\(\int \frac {e^{2 x+\frac {\log (x^3)}{e^3}} (\frac {3 \log (x^3)}{e^3}+2 x \log (x^3))}{x \log (x^3)} \, dx\) [1891]

Optimal result

Integrand size = 41, antiderivative size = 14 \[ \int \frac {e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \left (\frac {3 \log \left (x^3\right )}{e^3}+2 x \log \left (x^3\right )\right )}{x \log \left (x^3\right )} \, dx=e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {2306, 15, 2228} \[ \int \frac {e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \left (\frac {3 \log \left (x^3\right )}{e^3}+2 x \log \left (x^3\right )\right )}{x \log \left (x^3\right )} \, dx=e^{2 x} \left (x^3\right )^{\frac {1}{e^3}} \]

[In]

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

[Out]

E^(2*x)*(x^3)^E^(-3)

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \left (\frac {3 \log \left (x^3\right )}{e^3}+2 x \log \left (x^3\right )\right )}{x \log \left (x^3\right )} \, dx=e^{2 x} \left (x^3\right )^{\frac {1}{e^3}} \]

[In]

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

[Out]

E^(2*x)*(x^3)^E^(-3)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\({\mathrm e}^{\ln \left (x^{3}\right ) {\mathrm e}^{-3}+2 x}\)	\(13\)
norman	\({\mathrm e}^{\ln \left (x^{3}\right ) {\mathrm e}^{-3}+2 x}\)	\(13\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{\ln \left (\ln \left (x^{3}\right )\right )-3}+2 x}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \left (\frac {3 \log \left (x^3\right )}{e^3}+2 x \log \left (x^3\right )\right )}{x \log \left (x^3\right )} \, dx=e^{\left ({\left (2 \, x e^{3} + \log \left (x^{3}\right )\right )} e^{\left (-3\right )}\right )} \]

[In]

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

[Out]

e^((2*x*e^3 + log(x^3))*e^(-3))

Sympy [A] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \left (\frac {3 \log \left (x^3\right )}{e^3}+2 x \log \left (x^3\right )\right )}{x \log \left (x^3\right )} \, dx=\left (x^{3}\right )^{e^{-3}} e^{2 x} \]

[In]

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

[Out]

(x**3)**exp(-3)*exp(2*x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \left (\frac {3 \log \left (x^3\right )}{e^3}+2 x \log \left (x^3\right )\right )}{x \log \left (x^3\right )} \, dx=e^{\left (3 \, e^{\left (-3\right )} \log \left (x\right ) + 2 \, x\right )} \]

[In]

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

[Out]

e^(3*e^(-3)*log(x) + 2*x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \left (\frac {3 \log \left (x^3\right )}{e^3}+2 x \log \left (x^3\right )\right )}{x \log \left (x^3\right )} \, dx=e^{\left ({\left (2 \, x e^{3} + 3 \, \log \left (x\right )\right )} e^{\left (-3\right )}\right )} \]

[In]

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

[Out]

e^((2*x*e^3 + 3*log(x))*e^(-3))

Mupad [B] (verification not implemented)

Time = 8.82 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 x+\frac {\log \left (x^3\right )}{e^3}} \left (\frac {3 \log \left (x^3\right )}{e^3}+2 x \log \left (x^3\right )\right )}{x \log \left (x^3\right )} \, dx={\mathrm {e}}^{2\,x}\,{\left (x^3\right )}^{{\mathrm {e}}^{-3}} \]

[In]

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

[Out]

exp(2*x)*(x^3)^exp(-3)