3.28.46 \(\int \frac {-2 e^{e^2}+e^{\frac {1}{2} e^{-e^2} (2 e^{e^2} x+3 x^2+x^3)} (2 e^{e^2}+6 x+3 x^2)}{16 e^{e^2+\frac {1}{2} e^{-e^2} (2 e^{e^2} x+3 x^2+x^3)}+e^{e^2} (-16-16 x)+(2 e^{e^2+\frac {1}{2} e^{-e^2} (2 e^{e^2} x+3 x^2+x^3)}+e^{e^2} (-2-2 x)) \log (1-e^{\frac {1}{2} e^{-e^2} (2 e^{e^2} x+3 x^2+x^3)}+x)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (8+\log \left (1-e^{x+\frac {1}{2} e^{-e^2} x^2 (3+x)}+x\right )\right ) \]

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Rubi [A]  time = 1.49, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 191, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6688, 12, 6684} \begin {gather*} \log \left (\log \left (-e^{\frac {1}{2} e^{-e^2} (x+3) x^2+x}+x+1\right )+8\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^E^2 + E^((2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2))*(2*E^E^2 + 6*x + 3*x^2))/(16*E^(E^2 + (2*E^E^2*x + 3*x
^2 + x^3)/(2*E^E^2)) + E^E^2*(-16 - 16*x) + (2*E^(E^2 + (2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + E^E^2*(-2 - 2*x
))*Log[1 - E^((2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + x]),x]

[Out]

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

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Mathematica [A]  time = 0.27, size = 30, normalized size = 1.00 \begin {gather*} \log \left (8+\log \left (1-e^{x+\frac {1}{2} e^{-e^2} x^2 (3+x)}+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^E^2 + E^((2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2))*(2*E^E^2 + 6*x + 3*x^2))/(16*E^(E^2 + (2*E^E^2*x
+ 3*x^2 + x^3)/(2*E^E^2)) + E^E^2*(-16 - 16*x) + (2*E^(E^2 + (2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + E^E^2*(-2
- 2*x))*Log[1 - E^((2*E^E^2*x + 3*x^2 + x^3)/(2*E^E^2)) + x]),x]

[Out]

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

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fricas [A]  time = 0.57, size = 46, normalized size = 1.53 \begin {gather*} \log \left (\log \left ({\left ({\left (x + 1\right )} e^{\left (e^{2}\right )} - e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, {\left (x + e^{2}\right )} e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )}\right )}\right )} e^{\left (-e^{2}\right )}\right ) + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))-2*exp(exp(2)))/((2*exp(e
xp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))+(-2*x-2)*exp(exp(2)))*log(-exp(1/2*(2*x*exp(exp(2))+x^
3+3*x^2)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))+(-16*x-16)*exp(exp(
2))),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (3 \, x^{2} + 6 \, x + 2 \, e^{\left (e^{2}\right )}\right )} e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, x e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )}\right )} - 2 \, e^{\left (e^{2}\right )}}{2 \, {\left (8 \, {\left (x + 1\right )} e^{\left (e^{2}\right )} + {\left ({\left (x + 1\right )} e^{\left (e^{2}\right )} - e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, x e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )} + e^{2}\right )}\right )} \log \left (x - e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, x e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )}\right )} + 1\right ) - 8 \, e^{\left (\frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 2 \, x e^{\left (e^{2}\right )}\right )} e^{\left (-e^{2}\right )} + e^{2}\right )}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))-2*exp(exp(2)))/((2*exp(e
xp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))+(-2*x-2)*exp(exp(2)))*log(-exp(1/2*(2*x*exp(exp(2))+x^
3+3*x^2)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))+(-16*x-16)*exp(exp(
2))),x, algorithm="giac")

[Out]

integrate(-1/2*((3*x^2 + 6*x + 2*e^(e^2))*e^(1/2*(x^3 + 3*x^2 + 2*x*e^(e^2))*e^(-e^2)) - 2*e^(e^2))/(8*(x + 1)
*e^(e^2) + ((x + 1)*e^(e^2) - e^(1/2*(x^3 + 3*x^2 + 2*x*e^(e^2))*e^(-e^2) + e^2))*log(x - e^(1/2*(x^3 + 3*x^2
+ 2*x*e^(e^2))*e^(-e^2)) + 1) - 8*e^(1/2*(x^3 + 3*x^2 + 2*x*e^(e^2))*e^(-e^2) + e^2)), x)

