3.46.29 \(\int \frac {e^{e^{\frac {1}{x^2 \log ^2(2+\log (4))}}} (-2 e^{x+\frac {1}{x^2 \log ^2(2+\log (4))}}+e^x (x^2+x^3) \log ^2(2+\log (4)))}{8 x^2 \log ^2(2+\log (4))} \, dx\)

Optimal. Leaf size=24 \[ 2+\frac {1}{8} e^{e^{\frac {1}{x^2 \log ^2(2+\log (4))}}+x} x \]

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Rubi [A]  time = 0.11, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 2288} \begin {gather*} \frac {1}{8} x e^{e^{\frac {1}{x^2 \log ^2(2+\log (4))}}+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^(1/(x^2*Log[2 + Log[4]]^2))*(-2*E^(x + 1/(x^2*Log[2 + Log[4]]^2)) + E^x*(x^2 + x^3)*Log[2 + Log[4]]^2
))/(8*x^2*Log[2 + Log[4]]^2),x]

[Out]

(E^(E^(1/(x^2*Log[2 + Log[4]]^2)) + x)*x)/8

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 {\int \frac {e^{e^{\frac {1}{x^2 \log ^2(2+\log (4))}}} \left (-2 e^{x+\frac {1}{x^2 \log ^2(2+\log (4))}}+e^x \left (x^2+x^3\right ) \log ^2(2+\log (4))\right )}{x^2} \, dx}{8 \log ^2(2+\log (4))}\\ &=\frac {1}{8} e^{e^{\frac {1}{x^2 \log ^2(2+\log (4))}}+x} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 22, normalized size = 0.92 \begin {gather*} \frac {1}{8} e^{e^{\frac {1}{x^2 \log ^2(2+\log (4))}}+x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^(1/(x^2*Log[2 + Log[4]]^2))*(-2*E^(x + 1/(x^2*Log[2 + Log[4]]^2)) + E^x*(x^2 + x^3)*Log[2 + Log
[4]]^2))/(8*x^2*Log[2 + Log[4]]^2),x]

[Out]

(E^(E^(1/(x^2*Log[2 + Log[4]]^2)) + x)*x)/8

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fricas [A]  time = 0.45, size = 39, normalized size = 1.62 \begin {gather*} \frac {1}{8} \, x e^{\left (x + e^{\left (-x + \frac {x^{3} \log \left (2 \, \log \relax (2) + 2\right )^{2} + 1}{x^{2} \log \left (2 \, \log \relax (2) + 2\right )^{2}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-2*exp(x)*exp(1/x^2/log(2+2*log(2))^2)+(x^3+x^2)*exp(x)*log(2+2*log(2))^2)*exp(exp(1/x^2/log(2+
2*log(2))^2))/x^2/log(2+2*log(2))^2,x, algorithm="fricas")

[Out]

1/8*x*e^(x + e^(-x + (x^3*log(2*log(2) + 2)^2 + 1)/(x^2*log(2*log(2) + 2)^2)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-2*exp(x)*exp(1/x^2/log(2+2*log(2))^2)+(x^3+x^2)*exp(x)*log(2+2*log(2))^2)*exp(exp(1/x^2/log(2+
2*log(2))^2))/x^2/log(2+2*log(2))^2,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.10, size = 22, normalized size = 0.92

method	result	size
risch	\(\frac {x \,{\mathrm e}^{x +{\mathrm e}^{\frac {1}{x^{2} \left (\ln \relax (2)+\ln \left (1+\ln \relax (2)\right )\right )^{2}}}}}{8}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(-2*exp(x)*exp(1/x^2/ln(2+2*ln(2))^2)+(x^3+x^2)*exp(x)*ln(2+2*ln(2))^2)*exp(exp(1/x^2/ln(2+2*ln(2))^2)
)/x^2/ln(2+2*ln(2))^2,x,method=_RETURNVERBOSE)

[Out]

1/8*x*exp(x+exp(1/x^2/(ln(2)+ln(1+ln(2)))^2))

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maxima [B]  time = 0.58, size = 64, normalized size = 2.67 \begin {gather*} \frac {{\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2) + 1\right ) + \log \left (\log \relax (2) + 1\right )^{2}\right )} x e^{\left (x + e^{\left (\frac {1}{{\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2) + 1\right ) + \log \left (\log \relax (2) + 1\right )^{2}\right )} x^{2}}\right )}\right )}}{8 \, \log \left (2 \, \log \relax (2) + 2\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-2*exp(x)*exp(1/x^2/log(2+2*log(2))^2)+(x^3+x^2)*exp(x)*log(2+2*log(2))^2)*exp(exp(1/x^2/log(2+
2*log(2))^2))/x^2/log(2+2*log(2))^2,x, algorithm="maxima")

[Out]

1/8*(log(2)^2 + 2*log(2)*log(log(2) + 1) + log(log(2) + 1)^2)*x*e^(x + e^(1/((log(2)^2 + 2*log(2)*log(log(2) +
1) + log(log(2) + 1)^2)*x^2)))/log(2*log(2) + 2)^2

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mupad [B]  time = 3.42, size = 18, normalized size = 0.75 \begin {gather*} \frac {x\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {1}{x^2\,{\ln \left (\ln \relax (4)+2\right )}^2}}}\,{\mathrm {e}}^x}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(1/(x^2*log(2*log(2) + 2)^2)))*(2*exp(1/(x^2*log(2*log(2) + 2)^2))*exp(x) - exp(x)*log(2*log(2) +
2)^2*(x^2 + x^3)))/(8*x^2*log(2*log(2) + 2)^2),x)

[Out]

(x*exp(exp(1/(x^2*log(log(4) + 2)^2)))*exp(x))/8

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

[Out]

Timed out

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