3.11.82 \(\int \frac {e^{\frac {1}{25} (25 x^2+10 x \log (\log (x^2))+\log ^2(\log (x^2)))} (60 x+20 x^2+(-25 x+150 x^2+50 x^3) \log (x^2)+(12+4 x+(30 x+10 x^2) \log (x^2)) \log (\log (x^2)))}{(225 x+150 x^2+25 x^3) \log (x^2)} \, dx\) [1082]

3.11.82.1 Optimal result

Integrand size = 103, antiderivative size = 21 \begin {dmath*} \int \frac {e^{\frac {1}{25} \left (25 x^2+10 x \log \left (\log \left (x^2\right )\right )+\log ^2\left (\log \left (x^2\right )\right )\right )} \left (60 x+20 x^2+\left (-25 x+150 x^2+50 x^3\right ) \log \left (x^2\right )+\left (12+4 x+\left (30 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (225 x+150 x^2+25 x^3\right ) \log \left (x^2\right )} \, dx=\frac {e^{\left (x+\frac {1}{5} \log \left (\log \left (x^2\right )\right )\right )^2}}{3+x} \end {dmath*}

output

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

3.11.82.2 Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \begin {dmath*} \int \frac {e^{\frac {1}{25} \left (25 x^2+10 x \log \left (\log \left (x^2\right )\right )+\log ^2\left (\log \left (x^2\right )\right )\right )} \left (60 x+20 x^2+\left (-25 x+150 x^2+50 x^3\right ) \log \left (x^2\right )+\left (12+4 x+\left (30 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (225 x+150 x^2+25 x^3\right ) \log \left (x^2\right )} \, dx=\frac {e^{x^2+\frac {1}{25} \log ^2\left (\log \left (x^2\right )\right )} \log ^{\frac {2 x}{5}}\left (x^2\right )}{3+x} \end {dmath*}

input

Integrate[(E^((25*x^2 + 10*x*Log[Log[x^2]] + Log[Log[x^2]]^2)/25)*(60*x + 
20*x^2 + (-25*x + 150*x^2 + 50*x^3)*Log[x^2] + (12 + 4*x + (30*x + 10*x^2) 
*Log[x^2])*Log[Log[x^2]]))/((225*x + 150*x^2 + 25*x^3)*Log[x^2]),x]
 

output

(E^(x^2 + Log[Log[x^2]]^2/25)*Log[x^2]^((2*x)/5))/(3 + x)
 

3.11.82.3 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.

\(\displaystyle \int \frac {\left (20 x^2+\left (\left (10 x^2+30 x\right ) \log \left (x^2\right )+4 x+12\right ) \log \left (\log \left (x^2\right )\right )+\left (50 x^3+150 x^2-25 x\right ) \log \left (x^2\right )+60 x\right ) \exp \left (\frac {1}{25} \left (25 x^2+\log ^2\left (\log \left (x^2\right )\right )+10 x \log \left (\log \left (x^2\right )\right )\right )\right )}{\left (25 x^3+150 x^2+225 x\right ) \log \left (x^2\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (20 x^2+\left (\left (10 x^2+30 x\right ) \log \left (x^2\right )+4 x+12\right ) \log \left (\log \left (x^2\right )\right )+\left (50 x^3+150 x^2-25 x\right ) \log \left (x^2\right )+60 x\right ) \exp \left (\frac {1}{25} \left (25 x^2+\log ^2\left (\log \left (x^2\right )\right )+10 x \log \left (\log \left (x^2\right )\right )\right )\right )}{x \left (25 x^2+150 x+225\right ) \log \left (x^2\right )}dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {\left (20 x^2+\left (\left (10 x^2+30 x\right ) \log \left (x^2\right )+4 x+12\right ) \log \left (\log \left (x^2\right )\right )+\left (50 x^3+150 x^2-25 x\right ) \log \left (x^2\right )+60 x\right ) \exp \left (\frac {1}{25} \left (25 x^2+\log ^2\left (\log \left (x^2\right )\right )+10 x \log \left (\log \left (x^2\right )\right )\right )\right )}{x (5 x+15)^2 \log \left (x^2\right )}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {1}{25} \left (\log \left (\log \left (x^2\right )\right )+5 x\right )^2} \left (20 x^2+\left (\left (10 x^2+30 x\right ) \log \left (x^2\right )+4 x+12\right ) \log \left (\log \left (x^2\right )\right )+\left (50 x^3+150 x^2-25 x\right ) \log \left (x^2\right )+60 x\right )}{x (5 x+15)^2 \log \left (x^2\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^{\frac {1}{25} \left (\log \left (\log \left (x^2\right )\right )+5 x\right )^2} \left (10 x^2 \log \left (x^2\right )+30 x \log \left (x^2\right )-5 \log \left (x^2\right )+4 x+12\right )}{5 (x+3)^2 \log \left (x^2\right )}+\frac {2 e^{\frac {1}{25} \left (\log \left (\log \left (x^2\right )\right )+5 x\right )^2} \left (5 x \log \left (x^2\right )+2\right ) \log \left (\log \left (x^2\right )\right )}{25 x (x+3) \log \left (x^2\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \int e^{\frac {1}{25} \left (5 x+\log \left (\log \left (x^2\right )\right )\right )^2}dx-\int \frac {e^{\frac {1}{25} \left (5 x+\log \left (\log \left (x^2\right )\right )\right )^2}}{(x+3)^2}dx-6 \int \frac {e^{\frac {1}{25} \left (5 x+\log \left (\log \left (x^2\right )\right )\right )^2}}{x+3}dx+\frac {4}{5} \int \frac {e^{\frac {1}{25} \left (5 x+\log \left (\log \left (x^2\right )\right )\right )^2}}{(x+3) \log \left (x^2\right )}dx+\frac {2}{5} \int \frac {e^{\frac {1}{25} \left (5 x+\log \left (\log \left (x^2\right )\right )\right )^2} \log \left (\log \left (x^2\right )\right )}{x+3}dx+\frac {4}{75} \int \frac {e^{\frac {1}{25} \left (5 x+\log \left (\log \left (x^2\right )\right )\right )^2} \log \left (\log \left (x^2\right )\right )}{x \log \left (x^2\right )}dx-\frac {4}{75} \int \frac {e^{\frac {1}{25} \left (5 x+\log \left (\log \left (x^2\right )\right )\right )^2} \log \left (\log \left (x^2\right )\right )}{(x+3) \log \left (x^2\right )}dx\)

