3.26.91 \(\int \frac {e^{-\frac {e^{\frac {1}{4} (17+e^4-4 x)}}{-5+3 x}} (-25+30 x-9 x^2+e^{\frac {1}{4} (17+e^4-4 x)} (-2 x+3 x^2))}{25 x^2-30 x^3+9 x^4} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^{\frac {e^{3+\frac {1}{4} \left (5+e^4\right )-x}}{5-3 x}}}{x} \]

________________________________________________________________________________________

Rubi [B]  time = 1.80, antiderivative size = 107, normalized size of antiderivative = 3.57, number of steps used = 3, number of rules used = 3, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1594, 27, 2288} \begin {gather*} -\frac {\left (2 x-3 x^2\right ) \exp \left (\frac {1}{4} \left (-4 x+e^4+17\right )+\frac {e^{\frac {1}{4} \left (-4 x+e^4+17\right )}}{5-3 x}\right )}{\left (\frac {3 e^{\frac {1}{4} \left (-4 x+e^4+17\right )}}{(5-3 x)^2}-\frac {e^{\frac {1}{4} \left (-4 x+e^4+17\right )}}{5-3 x}\right ) (5-3 x)^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 + 30*x - 9*x^2 + E^((17 + E^4 - 4*x)/4)*(-2*x + 3*x^2))/(E^(E^((17 + E^4 - 4*x)/4)/(-5 + 3*x))*(25*x^
2 - 30*x^3 + 9*x^4)),x]

[Out]

-((E^((17 + E^4 - 4*x)/4 + E^((17 + E^4 - 4*x)/4)/(5 - 3*x))*(2*x - 3*x^2))/(((3*E^((17 + E^4 - 4*x)/4))/(5 -
3*x)^2 - E^((17 + E^4 - 4*x)/4)/(5 - 3*x))*(5 - 3*x)^2*x^2))

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {e^{\frac {1}{4} \left (17+e^4-4 x\right )}}{-5+3 x}} \left (-25+30 x-9 x^2+e^{\frac {1}{4} \left (17+e^4-4 x\right )} \left (-2 x+3 x^2\right )\right )}{x^2 \left (25-30 x+9 x^2\right )} \, dx\\ &=\int \frac {e^{-\frac {e^{\frac {1}{4} \left (17+e^4-4 x\right )}}{-5+3 x}} \left (-25+30 x-9 x^2+e^{\frac {1}{4} \left (17+e^4-4 x\right )} \left (-2 x+3 x^2\right )\right )}{x^2 (-5+3 x)^2} \, dx\\ &=-\frac {\exp \left (\frac {1}{4} \left (17+e^4-4 x\right )+\frac {e^{\frac {1}{4} \left (17+e^4-4 x\right )}}{5-3 x}\right ) \left (2 x-3 x^2\right )}{\left (\frac {3 e^{\frac {1}{4} \left (17+e^4-4 x\right )}}{(5-3 x)^2}-\frac {e^{\frac {1}{4} \left (17+e^4-4 x\right )}}{5-3 x}\right ) (5-3 x)^2 x^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 1.55, size = 30, normalized size = 1.00 \begin {gather*} \frac {e^{-\frac {e^{\frac {1}{4} \left (17+e^4\right )-x}}{-5+3 x}}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 + 30*x - 9*x^2 + E^((17 + E^4 - 4*x)/4)*(-2*x + 3*x^2))/(E^(E^((17 + E^4 - 4*x)/4)/(-5 + 3*x))*
(25*x^2 - 30*x^3 + 9*x^4)),x]

[Out]

1/(E^(E^((17 + E^4)/4 - x)/(-5 + 3*x))*x)

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 24, normalized size = 0.80 \begin {gather*} \frac {e^{\left (-\frac {e^{\left (-x + \frac {1}{4} \, e^{4} + \frac {17}{4}\right )}}{3 \, x - 5}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-2*x)*exp(1/4*exp(4)-x+17/4)-9*x^2+30*x-25)*exp(-exp(1/4*exp(4)-x+17/4)/(3*x-5))/(9*x^4-30*x^
3+25*x^2),x, algorithm="fricas")

