\(\int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx\) [6406]

Optimal result

Integrand size = 39, antiderivative size = 12 \[ \int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx=e^{\frac {3 x}{\log (\log (x))}}+x \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6874, 6838} \[ \int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx=x+e^{\frac {3 x}{\log (\log (x))}} \]

[In]

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

[Out]

E^((3*x)/Log[Log[x]]) + x

Rule 6838

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {3 e^{\frac {3 x}{\log (\log (x))}} (-1+\log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))}\right ) \, dx \\ & = x+3 \int \frac {e^{\frac {3 x}{\log (\log (x))}} (-1+\log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx \\ & = e^{\frac {3 x}{\log (\log (x))}}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx=e^{\frac {3 x}{\log (\log (x))}}+x \]

[In]

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

[Out]

E^((3*x)/Log[Log[x]]) + x

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00

method	result	size
risch	\({\mathrm e}^{\frac {3 x}{\ln \left (\ln \left (x \right )\right )}}+x\)	\(12\)
parallelrisch	\({\mathrm e}^{\frac {3 x}{\ln \left (\ln \left (x \right )\right )}}+x\)	\(14\)

[In]

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

[Out]

exp(3*x/ln(ln(x)))+x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx=x + e^{\left (\frac {3 \, x}{\log \left (\log \left (x\right )\right )}\right )} \]

[In]

integrate(((3*log(x)*log(log(x))-3)*exp(3/2*x/log(log(x)))^2+log(x)*log(log(x))^2)/log(x)/log(log(x))^2,x, alg
orithm="fricas")

[Out]

x + e^(3*x/log(log(x)))

Sympy [F(-2)]

Exception generated. \[ \int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((3*ln(x)*ln(ln(x))-3)*exp(3/2*x/ln(ln(x)))**2+ln(x)*ln(ln(x))**2)/ln(x)/ln(ln(x))**2,x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [F(-2)]

Exception generated. \[ \int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(((3*log(x)*log(log(x))-3)*exp(3/2*x/log(log(x)))^2+log(x)*log(log(x))^2)/log(x)/log(log(x))^2,x, alg
orithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [F]

\[ \int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx=\int { \frac {\log \left (x\right ) \log \left (\log \left (x\right )\right )^{2} + 3 \, {\left (\log \left (x\right ) \log \left (\log \left (x\right )\right ) - 1\right )} e^{\left (\frac {3 \, x}{\log \left (\log \left (x\right )\right )}\right )}}{\log \left (x\right ) \log \left (\log \left (x\right )\right )^{2}} \,d x } \]

[In]

integrate(((3*log(x)*log(log(x))-3)*exp(3/2*x/log(log(x)))^2+log(x)*log(log(x))^2)/log(x)/log(log(x))^2,x, alg
orithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 12.49 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {\log (x) \log ^2(\log (x))+e^{\frac {3 x}{\log (\log (x))}} (-3+3 \log (x) \log (\log (x)))}{\log (x) \log ^2(\log (x))} \, dx=x+{\mathrm {e}}^{\frac {3\,x}{\ln \left (\ln \left (x\right )\right )}} \]

[In]

int((exp((3*x)/log(log(x)))*(3*log(log(x))*log(x) - 3) + log(log(x))^2*log(x))/(log(log(x))^2*log(x)),x)

[Out]

x + exp((3*x)/log(log(x)))