 
p(3) + x^2 - 1) - x*exp(x))/x)*(exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp( 
x))),x)
 

output

x*log(log((exp(5)*(4*x - x*exp(3) + x^2 - 1) - x*exp(x))/x))