input

Int[(E^((25*x^2 + 10*x*Log[Log[x^2]] + Log[Log[x^2]]^2)/25)*(60*x + 20*x^2 
 + (-25*x + 150*x^2 + 50*x^3)*Log[x^2] + (12 + 4*x + (30*x + 10*x^2)*Log[x 
^2])*Log[Log[x^2]]))/((225*x + 150*x^2 + 25*x^3)*Log[x^2]),x]
 

output

$Aborted
 

3.11.82.3.1 Defintions of rubi rules used

rule 2007

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, 
x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex 
pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol 
yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
 

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 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] 
]
 

3.11.82.4 Maple [A] (verified)

Time = 1.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{\frac {{\ln \left (\ln \left (x^{2}\right )\right )}^{2}}{25}+\frac {2 x \ln \left (\ln \left (x^{2}\right )\right )}{5}+x^{2}}}{3+x}\)	\(29\)

input

int((((10*x^2+30*x)*ln(x^2)+4*x+12)*ln(ln(x^2))+(50*x^3+150*x^2-25*x)*ln(x 
^2)+20*x^2+60*x)*exp(1/25*ln(ln(x^2))^2+2/5*x*ln(ln(x^2))+x^2)/(25*x^3+150 
*x^2+225*x)/ln(x^2),x,method=_RETURNVERBOSE)
 

output

exp(1/25*ln(ln(x^2))^2+2/5*x*ln(ln(x^2))+x^2)/(3+x)
 

3.11.82.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {e^{\frac {1}{25} \left (25 x^2+10 x \log \left (\log \left (x^2\right )\right )+\log ^2\left (\log \left (x^2\right )\right )\right )} \left (60 x+20 x^2+\left (-25 x+150 x^2+50 x^3\right ) \log \left (x^2\right )+\left (12+4 x+\left (30 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (225 x+150 x^2+25 x^3\right ) \log \left (x^2\right )} \, dx=\frac {e^{\left (x^{2} + \frac {2}{5} \, x \log \left (\log \left (x^{2}\right )\right ) + \frac {1}{25} \, \log \left (\log \left (x^{2}\right )\right )^{2}\right )}}{x + 3} \end {dmath*}

input

integrate((((10*x^2+30*x)*log(x^2)+4*x+12)*log(log(x^2))+(50*x^3+150*x^2-2 
5*x)*log(x^2)+20*x^2+60*x)*exp(1/25*log(log(x^2))^2+2/5*x*log(log(x^2))+x^ 
2)/(25*x^3+150*x^2+225*x)/log(x^2),x, algorithm=\
 

output

e^(x^2 + 2/5*x*log(log(x^2)) + 1/25*log(log(x^2))^2)/(x + 3)
 

3.11.82.6 Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \begin {dmath*} \int \frac {e^{\frac {1}{25} \left (25 x^2+10 x \log \left (\log \left (x^2\right )\right )+\log ^2\left (\log \left (x^2\right )\right )\right )} \left (60 x+20 x^2+\left (-25 x+150 x^2+50 x^3\right ) \log \left (x^2\right )+\left (12+4 x+\left (30 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (225 x+150 x^2+25 x^3\right ) \log \left (x^2\right )} \, dx=\frac {e^{x^{2} + \frac {2 x \log {\left (\log {\left (x^{2} \right )} \right )}}{5} + \frac {\log {\left (\log {\left (x^{2} \right )} \right )}^{2}}{25}}}{x + 3} \end {dmath*}

input

integrate((((10*x**2+30*x)*ln(x**2)+4*x+12)*ln(ln(x**2))+(50*x**3+150*x**2 
-25*x)*ln(x**2)+20*x**2+60*x)*exp(1/25*ln(ln(x**2))**2+2/5*x*ln(ln(x**2))+ 
x**2)/(25*x**3+150*x**2+225*x)/ln(x**2),x)
 

output

exp(x**2 + 2*x*log(log(x**2))/5 + log(log(x**2))**2/25)/(x + 3)
 

