3.9.33 \(\int \frac {(e^x \log (\log (e^{x+x^2}-e^x x)))^{2/9} (-4-4 x+e^{x^2} (4+8 x)+(4 e^{x^2}-4 x) \log (e^{x+x^2}-e^x x) \log (\log (e^{x+x^2}-e^x x)))}{(9 e^{x^2}-9 x) \log (e^{x+x^2}-e^x x) \log (\log (e^{x+x^2}-e^x x))} \, dx\) [833]

3.9.33.1 Optimal result

Integrand size = 132, antiderivative size = 25 \[ \int \frac {\left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9} \left (-4-4 x+e^{x^2} (4+8 x)+\left (4 e^{x^2}-4 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )}{\left (9 e^{x^2}-9 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx=2 \left (e^x \log \left (\log \left (e^x \left (e^{x^2}-x\right )\right )\right )\right )^{2/9} \]

output

2*(ln(ln((exp(x^2)-x)*exp(x)))*exp(x))^(2/9)
 

3.9.33.2 Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9} \left (-4-4 x+e^{x^2} (4+8 x)+\left (4 e^{x^2}-4 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )}{\left (9 e^{x^2}-9 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx=2 \left (e^x \log \left (\log \left (e^x \left (e^{x^2}-x\right )\right )\right )\right )^{2/9} \]

input

Integrate[((E^x*Log[Log[E^(x + x^2) - E^x*x]])^(2/9)*(-4 - 4*x + E^x^2*(4 
+ 8*x) + (4*E^x^2 - 4*x)*Log[E^(x + x^2) - E^x*x]*Log[Log[E^(x + x^2) - E^ 
x*x]]))/((9*E^x^2 - 9*x)*Log[E^(x + x^2) - E^x*x]*Log[Log[E^(x + x^2) - E^ 
x*x]]),x]
 

output

2*(E^x*Log[Log[E^x*(E^x^2 - x)]])^(2/9)
 

3.9.33.3 Rubi [A] (verified)

Time = 2.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {7270, 27, 2717, 2726}

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.

\(\displaystyle \int \frac {\left (e^x \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )\right )^{2/9} \left (e^{x^2} (8 x+4)+\left (4 e^{x^2}-4 x\right ) \log \left (e^{x^2+x}-e^x x\right ) \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )-4 x-4\right )}{\left (9 e^{x^2}-9 x\right ) \log \left (e^{x^2+x}-e^x x\right ) \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )} \, dx\)

\(\Big \downarrow \) 7270

\(\displaystyle \frac {\left (e^x \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )\right )^{2/9} \int -\frac {4 \left (e^x\right )^{2/9} \left (x-e^{x^2} (2 x+1)-\left (e^{x^2}-x\right ) \log \left (e^{x^2+x}-e^x x\right ) \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )+1\right )}{9 \left (e^{x^2}-x\right ) \log \left (e^{x^2+x}-e^x x\right ) \log ^{\frac {7}{9}}\left (\log \left (e^{x^2+x}-e^x x\right )\right )}dx}{\left (e^x\right )^{2/9} \log ^{\frac {2}{9}}\left (\log \left (e^{x^2+x}-e^x x\right )\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {4 \left (e^x \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )\right )^{2/9} \int \frac {\left (e^x\right )^{2/9} \left (x-e^{x^2} (2 x+1)-\left (e^{x^2}-x\right ) \log \left (e^{x^2+x}-e^x x\right ) \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )+1\right )}{\left (e^{x^2}-x\right ) \log \left (e^{x^2+x}-e^x x\right ) \log ^{\frac {7}{9}}\left (\log \left (e^{x^2+x}-e^x x\right )\right )}dx}{9 \left (e^x\right )^{2/9} \log ^{\frac {2}{9}}\left (\log \left (e^{x^2+x}-e^x x\right )\right )}\)

\(\Big \downarrow \) 2717

\(\displaystyle -\frac {4 e^{-2 x/9} \left (e^x \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )\right )^{2/9} \int \frac {e^{2 x/9} \left (x-e^{x^2} (2 x+1)-\left (e^{x^2}-x\right ) \log \left (e^{x^2+x}-e^x x\right ) \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )+1\right )}{\left (e^{x^2}-x\right ) \log \left (e^{x^2+x}-e^x x\right ) \log ^{\frac {7}{9}}\left (\log \left (e^{x^2+x}-e^x x\right )\right )}dx}{9 \log ^{\frac {2}{9}}\left (\log \left (e^{x^2+x}-e^x x\right )\right )}\)

