3.93.89 \(\int \frac {1}{6} e^{\frac {1}{2} (e^x-4 e^{\frac {1}{3} (4 x+6 x^2)}+e^5 x)} (3 e^5+3 e^x+e^{\frac {1}{3} (4 x+6 x^2)} (-16-48 x)) \, dx\)

Optimal. Leaf size=27 \[ e^{\frac {1}{2} \left (e^x-4 e^{2 x \left (\frac {2}{3}+x\right )}+e^5 x\right )} \]

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Rubi [A]  time = 0.29, antiderivative size = 32, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {12, 6706} \begin {gather*} \exp \left (\frac {1}{2} \left (-4 e^{\frac {2}{3} \left (3 x^2+2 x\right )}+e^5 x+e^x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((E^x - 4*E^((4*x + 6*x^2)/3) + E^5*x)/2)*(3*E^5 + 3*E^x + E^((4*x + 6*x^2)/3)*(-16 - 48*x)))/6,x]

[Out]

E^((E^x - 4*E^((2*(2*x + 3*x^2))/3) + E^5*x)/2)

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}{6} \int \exp \left (\frac {1}{2} \left (e^x-4 e^{\frac {1}{3} \left (4 x+6 x^2\right )}+e^5 x\right )\right ) \left (3 e^5+3 e^x+e^{\frac {1}{3} \left (4 x+6 x^2\right )} (-16-48 x)\right ) \, dx\\ &=\exp \left (\frac {1}{2} \left (e^x-4 e^{\frac {2}{3} \left (2 x+3 x^2\right )}+e^5 x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.76, size = 29, normalized size = 1.07 \begin {gather*} e^{\frac {1}{2} \left (e^x-4 e^{\frac {2}{3} x (2+3 x)}+e^5 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((E^x - 4*E^((4*x + 6*x^2)/3) + E^5*x)/2)*(3*E^5 + 3*E^x + E^((4*x + 6*x^2)/3)*(-16 - 48*x)))/6,x
]

[Out]

E^((E^x - 4*E^((2*x*(2 + 3*x))/3) + E^5*x)/2)

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fricas [A]  time = 0.78, size = 23, normalized size = 0.85 \begin {gather*} e^{\left (\frac {1}{2} \, x e^{5} - 2 \, e^{\left (2 \, x^{2} + \frac {4}{3} \, x\right )} + \frac {1}{2} \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((-48*x-16)*exp(2*x^2+4/3*x)+3*exp(x)+3*exp(5))*exp(-exp(2*x^2+4/3*x)+1/4*exp(x)+1/4*x*exp(5))^2
,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 23, normalized size = 0.85 \begin {gather*} e^{\left (\frac {1}{2} \, x e^{5} - 2 \, e^{\left (2 \, x^{2} + \frac {4}{3} \, x\right )} + \frac {1}{2} \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((-48*x-16)*exp(2*x^2+4/3*x)+3*exp(x)+3*exp(5))*exp(-exp(2*x^2+4/3*x)+1/4*exp(x)+1/4*x*exp(5))^2
,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.16, size = 23, normalized size = 0.85

method	result	size
risch	\({\mathrm e}^{-2 \,{\mathrm e}^{\frac {2 x \left (3 x +2\right )}{3}}+\frac {{\mathrm e}^{x}}{2}+\frac {x \,{\mathrm e}^{5}}{2}}\)	\(23\)
norman	\({\mathrm e}^{-2 \,{\mathrm e}^{2 x^{2}+\frac {4}{3} x}+\frac {{\mathrm e}^{x}}{2}+\frac {x \,{\mathrm e}^{5}}{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*((-48*x-16)*exp(2*x^2+4/3*x)+3*exp(x)+3*exp(5))*exp(-exp(2*x^2+4/3*x)+1/4*exp(x)+1/4*x*exp(5))^2,x,met
hod=_RETURNVERBOSE)

[Out]

exp(-2*exp(2/3*x*(3*x+2))+1/2*exp(x)+1/2*x*exp(5))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{6} \, \int {\left (16 \, {\left (3 \, x + 1\right )} e^{\left (2 \, x^{2} + \frac {4}{3} \, x\right )} - 3 \, e^{5} - 3 \, e^{x}\right )} e^{\left (\frac {1}{2} \, x e^{5} - 2 \, e^{\left (2 \, x^{2} + \frac {4}{3} \, x\right )} + \frac {1}{2} \, e^{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((-48*x-16)*exp(2*x^2+4/3*x)+3*exp(x)+3*exp(5))*exp(-exp(2*x^2+4/3*x)+1/4*exp(x)+1/4*x*exp(5))^2
,x, algorithm="maxima")

[Out]

-1/6*integrate((16*(3*x + 1)*e^(2*x^2 + 4/3*x) - 3*e^5 - 3*e^x)*e^(1/2*x*e^5 - 2*e^(2*x^2 + 4/3*x) + 1/2*e^x),
x)

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mupad [B]  time = 0.28, size = 25, normalized size = 0.93 \begin {gather*} {\mathrm {e}}^{-2\,{\mathrm {e}}^{\frac {4\,x}{3}}\,{\mathrm {e}}^{2\,x^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{2}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x)/2 - 2*exp((4*x)/3 + 2*x^2) + (x*exp(5))/2)*(3*exp(5) + 3*exp(x) - exp((4*x)/3 + 2*x^2)*(48*x +
16)))/6,x)

[Out]

exp(-2*exp((4*x)/3)*exp(2*x^2))*exp(exp(x)/2)*exp((x*exp(5))/2)

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sympy [A]  time = 0.35, size = 26, normalized size = 0.96 \begin {gather*} e^{\frac {x e^{5}}{2} + \frac {e^{x}}{2} - 2 e^{2 x^{2} + \frac {4 x}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((-48*x-16)*exp(2*x**2+4/3*x)+3*exp(x)+3*exp(5))*exp(-exp(2*x**2+4/3*x)+1/4*exp(x)+1/4*x*exp(5))
**2,x)

[Out]

exp(x*exp(5)/2 + exp(x)/2 - 2*exp(2*x**2 + 4*x/3))

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