3.35.81 \(\int \frac {e^{-\frac {2}{-45 x^3+9 x^4+e^6 (-45 x+9 x^2)+e^3 (90 x^2-18 x^3)}} (30 x-8 x^2+e^3 (-10+4 x))}{-225 x^5+90 x^6-9 x^7+e^9 (225 x^2-90 x^3+9 x^4)+e^6 (-675 x^3+270 x^4-27 x^5)+e^3 (675 x^4-270 x^5+27 x^6)} \, dx\)

Optimal. Leaf size=23 \[ e^{-\frac {2}{9 \left (e^3-x\right )^2 (-5+x) x}} \]

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Rubi [A]  time = 1.53, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, integrand size = 142, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6688, 12, 6706} \begin {gather*} e^{\frac {2}{9 (5-x) \left (e^3-x\right )^2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(30*x - 8*x^2 + E^3*(-10 + 4*x))/(E^(2/(-45*x^3 + 9*x^4 + E^6*(-45*x + 9*x^2) + E^3*(90*x^2 - 18*x^3)))*(-
225*x^5 + 90*x^6 - 9*x^7 + E^9*(225*x^2 - 90*x^3 + 9*x^4) + E^6*(-675*x^3 + 270*x^4 - 27*x^5) + E^3*(675*x^4 -
270*x^5 + 27*x^6))),x]

[Out]

E^(2/(9*(5 - x)*(E^3 - x)^2*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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} &=\int \frac {e^{-\frac {2}{9 \left (e^3-x\right )^2 (-5+x) x}} \left (-10 e^3+2 \left (15+2 e^3\right ) x-8 x^2\right )}{9 (5-x)^2 \left (e^3-x\right )^3 x^2} \, dx\\ &=\frac {1}{9} \int \frac {e^{-\frac {2}{9 \left (e^3-x\right )^2 (-5+x) x}} \left (-10 e^3+2 \left (15+2 e^3\right ) x-8 x^2\right )}{(5-x)^2 \left (e^3-x\right )^3 x^2} \, dx\\ &=e^{\frac {2}{9 (5-x) \left (e^3-x\right )^2 x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 23, normalized size = 1.00 \begin {gather*} e^{-\frac {2}{9 \left (e^3-x\right )^2 (-5+x) x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(30*x - 8*x^2 + E^3*(-10 + 4*x))/(E^(2/(-45*x^3 + 9*x^4 + E^6*(-45*x + 9*x^2) + E^3*(90*x^2 - 18*x^3
)))*(-225*x^5 + 90*x^6 - 9*x^7 + E^9*(225*x^2 - 90*x^3 + 9*x^4) + E^6*(-675*x^3 + 270*x^4 - 27*x^5) + E^3*(675
*x^4 - 270*x^5 + 27*x^6))),x]

[Out]

E^(-2/(9*(E^3 - x)^2*(-5 + x)*x))

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fricas [A]  time = 0.53, size = 37, normalized size = 1.61 \begin {gather*} e^{\left (-\frac {2}{9 \, {\left (x^{4} - 5 \, x^{3} + {\left (x^{2} - 5 \, x\right )} e^{6} - 2 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{3}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-10)*exp(3)-8*x^2+30*x)*exp(-2/((9*x^2-45*x)*exp(3)^2+(-18*x^3+90*x^2)*exp(3)+9*x^4-45*x^3))/((
9*x^4-90*x^3+225*x^2)*exp(3)^3+(-27*x^5+270*x^4-675*x^3)*exp(3)^2+(27*x^6-270*x^5+675*x^4)*exp(3)-9*x^7+90*x^6
-225*x^5),x, algorithm="fricas")

[Out]

e^(-2/9/(x^4 - 5*x^3 + (x^2 - 5*x)*e^6 - 2*(x^3 - 5*x^2)*e^3))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-10)*exp(3)-8*x^2+30*x)*exp(-2/((9*x^2-45*x)*exp(3)^2+(-18*x^3+90*x^2)*exp(3)+9*x^4-45*x^3))/((
9*x^4-90*x^3+225*x^2)*exp(3)^3+(-27*x^5+270*x^4-675*x^3)*exp(3)^2+(27*x^6-270*x^5+675*x^4)*exp(3)-9*x^7+90*x^6
-225*x^5),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.64, size = 29, normalized size = 1.26

method	result	size
risch	\({\mathrm e}^{\frac {2}{9 x \left (x -5\right ) \left (2 x \,{\mathrm e}^{3}-x^{2}-{\mathrm e}^{6}\right )}}\)	\(29\)
gosper	\({\mathrm e}^{-\frac {2}{9 x \left (x \,{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{3}-5 \,{\mathrm e}^{6}+10 x \,{\mathrm e}^{3}-5 x^{2}\right )}}\)	\(42\)
norman	\(\frac {x^{4} {\mathrm e}^{-\frac {2}{\left (9 x^{2}-45 x \right ) {\mathrm e}^{6}+\left (-18 x^{3}+90 x^{2}\right ) {\mathrm e}^{3}+9 x^{4}-45 x^{3}}}+\left ({\mathrm e}^{6}+10 \,{\mathrm e}^{3}\right ) x^{2} {\mathrm e}^{-\frac {2}{\left (9 x^{2}-45 x \right ) {\mathrm e}^{6}+\left (-18 x^{3}+90 x^{2}\right ) {\mathrm e}^{3}+9 x^{4}-45 x^{3}}}+\left (-2 \,{\mathrm e}^{3}-5\right ) x^{3} {\mathrm e}^{-\frac {2}{\left (9 x^{2}-45 x \right ) {\mathrm e}^{6}+\left (-18 x^{3}+90 x^{2}\right ) {\mathrm e}^{3}+9 x^{4}-45 x^{3}}}-5 x \,{\mathrm e}^{6} {\mathrm e}^{-\frac {2}{\left (9 x^{2}-45 x \right ) {\mathrm e}^{6}+\left (-18 x^{3}+90 x^{2}\right ) {\mathrm e}^{3}+9 x^{4}-45 x^{3}}}}{x \left (x -5\right ) \left (-x +{\mathrm e}^{3}\right )^{2}}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x-10)*exp(3)-8*x^2+30*x)*exp(-2/((9*x^2-45*x)*exp(3)^2+(-18*x^3+90*x^2)*exp(3)+9*x^4-45*x^3))/((9*x^4-
90*x^3+225*x^2)*exp(3)^3+(-27*x^5+270*x^4-675*x^3)*exp(3)^2+(27*x^6-270*x^5+675*x^4)*exp(3)-9*x^7+90*x^6-225*x
^5),x,method=_RETURNVERBOSE)