[Out]

e^(-e^(-x + 1/4*e^4 + 17/4)/(3*x - 5))/x

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (9 \, x^{2} - {\left (3 \, x^{2} - 2 \, x\right )} e^{\left (-x + \frac {1}{4} \, e^{4} + \frac {17}{4}\right )} - 30 \, x + 25\right )} e^{\left (-\frac {e^{\left (-x + \frac {1}{4} \, e^{4} + \frac {17}{4}\right )}}{3 \, x - 5}\right )}}{9 \, x^{4} - 30 \, x^{3} + 25 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-2*x)*exp(1/4*exp(4)-x+17/4)-9*x^2+30*x-25)*exp(-exp(1/4*exp(4)-x+17/4)/(3*x-5))/(9*x^4-30*x^
3+25*x^2),x, algorithm="giac")

[Out]

integrate(-(9*x^2 - (3*x^2 - 2*x)*e^(-x + 1/4*e^4 + 17/4) - 30*x + 25)*e^(-e^(-x + 1/4*e^4 + 17/4)/(3*x - 5))/
(9*x^4 - 30*x^3 + 25*x^2), x)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 25, normalized size = 0.83

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{4}-x +\frac {17}{4}}}{3 x -5}}}{x}\)	\(25\)
norman	\(\frac {3 x \,{\mathrm e}^{-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{4}-x +\frac {17}{4}}}{3 x -5}}-5 \,{\mathrm e}^{-\frac {{\mathrm e}^{\frac {{\mathrm e}^{4}}{4}-x +\frac {17}{4}}}{3 x -5}}}{x \left (3 x -5\right )}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2-2*x)*exp(1/4*exp(4)-x+17/4)-9*x^2+30*x-25)*exp(-exp(1/4*exp(4)-x+17/4)/(3*x-5))/(9*x^4-30*x^3+25*x
^2),x,method=_RETURNVERBOSE)

[Out]

1/x*exp(-exp(1/4*exp(4)-x+17/4)/(3*x-5))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (9 \, x^{2} - {\left (3 \, x^{2} - 2 \, x\right )} e^{\left (-x + \frac {1}{4} \, e^{4} + \frac {17}{4}\right )} - 30 \, x + 25\right )} e^{\left (-\frac {e^{\left (-x + \frac {1}{4} \, e^{4} + \frac {17}{4}\right )}}{3 \, x - 5}\right )}}{9 \, x^{4} - 30 \, x^{3} + 25 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-2*x)*exp(1/4*exp(4)-x+17/4)-9*x^2+30*x-25)*exp(-exp(1/4*exp(4)-x+17/4)/(3*x-5))/(9*x^4-30*x^
3+25*x^2),x, algorithm="maxima")

[Out]

-integrate((9*x^2 - (3*x^2 - 2*x)*e^(-x + 1/4*e^4 + 17/4) - 30*x + 25)*e^(-e^(-x + 1/4*e^4 + 17/4)/(3*x - 5))/
(9*x^4 - 30*x^3 + 25*x^2), x)

________________________________________________________________________________________

mupad [B]  time = 0.41, size = 25, normalized size = 0.83 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^4}{4}}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{17/4}}{3\,x-5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-exp(exp(4)/4 - x + 17/4)/(3*x - 5))*(exp(exp(4)/4 - x + 17/4)*(2*x - 3*x^2) - 30*x + 9*x^2 + 25))/(
25*x^2 - 30*x^3 + 9*x^4),x)

[Out]

exp(-(exp(exp(4)/4)*exp(-x)*exp(17/4))/(3*x - 5))/x

________________________________________________________________________________________

sympy [A]  time = 0.23, size = 20, normalized size = 0.67 \begin {gather*} \frac {e^{- \frac {e^{- x + \frac {17}{4} + \frac {e^{4}}{4}}}{3 x - 5}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2-2*x)*exp(1/4*exp(4)-x+17/4)-9*x**2+30*x-25)*exp(-exp(1/4*exp(4)-x+17/4)/(3*x-5))/(9*x**4-30
*x**3+25*x**2),x)

[Out]

exp(-exp(-x + 17/4 + exp(4)/4)/(3*x - 5))/x

________________________________________________________________________________________