3.16.89 \(\int \frac {e^7 (-8-8 x)+e^{2 e^{-2+x}} (2 e^2+4 e^x x)}{(e^{2+2 e^{-2+x}} x+e^7 (-4 x-2 x^2)) \log (\frac {e^{4 e^{-2+x}} x^2+e^{5+2 e^{-2+x}} (-8 x^2-4 x^3)+e^{10} (16 x^2+16 x^3+4 x^4)}{e^{10}})} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\log \left (\left (x+x \left (3-e^{-5+2 e^{-2+x}}+2 x\right )\right )^2\right )\right ) \]

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Rubi [A]  time = 0.73, antiderivative size = 29, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, integrand size = 126, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6688, 12, 6684} \begin {gather*} \log \left (\log \left (\frac {x^2 \left (e^{2 e^{x-2}}-2 e^5 (x+2)\right )^2}{e^{10}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^7*(-8 - 8*x) + E^(2*E^(-2 + x))*(2*E^2 + 4*E^x*x))/((E^(2 + 2*E^(-2 + x))*x + E^7*(-4*x - 2*x^2))*Log[(
E^(4*E^(-2 + x))*x^2 + E^(5 + 2*E^(-2 + x))*(-8*x^2 - 4*x^3) + E^10*(16*x^2 + 16*x^3 + 4*x^4))/E^10]),x]

[Out]

Log[Log[(x^2*(E^(2*E^(-2 + x)) - 2*E^5*(2 + x))^2)/E^10]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{2+2 e^{-2+x}}+4 e^{2 e^{-2+x}+x} x-8 e^7 (1+x)}{e^2 x \left (e^{2 e^{-2+x}}-2 e^5 (2+x)\right ) \log \left (\frac {x^2 \left (e^{2 e^{-2+x}}-2 e^5 (2+x)\right )^2}{e^{10}}\right )} \, dx\\ &=\frac {\int \frac {2 e^{2+2 e^{-2+x}}+4 e^{2 e^{-2+x}+x} x-8 e^7 (1+x)}{x \left (e^{2 e^{-2+x}}-2 e^5 (2+x)\right ) \log \left (\frac {x^2 \left (e^{2 e^{-2+x}}-2 e^5 (2+x)\right )^2}{e^{10}}\right )} \, dx}{e^2}\\ &=\log \left (\log \left (\frac {x^2 \left (e^{2 e^{-2+x}}-2 e^5 (2+x)\right )^2}{e^{10}}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 29, normalized size = 1.12 \begin {gather*} \log \left (\log \left (\frac {x^2 \left (e^{2 e^{-2+x}}-2 e^5 (2+x)\right )^2}{e^{10}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^7*(-8 - 8*x) + E^(2*E^(-2 + x))*(2*E^2 + 4*E^x*x))/((E^(2 + 2*E^(-2 + x))*x + E^7*(-4*x - 2*x^2))
*Log[(E^(4*E^(-2 + x))*x^2 + E^(5 + 2*E^(-2 + x))*(-8*x^2 - 4*x^3) + E^10*(16*x^2 + 16*x^3 + 4*x^4))/E^10]),x]

[Out]

Log[Log[(x^2*(E^(2*E^(-2 + x)) - 2*E^5*(2 + x))^2)/E^10]]

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fricas [B]  time = 0.79, size = 68, normalized size = 2.62 \begin {gather*} \log \left (\log \left ({\left (x^{2} e^{\left (2 \, {\left (5 \, e^{2} + 2 \, e^{x}\right )} e^{\left (-2\right )}\right )} + 4 \, {\left (x^{4} + 4 \, x^{3} + 4 \, x^{2}\right )} e^{20} - 4 \, {\left (x^{3} + 2 \, x^{2}\right )} e^{\left ({\left (5 \, e^{2} + 2 \, e^{x}\right )} e^{\left (-2\right )} + 10\right )}\right )} e^{\left (-20\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x+2*exp(2))*exp(exp(x)/exp(2))^2+(-8*x-8)*exp(2)*exp(5))/(x*exp(2)*exp(exp(x)/exp(2))^2+(
-2*x^2-4*x)*exp(2)*exp(5))/log((x^2*exp(exp(x)/exp(2))^4+(-4*x^3-8*x^2)*exp(5)*exp(exp(x)/exp(2))^2+(4*x^4+16*
x^3+16*x^2)*exp(5)^2)/exp(5)^2),x, algorithm="fricas")

