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

Optimal. Leaf size=26 \[ \frac {x}{16 \left (\frac {25 e^{5/2} (1-x)}{x}-x\right )} \]

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Rubi [A]  time = 0.11, antiderivative size = 37, normalized size of antiderivative = 1.42, number of steps used = 5, number of rules used = 4, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 1593, 1680, 776} \begin {gather*} \frac {25 e^{5/2} (1-x)}{16 \left (-x^2-25 e^{5/2} x+25 e^{5/2}\right )} \end {gather*}

Antiderivative was successfully verified.

[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 776

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 1593

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 1680

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 {gather*} \begin {aligned} \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} \operatorname {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 ) \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 30, normalized size = 1.15 \begin {gather*} -\frac {25 e^{5/2} (1-x)}{16 \left (25 e^{5/2} (-1+x)+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.45, size = 20, normalized size = 0.77 \begin {gather*} \frac {25 \, {\left (x - 1\right )} e^{\frac {5}{2}}}{16 \, {\left (x^{2} + 25 \, {\left (x - 1\right )} e^{\frac {5}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

undef

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maple [A]  time = 0.11, size = 23, normalized size = 0.88

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

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {25}{16} \, e^{\frac {5}{2}} \int \frac {x^{2} - 2 \, x}{x^{4} + 625 \, {\left (x^{2} - 2 \, x + 1\right )} e^{5} + 50 \, {\left (x^{3} - x^{2}\right )} e^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.12, size = 24, normalized size = 0.92 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [B]  time = 1.05, size = 100, normalized size = 3.85 \begin {gather*} - \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} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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