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maple [A]  time = 0.45, size = 31, normalized size = 1.03

method	result	size
risch	\(\ln \left (\ln \left (-{\mathrm e}^{\frac {x \left (x^{2}+2 \,{\mathrm e}^{{\mathrm e}^{2}}+3 x \right ) {\mathrm e}^{-{\mathrm e}^{2}}}{2}}+x +1\right )+8\right )\)	\(31\)
norman	\(\ln \left (\ln \left (-{\mathrm e}^{\frac {\left (2 x \,{\mathrm e}^{{\mathrm e}^{2}}+x^{3}+3 x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{2}}}{2}}+x +1\right )+8\right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))-2*exp(exp(2)))/((2*exp(exp(2))
*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))+(-2*x-2)*exp(exp(2)))*ln(-exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2
)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))+(-16*x-16)*exp(exp(2))),x,
method=_RETURNVERBOSE)

[Out]

ln(ln(-exp(1/2*x*(x^2+2*exp(exp(2))+3*x)*exp(-exp(2)))+x+1)+8)

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maxima [A]  time = 0.57, size = 32, normalized size = 1.07 \begin {gather*} \log \left (\log \left (x - e^{\left (\frac {1}{2} \, x^{3} e^{\left (-e^{2}\right )} + \frac {3}{2} \, x^{2} e^{\left (-e^{2}\right )} + x\right )} + 1\right ) + 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(exp(2))+3*x^2+6*x)*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))-2*exp(exp(2)))/((2*exp(e
xp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))+(-2*x-2)*exp(exp(2)))*log(-exp(1/2*(2*x*exp(exp(2))+x^
3+3*x^2)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x^3+3*x^2)/exp(exp(2)))+(-16*x-16)*exp(exp(
2))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.42, size = 32, normalized size = 1.07 \begin {gather*} \ln \left (\ln \left (x-{\mathrm {e}}^{\frac {{\mathrm {e}}^{-{\mathrm {e}}^2}\,x^3}{2}+\frac {3\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,x^2}{2}+x}+1\right )+8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(exp(2)) - exp(exp(-exp(2))*(x*exp(exp(2)) + (3*x^2)/2 + x^3/2))*(6*x + 2*exp(exp(2)) + 3*x^2))/(exp
(exp(2))*(16*x + 16) - 16*exp(exp(-exp(2))*(x*exp(exp(2)) + (3*x^2)/2 + x^3/2))*exp(exp(2)) + log(x - exp(exp(
-exp(2))*(x*exp(exp(2)) + (3*x^2)/2 + x^3/2)) + 1)*(exp(exp(2))*(2*x + 2) - 2*exp(exp(-exp(2))*(x*exp(exp(2))
+ (3*x^2)/2 + x^3/2))*exp(exp(2)))),x)

[Out]

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

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sympy [A]  time = 1.10, size = 32, normalized size = 1.07 \begin {gather*} \log {\left (\log {\left (x - e^{\frac {\frac {x^{3}}{2} + \frac {3 x^{2}}{2} + x e^{e^{2}}}{e^{e^{2}}}} + 1 \right )} + 8 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(exp(2))+3*x**2+6*x)*exp(1/2*(2*x*exp(exp(2))+x**3+3*x**2)/exp(exp(2)))-2*exp(exp(2)))/((2*ex
p(exp(2))*exp(1/2*(2*x*exp(exp(2))+x**3+3*x**2)/exp(exp(2)))+(-2*x-2)*exp(exp(2)))*ln(-exp(1/2*(2*x*exp(exp(2)
)+x**3+3*x**2)/exp(exp(2)))+x+1)+16*exp(exp(2))*exp(1/2*(2*x*exp(exp(2))+x**3+3*x**2)/exp(exp(2)))+(-16*x-16)*
exp(exp(2))),x)

[Out]

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

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