Integrand size = 54, antiderivative size = 26 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {x}{16 \left (\frac {25 e^{5/2} (1-x)}{x}-x\right )} \]
Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=-\frac {25 e^{5/2} (1-x)}{16 \left (25 e^{5/2} (-1+x)+x^2\right )} \]
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]
Leaf count is larger than twice the leaf count of optimal. \(65\) vs. \(2(26)=52\).
Time = 0.38 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.50, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {27, 27, 2027, 2459, 1380, 27, 2345, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000 x^2-20000 x+10000\right )+e^{5/2} \left (800 x^3-800 x^2\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{5/2} \int \frac {25 \left (2 x-x^2\right )}{16 \left (x^4+625 e^5 \left (x^2-2 x+1\right )-50 e^{5/2} \left (x^2-x^3\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {25}{16} e^{5/2} \int \frac {2 x-x^2}{x^4+625 e^5 \left (x^2-2 x+1\right )-50 e^{5/2} \left (x^2-x^3\right )}dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \frac {25}{16} e^{5/2} \int \frac {(2-x) x}{x^4+625 e^5 \left (x^2-2 x+1\right )-50 e^{5/2} \left (x^2-x^3\right )}dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle \frac {25}{16} e^{5/2} \int \frac {-\left (x+\frac {25 e^{5/2}}{2}\right )^2+\left (2+25 e^{5/2}\right ) \left (x+\frac {25 e^{5/2}}{2}\right )-\frac {25}{4} e^{5/2} \left (4+25 e^{5/2}\right )}{\left (x+\frac {25 e^{5/2}}{2}\right )^4-\frac {25}{2} e^{5/2} \left (4+25 e^{5/2}\right ) \left (x+\frac {25 e^{5/2}}{2}\right )^2+\frac {625}{16} e^5 \left (4+25 e^{5/2}\right )^2}d\left (x+\frac {25 e^{5/2}}{2}\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \frac {25}{16} e^{5/2} \int -\frac {4 \left (4 \left (x+\frac {25 e^{5/2}}{2}\right )^2-4 \left (2+25 e^{5/2}\right ) \left (x+\frac {25 e^{5/2}}{2}\right )+25 e^{5/2} \left (4+25 e^{5/2}\right )\right )}{\left (25 e^{5/2} \left (4+25 e^{5/2}\right )-4 \left (x+\frac {25 e^{5/2}}{2}\right )^2\right )^2}d\left (x+\frac {25 e^{5/2}}{2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {25}{4} e^{5/2} \int \frac {4 \left (x+\frac {25 e^{5/2}}{2}\right )^2-4 \left (2+25 e^{5/2}\right ) \left (x+\frac {25 e^{5/2}}{2}\right )+25 e^{5/2} \left (4+25 e^{5/2}\right )}{\left (25 e^{5/2} \left (4+25 e^{5/2}\right )-4 \left (x+\frac {25 e^{5/2}}{2}\right )^2\right )^2}d\left (x+\frac {25 e^{5/2}}{2}\right )\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {25}{4} e^{5/2} \left (-\frac {\int 0d\left (x+\frac {25 e^{5/2}}{2}\right )}{50 e^{5/2} \left (4+25 e^{5/2}\right )}-\frac {-2 \left (x+\frac {25 e^{5/2}}{2}\right )+25 e^{5/2}+2}{2 \left (25 e^{5/2} \left (4+25 e^{5/2}\right )-4 \left (x+\frac {25 e^{5/2}}{2}\right )^2\right )}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {25 e^{5/2} \left (-2 \left (x+\frac {25 e^{5/2}}{2}\right )+25 e^{5/2}+2\right )}{8 \left (25 e^{5/2} \left (4+25 e^{5/2}\right )-4 \left (x+\frac {25 e^{5/2}}{2}\right )^2\right )}\) |
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]
