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

Optimal. Leaf size=25 \[ 2 \left (e^x \log \left (\log \left (e^x \left (e^{x^2}-x\right )\right )\right )\right )^{2/9} \]

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Rubi [A]  time = 2.72, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, integrand size = 132, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6719, 2281, 2288} \begin {gather*} 2 \left (e^x \log \left (\log \left (e^{x^2+x}-e^x x\right )\right )\right )^{2/9} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

Rule 2281

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

Rule 2288

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 6719

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9} \int \frac {\left (e^x\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 ^{\frac {7}{9}}\left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx}{\left (e^x\right )^{2/9} \log ^{\frac {2}{9}}\left (\log \left (e^{x+x^2}-e^x x\right )\right )}\\ &=\frac {\left (e^{-2 x/9} \left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9}\right ) \int \frac {e^{2 x/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 ^{\frac {7}{9}}\left (\log \left (e^{x+x^2}-e^x x\right )\right )} \, dx}{\log ^{\frac {2}{9}}\left (\log \left (e^{x+x^2}-e^x x\right )\right )}\\ &=2 \left (e^x \log \left (\log \left (e^{x+x^2}-e^x x\right )\right )\right )^{2/9}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 25, normalized size = 1.00 \begin {gather*} 2 \left (e^x \log \left (\log \left (e^x \left (e^{x^2}-x\right )\right )\right )\right )^{2/9} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x^2)-4*x)*log(exp(x)*exp(x^2)-exp(x)*x)*log(log(exp(x)*exp(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(lo
g(exp(x)*exp(x^2)-exp(x)*x)),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x^2)-4*x)*log(exp(x)*exp(x^2)-exp(x)*x)*log(log(exp(x)*exp(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(lo
g(exp(x)*exp(x^2)-exp(x)*x)),x, algorithm="giac")

[Out]

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)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (4 \,{\mathrm e}^{x^{2}}-4 x \right ) \ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right ) \ln \left (\ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right )\right )+\left (8 x +4\right ) {\mathrm e}^{x^{2}}-4 x -4\right ) \left ({\mathrm e}^{x} \ln \left (\ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right )\right )\right )^{\frac {2}{9}}}{\left (9 \,{\mathrm e}^{x^{2}}-9 x \right ) \ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right ) \ln \left (\ln \left ({\mathrm e}^{x} {\mathrm e}^{x^{2}}-{\mathrm e}^{x} x \right )\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {4}{9} \, \int \frac {{\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 )}}{{\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x^2)-4*x)*log(exp(x)*exp(x^2)-exp(x)*x)*log(log(exp(x)*exp(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(lo
g(exp(x)*exp(x^2)-exp(x)*x)),x, algorithm="maxima")

[Out]

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)

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mupad [B]  time = 1.09, size = 23, normalized size = 0.92 \begin {gather*} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*exp(x))*(4*x - 4*exp(x^2)) + 4))/(log(log(exp(x^2)*exp(x) - x*exp(x)))*log
(exp(x^2)*exp(x) - x*exp(x))*(9*x - 9*exp(x^2))),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x**2)-4*x)*ln(exp(x)*exp(x**2)-exp(x)*x)*ln(ln(exp(x)*exp(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(l
n(exp(x)*exp(x**2)-exp(x)*x)),x)

[Out]

Timed out

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