3.91.17 \(\int \frac {e^{\frac {1}{25} (-4775-560 x-16 x^2)} (-1680-96 x+e^5 (1145+624 x+32 x^2))}{225+e^5 (-300-150 x)+e^{10} (100+100 x+25 x^2)} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^{5-\left (-14-\frac {4 x}{5}\right )^2}}{3-e^5 (2+x)} \]

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Rubi [A]  time = 0.34, antiderivative size = 32, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6741, 27, 12, 2289} \begin {gather*} \frac {e^{\frac {1}{25} \left (-16 x^2-560 x-4775\right )}}{-e^5 x-2 e^5+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-4775 - 560*x - 16*x^2)/25)*(-1680 - 96*x + E^5*(1145 + 624*x + 32*x^2)))/(225 + E^5*(-300 - 150*x) +
E^10*(100 + 100*x + 25*x^2)),x]

[Out]

E^((-4775 - 560*x - 16*x^2)/25)/(3 - 2*E^5 - E^5*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 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 2289

Int[(F_)^(u_)*(v_)^(n_.)*(w_), x_Symbol] :> With[{z = Log[F]*v*D[u, x] + (n + 1)*D[v, x]}, Simp[(Coefficient[w
, x, Exponent[w, x]]*F^u*v^(n + 1))/Coefficient[z, x, Exponent[z, x]], x] /; EqQ[Exponent[w, x], Exponent[z, x
]] && EqQ[w*Coefficient[z, x, Exponent[z, x]], z*Coefficient[w, x, Exponent[w, x]]]] /; FreeQ[{F, n}, x] && Po
lynomialQ[u, x] && PolynomialQ[v, x] && PolynomialQ[w, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{25} \left (-4775-560 x-16 x^2\right )} \left (-5 \left (336-229 e^5\right )-48 \left (2-13 e^5\right ) x+32 e^5 x^2\right )}{25 \left (3-2 e^5\right )^2-50 e^5 \left (3-2 e^5\right ) x+25 e^{10} x^2} \, dx\\ &=\int \frac {e^{\frac {1}{25} \left (-4775-560 x-16 x^2\right )} \left (-5 \left (336-229 e^5\right )-48 \left (2-13 e^5\right ) x+32 e^5 x^2\right )}{25 \left (-3+2 e^5+e^5 x\right )^2} \, dx\\ &=\frac {1}{25} \int \frac {e^{\frac {1}{25} \left (-4775-560 x-16 x^2\right )} \left (-5 \left (336-229 e^5\right )-48 \left (2-13 e^5\right ) x+32 e^5 x^2\right )}{\left (-3+2 e^5+e^5 x\right )^2} \, dx\\ &=\frac {e^{\frac {1}{25} \left (-4775-560 x-16 x^2\right )}}{3-2 e^5-e^5 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 29, normalized size = 1.04 \begin {gather*} -\frac {e^{-191-\frac {112 x}{5}-\frac {16 x^2}{25}}}{-3+e^5 (2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-4775 - 560*x - 16*x^2)/25)*(-1680 - 96*x + E^5*(1145 + 624*x + 32*x^2)))/(225 + E^5*(-300 - 15
0*x) + E^10*(100 + 100*x + 25*x^2)),x]

[Out]

-(E^(-191 - (112*x)/5 - (16*x^2)/25)/(-3 + E^5*(2 + x)))

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fricas [A]  time = 0.56, size = 23, normalized size = 0.82 \begin {gather*} -\frac {e^{\left (-\frac {16}{25} \, x^{2} - \frac {112}{5} \, x - 191\right )}}{{\left (x + 2\right )} e^{5} - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2+624*x+1145)*exp(5)-96*x-1680)*exp(-16/25*x^2-112/5*x-191)/((25*x^2+100*x+100)*exp(5)^2+(-15
0*x-300)*exp(5)+225),x, algorithm="fricas")

[Out]

-e^(-16/25*x^2 - 112/5*x - 191)/((x + 2)*e^5 - 3)