[Out]

log(log((x^2*e^(2*(5*e^2 + 2*e^x)*e^(-2)) + 4*(x^4 + 4*x^3 + 4*x^2)*e^20 - 4*(x^3 + 2*x^2)*e^((5*e^2 + 2*e^x)*
e^(-2) + 10))*e^(-20)))

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giac [B]  time = 0.65, size = 60, normalized size = 2.31 \begin {gather*} \log \left (-\log \left (4 \, x^{2} e^{10} + 16 \, x e^{10} - 4 \, x e^{\left (2 \, e^{\left (x - 2\right )} + 5\right )} + 16 \, e^{10} + e^{\left (4 \, e^{\left (x - 2\right )}\right )} - 8 \, e^{\left (2 \, e^{\left (x - 2\right )} + 5\right )}\right ) - 2 \, \log \left (x \mathrm {sgn}\relax (x)\right ) + 10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x+2*exp(2))*exp(exp(x)/exp(2))^2+(-8*x-8)*exp(2)*exp(5))/(x*exp(2)*exp(exp(x)/exp(2))^2+(
-2*x^2-4*x)*exp(2)*exp(5))/log((x^2*exp(exp(x)/exp(2))^4+(-4*x^3-8*x^2)*exp(5)*exp(exp(x)/exp(2))^2+(4*x^4+16*
x^3+16*x^2)*exp(5)^2)/exp(5)^2),x, algorithm="giac")

[Out]

log(-log(4*x^2*e^10 + 16*x*e^10 - 4*x*e^(2*e^(x - 2) + 5) + 16*e^10 + e^(4*e^(x - 2)) - 8*e^(2*e^(x - 2) + 5))
- 2*log(x*sgn(x)) + 10)

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maple [C]  time = 0.34, size = 373, normalized size = 14.35

method	result	size
risch	\(\ln \left (\ln \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )-\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right ) \mathrm {csgn}\left (i x^{2} \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right )-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{2} \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )\right )^{2} \mathrm {csgn}\left (i \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right )-2 \pi \,\mathrm {csgn}\left (i \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )\right ) \mathrm {csgn}\left (i \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right ) \mathrm {csgn}\left (i x^{2} \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2} \left (\left (2+x \right ) {\mathrm e}^{5}-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x -2}}}{2}\right )^{2}\right )^{3}+4 i \ln \relax (x )+4 i \ln \relax (2)-20 i\right )}{4}\right )\)	\(373\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(x)*x+2*exp(2))*exp(exp(x)/exp(2))^2+(-8*x-8)*exp(2)*exp(5))/(x*exp(2)*exp(exp(x)/exp(2))^2+(-2*x^2
-4*x)*exp(2)*exp(5))/ln((x^2*exp(exp(x)/exp(2))^4+(-4*x^3-8*x^2)*exp(5)*exp(exp(x)/exp(2))^2+(4*x^4+16*x^3+16*
x^2)*exp(5)^2)/exp(5)^2),x,method=_RETURNVERBOSE)

[Out]