\(\Big \downarrow \) 2726

\(\displaystyle 2 \left (e^x \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )\right )^{2/9}\)

input

Int[((E^x*Log[Log[E^(x + x^2) - E^x*x]])^(2/9)*(-4 - 4*x + E^x^2*(4 + 8*x) 
 + (4*E^x^2 - 4*x)*Log[E^(x + x^2) - E^x*x]*Log[Log[E^(x + x^2) - E^x*x]]) 
)/((9*E^x^2 - 9*x)*Log[E^(x + x^2) - E^x*x]*Log[Log[E^(x + x^2) - E^x*x]]) 
,x]
 

output

2*(E^x*Log[Log[E^(x + x^2) - E^x*x]])^(2/9)
 

3.9.33.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2717

Int[(u_.)*((a_.)*(F_)^(v_))^(n_), x_Symbol] :> Simp[(a*F^v)^n/F^(n*v)   Int 
[u*F^(n*v), x], x] /; FreeQ[{F, a, n}, x] &&  !IntegerQ[n]
 

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

rule 7270

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p 
]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p])))   Int[u*v 
^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !Free 
Q[v, x] &&  !FreeQ[w, x]
 

3.9.33.4 Maple [F]

\[\int \frac {\left (\left (4 \,{\mathrm e}^{x^{2}}-4 x \right ) \ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right ) \ln \left (\ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right )\right )+\left (8 x +4\right ) {\mathrm e}^{x^{2}}-4 x -4\right ) \left ({\mathrm e}^{x} \ln \left (\ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right )\right )\right )^{\frac {2}{9}}}{\left (9 \,{\mathrm e}^{x^{2}}-9 x \right ) \ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right ) \ln \left (\ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right )\right )}d x\]

input

int(((4*exp(x^2)-4*x)*ln(exp(x)*exp(x^2)-exp(x)*x)*ln(ln(exp(x)*exp(x^2)-e 
xp(x)*x))+(8*x+4)*exp(x^2)-4*x-4)*(exp(x)*ln(ln(exp(x)*exp(x^2)-exp(x)*x)) 
)^(2/9)/(9*exp(x^2)-9*x)/ln(exp(x)*exp(x^2)-exp(x)*x)/ln(ln(exp(x)*exp(x^2 
)-exp(x)*x)),x)
 

output

int(((4*exp(x^2)-4*x)*ln(exp(x)*exp(x^2)-exp(x)*x)*ln(ln(exp(x)*exp(x^2)-e 
xp(x)*x))+(8*x+4)*exp(x^2)-4*x-4)*(exp(x)*ln(ln(exp(x)*exp(x^2)-exp(x)*x)) 
)^(2/9)/(9*exp(x^2)-9*x)/ln(exp(x)*exp(x^2)-exp(x)*x)/ln(ln(exp(x)*exp(x^2 
)-exp(x)*x)),x)
 

3.9.33.5 Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9} \left (-4-4 x+e^{x^2} (4+8 x)+\left (4 e^{x^2}-4 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )}{\left (9 e^{x^2}-9 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx=\text {Exception raised: TypeError} \]

input

integrate(((4*exp(x^2)-4*x)*log(exp(x)*exp(x^2)-exp(x)*x)*log(log(exp(x)*e 
xp(x^2)-exp(x)*x))+(8*x+4)*exp(x^2)-4*x-4)*(exp(x)*log(log(exp(x)*exp(x^2) 
-exp(x)*x)))^(2/9)/(9*exp(x^2)-9*x)/log(exp(x)*exp(x^2)-exp(x)*x)/log(log( 
exp(x)*exp(x^2)-exp(x)*x)),x, algorithm=\
 

output

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

3.9.33.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9} \left (-4-4 x+e^{x^2} (4+8 x)+\left (4 e^{x^2}-4 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )}{\left (9 e^{x^2}-9 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx=\text {Timed out} \]

input

integrate(((4*exp(x**2)-4*x)*ln(exp(x)*exp(x**2)-exp(x)*x)*ln(ln(exp(x)*ex 
p(x**2)-exp(x)*x))+(8*x+4)*exp(x**2)-4*x-4)*(exp(x)*ln(ln(exp(x)*exp(x**2) 
-exp(x)*x)))**(2/9)/(9*exp(x**2)-9*x)/ln(exp(x)*exp(x**2)-exp(x)*x)/ln(ln( 
exp(x)*exp(x**2)-exp(x)*x)),x)
 