(25*E^(5/2)*(2 + 25*E^(5/2) - 2*((25*E^(5/2))/2 + x)))/(8*(25*E^(5/2)*(4 + 25*E^(5/2)) - 4*((25*E^(5/2))/2 + x)^2))
3.8.25.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
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 | \(-\frac {25 \,{\mathrm e}^{\frac {5}{2}} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+50 \textit {\_Z}^{3} {\mathrm e}^{\frac {5}{2}}+\left (625 \,{\mathrm e}^{5}-50 \,{\mathrm e}^{\frac {5}{2}}\right ) \textit {\_Z}^{2}-1250 \textit {\_Z} \,{\mathrm e}^{5}+625 \,{\mathrm e}^{5}\right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{625 \textit {\_R} \,{\mathrm e}^{5}+75 \textit {\_R}^{2} {\mathrm e}^{\frac {5}{2}}+2 \textit {\_R}^{3}-625 \,{\mathrm e}^{5}-50 \textit {\_R} \,{\mathrm e}^{\frac {5}{2}}}\right )}{32}\) | \(85\) |
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)
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {25 \, {\left (x - 1\right )} e^{\frac {5}{2}}}{16 \, {\left (x^{2} + 25 \, {\left (x - 1\right )} e^{\frac {5}{2}}\right )}} \]
integrate((-25*x^2+50*x)*exp(5/2)/((10000*x^2-20000*x+10000)*exp(5/2)^2+(8 00*x^3-800*x^2)*exp(5/2)+16*x^4),x, algorithm=\
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 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=- \frac {x \left (- 15625 e^{\frac {25}{2}} - 5000 e^{10} - 400 e^{\frac {15}{2}}\right ) + 400 e^{\frac {15}{2}} + 5000 e^{10} + 15625 e^{\frac {25}{2}}}{x^{2} \cdot \left (256 e^{5} + 3200 e^{\frac {15}{2}} + 10000 e^{10}\right ) + x \left (6400 e^{\frac {15}{2}} + 80000 e^{10} + 250000 e^{\frac {25}{2}}\right ) - 250000 e^{\frac {25}{2}} - 80000 e^{10} - 6400 e^{\frac {15}{2}}} \]
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)
-(x*(-15625*exp(25/2) - 5000*exp(10) - 400*exp(15/2)) + 400*exp(15/2) + 50 00*exp(10) + 15625*exp(25/2))/(x**2*(256*exp(5) + 3200*exp(15/2) + 10000*e xp(10)) + x*(6400*exp(15/2) + 80000*exp(10) + 250000*exp(25/2)) - 250000*e xp(25/2) - 80000*exp(10) - 6400*exp(15/2))
\[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\int { -\frac {25 \, {\left (x^{2} - 2 \, x\right )} e^{\frac {5}{2}}}{16 \, {\left (x^{4} + 625 \, {\left (x^{2} - 2 \, x + 1\right )} e^{5} + 50 \, {\left (x^{3} - x^{2}\right )} e^{\frac {5}{2}}\right )}} \,d x } \]
integrate((-25*x^2+50*x)*exp(5/2)/((10000*x^2-20000*x+10000)*exp(5/2)^2+(8 00*x^3-800*x^2)*exp(5/2)+16*x^4),x, algorithm=\
-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)
Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {25 \, {\left (x - 1\right )} e^{\frac {5}{2}}}{16 \, {\left (x^{2} + 25 \, x e^{\frac {5}{2}} - 25 \, e^{\frac {5}{2}}\right )}} \]
integrate((-25*x^2+50*x)*exp(5/2)/((10000*x^2-20000*x+10000)*exp(5/2)^2+(8 00*x^3-800*x^2)*exp(5/2)+16*x^4),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{5/2} \left (50 x-25 x^2\right )}{16 x^4+e^5 \left (10000-20000 x+10000 x^2\right )+e^{5/2} \left (-800 x^2+800 x^3\right )} \, dx=\frac {25\,{\mathrm {e}}^{5/2}\,\left (x-1\right )}{16\,\left (x^2+25\,{\mathrm {e}}^{5/2}\,x-25\,{\mathrm {e}}^{5/2}\right )} \]