\(\int \frac {e^5 (-4 x+2 x^2)}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx\) [3216]

Optimal result

Integrand size = 35, antiderivative size = 26 \[ \int \frac {e^5 \left (-4 x+2 x^2\right )}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx=\frac {e^5 x}{3 \left (5-\frac {5}{2 x}\right ) (3+3 x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 1607, 1694, 790} \[ \int \frac {e^5 \left (-4 x+2 x^2\right )}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx=-\frac {8 e^5 (1-x)}{45 \left (9-(4 x+1)^2\right )} \]

[In]

Int[(E^5*(-4*x + 2*x^2))/(45 - 90*x - 135*x^2 + 180*x^3 + 180*x^4),x]

[Out]

(-8*E^5*(1 - x))/(45*(9 - (1 + 4*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 790

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] /; FreeQ[{a, c, d, e, f, g, p}, x] &
& EqQ[a*e*g - c*d*f*(2*p + 3), 0] && NeQ[p, -1]

Rule 1607

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

Rule 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = e^5 \int \frac {-4 x+2 x^2}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx \\ & = e^5 \int \frac {x (-4+2 x)}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx \\ & = e^5 \text {Subst}\left (\int \frac {8 (1-4 x) (9-4 x)}{45 \left (9-16 x^2\right )^2} \, dx,x,\frac {1}{4}+x\right ) \\ & = \frac {1}{45} \left (8 e^5\right ) \text {Subst}\left (\int \frac {(1-4 x) (9-4 x)}{\left (9-16 x^2\right )^2} \, dx,x,\frac {1}{4}+x\right ) \\ & = -\frac {8 e^5 (1-x)}{45 \left (9-(1+4 x)^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^5 \left (-4 x+2 x^2\right )}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx=\frac {e^5 (1-x)}{45 \left (-1+x+2 x^2\right )} \]

[In]

Integrate[(E^5*(-4*x + 2*x^2))/(45 - 90*x - 135*x^2 + 180*x^3 + 180*x^4),x]

[Out]

(E^5*(1 - x))/(45*(-1 + x + 2*x^2))

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69

method	result	size
gosper	\(-\frac {\left (-1+x \right ) {\mathrm e}^{5}}{45 \left (2 x^{2}+x -1\right )}\)	\(18\)
norman	\(\frac {2 \,{\mathrm e}^{5} x^{2}}{45 \left (2 x^{2}+x -1\right )}\)	\(18\)
parallelrisch	\(\frac {2 \,{\mathrm e}^{5} x^{2}}{45 \left (2 x^{2}+x -1\right )}\)	\(18\)
risch	\(\frac {{\mathrm e}^{5} \left (-\frac {x}{90}+\frac {1}{90}\right )}{x^{2}+\frac {1}{2} x -\frac {1}{2}}\)	\(19\)
default	\(\frac {2 \,{\mathrm e}^{5} \left (\frac {1}{12 x -6}-\frac {1}{3 \left (1+x \right )}\right )}{45}\)	\(22\)

[In]

int((2*x^2-4*x)*exp(5)/(180*x^4+180*x^3-135*x^2-90*x+45),x,method=_RETURNVERBOSE)

[Out]

-1/45*(-1+x)*exp(5)/(2*x^2+x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e^5 \left (-4 x+2 x^2\right )}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx=-\frac {{\left (x - 1\right )} e^{5}}{45 \, {\left (2 \, x^{2} + x - 1\right )}} \]

[In]

integrate((2*x^2-4*x)*exp(5)/(180*x^4+180*x^3-135*x^2-90*x+45),x, algorithm="fricas")

[Out]

-1/45*(x - 1)*e^5/(2*x^2 + x - 1)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e^5 \left (-4 x+2 x^2\right )}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx=\frac {- x e^{5} + e^{5}}{90 x^{2} + 45 x - 45} \]

[In]

integrate((2*x**2-4*x)*exp(5)/(180*x**4+180*x**3-135*x**2-90*x+45),x)

[Out]

(-x*exp(5) + exp(5))/(90*x**2 + 45*x - 45)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e^5 \left (-4 x+2 x^2\right )}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx=-\frac {{\left (x - 1\right )} e^{5}}{45 \, {\left (2 \, x^{2} + x - 1\right )}} \]

[In]

integrate((2*x^2-4*x)*exp(5)/(180*x^4+180*x^3-135*x^2-90*x+45),x, algorithm="maxima")

[Out]

-1/45*(x - 1)*e^5/(2*x^2 + x - 1)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e^5 \left (-4 x+2 x^2\right )}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx=-\frac {{\left (x - 1\right )} e^{5}}{45 \, {\left (2 \, x^{2} + x - 1\right )}} \]

[In]

integrate((2*x^2-4*x)*exp(5)/(180*x^4+180*x^3-135*x^2-90*x+45),x, algorithm="giac")

[Out]

-1/45*(x - 1)*e^5/(2*x^2 + x - 1)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^5 \left (-4 x+2 x^2\right )}{45-90 x-135 x^2+180 x^3+180 x^4} \, dx=\frac {{\mathrm {e}}^5}{135\,\left (2\,x-1\right )}-\frac {2\,{\mathrm {e}}^5}{135\,\left (x+1\right )} \]

[In]

int(-(exp(5)*(4*x - 2*x^2))/(180*x^3 - 135*x^2 - 90*x + 180*x^4 + 45),x)

[Out]

exp(5)/(135*(2*x - 1)) - (2*exp(5))/(135*(x + 1))