3.29.81 \(\int \frac {e^{-1-e^x+x+5 x^2+x \log (\log ^2(\log (x)))} (2+(1-e^x+10 x) \log (x) \log (\log (x))+\log (x) \log (\log (x)) \log (\log ^2(\log (x))))}{\log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=25 \[ e^{-1-e^x+x+x^2+x \left (4 x+\log \left (\log ^2(\log (x))\right )\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.80, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 1, number of rules used = 1, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6706} \begin {gather*} e^{5 x^2+x-e^x-1} \log ^2(\log (x))^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-1 - E^x + x + 5*x^2 + x*Log[Log[Log[x]]^2])*(2 + (1 - E^x + 10*x)*Log[x]*Log[Log[x]] + Log[x]*Log[Log
[x]]*Log[Log[Log[x]]^2]))/(Log[x]*Log[Log[x]]),x]

[Out]

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-1-e^x+x+5 x^2} \log ^2(\log (x))^x\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 1.04, size = 23, normalized size = 0.92 \begin {gather*} e^{-1-e^x+x+5 x^2} \log ^2(\log (x))^x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 - E^x + x + 5*x^2 + x*Log[Log[Log[x]]^2])*(2 + (1 - E^x + 10*x)*Log[x]*Log[Log[x]] + Log[x]*L
og[Log[x]]*Log[Log[Log[x]]^2]))/(Log[x]*Log[Log[x]]),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 21, normalized size = 0.84 \begin {gather*} e^{\left (5 \, x^{2} + x \log \left (\log \left (\log \relax (x)\right )^{2}\right ) + x - e^{x} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)*log(log(x))*log(log(log(x))^2)+(-exp(x)+10*x+1)*log(x)*log(log(x))+2)*exp(x*log(log(log(x))^
2)-exp(x)+5*x^2+x-1)/log(x)/log(log(x)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)*log(log(x))*log(log(log(x))^2)+(-exp(x)+10*x+1)*log(x)*log(log(x))+2)*exp(x*log(log(log(x))^
2)-exp(x)+5*x^2+x-1)/log(x)/log(log(x)),x, algorithm="giac")

[Out]

undef

________________________________________________________________________________________

maple [C]  time = 0.53, size = 84, normalized size = 3.36

method	result	size
risch	\(\ln \left (\ln \relax (x )\right )^{2 x} {\mathrm e}^{-1-\frac {i x \pi \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )^{3}}{2}+i x \pi \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right )^{2} \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )\right )-\frac {i x \pi \,\mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )^{2}\right ) \mathrm {csgn}\left (i \ln \left (\ln \relax (x )\right )\right )^{2}}{2}-{\mathrm e}^{x}+5 x^{2}+x}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)*ln(ln(x))*ln(ln(ln(x))^2)+(-exp(x)+10*x+1)*ln(x)*ln(ln(x))+2)*exp(x*ln(ln(ln(x))^2)-exp(x)+5*x^2+x-
1)/ln(x)/ln(ln(x)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x))^(2*x)*exp(-1-1/2*I*x*Pi*csgn(I*ln(ln(x))^2)^3+I*x*Pi*csgn(I*ln(ln(x))^2)^2*csgn(I*ln(ln(x)))-1/2*I*x
*Pi*csgn(I*ln(ln(x))^2)*csgn(I*ln(ln(x)))^2-exp(x)+5*x^2+x)

________________________________________________________________________________________

maxima [A]  time = 0.56, size = 20, normalized size = 0.80 \begin {gather*} e^{\left (5 \, x^{2} + 2 \, x \log \left (\log \left (\log \relax (x)\right )\right ) + x - e^{x} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)*log(log(x))*log(log(log(x))^2)+(-exp(x)+10*x+1)*log(x)*log(log(x))+2)*exp(x*log(log(log(x))^
2)-exp(x)+5*x^2+x-1)/log(x)/log(log(x)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.84, size = 23, normalized size = 0.92 \begin {gather*} {\mathrm {e}}^{-1}\,{\mathrm {e}}^{5\,x^2}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^x\,{\left ({\ln \left (\ln \relax (x)\right )}^2\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - exp(x) + 5*x^2 + x*log(log(log(x))^2) - 1)*(log(log(x))*log(x)*(10*x - exp(x) + 1) + log(log(log(
x))^2)*log(log(x))*log(x) + 2))/(log(log(x))*log(x)),x)

[Out]

exp(-1)*exp(5*x^2)*exp(-exp(x))*exp(x)*(log(log(x))^2)^x

________________________________________________________________________________________

sympy [A]  time = 4.55, size = 22, normalized size = 0.88 \begin {gather*} e^{5 x^{2} + x \log {\left (\log {\left (\log {\relax (x )} \right )}^{2} \right )} + x - e^{x} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)*ln(ln(x))*ln(ln(ln(x))**2)+(-exp(x)+10*x+1)*ln(x)*ln(ln(x))+2)*exp(x*ln(ln(ln(x))**2)-exp(x)+
5*x**2+x-1)/ln(x)/ln(ln(x)),x)

[Out]

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

________________________________________________________________________________________