3.3.14 \(\int \frac {e^{\frac {2}{3} (5 e^{9+6 x+x^2}-3 x-3 x^2 \log (e^2+2 x))} (-6 e^2-12 x-12 x^2+e^{9+6 x+x^2} (120 x+40 x^2+e^2 (60+20 x))+(-12 e^2 x-24 x^2) \log (e^2+2 x))}{3 e^2+6 x} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {10}{3} e^{(3+x)^2}-2 x-2 x^2 \log \left (e^2+2 x\right )} \]

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Rubi [A]  time = 2.69, antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 1, number of rules used = 1, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6706} \begin {gather*} e^{\frac {2}{3} \left (5 e^{x^2+6 x+9}-3 x\right )} \left (2 x+e^2\right )^{-2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*(5*E^(9 + 6*x + x^2) - 3*x - 3*x^2*Log[E^2 + 2*x]))/3)*(-6*E^2 - 12*x - 12*x^2 + E^(9 + 6*x + x^2)*
(120*x + 40*x^2 + E^2*(60 + 20*x)) + (-12*E^2*x - 24*x^2)*Log[E^2 + 2*x]))/(3*E^2 + 6*x),x]

[Out]

E^((2*(5*E^(9 + 6*x + x^2) - 3*x))/3)/(E^2 + 2*x)^(2*x^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} &=e^{\frac {2}{3} \left (5 e^{9+6 x+x^2}-3 x\right )} \left (e^2+2 x\right )^{-2 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 31, normalized size = 1.03 \begin {gather*} e^{\frac {10}{3} e^{(3+x)^2}-2 x} \left (e^2+2 x\right )^{-2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(5*E^(9 + 6*x + x^2) - 3*x - 3*x^2*Log[E^2 + 2*x]))/3)*(-6*E^2 - 12*x - 12*x^2 + E^(9 + 6*x +
x^2)*(120*x + 40*x^2 + E^2*(60 + 20*x)) + (-12*E^2*x - 24*x^2)*Log[E^2 + 2*x]))/(3*E^2 + 6*x),x]

[Out]

E^((10*E^(3 + x)^2)/3 - 2*x)/(E^2 + 2*x)^(2*x^2)

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fricas [A]  time = 0.61, size = 28, normalized size = 0.93 \begin {gather*} e^{\left (-2 \, x^{2} \log \left (2 \, x + e^{2}\right ) - 2 \, x + \frac {10}{3} \, e^{\left (x^{2} + 6 \, x + 9\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*exp(2)*x-24*x^2)*log(exp(2)+2*x)+((20*x+60)*exp(2)+40*x^2+120*x)*exp(x^2+6*x+9)-6*exp(2)-12*x^
2-12*x)*exp(-x^2*log(exp(2)+2*x)+5/3*exp(x^2+6*x+9)-x)^2/(3*exp(2)+6*x),x, algorithm="fricas")

[Out]

e^(-2*x^2*log(2*x + e^2) - 2*x + 10/3*e^(x^2 + 6*x + 9))

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giac [A]  time = 1.21, size = 28, normalized size = 0.93 \begin {gather*} e^{\left (-2 \, x^{2} \log \left (2 \, x + e^{2}\right ) - 2 \, x + \frac {10}{3} \, e^{\left (x^{2} + 6 \, x + 9\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*exp(2)*x-24*x^2)*log(exp(2)+2*x)+((20*x+60)*exp(2)+40*x^2+120*x)*exp(x^2+6*x+9)-6*exp(2)-12*x^
2-12*x)*exp(-x^2*log(exp(2)+2*x)+5/3*exp(x^2+6*x+9)-x)^2/(3*exp(2)+6*x),x, algorithm="giac")

[Out]

e^(-2*x^2*log(2*x + e^2) - 2*x + 10/3*e^(x^2 + 6*x + 9))

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maple [A]  time = 0.12, size = 29, normalized size = 0.97

method	result	size
risch	\(\left ({\mathrm e}^{2}+2 x \right )^{-2 x^{2}} {\mathrm e}^{\frac {10 \,{\mathrm e}^{\left (3+x \right )^{2}}}{3}-2 x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*exp(2)*x-24*x^2)*ln(exp(2)+2*x)+((20*x+60)*exp(2)+40*x^2+120*x)*exp(x^2+6*x+9)-6*exp(2)-12*x^2-12*x)
*exp(-x^2*ln(exp(2)+2*x)+5/3*exp(x^2+6*x+9)-x)^2/(3*exp(2)+6*x),x,method=_RETURNVERBOSE)

[Out]

((exp(2)+2*x)^(-x^2))^2*exp(10/3*exp((3+x)^2)-2*x)

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maxima [A]  time = 0.75, size = 28, normalized size = 0.93 \begin {gather*} e^{\left (-2 \, x^{2} \log \left (2 \, x + e^{2}\right ) - 2 \, x + \frac {10}{3} \, e^{\left (x^{2} + 6 \, x + 9\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*exp(2)*x-24*x^2)*log(exp(2)+2*x)+((20*x+60)*exp(2)+40*x^2+120*x)*exp(x^2+6*x+9)-6*exp(2)-12*x^
2-12*x)*exp(-x^2*log(exp(2)+2*x)+5/3*exp(x^2+6*x+9)-x)^2/(3*exp(2)+6*x),x, algorithm="maxima")

[Out]

e^(-2*x^2*log(2*x + e^2) - 2*x + 10/3*e^(x^2 + 6*x + 9))

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mupad [B]  time = 0.72, size = 32, normalized size = 1.07 \begin {gather*} \frac {{\mathrm {e}}^{\frac {10\,{\mathrm {e}}^{6\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^9}{3}-2\,x}}{{\left (2\,x+{\mathrm {e}}^2\right )}^{2\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((10*exp(6*x + x^2 + 9))/3 - 2*x - 2*x^2*log(2*x + exp(2)))*(12*x + 6*exp(2) + log(2*x + exp(2))*(12*
x*exp(2) + 24*x^2) - exp(6*x + x^2 + 9)*(120*x + 40*x^2 + exp(2)*(20*x + 60)) + 12*x^2))/(6*x + 3*exp(2)),x)

[Out]

exp((10*exp(6*x)*exp(x^2)*exp(9))/3 - 2*x)/(2*x + exp(2))^(2*x^2)

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sympy [A]  time = 1.01, size = 31, normalized size = 1.03 \begin {gather*} e^{- 2 x^{2} \log {\left (2 x + e^{2} \right )} - 2 x + \frac {10 e^{x^{2} + 6 x + 9}}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*exp(2)*x-24*x**2)*ln(exp(2)+2*x)+((20*x+60)*exp(2)+40*x**2+120*x)*exp(x**2+6*x+9)-6*exp(2)-12*
x**2-12*x)*exp(-x**2*ln(exp(2)+2*x)+5/3*exp(x**2+6*x+9)-x)**2/(3*exp(2)+6*x),x)

[Out]

exp(-2*x**2*log(2*x + exp(2)) - 2*x + 10*exp(x**2 + 6*x + 9)/3)

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