3.3.46 \(\int \frac {1}{3} e^{\frac {1}{3} (15-52 x+9 x^2-5 x^3+e^3 (-21+15 x))} (-52+15 e^3+18 x-15 x^2) \, dx\)

Optimal. Leaf size=35 \[ e^{x-(3+x)^2+(7-5 x) \left (2-e^3+\frac {1}{3} \left (-x+x^2\right )\right )} \]

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Rubi [A]  time = 0.17, antiderivative size = 31, normalized size of antiderivative = 0.89, number of steps used = 2, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {12, 6706} \begin {gather*} \exp \left (\frac {1}{3} \left (-5 x^3+9 x^2-52 x-3 e^3 (7-5 x)+15\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((15 - 52*x + 9*x^2 - 5*x^3 + E^3*(-21 + 15*x))/3)*(-52 + 15*E^3 + 18*x - 15*x^2))/3,x]

[Out]

E^((15 - 3*E^3*(7 - 5*x) - 52*x + 9*x^2 - 5*x^3)/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 (15-52 x+9 x^2-5 x^3+e^3 (-21+15 x)\right )\right ) \left (-52+15 e^3+18 x-15 x^2\right ) \, dx\\ &=\exp \left (\frac {1}{3} \left (15-3 e^3 (7-5 x)-52 x+9 x^2-5 x^3\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 30, normalized size = 0.86 \begin {gather*} e^{5-\frac {52 x}{3}+3 x^2-\frac {5 x^3}{3}+e^3 (-7+5 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((15 - 52*x + 9*x^2 - 5*x^3 + E^3*(-21 + 15*x))/3)*(-52 + 15*E^3 + 18*x - 15*x^2))/3,x]

[Out]

E^(5 - (52*x)/3 + 3*x^2 - (5*x^3)/3 + E^3*(-7 + 5*x))

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fricas [A]  time = 0.64, size = 24, normalized size = 0.69 \begin {gather*} e^{\left (-\frac {5}{3} \, x^{3} + 3 \, x^{2} + {\left (5 \, x - 7\right )} e^{3} - \frac {52}{3} \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(15*exp(3)-15*x^2+18*x-52)*exp(1/3*(15*x-21)*exp(3)-5/3*x^3+3*x^2-52/3*x+5),x, algorithm="fricas
")

[Out]

e^(-5/3*x^3 + 3*x^2 + (5*x - 7)*e^3 - 52/3*x + 5)

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giac [A]  time = 0.33, size = 25, normalized size = 0.71 \begin {gather*} e^{\left (-\frac {5}{3} \, x^{3} + 3 \, x^{2} + 5 \, x e^{3} - \frac {52}{3} \, x - 7 \, e^{3} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(15*exp(3)-15*x^2+18*x-52)*exp(1/3*(15*x-21)*exp(3)-5/3*x^3+3*x^2-52/3*x+5),x, algorithm="giac")

[Out]

e^(-5/3*x^3 + 3*x^2 + 5*x*e^3 - 52/3*x - 7*e^3 + 5)

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maple [A]  time = 0.13, size = 26, normalized size = 0.74

method	result	size
gosper	\({\mathrm e}^{5 x \,{\mathrm e}^{3}-7 \,{\mathrm e}^{3}-\frac {5 x^{3}}{3}+3 x^{2}-\frac {52 x}{3}+5}\)	\(26\)
norman	\({\mathrm e}^{\frac {\left (15 x -21\right ) {\mathrm e}^{3}}{3}-\frac {5 x^{3}}{3}+3 x^{2}-\frac {52 x}{3}+5}\)	\(26\)
risch	\({\mathrm e}^{5 x \,{\mathrm e}^{3}-7 \,{\mathrm e}^{3}-\frac {5 x^{3}}{3}+3 x^{2}-\frac {52 x}{3}+5}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(15*exp(3)-15*x^2+18*x-52)*exp(1/3*(15*x-21)*exp(3)-5/3*x^3+3*x^2-52/3*x+5),x,method=_RETURNVERBOSE)

[Out]

exp(5*x*exp(3)-7*exp(3)-5/3*x^3+3*x^2-52/3*x+5)

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maxima [A]  time = 0.44, size = 24, normalized size = 0.69 \begin {gather*} e^{\left (-\frac {5}{3} \, x^{3} + 3 \, x^{2} + {\left (5 \, x - 7\right )} e^{3} - \frac {52}{3} \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(15*exp(3)-15*x^2+18*x-52)*exp(1/3*(15*x-21)*exp(3)-5/3*x^3+3*x^2-52/3*x+5),x, algorithm="maxima
")

[Out]

e^(-5/3*x^3 + 3*x^2 + (5*x - 7)*e^3 - 52/3*x + 5)

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mupad [B]  time = 0.37, size = 30, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^{-7\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-\frac {52\,x}{3}}\,{\mathrm {e}}^5\,{\mathrm {e}}^{3\,x^2}\,{\mathrm {e}}^{-\frac {5\,x^3}{3}}\,{\mathrm {e}}^{5\,x\,{\mathrm {e}}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*x^2 - (52*x)/3 - (5*x^3)/3 + (exp(3)*(15*x - 21))/3 + 5)*(18*x + 15*exp(3) - 15*x^2 - 52))/3,x)

[Out]

exp(-7*exp(3))*exp(-(52*x)/3)*exp(5)*exp(3*x^2)*exp(-(5*x^3)/3)*exp(5*x*exp(3))

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sympy [A]  time = 0.15, size = 27, normalized size = 0.77 \begin {gather*} e^{- \frac {5 x^{3}}{3} + 3 x^{2} - \frac {52 x}{3} + \left (5 x - 7\right ) e^{3} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(15*exp(3)-15*x**2+18*x-52)*exp(1/3*(15*x-21)*exp(3)-5/3*x**3+3*x**2-52/3*x+5),x)

[Out]

exp(-5*x**3/3 + 3*x**2 - 52*x/3 + (5*x - 7)*exp(3) + 5)

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