\(\int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log (\frac {x}{3})}{3 x}} (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log (\frac {x}{3}))}{4 x^2} \, dx\) [6327]

Optimal result

Integrand size = 68, antiderivative size = 35 \[ \int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}} \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{4 x^2} \, dx=3 \left (-4+e^{3+\frac {16 e^{\frac {1}{4} (-3+x)}-\log \left (\frac {x}{3}\right )}{3 x}}\right ) \]

[Out]

3*exp(3+1/3*(exp(4*ln(2)+1/4*x-3/4)-ln(1/3*x))/x)-12

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {12, 6838} \[ \int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}} \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{4 x^2} \, dx=3^{\frac {1}{3 x}+1} e^{\frac {9 x+16 e^{\frac {x-3}{4}}}{3 x}} x^{\left .-\frac {1}{3}\right /x} \]

[In]

Int[(E^((E^((-3 + x + 16*Log[2])/4) + 9*x - Log[x/3])/(3*x))*(-4 + E^((-3 + x + 16*Log[2])/4)*(-4 + x) + 4*Log
[x/3]))/(4*x^2),x]

[Out]

(3^(1 + 1/(3*x))*E^((16*E^((-3 + x)/4) + 9*x)/(3*x)))/x^(1/(3*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {\exp \left (\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}\right ) \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{x^2} \, dx \\ & = 3^{1+\frac {1}{3 x}} e^{\frac {16 e^{\frac {1}{4} (-3+x)}+9 x}{3 x}} x^{\left .-\frac {1}{3}\right /x} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}} \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{4 x^2} \, dx=3^{1+\frac {1}{3 x}} e^{3+\frac {16 e^{\frac {1}{4} (-3+x)}}{3 x}} x^{\left .-\frac {1}{3}\right /x} \]

[In]

Integrate[(E^((E^((-3 + x + 16*Log[2])/4) + 9*x - Log[x/3])/(3*x))*(-4 + E^((-3 + x + 16*Log[2])/4)*(-4 + x) +
4*Log[x/3]))/(4*x^2),x]

[Out]

(3^(1 + 1/(3*x))*E^(3 + (16*E^((-3 + x)/4))/(3*x)))/x^(1/(3*x))

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77

method	result	size
risch	\(3 \,{\mathrm e}^{\frac {-\ln \left (\frac {x}{3}\right )+16 \,{\mathrm e}^{\frac {x}{4}-\frac {3}{4}}+9 x}{3 x}}\)	\(27\)
parallelrisch	\(3 \,{\mathrm e}^{-\frac {\ln \left (\frac {x}{3}\right )-{\mathrm e}^{4 \ln \left (2\right )+\frac {x}{4}-\frac {3}{4}}-9 x}{3 x}}\)	\(29\)

[In]

int(1/4*(4*ln(1/3*x)+(x-4)*exp(4*ln(2)+1/4*x-3/4)-4)*exp(1/3*(-ln(1/3*x)+exp(4*ln(2)+1/4*x-3/4)+9*x)/x)/x^2,x,
method=_RETURNVERBOSE)

[Out]

3*exp(1/3*(-ln(1/3*x)+16*exp(1/4*x-3/4)+9*x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}} \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{4 x^2} \, dx=3 \, e^{\left (\frac {9 \, x + e^{\left (\frac {1}{4} \, x + 4 \, \log \left (2\right ) - \frac {3}{4}\right )} - \log \left (\frac {1}{3} \, x\right )}{3 \, x}\right )} \]

[In]

integrate(1/4*(4*log(1/3*x)+(x-4)*exp(4*log(2)+1/4*x-3/4)-4)*exp(1/3*(-log(1/3*x)+exp(4*log(2)+1/4*x-3/4)+9*x)
/x)/x^2,x, algorithm="fricas")

[Out]

3*e^(1/3*(9*x + e^(1/4*x + 4*log(2) - 3/4) - log(1/3*x))/x)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}} \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{4 x^2} \, dx=3 e^{\frac {3 x + \frac {16 e^{\frac {x}{4} - \frac {3}{4}}}{3} - \frac {\log {\left (\frac {x}{3} \right )}}{3}}{x}} \]

[In]

integrate(1/4*(4*ln(1/3*x)+(x-4)*exp(4*ln(2)+1/4*x-3/4)-4)*exp(1/3*(-ln(1/3*x)+exp(4*ln(2)+1/4*x-3/4)+9*x)/x)/
x**2,x)

[Out]

3*exp((3*x + 16*exp(x/4 - 3/4)/3 - log(x/3)/3)/x)

Maxima [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}} \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{4 x^2} \, dx=3 \, e^{\left (\frac {16 \, e^{\left (\frac {1}{4} \, x - \frac {3}{4}\right )}}{3 \, x} + \frac {\log \left (3\right )}{3 \, x} - \frac {\log \left (x\right )}{3 \, x} + 3\right )} \]

[In]

integrate(1/4*(4*log(1/3*x)+(x-4)*exp(4*log(2)+1/4*x-3/4)-4)*exp(1/3*(-log(1/3*x)+exp(4*log(2)+1/4*x-3/4)+9*x)
/x)/x^2,x, algorithm="maxima")

[Out]

3*e^(16/3*e^(1/4*x - 3/4)/x + 1/3*log(3)/x - 1/3*log(x)/x + 3)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}} \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{4 x^2} \, dx=3 \, e^{\left (\frac {e^{\left (\frac {1}{4} \, x + 4 \, \log \left (2\right ) - \frac {3}{4}\right )}}{3 \, x} - \frac {\log \left (\frac {1}{3} \, x\right )}{3 \, x} + 3\right )} \]

[In]

integrate(1/4*(4*log(1/3*x)+(x-4)*exp(4*log(2)+1/4*x-3/4)-4)*exp(1/3*(-log(1/3*x)+exp(4*log(2)+1/4*x-3/4)+9*x)
/x)/x^2,x, algorithm="giac")

[Out]

3*e^(1/3*e^(1/4*x + 4*log(2) - 3/4)/x - 1/3*log(1/3*x)/x + 3)

Mupad [B] (verification not implemented)

Time = 12.49 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {e^{\frac {1}{4} (-3+x+16 \log (2))}+9 x-\log \left (\frac {x}{3}\right )}{3 x}} \left (-4+e^{\frac {1}{4} (-3+x+16 \log (2))} (-4+x)+4 \log \left (\frac {x}{3}\right )\right )}{4 x^2} \, dx=\frac {3^{\frac {1}{3\,x}+1}\,{\mathrm {e}}^{\frac {16\,{\mathrm {e}}^{x/4}\,{\mathrm {e}}^{-\frac {3}{4}}}{3\,x}+3}}{x^{\frac {1}{3\,x}}} \]

[In]

int((exp((3*x - log(x/3)/3 + exp(x/4 + 4*log(2) - 3/4)/3)/x)*(4*log(x/3) + exp(x/4 + 4*log(2) - 3/4)*(x - 4) -
4))/(4*x^2),x)

[Out]

(3^(1/(3*x) + 1)*exp((16*exp(x/4)*exp(-3/4))/(3*x) + 3))/x^(1/(3*x))