3.86.59 \(\int \frac {e^{x-x^2 \log ^2(\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))})} (4+\log (x-x \log (4))+4 x \log (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))})+(-8 x-2 x \log (x-x \log (4))) \log ^2(\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}))}{4+\log (x-x \log (4))} \, dx\)

Optimal. Leaf size=26 \[ 1+e^{x-x^2 \log ^2\left (\frac {1}{(4+\log (x-x \log (4)))^2}\right )} \]

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Rubi [A]  time = 0.63, antiderivative size = 38, normalized size of antiderivative = 1.46, number of steps used = 2, number of rules used = 2, integrand size = 130, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {2444, 6706} \begin {gather*} \exp \left (x-x^2 \log ^2\left (\frac {1}{\log ^2(x (1-\log (4)))+8 \log (x (1-\log (4)))+16}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x - x^2*Log[(16 + 8*Log[x - x*Log[4]] + Log[x - x*Log[4]]^2)^(-1)]^2)*(4 + Log[x - x*Log[4]] + 4*x*Log
[(16 + 8*Log[x - x*Log[4]] + Log[x - x*Log[4]]^2)^(-1)] + (-8*x - 2*x*Log[x - x*Log[4]])*Log[(16 + 8*Log[x - x
*Log[4]] + Log[x - x*Log[4]]^2)^(-1)]^2))/(4 + Log[x - x*Log[4]]),x]

[Out]

E^(x - x^2*Log[(16 + 8*Log[x*(1 - Log[4])] + Log[x*(1 - Log[4])]^2)^(-1)]^2)

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 6706

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (x-x^2 \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right ) \left (4+\log (x-x \log (4))+4 x \log \left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )+(-8 x-2 x \log (x-x \log (4))) \log ^2\left (\frac {1}{16+8 \log (x-x \log (4))+\log ^2(x-x \log (4))}\right )\right )}{4+\log (x (1-\log (4)))} \, dx\\ &=\exp \left (x-x^2 \log ^2\left (\frac {1}{16+8 \log (x (1-\log (4)))+\log ^2(x (1-\log (4)))}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 24, normalized size = 0.92 \begin {gather*} e^{x-x^2 \log ^2\left (\frac {1}{(4+\log (x-x \log (4)))^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x - x^2*Log[(16 + 8*Log[x - x*Log[4]] + Log[x - x*Log[4]]^2)^(-1)]^2)*(4 + Log[x - x*Log[4]] + 4
*x*Log[(16 + 8*Log[x - x*Log[4]] + Log[x - x*Log[4]]^2)^(-1)] + (-8*x - 2*x*Log[x - x*Log[4]])*Log[(16 + 8*Log
[x - x*Log[4]] + Log[x - x*Log[4]]^2)^(-1)]^2))/(4 + Log[x - x*Log[4]]),x]

[Out]

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

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fricas [A]  time = 0.62, size = 35, normalized size = 1.35 \begin {gather*} e^{\left (-x^{2} \log \left (\frac {1}{\log \left (-2 \, x \log \relax (2) + x\right )^{2} + 8 \, \log \left (-2 \, x \log \relax (2) + x\right ) + 16}\right )^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x-2*x*log(2))-8*x)*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log(2))+16))^2+4*x*log(1/(log(x
-2*x*log(2))^2+8*log(x-2*x*log(2))+16))+log(x-2*x*log(2))+4)*exp(-x^2*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*l
og(2))+16))^2+x)/(log(x-2*x*log(2))+4),x, algorithm="fricas")

[Out]

e^(-x^2*log(1/(log(-2*x*log(2) + x)^2 + 8*log(-2*x*log(2) + x) + 16))^2 + x)

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giac [A]  time = 2.10, size = 33, normalized size = 1.27 \begin {gather*} e^{\left (-x^{2} \log \left (\log \left (-2 \, x \log \relax (2) + x\right )^{2} + 8 \, \log \left (-2 \, x \log \relax (2) + x\right ) + 16\right )^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x-2*x*log(2))-8*x)*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log(2))+16))^2+4*x*log(1/(log(x
-2*x*log(2))^2+8*log(x-2*x*log(2))+16))+log(x-2*x*log(2))+4)*exp(-x^2*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*l
og(2))+16))^2+x)/(log(x-2*x*log(2))+4),x, algorithm="giac")

[Out]

e^(-x^2*log(log(-2*x*log(2) + x)^2 + 8*log(-2*x*log(2) + x) + 16)^2 + x)

