\(\int \frac {e^{e^{-4+e^{\frac {1}{4} (e^x+8 x)}}} (e^{-4+e^{\frac {1}{4} (e^x+8 x)}+\frac {1}{4} (e^x+8 x)+\log ^2(x)} (8 x+e^x x)+8 e^{\log ^2(x)} \log (x))}{8 x} \, dx\) [4349]

Optimal result

Integrand size = 79, antiderivative size = 28 \[ \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)} \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{8 x} \, dx=\frac {1}{2} e^{e^{-4+e^{\frac {e^x}{4}+2 x}}+\log ^2(x)} \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(71\) vs. \(2(28)=56\).

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.54, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {12, 2326} \[ \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)} \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{8 x} \, dx=\frac {\left (e^x x+8 x\right ) \exp \left (e^{e^{\frac {1}{4} \left (8 x+e^x\right )}-4}+\frac {1}{4} \left (-8 x-e^x\right )+\frac {1}{4} \left (8 x+e^x\right )+\log ^2(x)\right )}{2 \left (e^x+8\right ) x} \]

[In]

Int[(E^E^(-4 + E^((E^x + 8*x)/4))*(E^(-4 + E^((E^x + 8*x)/4) + (E^x + 8*x)/4 + Log[x]^2)*(8*x + E^x*x) + 8*E^L
og[x]^2*Log[x]))/(8*x),x]

[Out]

(E^(E^(-4 + E^((E^x + 8*x)/4)) + (-E^x - 8*x)/4 + (E^x + 8*x)/4 + Log[x]^2)*(8*x + E^x*x))/(2*(8 + E^x)*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 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}& = \frac {1}{8} \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (\exp \left (-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)\right ) \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{x} \, dx \\ & = \frac {\exp \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}+\frac {1}{4} \left (-e^x-8 x\right )+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)\right ) \left (8 x+e^x x\right )}{2 \left (8+e^x\right ) x} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)} \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{8 x} \, dx=\frac {1}{2} e^{e^{-4+e^{\frac {e^x}{4}+2 x}}+\log ^2(x)} \]

[In]

Integrate[(E^E^(-4 + E^((E^x + 8*x)/4))*(E^(-4 + E^((E^x + 8*x)/4) + (E^x + 8*x)/4 + Log[x]^2)*(8*x + E^x*x) +
8*E^Log[x]^2*Log[x]))/(8*x),x]

[Out]

E^(E^(-4 + E^(E^x/4 + 2*x)) + Log[x]^2)/2

Maple [A] (verified)

Time = 9.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{x}}{4}+2 x}-4}+\ln \left (x \right )^{2}}}{2}\)	\(21\)
parallelrisch	\(\frac {{\mathrm e}^{\ln \left (x \right )^{2}} {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{x}}{4}+2 x}-4}}}{2}\)	\(21\)

[In]

int(1/8*((exp(x)*x+8*x)*exp(1/4*exp(x)+2*x)*exp(ln(x)^2)*exp(exp(1/4*exp(x)+2*x)-4)+8*ln(x)*exp(ln(x)^2))*exp(
exp(exp(1/4*exp(x)+2*x)-4))/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)} \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{8 x} \, dx=\frac {1}{2} \, e^{\left (\log \left (x\right )^{2} + e^{\left (e^{\left (2 \, x + \frac {1}{4} \, e^{x}\right )} - 4\right )}\right )} \]

[In]

integrate(1/8*((exp(x)*x+8*x)*exp(1/4*exp(x)+2*x)*exp(log(x)^2)*exp(exp(1/4*exp(x)+2*x)-4)+8*log(x)*exp(log(x)
^2))*exp(exp(exp(1/4*exp(x)+2*x)-4))/x,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)} \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{8 x} \, dx=\text {Timed out} \]

[In]

integrate(1/8*((exp(x)*x+8*x)*exp(1/4*exp(x)+2*x)*exp(ln(x)**2)*exp(exp(1/4*exp(x)+2*x)-4)+8*ln(x)*exp(ln(x)**
2))*exp(exp(exp(1/4*exp(x)+2*x)-4))/x,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)} \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{8 x} \, dx=\frac {1}{2} \, e^{\left (\log \left (x\right )^{2} + e^{\left (e^{\left (2 \, x + \frac {1}{4} \, e^{x}\right )} - 4\right )}\right )} \]

[In]

integrate(1/8*((exp(x)*x+8*x)*exp(1/4*exp(x)+2*x)*exp(log(x)^2)*exp(exp(1/4*exp(x)+2*x)-4)+8*log(x)*exp(log(x)
^2))*exp(exp(exp(1/4*exp(x)+2*x)-4))/x,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)} \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{8 x} \, dx=\int { \frac {{\left ({\left (x e^{x} + 8 \, x\right )} e^{\left (\log \left (x\right )^{2} + 2 \, x + e^{\left (2 \, x + \frac {1}{4} \, e^{x}\right )} + \frac {1}{4} \, e^{x} - 4\right )} + 8 \, e^{\left (\log \left (x\right )^{2}\right )} \log \left (x\right )\right )} e^{\left (e^{\left (e^{\left (2 \, x + \frac {1}{4} \, e^{x}\right )} - 4\right )}\right )}}{8 \, x} \,d x } \]

[In]

integrate(1/8*((exp(x)*x+8*x)*exp(1/4*exp(x)+2*x)*exp(log(x)^2)*exp(exp(1/4*exp(x)+2*x)-4)+8*log(x)*exp(log(x)
^2))*exp(exp(exp(1/4*exp(x)+2*x)-4))/x,x, algorithm="giac")

[Out]

integrate(1/8*((x*e^x + 8*x)*e^(log(x)^2 + 2*x + e^(2*x + 1/4*e^x) + 1/4*e^x - 4) + 8*e^(log(x)^2)*log(x))*e^(
e^(e^(2*x + 1/4*e^x) - 4))/x, x)

Mupad [B] (verification not implemented)

Time = 10.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}}} \left (e^{-4+e^{\frac {1}{4} \left (e^x+8 x\right )}+\frac {1}{4} \left (e^x+8 x\right )+\log ^2(x)} \left (8 x+e^x x\right )+8 e^{\log ^2(x)} \log (x)\right )}{8 x} \, dx=\frac {{\mathrm {e}}^{{\ln \left (x\right )}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{-4}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{4}}}}}{2} \]

[In]

int((exp(exp(exp(2*x + exp(x)/4) - 4))*(8*exp(log(x)^2)*log(x) + exp(exp(2*x + exp(x)/4) - 4)*exp(log(x)^2)*ex
p(2*x + exp(x)/4)*(8*x + x*exp(x))))/(8*x),x)

[Out]

(exp(log(x)^2)*exp(exp(-4)*exp(exp(2*x)*exp(exp(x)/4))))/2