ln(ln((2+x)*exp(5)-1/2*exp(2*exp(x-2)))-1/4*I*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn
(I*x^2)^3+Pi*csgn(I*x^2)*csgn(I*((2+x)*exp(5)-1/2*exp(2*exp(x-2)))^2)*csgn(I*x^2*((2+x)*exp(5)-1/2*exp(2*exp(x
-2)))^2)-Pi*csgn(I*x^2)*csgn(I*x^2*((2+x)*exp(5)-1/2*exp(2*exp(x-2)))^2)^2+Pi*csgn(I*((2+x)*exp(5)-1/2*exp(2*e
xp(x-2))))^2*csgn(I*((2+x)*exp(5)-1/2*exp(2*exp(x-2)))^2)-2*Pi*csgn(I*((2+x)*exp(5)-1/2*exp(2*exp(x-2))))*csgn
(I*((2+x)*exp(5)-1/2*exp(2*exp(x-2)))^2)^2+Pi*csgn(I*((2+x)*exp(5)-1/2*exp(2*exp(x-2)))^2)^3-Pi*csgn(I*((2+x)*
exp(5)-1/2*exp(2*exp(x-2)))^2)*csgn(I*x^2*((2+x)*exp(5)-1/2*exp(2*exp(x-2)))^2)^2+Pi*csgn(I*x^2*((2+x)*exp(5)-
1/2*exp(2*exp(x-2)))^2)^3+4*I*ln(x)+4*I*ln(2)-20*I))

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maxima [A]  time = 0.55, size = 25, normalized size = 0.96 \begin {gather*} \log \left (\log \left (2 \, x e^{5} + 4 \, e^{5} - e^{\left (2 \, e^{\left (x - 2\right )}\right )}\right ) + \log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x+2*exp(2))*exp(exp(x)/exp(2))^2+(-8*x-8)*exp(2)*exp(5))/(x*exp(2)*exp(exp(x)/exp(2))^2+(
-2*x^2-4*x)*exp(2)*exp(5))/log((x^2*exp(exp(x)/exp(2))^4+(-4*x^3-8*x^2)*exp(5)*exp(exp(x)/exp(2))^2+(4*x^4+16*
x^3+16*x^2)*exp(5)^2)/exp(5)^2),x, algorithm="maxima")

[Out]

log(log(2*x*e^5 + 4*e^5 - e^(2*e^(x - 2))) + log(x) - 5)

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mupad [B]  time = 2.22, size = 57, normalized size = 2.19 \begin {gather*} \ln \left (\ln \left (x^2\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}+{\mathrm {e}}^{10}\,\left (4\,x^4+16\,x^3+16\,x^2\right )-{\mathrm {e}}^{2\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^5\,\left (4\,x^3+8\,x^2\right )\right )-10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*exp(-2)*exp(x))*(2*exp(2) + 4*x*exp(x)) - exp(7)*(8*x + 8))/(log(exp(-10)*(x^2*exp(4*exp(-2)*exp(x
)) + exp(10)*(16*x^2 + 16*x^3 + 4*x^4) - exp(2*exp(-2)*exp(x))*exp(5)*(8*x^2 + 4*x^3)))*(exp(7)*(4*x + 2*x^2)
- x*exp(2*exp(-2)*exp(x))*exp(2))),x)

[Out]

log(log(x^2*exp(4*exp(-2)*exp(x)) + exp(10)*(16*x^2 + 16*x^3 + 4*x^4) - exp(2*exp(-2)*exp(x))*exp(5)*(8*x^2 +
4*x^3)) - 10)

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sympy [B]  time = 1.06, size = 63, normalized size = 2.42 \begin {gather*} \log {\left (\log {\left (\frac {x^{2} e^{\frac {4 e^{x}}{e^{2}}} + \left (- 4 x^{3} - 8 x^{2}\right ) e^{5} e^{\frac {2 e^{x}}{e^{2}}} + \left (4 x^{4} + 16 x^{3} + 16 x^{2}\right ) e^{10}}{e^{10}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(x)*x+2*exp(2))*exp(exp(x)/exp(2))**2+(-8*x-8)*exp(2)*exp(5))/(x*exp(2)*exp(exp(x)/exp(2))**2
+(-2*x**2-4*x)*exp(2)*exp(5))/ln((x**2*exp(exp(x)/exp(2))**4+(-4*x**3-8*x**2)*exp(5)*exp(exp(x)/exp(2))**2+(4*
x**4+16*x**3+16*x**2)*exp(5)**2)/exp(5)**2),x)

[Out]

log(log((x**2*exp(4*exp(-2)*exp(x)) + (-4*x**3 - 8*x**2)*exp(5)*exp(2*exp(-2)*exp(x)) + (4*x**4 + 16*x**3 + 16
*x**2)*exp(10))*exp(-10)))

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