[Out]

exp(2/9/x/(x-5)/(2*x*exp(3)-x^2-exp(6)))

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maxima [B]  time = 1.05, size = 125, normalized size = 5.43 \begin {gather*} e^{\left (\frac {2 \, e^{\left (-6\right )}}{45 \, x} - \frac {2}{9 \, {\left (x^{2} {\left (e^{6} - 5 \, e^{3}\right )} - 2 \, x {\left (e^{9} - 5 \, e^{6}\right )} + e^{12} - 5 \, e^{9}\right )}} - \frac {10}{9 \, {\left (x {\left (e^{12} - 10 \, e^{9} + 25 \, e^{6}\right )} - e^{15} + 10 \, e^{12} - 25 \, e^{9}\right )}} + \frac {4}{9 \, {\left (x {\left (e^{9} - 10 \, e^{6} + 25 \, e^{3}\right )} - e^{12} + 10 \, e^{9} - 25 \, e^{6}\right )}} - \frac {2}{45 \, {\left (x {\left (e^{6} - 10 \, e^{3} + 25\right )} - 5 \, e^{6} + 50 \, e^{3} - 125\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-10)*exp(3)-8*x^2+30*x)*exp(-2/((9*x^2-45*x)*exp(3)^2+(-18*x^3+90*x^2)*exp(3)+9*x^4-45*x^3))/((
9*x^4-90*x^3+225*x^2)*exp(3)^3+(-27*x^5+270*x^4-675*x^3)*exp(3)^2+(27*x^6-270*x^5+675*x^4)*exp(3)-9*x^7+90*x^6
-225*x^5),x, algorithm="maxima")

[Out]

e^(2/45*e^(-6)/x - 2/9/(x^2*(e^6 - 5*e^3) - 2*x*(e^9 - 5*e^6) + e^12 - 5*e^9) - 10/9/(x*(e^12 - 10*e^9 + 25*e^
6) - e^15 + 10*e^12 - 25*e^9) + 4/9/(x*(e^9 - 10*e^6 + 25*e^3) - e^12 + 10*e^9 - 25*e^6) - 2/45/(x*(e^6 - 10*e
^3 + 25) - 5*e^6 + 50*e^3 - 125))

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mupad [B]  time = 4.52, size = 62, normalized size = 2.70 \begin {gather*} {\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-6}}{45\,x}-\frac {2}{45\,{\left ({\mathrm {e}}^3-5\right )}^2\,\left (x-5\right )}-\frac {2\,{\mathrm {e}}^{-6}\,\left (5\,x-10\,{\mathrm {e}}^3+3\,{\mathrm {e}}^6-2\,x\,{\mathrm {e}}^3\right )}{9\,{\left ({\mathrm {e}}^3-5\right )}^2\,\left (x^2-2\,{\mathrm {e}}^3\,x+{\mathrm {e}}^6\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2/(exp(6)*(45*x - 9*x^2) - exp(3)*(90*x^2 - 18*x^3) + 45*x^3 - 9*x^4))*(30*x - 8*x^2 + exp(3)*(4*x -
10)))/(exp(9)*(225*x^2 - 90*x^3 + 9*x^4) + exp(3)*(675*x^4 - 270*x^5 + 27*x^6) - exp(6)*(675*x^3 - 270*x^4 + 2
7*x^5) - 225*x^5 + 90*x^6 - 9*x^7),x)

[Out]

exp((2*exp(-6))/(45*x) - 2/(45*(exp(3) - 5)^2*(x - 5)) - (2*exp(-6)*(5*x - 10*exp(3) + 3*exp(6) - 2*x*exp(3)))
/(9*(exp(3) - 5)^2*(exp(6) - 2*x*exp(3) + x^2)))

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sympy [A]  time = 1.58, size = 39, normalized size = 1.70 \begin {gather*} e^{- \frac {2}{9 x^{4} - 45 x^{3} + \left (9 x^{2} - 45 x\right ) e^{6} + \left (- 18 x^{3} + 90 x^{2}\right ) e^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x-10)*exp(3)-8*x**2+30*x)*exp(-2/((9*x**2-45*x)*exp(3)**2+(-18*x**3+90*x**2)*exp(3)+9*x**4-45*x*
*3))/((9*x**4-90*x**3+225*x**2)*exp(3)**3+(-27*x**5+270*x**4-675*x**3)*exp(3)**2+(27*x**6-270*x**5+675*x**4)*e
xp(3)-9*x**7+90*x**6-225*x**5),x)

[Out]

exp(-2/(9*x**4 - 45*x**3 + (9*x**2 - 45*x)*exp(6) + (-18*x**3 + 90*x**2)*exp(3)))

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