output

Timed out
 

3.9.33.7 Maxima [F]

\[ \int \frac {\left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9} \left (-4-4 x+e^{x^2} (4+8 x)+\left (4 e^{x^2}-4 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )}{\left (9 e^{x^2}-9 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx=\int { \frac {4 \, {\left ({\left (x - e^{\left (x^{2}\right )}\right )} \log \left (-x e^{x} + e^{\left (x^{2} + x\right )}\right ) \log \left (\log \left (-x e^{x} + e^{\left (x^{2} + x\right )}\right )\right ) - {\left (2 \, x + 1\right )} e^{\left (x^{2}\right )} + x + 1\right )} e^{\left (\frac {2}{9} \, x\right )}}{9 \, {\left (x - e^{\left (x^{2}\right )}\right )} \log \left (-x e^{x} + e^{\left (x^{2} + x\right )}\right ) \log \left (\log \left (-x e^{x} + e^{\left (x^{2} + x\right )}\right )\right )^{\frac {7}{9}}} \,d x } \]

input

integrate(((4*exp(x^2)-4*x)*log(exp(x)*exp(x^2)-exp(x)*x)*log(log(exp(x)*e 
xp(x^2)-exp(x)*x))+(8*x+4)*exp(x^2)-4*x-4)*(exp(x)*log(log(exp(x)*exp(x^2) 
-exp(x)*x)))^(2/9)/(9*exp(x^2)-9*x)/log(exp(x)*exp(x^2)-exp(x)*x)/log(log( 
exp(x)*exp(x^2)-exp(x)*x)),x, algorithm=\
 

output

4/9*integrate(((x - e^(x^2))*log(-x*e^x + e^(x^2 + x))*log(log(-x*e^x + e^ 
(x^2 + x))) - (2*x + 1)*e^(x^2) + x + 1)*e^(2/9*x)/((x - e^(x^2))*log(-x*e 
^x + e^(x^2 + x))*log(log(-x*e^x + e^(x^2 + x)))^(7/9)), x)
 

3.9.33.8 Giac [F]

\[ \int \frac {\left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9} \left (-4-4 x+e^{x^2} (4+8 x)+\left (4 e^{x^2}-4 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )}{\left (9 e^{x^2}-9 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx=\int { \frac {4 \, {\left ({\left (x - e^{\left (x^{2}\right )}\right )} \log \left (-x e^{x} + e^{\left (x^{2} + x\right )}\right ) \log \left (\log \left (-x e^{x} + e^{\left (x^{2} + x\right )}\right )\right ) - {\left (2 \, x + 1\right )} e^{\left (x^{2}\right )} + x + 1\right )} e^{\left (\frac {2}{9} \, x\right )}}{9 \, {\left (x - e^{\left (x^{2}\right )}\right )} \log \left (-x e^{x} + e^{\left (x^{2} + x\right )}\right ) \log \left (\log \left (-x e^{x} + e^{\left (x^{2} + x\right )}\right )\right )^{\frac {7}{9}}} \,d x } \]

input

integrate(((4*exp(x^2)-4*x)*log(exp(x)*exp(x^2)-exp(x)*x)*log(log(exp(x)*e 
xp(x^2)-exp(x)*x))+(8*x+4)*exp(x^2)-4*x-4)*(exp(x)*log(log(exp(x)*exp(x^2) 
-exp(x)*x)))^(2/9)/(9*exp(x^2)-9*x)/log(exp(x)*exp(x^2)-exp(x)*x)/log(log( 
exp(x)*exp(x^2)-exp(x)*x)),x, algorithm=\
 

output

integrate(4/9*((x - e^(x^2))*log(-x*e^x + e^(x^2 + x))*log(log(-x*e^x + e^ 
(x^2 + x))) - (2*x + 1)*e^(x^2) + x + 1)*e^(2/9*x)/((x - e^(x^2))*log(-x*e 
^x + e^(x^2 + x))*log(log(-x*e^x + e^(x^2 + x)))^(7/9)), x)
 

3.9.33.9 Mupad [B] (verification not implemented)

Time = 9.61 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {\left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9} \left (-4-4 x+e^{x^2} (4+8 x)+\left (4 e^{x^2}-4 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )}{\left (9 e^{x^2}-9 x\right ) \log \left (e^{x+x^2}-e^x x\right ) \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx=2\,{\mathrm {e}}^{\frac {2\,x}{9}}\,{\ln \left (\ln \left ({\mathrm {e}}^{x^2}\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x\right )\right )}^{2/9} \]

input

int(((exp(x)*log(log(exp(x^2)*exp(x) - x*exp(x))))^(2/9)*(4*x - exp(x^2)*( 
8*x + 4) + log(log(exp(x^2)*exp(x) - x*exp(x)))*log(exp(x^2)*exp(x) - x*ex 
p(x))*(4*x - 4*exp(x^2)) + 4))/(log(log(exp(x^2)*exp(x) - x*exp(x)))*log(e 
xp(x^2)*exp(x) - x*exp(x))*(9*x - 9*exp(x^2))),x)
 

output

2*exp((2*x)/9)*log(log(exp(x^2)*exp(x) - x*exp(x)))^(2/9)