3.30.76 \(\int \frac {1}{3} e^{\frac {1}{3} (3 e^{x^2-2 x \log (5)+\log ^2(5)}-e^{4 e^3} x^2+4 e^{3 e^3} x^3-6 e^{2 e^3} x^4+4 e^{e^3} x^5-x^6)} (-2 e^{4 e^3} x+12 e^{3 e^3} x^2-24 e^{2 e^3} x^3+20 e^{e^3} x^4-6 x^5+e^{x^2-2 x \log (5)+\log ^2(5)} (6 x-6 \log (5))) \, dx\)

Optimal. Leaf size=31 \[ e^{e^{(x-\log (5))^2}-\frac {1}{3} x^2 \left (-e^{e^3}+x\right )^4} \]

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Rubi [B]  time = 7.64, antiderivative size = 75, normalized size of antiderivative = 2.42, number of steps used = 2, number of rules used = 2, integrand size = 153, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {12, 6706} \begin {gather*} \exp \left (\frac {1}{3} \left (-x^6+4 e^{e^3} x^5-6 e^{2 e^3} x^4+4 e^{3 e^3} x^3-e^{4 e^3} x^2+3\ 5^{-2 x} e^{x^2+\log ^2(5)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((3*E^(x^2 - 2*x*Log[5] + Log[5]^2) - E^(4*E^3)*x^2 + 4*E^(3*E^3)*x^3 - 6*E^(2*E^3)*x^4 + 4*E^E^3*x^5 -
x^6)/3)*(-2*E^(4*E^3)*x + 12*E^(3*E^3)*x^2 - 24*E^(2*E^3)*x^3 + 20*E^E^3*x^4 - 6*x^5 + E^(x^2 - 2*x*Log[5] +
Log[5]^2)*(6*x - 6*Log[5])))/3,x]

[Out]

E^(((3*E^(x^2 + Log[5]^2))/5^(2*x) - E^(4*E^3)*x^2 + 4*E^(3*E^3)*x^3 - 6*E^(2*E^3)*x^4 + 4*E^E^3*x^5 - x^6)/3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{3} \int \exp \left (\frac {1}{3} \left (3 e^{x^2-2 x \log (5)+\log ^2(5)}-e^{4 e^3} x^2+4 e^{3 e^3} x^3-6 e^{2 e^3} x^4+4 e^{e^3} x^5-x^6\right )\right ) \left (-2 e^{4 e^3} x+12 e^{3 e^3} x^2-24 e^{2 e^3} x^3+20 e^{e^3} x^4-6 x^5+e^{x^2-2 x \log (5)+\log ^2(5)} (6 x-6 \log (5))\right ) \, dx\\ &=\exp \left (\frac {1}{3} \left (3\ 5^{-2 x} e^{x^2+\log ^2(5)}-e^{4 e^3} x^2+4 e^{3 e^3} x^3-6 e^{2 e^3} x^4+4 e^{e^3} x^5-x^6\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.27, size = 75, normalized size = 2.42 \begin {gather*} e^{\frac {1}{3} \left (3\ 5^{-2 x} e^{x^2+\log ^2(5)}-e^{4 e^3} x^2+4 e^{3 e^3} x^3-6 e^{2 e^3} x^4+4 e^{e^3} x^5-x^6\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3*E^(x^2 - 2*x*Log[5] + Log[5]^2) - E^(4*E^3)*x^2 + 4*E^(3*E^3)*x^3 - 6*E^(2*E^3)*x^4 + 4*E^E^3
*x^5 - x^6)/3)*(-2*E^(4*E^3)*x + 12*E^(3*E^3)*x^2 - 24*E^(2*E^3)*x^3 + 20*E^E^3*x^4 - 6*x^5 + E^(x^2 - 2*x*Log
[5] + Log[5]^2)*(6*x - 6*Log[5])))/3,x]

[Out]

E^(((3*E^(x^2 + Log[5]^2))/5^(2*x) - E^(4*E^3)*x^2 + 4*E^(3*E^3)*x^3 - 6*E^(2*E^3)*x^4 + 4*E^E^3*x^5 - x^6)/3)

