\(\int \frac {e^{\frac {1}{16} (-48-e^{-12+64 e^x}+16 \log (x))} (-1+4 e^{-12+64 e^x+x} x)}{x} \, dx\) [3584]

Optimal result

Integrand size = 42, antiderivative size = 22 \[ \int \frac {e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}+16 \log (x)\right )} \left (-1+4 e^{-12+64 e^x+x} x\right )}{x} \, dx=5-e^{-3-\frac {1}{16} e^{-12+64 e^x}} x \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2306, 2326} \[ \int \frac {e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}+16 \log (x)\right )} \left (-1+4 e^{-12+64 e^x+x} x\right )}{x} \, dx=-e^{\frac {1}{16} \left (-e^{64 e^x-12}-48\right )} x \]

[In]

Int[(E^((-48 - E^(-12 + 64*E^x) + 16*Log[x])/16)*(-1 + 4*E^(-12 + 64*E^x + x)*x))/x,x]

[Out]

-(E^((-48 - E^(-12 + 64*E^x))/16)*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]

Rule 2326

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]

Rubi steps \begin{align*} \text {integral}& = \int e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}\right )} \left (-1+4 e^{-12+64 e^x+x} x\right ) \, dx \\ & = -e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}\right )} x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}+16 \log (x)\right )} \left (-1+4 e^{-12+64 e^x+x} x\right )}{x} \, dx=-e^{-3-\frac {1}{16} e^{-12+64 e^x}} x \]

[In]

Integrate[(E^((-48 - E^(-12 + 64*E^x) + 16*Log[x])/16)*(-1 + 4*E^(-12 + 64*E^x + x)*x))/x,x]

[Out]

-(E^(-3 - E^(-12 + 64*E^x)/16)*x)

Maple [A] (verified)

Time = 5.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73

method	result	size
risch	\(-x \,{\mathrm e}^{-\frac {{\mathrm e}^{64 \,{\mathrm e}^{x}-12}}{16}-3}\)	\(16\)
parallelrisch	\(-{\mathrm e}^{-\frac {{\mathrm e}^{64 \,{\mathrm e}^{x}-12}}{16}+\ln \left (x \right )-3}\)	\(19\)

[In]

int((4*x*exp(x)*exp(16*exp(x)-3)^4-1)*exp(-1/16*exp(16*exp(x)-3)^4+ln(x)-3)/x,x,method=_RETURNVERBOSE)

[Out]

-x*exp(-1/16*exp(64*exp(x)-12)-3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}+16 \log (x)\right )} \left (-1+4 e^{-12+64 e^x+x} x\right )}{x} \, dx=-e^{\left (\frac {1}{16} \, {\left (16 \, e^{x} \log \left (x\right ) - e^{\left (x + 64 \, e^{x} - 12\right )} - 48 \, e^{x}\right )} e^{\left (-x\right )}\right )} \]

[In]

integrate((4*x*exp(x)*exp(16*exp(x)-3)^4-1)*exp(-1/16*exp(16*exp(x)-3)^4+log(x)-3)/x,x, algorithm="fricas")

[Out]

-e^(1/16*(16*e^x*log(x) - e^(x + 64*e^x - 12) - 48*e^x)*e^(-x))

Sympy [A] (verification not implemented)

Time = 1.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}+16 \log (x)\right )} \left (-1+4 e^{-12+64 e^x+x} x\right )}{x} \, dx=- x e^{- \frac {e^{64 e^{x} - 12}}{16} - 3} \]

[In]

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

[Out]

-x*exp(-exp(64*exp(x) - 12)/16 - 3)

Maxima [A] (verification not implemented)

none

Time = 1.16 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}+16 \log (x)\right )} \left (-1+4 e^{-12+64 e^x+x} x\right )}{x} \, dx=-x e^{\left (-\frac {1}{16} \, e^{\left (64 \, e^{x} - 12\right )} - 3\right )} \]

[In]

integrate((4*x*exp(x)*exp(16*exp(x)-3)^4-1)*exp(-1/16*exp(16*exp(x)-3)^4+log(x)-3)/x,x, algorithm="maxima")

[Out]

-x*e^(-1/16*e^(64*e^x - 12) - 3)

Giac [F]

\[ \int \frac {e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}+16 \log (x)\right )} \left (-1+4 e^{-12+64 e^x+x} x\right )}{x} \, dx=\int { \frac {{\left (4 \, x e^{\left (x + 64 \, e^{x} - 12\right )} - 1\right )} e^{\left (-\frac {1}{16} \, e^{\left (64 \, e^{x} - 12\right )} + \log \left (x\right ) - 3\right )}}{x} \,d x } \]

[In]

integrate((4*x*exp(x)*exp(16*exp(x)-3)^4-1)*exp(-1/16*exp(16*exp(x)-3)^4+log(x)-3)/x,x, algorithm="giac")

[Out]

integrate((4*x*e^(x + 64*e^x - 12) - 1)*e^(-1/16*e^(64*e^x - 12) + log(x) - 3)/x, x)

Mupad [B] (verification not implemented)

Time = 8.82 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {1}{16} \left (-48-e^{-12+64 e^x}+16 \log (x)\right )} \left (-1+4 e^{-12+64 e^x+x} x\right )}{x} \, dx=-x\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-12}\,{\mathrm {e}}^{64\,{\mathrm {e}}^x}}{16}} \]

[In]

int((exp(log(x) - exp(64*exp(x) - 12)/16 - 3)*(4*x*exp(64*exp(x) - 12)*exp(x) - 1))/x,x)

[Out]

-x*exp(-3)*exp(-(exp(-12)*exp(64*exp(x)))/16)