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giac [A]  time = 0.17, size = 27, normalized size = 0.96 \begin {gather*} -\frac {e^{\left (-\frac {16}{25} \, x^{2} - \frac {112}{5} \, x\right )}}{x e^{196} + 2 \, e^{196} - 3 \, e^{191}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2+624*x+1145)*exp(5)-96*x-1680)*exp(-16/25*x^2-112/5*x-191)/((25*x^2+100*x+100)*exp(5)^2+(-15
0*x-300)*exp(5)+225),x, algorithm="giac")

[Out]

-e^(-16/25*x^2 - 112/5*x)/(x*e^196 + 2*e^196 - 3*e^191)

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maple [A]  time = 0.54, size = 26, normalized size = 0.93

method	result	size
gosper	\(-\frac {{\mathrm e}^{-\frac {16}{25} x^{2}-\frac {112}{5} x -191}}{x \,{\mathrm e}^{5}+2 \,{\mathrm e}^{5}-3}\)	\(26\)
norman	\(-\frac {{\mathrm e}^{-\frac {16}{25} x^{2}-\frac {112}{5} x -191}}{x \,{\mathrm e}^{5}+2 \,{\mathrm e}^{5}-3}\)	\(26\)
risch	\(-\frac {{\mathrm e}^{-\frac {16}{25} x^{2}-\frac {112}{5} x -191}}{x \,{\mathrm e}^{5}+2 \,{\mathrm e}^{5}-3}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((32*x^2+624*x+1145)*exp(5)-96*x-1680)*exp(-16/25*x^2-112/5*x-191)/((25*x^2+100*x+100)*exp(5)^2+(-150*x-30
0)*exp(5)+225),x,method=_RETURNVERBOSE)

[Out]

-exp(-16/25*x^2-112/5*x-191)/(x*exp(5)+2*exp(5)-3)

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maxima [A]  time = 0.42, size = 27, normalized size = 0.96 \begin {gather*} -\frac {e^{\left (-\frac {16}{25} \, x^{2} - \frac {112}{5} \, x\right )}}{x e^{196} + 2 \, e^{196} - 3 \, e^{191}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x^2+624*x+1145)*exp(5)-96*x-1680)*exp(-16/25*x^2-112/5*x-191)/((25*x^2+100*x+100)*exp(5)^2+(-15
0*x-300)*exp(5)+225),x, algorithm="maxima")

[Out]

-e^(-16/25*x^2 - 112/5*x)/(x*e^196 + 2*e^196 - 3*e^191)

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mupad [B]  time = 1.02, size = 26, normalized size = 0.93 \begin {gather*} -\frac {{\mathrm {e}}^{-\frac {112\,x}{5}}\,{\mathrm {e}}^{-191}\,{\mathrm {e}}^{-\frac {16\,x^2}{25}}}{2\,{\mathrm {e}}^5+x\,{\mathrm {e}}^5-3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- (112*x)/5 - (16*x^2)/25 - 191)*(96*x - exp(5)*(624*x + 32*x^2 + 1145) + 1680))/(exp(10)*(100*x + 2
5*x^2 + 100) - exp(5)*(150*x + 300) + 225),x)

[Out]

-(exp(-(112*x)/5)*exp(-191)*exp(-(16*x^2)/25))/(2*exp(5) + x*exp(5) - 3)

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sympy [A]  time = 0.20, size = 29, normalized size = 1.04 \begin {gather*} - \frac {e^{- \frac {16 x^{2}}{25} - \frac {112 x}{5} - 191}}{x e^{5} - 3 + 2 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((32*x**2+624*x+1145)*exp(5)-96*x-1680)*exp(-16/25*x**2-112/5*x-191)/((25*x**2+100*x+100)*exp(5)**2+
(-150*x-300)*exp(5)+225),x)

[Out]

-exp(-16*x**2/25 - 112*x/5 - 191)/(x*exp(5) - 3 + 2*exp(5))

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