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fricas [B]  time = 0.63, size = 59, normalized size = 1.90 \begin {gather*} e^{\left (-\frac {1}{3} \, x^{6} + \frac {4}{3} \, x^{5} e^{\left (e^{3}\right )} - 2 \, x^{4} e^{\left (2 \, e^{3}\right )} + \frac {4}{3} \, x^{3} e^{\left (3 \, e^{3}\right )} - \frac {1}{3} \, x^{2} e^{\left (4 \, e^{3}\right )} + e^{\left (x^{2} - 2 \, x \log \relax (5) + \log \relax (5)^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-6*log(5)+6*x)*exp(log(5)^2-2*x*log(5)+x^2)-2*x*exp(exp(3))^4+12*x^2*exp(exp(3))^3-24*x^3*exp(
exp(3))^2+20*x^4*exp(exp(3))-6*x^5)*exp(exp(log(5)^2-2*x*log(5)+x^2)-1/3*x^2*exp(exp(3))^4+4/3*x^3*exp(exp(3))
^3-2*x^4*exp(exp(3))^2+4/3*x^5*exp(exp(3))-1/3*x^6),x, algorithm="fricas")

[Out]

e^(-1/3*x^6 + 4/3*x^5*e^(e^3) - 2*x^4*e^(2*e^3) + 4/3*x^3*e^(3*e^3) - 1/3*x^2*e^(4*e^3) + e^(x^2 - 2*x*log(5)
+ log(5)^2))

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giac [B]  time = 0.42, size = 59, normalized size = 1.90 \begin {gather*} e^{\left (-\frac {1}{3} \, x^{6} + \frac {4}{3} \, x^{5} e^{\left (e^{3}\right )} - 2 \, x^{4} e^{\left (2 \, e^{3}\right )} + \frac {4}{3} \, x^{3} e^{\left (3 \, e^{3}\right )} - \frac {1}{3} \, x^{2} e^{\left (4 \, e^{3}\right )} + e^{\left (x^{2} - 2 \, x \log \relax (5) + \log \relax (5)^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-6*log(5)+6*x)*exp(log(5)^2-2*x*log(5)+x^2)-2*x*exp(exp(3))^4+12*x^2*exp(exp(3))^3-24*x^3*exp(
exp(3))^2+20*x^4*exp(exp(3))-6*x^5)*exp(exp(log(5)^2-2*x*log(5)+x^2)-1/3*x^2*exp(exp(3))^4+4/3*x^3*exp(exp(3))
^3-2*x^4*exp(exp(3))^2+4/3*x^5*exp(exp(3))-1/3*x^6),x, algorithm="giac")

[Out]

e^(-1/3*x^6 + 4/3*x^5*e^(e^3) - 2*x^4*e^(2*e^3) + 4/3*x^3*e^(3*e^3) - 1/3*x^2*e^(4*e^3) + e^(x^2 - 2*x*log(5)
+ log(5)^2))

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maple [B]  time = 0.16, size = 59, normalized size = 1.90

method	result	size
risch	\({\mathrm e}^{\left (\frac {1}{25}\right )^{x} {\mathrm e}^{\ln \relax (5)^{2}+x^{2}}-\frac {x^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}}}{3}+\frac {4 x^{3} {\mathrm e}^{3 \,{\mathrm e}^{3}}}{3}-2 x^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}+\frac {4 x^{5} {\mathrm e}^{{\mathrm e}^{3}}}{3}-\frac {x^{6}}{3}}\)	\(59\)
norman	\({\mathrm e}^{{\mathrm e}^{\ln \relax (5)^{2}-2 x \ln \relax (5)+x^{2}}-\frac {x^{2} {\mathrm e}^{4 \,{\mathrm e}^{3}}}{3}+\frac {4 x^{3} {\mathrm e}^{3 \,{\mathrm e}^{3}}}{3}-2 x^{4} {\mathrm e}^{2 \,{\mathrm e}^{3}}+\frac {4 x^{5} {\mathrm e}^{{\mathrm e}^{3}}}{3}-\frac {x^{6}}{3}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-6*ln(5)+6*x)*exp(ln(5)^2-2*x*ln(5)+x^2)-2*x*exp(exp(3))^4+12*x^2*exp(exp(3))^3-24*x^3*exp(exp(3))^2
+20*x^4*exp(exp(3))-6*x^5)*exp(exp(ln(5)^2-2*x*ln(5)+x^2)-1/3*x^2*exp(exp(3))^4+4/3*x^3*exp(exp(3))^3-2*x^4*ex
p(exp(3))^2+4/3*x^5*exp(exp(3))-1/3*x^6),x,method=_RETURNVERBOSE)

[Out]

exp((1/25)^x*exp(ln(5)^2+x^2)-1/3*x^2*exp(4*exp(3))+4/3*x^3*exp(3*exp(3))-2*x^4*exp(2*exp(3))+4/3*x^5*exp(exp(
3))-1/3*x^6)

