3.52.48 \(\int \frac {e^{-25/x} (6 e^{25/x} x^2+e^{2 e^{-25/x}} (-2250-3000 x-1000 x^2))}{45 x^2+60 x^3+20 x^4} \, dx\)

Optimal. Leaf size=26 \[ -e^{2 e^{-25/x}}+\frac {2 x}{5 (3+2 x)} \]

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Rubi [A]  time = 0.64, antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1594, 27, 12, 6742, 6715, 2282, 2194} \begin {gather*} -e^{2 e^{-25/x}}-\frac {3}{5 (2 x+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6*E^(25/x)*x^2 + E^(2/E^(25/x))*(-2250 - 3000*x - 1000*x^2))/(E^(25/x)*(45*x^2 + 60*x^3 + 20*x^4)),x]

[Out]

-E^(2/E^(25/x)) - 3/(5*(3 + 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 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-25/x} \left (6 e^{25/x} x^2+e^{2 e^{-25/x}} \left (-2250-3000 x-1000 x^2\right )\right )}{x^2 \left (45+60 x+20 x^2\right )} \, dx\\ &=\int \frac {e^{-25/x} \left (6 e^{25/x} x^2+e^{2 e^{-25/x}} \left (-2250-3000 x-1000 x^2\right )\right )}{5 x^2 (3+2 x)^2} \, dx\\ &=\frac {1}{5} \int \frac {e^{-25/x} \left (6 e^{25/x} x^2+e^{2 e^{-25/x}} \left (-2250-3000 x-1000 x^2\right )\right )}{x^2 (3+2 x)^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {250 e^{2 e^{-25/x}-\frac {25}{x}}}{x^2}+\frac {6}{(3+2 x)^2}\right ) \, dx\\ &=-\frac {3}{5 (3+2 x)}-50 \int \frac {e^{2 e^{-25/x}-\frac {25}{x}}}{x^2} \, dx\\ &=-\frac {3}{5 (3+2 x)}+50 \operatorname {Subst}\left (\int e^{2 e^{-25 x}-25 x} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {3}{5 (3+2 x)}-2 \operatorname {Subst}\left (\int e^{2 x} \, dx,x,e^{-25/x}\right )\\ &=-e^{2 e^{-25/x}}-\frac {3}{5 (3+2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 31, normalized size = 1.19 \begin {gather*} -\frac {2}{5} \left (\frac {5}{2} e^{2 e^{-25/x}}+\frac {3}{2 (3+2 x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*E^(25/x)*x^2 + E^(2/E^(25/x))*(-2250 - 3000*x - 1000*x^2))/(E^(25/x)*(45*x^2 + 60*x^3 + 20*x^4)),
x]

[Out]

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

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fricas [A]  time = 0.52, size = 27, normalized size = 1.04 \begin {gather*} -\frac {5 \, {\left (2 \, x + 3\right )} e^{\left (2 \, e^{\left (-\frac {25}{x}\right )}\right )} + 3}{5 \, {\left (2 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1000*x^2-3000*x-2250)*exp(2/exp(25/x))+6*x^2*exp(25/x))/(20*x^4+60*x^3+45*x^2)/exp(25/x),x, algor
ithm="fricas")

[Out]

-1/5*(5*(2*x + 3)*e^(2*e^(-25/x)) + 3)/(2*x + 3)

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giac [B]  time = 0.98, size = 68, normalized size = 2.62 \begin {gather*} -\frac {10 \, x e^{\left (\frac {2 \, x e^{\left (-\frac {25}{x}\right )} - 25}{x}\right )} + 15 \, e^{\left (\frac {2 \, x e^{\left (-\frac {25}{x}\right )} - 25}{x}\right )} + 3 \, e^{\left (-\frac {25}{x}\right )}}{5 \, {\left (2 \, x e^{\left (-\frac {25}{x}\right )} + 3 \, e^{\left (-\frac {25}{x}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1000*x^2-3000*x-2250)*exp(2/exp(25/x))+6*x^2*exp(25/x))/(20*x^4+60*x^3+45*x^2)/exp(25/x),x, algor
ithm="giac")

[Out]

-1/5*(10*x*e^((2*x*e^(-25/x) - 25)/x) + 15*e^((2*x*e^(-25/x) - 25)/x) + 3*e^(-25/x))/(2*x*e^(-25/x) + 3*e^(-25
/x))

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maple [A]  time = 0.10, size = 20, normalized size = 0.77

method	result	size
risch	\(-\frac {3}{10 \left (x +\frac {3}{2}\right )}-{\mathrm e}^{2 \,{\mathrm e}^{-\frac {25}{x}}}\)	\(20\)
norman	\(\frac {\left (-\frac {3 x \,{\mathrm e}^{\frac {25}{x}}}{5}-3 \,{\mathrm e}^{\frac {25}{x}} {\mathrm e}^{2 \,{\mathrm e}^{-\frac {25}{x}}} x -2 \,{\mathrm e}^{2 \,{\mathrm e}^{-\frac {25}{x}}} x^{2} {\mathrm e}^{\frac {25}{x}}\right ) {\mathrm e}^{-\frac {25}{x}}}{x \left (2 x +3\right )}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-1000*x^2-3000*x-2250)*exp(2/exp(25/x))+6*x^2*exp(25/x))/(20*x^4+60*x^3+45*x^2)/exp(25/x),x,method=_RETU
RNVERBOSE)

[Out]

-3/10/(x+3/2)-exp(2*exp(-25/x))

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maxima [A]  time = 0.45, size = 21, normalized size = 0.81 \begin {gather*} -\frac {3}{5 \, {\left (2 \, x + 3\right )}} - e^{\left (2 \, e^{\left (-\frac {25}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1000*x^2-3000*x-2250)*exp(2/exp(25/x))+6*x^2*exp(25/x))/(20*x^4+60*x^3+45*x^2)/exp(25/x),x, algor
ithm="maxima")

[Out]

-3/5/(2*x + 3) - e^(2*e^(-25/x))

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mupad [B]  time = 3.39, size = 21, normalized size = 0.81 \begin {gather*} -{\mathrm {e}}^{2\,{\mathrm {e}}^{-\frac {25}{x}}}-\frac {3}{5\,\left (2\,x+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-25/x)*(exp(2*exp(-25/x))*(3000*x + 1000*x^2 + 2250) - 6*x^2*exp(25/x)))/(45*x^2 + 60*x^3 + 20*x^4),
x)

[Out]

- exp(2*exp(-25/x)) - 3/(5*(2*x + 3))

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sympy [A]  time = 0.37, size = 15, normalized size = 0.58 \begin {gather*} - e^{2 e^{- \frac {25}{x}}} - \frac {6}{20 x + 30} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1000*x**2-3000*x-2250)*exp(2/exp(25/x))+6*x**2*exp(25/x))/(20*x**4+60*x**3+45*x**2)/exp(25/x),x)

[Out]

-exp(2*exp(-25/x)) - 6/(20*x + 30)

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