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maple [C]  time = 0.78, size = 280, normalized size = 10.77

method	result	size
risch	\(\left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )^{4 i \pi \,\mathrm {csgn}\left (i \left (\ln \left (-\left (2 \ln \relax (2)-1\right ) x \right )+4\right )^{2}\right ) x^{2}} \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )^{-4 i \pi \,\mathrm {csgn}\left (i \left (\ln \left (-x \right )+\ln \left (2 \ln \relax (2)-1\right )+4\right )\right ) x^{2}} {\mathrm e}^{-\frac {x \left (-x \,\pi ^{2} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )^{2}\right )^{6}+4 x \,\pi ^{2} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )^{2}\right )^{5} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )\right )-6 x \,\pi ^{2} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )^{2}\right )^{4} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )\right )^{2}+4 x \,\pi ^{2} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )^{2}\right )^{3} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )\right )^{3}-x \,\pi ^{2} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )\right )^{4}+16 x \ln \left (\ln \left (x -2 x \ln \relax (2)\right )+4\right )^{2}-4\right )}{4}}\)	\(280\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*ln(x-2*x*ln(2))-8*x)*ln(1/(ln(x-2*x*ln(2))^2+8*ln(x-2*x*ln(2))+16))^2+4*x*ln(1/(ln(x-2*x*ln(2))^2+8
*ln(x-2*x*ln(2))+16))+ln(x-2*x*ln(2))+4)*exp(-x^2*ln(1/(ln(x-2*x*ln(2))^2+8*ln(x-2*x*ln(2))+16))^2+x)/(ln(x-2*
x*ln(2))+4),x,method=_RETURNVERBOSE)

[Out]

((ln(x-2*x*ln(2))+4)^(2*I*Pi*csgn(I*(ln(-x)+ln(2*ln(2)-1)+4)^2)*x^2))^2*(ln(x-2*x*ln(2))+4)^(-4*I*Pi*csgn(I*(l
n(-x)+ln(2*ln(2)-1)+4))*x^2)*exp(-1/4*x*(-x*Pi^2*csgn(I*(ln(x-2*x*ln(2))+4)^2)^6+4*x*Pi^2*csgn(I*(ln(x-2*x*ln(
2))+4)^2)^5*csgn(I*(ln(x-2*x*ln(2))+4))-6*x*Pi^2*csgn(I*(ln(x-2*x*ln(2))+4)^2)^4*csgn(I*(ln(x-2*x*ln(2))+4))^2
+4*x*Pi^2*csgn(I*(ln(x-2*x*ln(2))+4)^2)^3*csgn(I*(ln(x-2*x*ln(2))+4))^3-x*Pi^2*csgn(I*(ln(x-2*x*ln(2))+4)^2)^2
*csgn(I*(ln(x-2*x*ln(2))+4))^4+16*x*ln(ln(x-2*x*ln(2))+4)^2-4))

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maxima [A]  time = 0.61, size = 22, normalized size = 0.85 \begin {gather*} e^{\left (-4 \, x^{2} \log \left (\log \relax (x) + \log \left (-2 \, \log \relax (2) + 1\right ) + 4\right )^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*log(x-2*x*log(2))-8*x)*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*log(2))+16))^2+4*x*log(1/(log(x
-2*x*log(2))^2+8*log(x-2*x*log(2))+16))+log(x-2*x*log(2))+4)*exp(-x^2*log(1/(log(x-2*x*log(2))^2+8*log(x-2*x*l
og(2))+16))^2+x)/(log(x-2*x*log(2))+4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.48, size = 36, normalized size = 1.38 \begin {gather*} {\mathrm {e}}^x\,{\mathrm {e}}^{-x^2\,{\ln \left (\frac {1}{{\ln \left (x-2\,x\,\ln \relax (2)\right )}^2+8\,\ln \left (x-2\,x\,\ln \relax (2)\right )+16}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x - x^2*log(1/(8*log(x - 2*x*log(2)) + log(x - 2*x*log(2))^2 + 16))^2)*(log(x - 2*x*log(2)) - log(1/(
8*log(x - 2*x*log(2)) + log(x - 2*x*log(2))^2 + 16))^2*(8*x + 2*x*log(x - 2*x*log(2))) + 4*x*log(1/(8*log(x -
2*x*log(2)) + log(x - 2*x*log(2))^2 + 16)) + 4))/(log(x - 2*x*log(2)) + 4),x)

[Out]

exp(x)*exp(-x^2*log(1/(8*log(x - 2*x*log(2)) + log(x - 2*x*log(2))^2 + 16))^2)

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sympy [A]  time = 0.67, size = 36, normalized size = 1.38 \begin {gather*} e^{- x^{2} \log {\left (\frac {1}{\log {\left (- 2 x \log {\relax (2 )} + x \right )}^{2} + 8 \log {\left (- 2 x \log {\relax (2 )} + x \right )} + 16} \right )}^{2} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*ln(x-2*x*ln(2))-8*x)*ln(1/(ln(x-2*x*ln(2))**2+8*ln(x-2*x*ln(2))+16))**2+4*x*ln(1/(ln(x-2*x*ln
(2))**2+8*ln(x-2*x*ln(2))+16))+ln(x-2*x*ln(2))+4)*exp(-x**2*ln(1/(ln(x-2*x*ln(2))**2+8*ln(x-2*x*ln(2))+16))**2
+x)/(ln(x-2*x*ln(2))+4),x)

[Out]

exp(-x**2*log(1/(log(-2*x*log(2) + x)**2 + 8*log(-2*x*log(2) + x) + 16))**2 + x)

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