3.28.83 \(\int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2(16-8 x+x^2)} (16-12 x+2 x^2+e^x (-8+6 x-x^2)+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log (16-8 x+x^2))}{8-6 x+x^2} \, dx\)

Optimal. Leaf size=30 \[ e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left ((-4+x)^2\right )} \]

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Rubi [A]  time = 4.55, antiderivative size = 33, normalized size of antiderivative = 1.10, number of steps used = 1, number of rules used = 1, integrand size = 101, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6706} \begin {gather*} e^{\log ^2\left (x^2-8 x+16\right )-e^x+2 x-e^{\log ^2(x-2)}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 - E^x - E^Log[-2 + x]^2 + 2*x + Log[16 - 8*x + x^2]^2)*(16 - 12*x + 2*x^2 + E^x*(-8 + 6*x - x^2) + E
^Log[-2 + x]^2*(8 - 2*x)*Log[-2 + x] + (-8 + 4*x)*Log[16 - 8*x + x^2]))/(8 - 6*x + x^2),x]

[Out]

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

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} &=\exp \left (1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 30, normalized size = 1.00 \begin {gather*} e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left ((-4+x)^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 - E^x - E^Log[-2 + x]^2 + 2*x + Log[16 - 8*x + x^2]^2)*(16 - 12*x + 2*x^2 + E^x*(-8 + 6*x - x^
2) + E^Log[-2 + x]^2*(8 - 2*x)*Log[-2 + x] + (-8 + 4*x)*Log[16 - 8*x + x^2]))/(8 - 6*x + x^2),x]

[Out]

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

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fricas [A]  time = 0.57, size = 30, normalized size = 1.00 \begin {gather*} e^{\left (\log \left (x^{2} - 8 \, x + 16\right )^{2} + 2 \, x - e^{\left (\log \left (x - 2\right )^{2}\right )} - e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x+8)*log(x-2)*exp(log(x-2)^2)+(4*x-8)*log(x^2-8*x+16)+(-x^2+6*x-8)*exp(x)+2*x^2-12*x+16)*exp(-e
xp(log(x-2)^2)+log(x^2-8*x+16)^2-exp(x)+2*x+1)/(x^2-6*x+8),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 30, normalized size = 1.00 \begin {gather*} e^{\left (\log \left (x^{2} - 8 \, x + 16\right )^{2} + 2 \, x - e^{\left (\log \left (x - 2\right )^{2}\right )} - e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x+8)*log(x-2)*exp(log(x-2)^2)+(4*x-8)*log(x^2-8*x+16)+(-x^2+6*x-8)*exp(x)+2*x^2-12*x+16)*exp(-e
xp(log(x-2)^2)+log(x^2-8*x+16)^2-exp(x)+2*x+1)/(x^2-6*x+8),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.64, size = 175, normalized size = 5.83

method	result	size
risch	\(\left (x -4\right )^{-4 i \pi \,\mathrm {csgn}\left (i \left (x -4\right )^{2}\right )} \left (x -4\right )^{4 i \pi \,\mathrm {csgn}\left (i \left (x -4\right )\right )} {\mathrm e}^{-{\mathrm e}^{\ln \left (x -2\right )^{2}}+4 \ln \left (x -4\right )^{2}+1-\frac {\pi ^{2} \mathrm {csgn}\left (i \left (x -4\right )^{2}\right )^{6}}{4}+\pi ^{2} \mathrm {csgn}\left (i \left (x -4\right )^{2}\right )^{5} \mathrm {csgn}\left (i \left (x -4\right )\right )-\frac {3 \pi ^{2} \mathrm {csgn}\left (i \left (x -4\right )^{2}\right )^{4} \mathrm {csgn}\left (i \left (x -4\right )\right )^{2}}{2}+\pi ^{2} \mathrm {csgn}\left (i \left (x -4\right )^{2}\right )^{3} \mathrm {csgn}\left (i \left (x -4\right )\right )^{3}-\frac {\pi ^{2} \mathrm {csgn}\left (i \left (x -4\right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (x -4\right )\right )^{4}}{4}-{\mathrm e}^{x}+2 x}\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x+8)*ln(x-2)*exp(ln(x-2)^2)+(4*x-8)*ln(x^2-8*x+16)+(-x^2+6*x-8)*exp(x)+2*x^2-12*x+16)*exp(-exp(ln(x-2
)^2)+ln(x^2-8*x+16)^2-exp(x)+2*x+1)/(x^2-6*x+8),x,method=_RETURNVERBOSE)

[Out]

((x-4)^(-2*I*Pi*csgn(I*(x-4)^2)))^2*(x-4)^(4*I*Pi*csgn(I*(x-4)))*exp(-exp(ln(x-2)^2)+4*ln(x-4)^2+1-1/4*Pi^2*cs
gn(I*(x-4)^2)^6+Pi^2*csgn(I*(x-4)^2)^5*csgn(I*(x-4))-3/2*Pi^2*csgn(I*(x-4)^2)^4*csgn(I*(x-4))^2+Pi^2*csgn(I*(x
-4)^2)^3*csgn(I*(x-4))^3-1/4*Pi^2*csgn(I*(x-4)^2)^2*csgn(I*(x-4))^4-exp(x)+2*x)

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maxima [A]  time = 0.55, size = 27, normalized size = 0.90 \begin {gather*} e^{\left (4 \, \log \left (x - 4\right )^{2} + 2 \, x - e^{\left (\log \left (x - 2\right )^{2}\right )} - e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x+8)*log(x-2)*exp(log(x-2)^2)+(4*x-8)*log(x^2-8*x+16)+(-x^2+6*x-8)*exp(x)+2*x^2-12*x+16)*exp(-e
xp(log(x-2)^2)+log(x^2-8*x+16)^2-exp(x)+2*x+1)/(x^2-6*x+8),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.27, size = 34, normalized size = 1.13 \begin {gather*} {\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\ln \left (x^2-8\,x+16\right )}^2}\,\mathrm {e}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^{-{\mathrm {e}}^{{\ln \left (x-2\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x - exp(log(x - 2)^2) - exp(x) + log(x^2 - 8*x + 16)^2 + 1)*(12*x - log(x^2 - 8*x + 16)*(4*x - 8)
+ exp(x)*(x^2 - 6*x + 8) - 2*x^2 + log(x - 2)*exp(log(x - 2)^2)*(2*x - 8) - 16))/(x^2 - 6*x + 8),x)

[Out]

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

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sympy [A]  time = 3.85, size = 29, normalized size = 0.97 \begin {gather*} e^{2 x - e^{x} - e^{\log {\left (x - 2 \right )}^{2}} + \log {\left (x^{2} - 8 x + 16 \right )}^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x+8)*ln(x-2)*exp(ln(x-2)**2)+(4*x-8)*ln(x**2-8*x+16)+(-x**2+6*x-8)*exp(x)+2*x**2-12*x+16)*exp(-
exp(ln(x-2)**2)+ln(x**2-8*x+16)**2-exp(x)+2*x+1)/(x**2-6*x+8),x)

[Out]

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

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