\(\int \frac {e^{5/2} (50 x-25 x^2)}{16 x^4+e^5 (10000-20000 x+10000 x^2)+e^{5/2} (-800 x^2+800 x^3)} \, dx\) [6725]

Optimal result

Integrand size = 54, antiderivative size = 26 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {x}{16 \left (\frac {25 e^{5/2} (1-x)}{x}-x\right )} \]

[Out]

x/(-16*x+400*(1-x)/x*exp(5/2))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.42, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 1607, 1694, 790} \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {25 e^{5/2} (1-x)}{16 \left (-x^2-25 e^{5/2} x+25 e^{5/2}\right )} \]

[In]

Int[(E^(5/2)*(50*x - 25*x^2))/(16*x^4 + E^5*(10000 - 20000*x + 10000*x^2) + E^(5/2)*(-800*x^2 + 800*x^3)),x]

[Out]

(25*E^(5/2)*(1 - x))/(16*(25*E^(5/2) - 25*E^(5/2)*x - 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^{5/2} \int \frac {50 x-25 x^2}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx \\ & = e^{5/2} \int \frac {(50-25 x) x}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx \\ & = e^{5/2} \text {Subst}\left (\int \frac {25 \left (25 e^{5/2}-2 x\right ) \left (-4-25 e^{5/2}+2 x\right )}{4 \left (100 e^{5/2}+625 e^5-4 x^2\right )^2} \, dx,x,\frac {25 e^{5/2}}{2}+x\right ) \\ & = \frac {1}{4} \left (25 e^{5/2}\right ) \text {Subst}\left (\int \frac {\left (25 e^{5/2}-2 x\right ) \left (-4-25 e^{5/2}+2 x\right )}{\left (100 e^{5/2}+625 e^5-4 x^2\right )^2} \, dx,x,\frac {25 e^{5/2}}{2}+x\right ) \\ & = \frac {25 e^{5/2} (1-x)}{16 \left (25 e^{5/2}-25 e^{5/2} x-x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=-\frac {25 e^{5/2} (1-x)}{16 \left (25 e^{5/2} (-1+x)+x^2\right )} \]

[In]

Integrate[(E^(5/2)*(50*x - 25*x^2))/(16*x^4 + E^5*(10000 - 20000*x + 10000*x^2) + E^(5/2)*(-800*x^2 + 800*x^3)
),x]

[Out]

(-25*E^(5/2)*(1 - x))/(16*(25*E^(5/2)*(-1 + x) + x^2))

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
gosper	\(\frac {25 \left (-1+x \right ) {\mathrm e}^{\frac {5}{2}}}{16 \left (25 x \,{\mathrm e}^{\frac {5}{2}}+x^{2}-25 \,{\mathrm e}^{\frac {5}{2}}\right )}\)	\(23\)
risch	\(\frac {{\mathrm e}^{\frac {5}{2}} \left (\frac {x}{16}-\frac {1}{16}\right )}{x \,{\mathrm e}^{\frac {5}{2}}+\frac {x^{2}}{25}-{\mathrm e}^{\frac {5}{2}}}\)	\(25\)
parallelrisch	\(\frac {{\mathrm e}^{\frac {5}{2}} \left (25 x -25\right )}{400 x \,{\mathrm e}^{\frac {5}{2}}+16 x^{2}-400 \,{\mathrm e}^{\frac {5}{2}}}\)	\(25\)
norman	\(\frac {\frac {25 x \,{\mathrm e}^{\frac {5}{2}}}{16}-\frac {25 \,{\mathrm e}^{\frac {5}{2}}}{16}}{25 x \,{\mathrm e}^{\frac {5}{2}}+x^{2}-25 \,{\mathrm e}^{\frac {5}{2}}}\)	\(27\)
default	\(-\frac {25 \,{\mathrm e}^{\frac {5}{2}} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+50 \textit {\_Z}^{3} {\mathrm e}^{\frac {5}{2}}+\left (625 \,{\mathrm e}^{5}-50 \,{\mathrm e}^{\frac {5}{2}}\right ) \textit {\_Z}^{2}-1250 \textit {\_Z} \,{\mathrm e}^{5}+625 \,{\mathrm e}^{5}\right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{625 \textit {\_R} \,{\mathrm e}^{5}+75 \textit {\_R}^{2} {\mathrm e}^{\frac {5}{2}}+2 \textit {\_R}^{3}-625 \,{\mathrm e}^{5}-50 \textit {\_R} \,{\mathrm e}^{\frac {5}{2}}}\right )}{32}\)	\(85\)