3.11.82.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.39 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00 \begin {dmath*} \int \frac {e^{\frac {1}{25} \left (25 x^2+10 x \log \left (\log \left (x^2\right )\right )+\log ^2\left (\log \left (x^2\right )\right )\right )} \left (60 x+20 x^2+\left (-25 x+150 x^2+50 x^3\right ) \log \left (x^2\right )+\left (12+4 x+\left (30 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (225 x+150 x^2+25 x^3\right ) \log \left (x^2\right )} \, dx=\frac {e^{\left (x^{2} + \frac {2}{5} \, x \log \left (2\right ) + \frac {1}{25} \, \log \left (2\right )^{2} + \frac {2}{5} \, x \log \left (\log \left (x\right )\right ) + \frac {2}{25} \, \log \left (2\right ) \log \left (\log \left (x\right )\right ) + \frac {1}{25} \, \log \left (\log \left (x\right )\right )^{2}\right )}}{x + 3} \end {dmath*}

input

integrate((((10*x^2+30*x)*log(x^2)+4*x+12)*log(log(x^2))+(50*x^3+150*x^2-2 
5*x)*log(x^2)+20*x^2+60*x)*exp(1/25*log(log(x^2))^2+2/5*x*log(log(x^2))+x^ 
2)/(25*x^3+150*x^2+225*x)/log(x^2),x, algorithm=\
 

output

e^(x^2 + 2/5*x*log(2) + 1/25*log(2)^2 + 2/5*x*log(log(x)) + 2/25*log(2)*lo 
g(log(x)) + 1/25*log(log(x))^2)/(x + 3)
 

3.11.82.8 Giac [F]

\begin {dmath*} \int \frac {e^{\frac {1}{25} \left (25 x^2+10 x \log \left (\log \left (x^2\right )\right )+\log ^2\left (\log \left (x^2\right )\right )\right )} \left (60 x+20 x^2+\left (-25 x+150 x^2+50 x^3\right ) \log \left (x^2\right )+\left (12+4 x+\left (30 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (225 x+150 x^2+25 x^3\right ) \log \left (x^2\right )} \, dx=\int { \frac {{\left (20 \, x^{2} + 25 \, {\left (2 \, x^{3} + 6 \, x^{2} - x\right )} \log \left (x^{2}\right ) + 2 \, {\left (5 \, {\left (x^{2} + 3 \, x\right )} \log \left (x^{2}\right ) + 2 \, x + 6\right )} \log \left (\log \left (x^{2}\right )\right ) + 60 \, x\right )} e^{\left (x^{2} + \frac {2}{5} \, x \log \left (\log \left (x^{2}\right )\right ) + \frac {1}{25} \, \log \left (\log \left (x^{2}\right )\right )^{2}\right )}}{25 \, {\left (x^{3} + 6 \, x^{2} + 9 \, x\right )} \log \left (x^{2}\right )} \,d x } \end {dmath*}

input

integrate((((10*x^2+30*x)*log(x^2)+4*x+12)*log(log(x^2))+(50*x^3+150*x^2-2 
5*x)*log(x^2)+20*x^2+60*x)*exp(1/25*log(log(x^2))^2+2/5*x*log(log(x^2))+x^ 
2)/(25*x^3+150*x^2+225*x)/log(x^2),x, algorithm=\
 

output

undef
 

3.11.82.9 Mupad [B] (verification not implemented)

Time = 17.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \begin {dmath*} \int \frac {e^{\frac {1}{25} \left (25 x^2+10 x \log \left (\log \left (x^2\right )\right )+\log ^2\left (\log \left (x^2\right )\right )\right )} \left (60 x+20 x^2+\left (-25 x+150 x^2+50 x^3\right ) \log \left (x^2\right )+\left (12+4 x+\left (30 x+10 x^2\right ) \log \left (x^2\right )\right ) \log \left (\log \left (x^2\right )\right )\right )}{\left (225 x+150 x^2+25 x^3\right ) \log \left (x^2\right )} \, dx=\frac {{\ln \left (x^2\right )}^{\frac {2\,x}{5}}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\frac {{\ln \left (\ln \left (x^2\right )\right )}^2}{25}}}{x+3} \end {dmath*}

input

int((exp(log(log(x^2))^2/25 + (2*x*log(log(x^2)))/5 + x^2)*(60*x + log(x^2 
)*(150*x^2 - 25*x + 50*x^3) + log(log(x^2))*(4*x + log(x^2)*(30*x + 10*x^2 
) + 12) + 20*x^2))/(log(x^2)*(225*x + 150*x^2 + 25*x^3)),x)
 

output

(log(x^2)^((2*x)/5)*exp(x^2)*exp(log(log(x^2))^2/25))/(x + 3)