3.23.16 \(\int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} (-2 e^8-5 e^4 x-2 x^2)+e^4 (12 x+8 x^2))}{2 e^8+4 e^4 x+2 x^2} \, dx\)

Optimal. Leaf size=28 \[ e^{x \left (3-e^{\frac {x}{2 \left (e^4+x\right )}}+\frac {5}{x}+x\right )} \]

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Rubi [A]  time = 2.53, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {27, 12, 6706} \begin {gather*} e^{x^2-e^{\frac {x}{2 \left (x+e^4\right )}} x+3 x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5 + 3*x - E^(x/(2*E^4 + 2*x))*x + x^2)*(6*x^2 + 4*x^3 + E^8*(6 + 4*x) + E^(x/(2*E^4 + 2*x))*(-2*E^8 -
5*E^4*x - 2*x^2) + E^4*(12*x + 8*x^2)))/(2*E^8 + 4*E^4*x + 2*x^2),x]

[Out]

E^(5 + 3*x - E^(x/(2*(E^4 + x)))*x + 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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} &=\int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{2 \left (e^4+x\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{5+3 x-e^{\frac {x}{2 e^4+2 x}} x+x^2} \left (6 x^2+4 x^3+e^8 (6+4 x)+e^{\frac {x}{2 e^4+2 x}} \left (-2 e^8-5 e^4 x-2 x^2\right )+e^4 \left (12 x+8 x^2\right )\right )}{\left (e^4+x\right )^2} \, dx\\ &=e^{5+3 x-e^{\frac {x}{2 \left (e^4+x\right )}} x+x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 26, normalized size = 0.93 \begin {gather*} e^{5-\left (-3+e^{\frac {x}{2 \left (e^4+x\right )}}\right ) x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + 3*x - E^(x/(2*E^4 + 2*x))*x + x^2)*(6*x^2 + 4*x^3 + E^8*(6 + 4*x) + E^(x/(2*E^4 + 2*x))*(-2*
E^8 - 5*E^4*x - 2*x^2) + E^4*(12*x + 8*x^2)))/(2*E^8 + 4*E^4*x + 2*x^2),x]

[Out]

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

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fricas [A]  time = 0.94, size = 22, normalized size = 0.79 \begin {gather*} e^{\left (x^{2} - x e^{\left (\frac {x}{2 \, {\left (x + e^{4}\right )}}\right )} + 3 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*exp(4)^2+(8*x^2+12*x)*exp(4)+4*x^3+6*x
^2)*exp(-x*exp(x/(2*exp(4)+2*x))+x^2+3*x+5)/(2*exp(4)^2+4*x*exp(4)+2*x^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, x^{3} + 6 \, x^{2} + 2 \, {\left (2 \, x + 3\right )} e^{8} + 4 \, {\left (2 \, x^{2} + 3 \, x\right )} e^{4} - {\left (2 \, x^{2} + 5 \, x e^{4} + 2 \, e^{8}\right )} e^{\left (\frac {x}{2 \, {\left (x + e^{4}\right )}}\right )}\right )} e^{\left (x^{2} - x e^{\left (\frac {x}{2 \, {\left (x + e^{4}\right )}}\right )} + 3 \, x + 5\right )}}{2 \, {\left (x^{2} + 2 \, x e^{4} + e^{8}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*exp(4)^2+(8*x^2+12*x)*exp(4)+4*x^3+6*x
^2)*exp(-x*exp(x/(2*exp(4)+2*x))+x^2+3*x+5)/(2*exp(4)^2+4*x*exp(4)+2*x^2),x, algorithm="giac")

[Out]

integrate(1/2*(4*x^3 + 6*x^2 + 2*(2*x + 3)*e^8 + 4*(2*x^2 + 3*x)*e^4 - (2*x^2 + 5*x*e^4 + 2*e^8)*e^(1/2*x/(x +
e^4)))*e^(x^2 - x*e^(1/2*x/(x + e^4)) + 3*x + 5)/(x^2 + 2*x*e^4 + e^8), x)

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maple [A]  time = 0.43, size = 23, normalized size = 0.82

method	result	size
risch	\({\mathrm e}^{-x \,{\mathrm e}^{\frac {x}{2 \,{\mathrm e}^{4}+2 x}}+x^{2}+3 x +5}\)	\(23\)
norman	\(\frac {x \,{\mathrm e}^{-x \,{\mathrm e}^{\frac {x}{2 \,{\mathrm e}^{4}+2 x}}+x^{2}+3 x +5}+{\mathrm e}^{4} {\mathrm e}^{-x \,{\mathrm e}^{\frac {x}{2 \,{\mathrm e}^{4}+2 x}}+x^{2}+3 x +5}}{x +{\mathrm e}^{4}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*exp(4)^2+(8*x^2+12*x)*exp(4)+4*x^3+6*x^2)*ex
p(-x*exp(x/(2*exp(4)+2*x))+x^2+3*x+5)/(2*exp(4)^2+4*x*exp(4)+2*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.57, size = 25, normalized size = 0.89 \begin {gather*} e^{\left (x^{2} - x e^{\left (-\frac {e^{4}}{2 \, {\left (x + e^{4}\right )}} + \frac {1}{2}\right )} + 3 \, x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4)^2-5*x*exp(4)-2*x^2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*exp(4)^2+(8*x^2+12*x)*exp(4)+4*x^3+6*x
^2)*exp(-x*exp(x/(2*exp(4)+2*x))+x^2+3*x+5)/(2*exp(4)^2+4*x*exp(4)+2*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.84, size = 28, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-x\,{\mathrm {e}}^{\frac {x}{2\,x+2\,{\mathrm {e}}^4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*x - x*exp(x/(2*x + 2*exp(4))) + x^2 + 5)*(exp(4)*(12*x + 8*x^2) - exp(x/(2*x + 2*exp(4)))*(2*exp(8)
+ 5*x*exp(4) + 2*x^2) + 6*x^2 + 4*x^3 + exp(8)*(4*x + 6)))/(2*exp(8) + 4*x*exp(4) + 2*x^2),x)

[Out]

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

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sympy [A]  time = 0.68, size = 22, normalized size = 0.79 \begin {gather*} e^{x^{2} - x e^{\frac {x}{2 x + 2 e^{4}}} + 3 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(4)**2-5*x*exp(4)-2*x**2)*exp(x/(2*exp(4)+2*x))+(4*x+6)*exp(4)**2+(8*x**2+12*x)*exp(4)+4*x**
3+6*x**2)*exp(-x*exp(x/(2*exp(4)+2*x))+x**2+3*x+5)/(2*exp(4)**2+4*x*exp(4)+2*x**2),x)

[Out]

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

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