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

Optimal. Leaf size=28 \[ \frac {1}{2} e^{e^{-4+e^{\frac {e^x}{4}+2 x}}+\log ^2(x)} \]

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Rubi [B]  time = 0.28, antiderivative size = 71, normalized size of antiderivative = 2.54, number of steps used = 2, number of rules used = 2, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {12, 2288} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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 2288

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.86, size = 28, normalized size = 1.00 \begin {gather*} \frac {1}{2} e^{e^{-4+e^{\frac {e^x}{4}+2 x}}+\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.64, size = 20, normalized size = 0.71 \begin {gather*} \frac {1}{2} \, e^{\left (\log \relax (x)^{2} + e^{\left (e^{\left (2 \, x + \frac {1}{4} \, e^{x}\right )} - 4\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (x e^{x} + 8 \, x\right )} e^{\left (\log \relax (x)^{2} + 2 \, x + e^{\left (2 \, x + \frac {1}{4} \, e^{x}\right )} + \frac {1}{4} \, e^{x} - 4\right )} + 8 \, e^{\left (\log \relax (x)^{2}\right )} \log \relax (x)\right )} e^{\left (e^{\left (e^{\left (2 \, x + \frac {1}{4} \, e^{x}\right )} - 4\right )}\right )}}{8 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.09, size = 21, normalized size = 0.75

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

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [A]  time = 0.80, size = 20, normalized size = 0.71 \begin {gather*} \frac {1}{2} \, e^{\left (\log \relax (x)^{2} + e^{\left (e^{\left (2 \, x + \frac {1}{4} \, e^{x}\right )} - 4\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 3.11, size = 22, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {e}}^{{\ln \relax (x)}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{-4}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{4}}}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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