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maxima [B]  time = 1.09, size = 59, normalized size = 1.90 \begin {gather*} e^{\left (-\frac {1}{3} \, x^{6} + \frac {4}{3} \, x^{5} e^{\left (e^{3}\right )} - 2 \, x^{4} e^{\left (2 \, e^{3}\right )} + \frac {4}{3} \, x^{3} e^{\left (3 \, e^{3}\right )} - \frac {1}{3} \, x^{2} e^{\left (4 \, e^{3}\right )} + e^{\left (x^{2} - 2 \, x \log \relax (5) + \log \relax (5)^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-6*log(5)+6*x)*exp(log(5)^2-2*x*log(5)+x^2)-2*x*exp(exp(3))^4+12*x^2*exp(exp(3))^3-24*x^3*exp(
exp(3))^2+20*x^4*exp(exp(3))-6*x^5)*exp(exp(log(5)^2-2*x*log(5)+x^2)-1/3*x^2*exp(exp(3))^4+4/3*x^3*exp(exp(3))
^3-2*x^4*exp(exp(3))^2+4/3*x^5*exp(exp(3))-1/3*x^6),x, algorithm="maxima")

[Out]

e^(-1/3*x^6 + 4/3*x^5*e^(e^3) - 2*x^4*e^(2*e^3) + 4/3*x^3*e^(3*e^3) - 1/3*x^2*e^(4*e^3) + e^(x^2 - 2*x*log(5)
+ log(5)^2))

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mupad [B]  time = 2.22, size = 63, normalized size = 2.03 \begin {gather*} {\mathrm {e}}^{{\left (\frac {1}{25}\right )}^x\,{\mathrm {e}}^{{\ln \relax (5)}^2}\,{\mathrm {e}}^{x^2}}\,{\mathrm {e}}^{\frac {4\,x^5\,{\mathrm {e}}^{{\mathrm {e}}^3}}{3}}\,{\mathrm {e}}^{-\frac {x^6}{3}}\,{\mathrm {e}}^{-2\,x^4\,{\mathrm {e}}^{2\,{\mathrm {e}}^3}}\,{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^{4\,{\mathrm {e}}^3}}{3}}\,{\mathrm {e}}^{\frac {4\,x^3\,{\mathrm {e}}^{3\,{\mathrm {e}}^3}}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(log(5)^2 - 2*x*log(5) + x^2) + (4*x^5*exp(exp(3)))/3 - (x^2*exp(4*exp(3)))/3 + (4*x^3*exp(3*exp(
3)))/3 - 2*x^4*exp(2*exp(3)) - x^6/3)*(2*x*exp(4*exp(3)) - 20*x^4*exp(exp(3)) - exp(log(5)^2 - 2*x*log(5) + x^
2)*(6*x - 6*log(5)) - 12*x^2*exp(3*exp(3)) + 24*x^3*exp(2*exp(3)) + 6*x^5))/3,x)

[Out]

exp((1/25)^x*exp(log(5)^2)*exp(x^2))*exp((4*x^5*exp(exp(3)))/3)*exp(-x^6/3)*exp(-2*x^4*exp(2*exp(3)))*exp(-(x^
2*exp(4*exp(3)))/3)*exp((4*x^3*exp(3*exp(3)))/3)

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sympy [B]  time = 0.55, size = 71, normalized size = 2.29 \begin {gather*} e^{- \frac {x^{6}}{3} + \frac {4 x^{5} e^{e^{3}}}{3} - 2 x^{4} e^{2 e^{3}} + \frac {4 x^{3} e^{3 e^{3}}}{3} - \frac {x^{2} e^{4 e^{3}}}{3} + e^{x^{2} - 2 x \log {\relax (5 )} + \log {\relax (5 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-6*ln(5)+6*x)*exp(ln(5)**2-2*x*ln(5)+x**2)-2*x*exp(exp(3))**4+12*x**2*exp(exp(3))**3-24*x**3*e
xp(exp(3))**2+20*x**4*exp(exp(3))-6*x**5)*exp(exp(ln(5)**2-2*x*ln(5)+x**2)-1/3*x**2*exp(exp(3))**4+4/3*x**3*ex
p(exp(3))**3-2*x**4*exp(exp(3))**2+4/3*x**5*exp(exp(3))-1/3*x**6),x)

[Out]

exp(-x**6/3 + 4*x**5*exp(exp(3))/3 - 2*x**4*exp(2*exp(3)) + 4*x**3*exp(3*exp(3))/3 - x**2*exp(4*exp(3))/3 + ex
p(x**2 - 2*x*log(5) + log(5)**2))

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