[In]

int((-25*x^2+50*x)*exp(5/2)/((10000*x^2-20000*x+10000)*exp(5/2)^2+(800*x^3-800*x^2)*exp(5/2)+16*x^4),x,method=
_RETURNVERBOSE)

[Out]

25/16*(-1+x)*exp(5/2)/(25*x*exp(5/2)+x^2-25*exp(5/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {25 \, {\left (x - 1\right )} e^{\frac {5}{2}}}{16 \, {\left (x^{2} + 25 \, {\left (x - 1\right )} e^{\frac {5}{2}}\right )}} \]

[In]

integrate((-25*x^2+50*x)*exp(5/2)/((10000*x^2-20000*x+10000)*exp(5/2)^2+(800*x^3-800*x^2)*exp(5/2)+16*x^4),x,
algorithm="fricas")

[Out]

25/16*(x - 1)*e^(5/2)/(x^2 + 25*(x - 1)*e^(5/2))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (15) = 30\).

Time = 0.77 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.85 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=- \frac {x \left (- 15625 e^{\frac {25}{2}} - 5000 e^{10} - 400 e^{\frac {15}{2}}\right ) + 400 e^{\frac {15}{2}} + 5000 e^{10} + 15625 e^{\frac {25}{2}}}{x^{2} \cdot \left (256 e^{5} + 3200 e^{\frac {15}{2}} + 10000 e^{10}\right ) + x \left (6400 e^{\frac {15}{2}} + 80000 e^{10} + 250000 e^{\frac {25}{2}}\right ) - 250000 e^{\frac {25}{2}} - 80000 e^{10} - 6400 e^{\frac {15}{2}}} \]

[In]

integrate((-25*x**2+50*x)*exp(5/2)/((10000*x**2-20000*x+10000)*exp(5/2)**2+(800*x**3-800*x**2)*exp(5/2)+16*x**
4),x)

[Out]

-(x*(-15625*exp(25/2) - 5000*exp(10) - 400*exp(15/2)) + 400*exp(15/2) + 5000*exp(10) + 15625*exp(25/2))/(x**2*
(256*exp(5) + 3200*exp(15/2) + 10000*exp(10)) + x*(6400*exp(15/2) + 80000*exp(10) + 250000*exp(25/2)) - 250000
*exp(25/2) - 80000*exp(10) - 6400*exp(15/2))

Maxima [F]

\[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\int { -\frac {25 \, {\left (x^{2} - 2 \, x\right )} e^{\frac {5}{2}}}{16 \, {\left (x^{4} + 625 \, {\left (x^{2} - 2 \, x + 1\right )} e^{5} + 50 \, {\left (x^{3} - x^{2}\right )} e^{\frac {5}{2}}\right )}} \,d x } \]

[In]

integrate((-25*x^2+50*x)*exp(5/2)/((10000*x^2-20000*x+10000)*exp(5/2)^2+(800*x^3-800*x^2)*exp(5/2)+16*x^4),x,
algorithm="maxima")

[Out]

-25/16*e^(5/2)*integrate((x^2 - 2*x)/(x^4 + 625*(x^2 - 2*x + 1)*e^5 + 50*(x^3 - x^2)*e^(5/2)), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {25 \, {\left (x - 1\right )} e^{\frac {5}{2}}}{16 \, {\left (x^{2} + 25 \, x e^{\frac {5}{2}} - 25 \, e^{\frac {5}{2}}\right )}} \]

[In]

integrate((-25*x^2+50*x)*exp(5/2)/((10000*x^2-20000*x+10000)*exp(5/2)^2+(800*x^3-800*x^2)*exp(5/2)+16*x^4),x,
algorithm="giac")

[Out]

25/16*(x - 1)*e^(5/2)/(x^2 + 25*x*e^(5/2) - 25*e^(5/2))

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {25\,{\mathrm {e}}^{5/2}\,\left (x-1\right )}{16\,\left (x^2+25\,{\mathrm {e}}^{5/2}\,x-25\,{\mathrm {e}}^{5/2}\right )} \]

[In]

int((exp(5/2)*(50*x - 25*x^2))/(exp(5)*(10000*x^2 - 20000*x + 10000) - exp(5/2)*(800*x^2 - 800*x^3) + 16*x^4),
x)

[Out]

(25*exp(5/2)*(x - 1))/(16*(25*x*exp(5/2) - 25*exp(5/2) + x^2))