Integrand size = 70, antiderivative size = 32 \[ \int \frac {e^{-4+2 x} \left (4-x^2\right )^2 \left (4000-4300 x+200 x^2+350 x^3-50 x^4+\frac {e^{4-2 x} \left (324-81 x^2\right )}{\left (4-x^2\right )^2}\right )}{-324+81 x^2} \, dx=5-x-\frac {25}{81} e^{-4+2 x} (5-x)^2 \left (4-x^2\right )^2 \]
Time = 0.85 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-4+2 x} \left (4-x^2\right )^2 \left (4000-4300 x+200 x^2+350 x^3-50 x^4+\frac {e^{4-2 x} \left (324-81 x^2\right )}{\left (4-x^2\right )^2}\right )}{-324+81 x^2} \, dx=-\frac {81 e^4 x+25 e^{2 x} \left (20-4 x-5 x^2+x^3\right )^2}{81 e^4} \]
Integrate[(E^(-4 + 2*x)*(4 - x^2)^2*(4000 - 4300*x + 200*x^2 + 350*x^3 - 5 0*x^4 + (E^(4 - 2*x)*(324 - 81*x^2))/(4 - x^2)^2))/(-324 + 81*x^2),x]
Leaf count is larger than twice the leaf count of optimal. \(87\) vs. \(2(32)=64\).
Time = 0.92 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.72, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {281, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{2 x-4} \left (4-x^2\right )^2 \left (-50 x^4+350 x^3+200 x^2+\frac {e^{4-2 x} \left (324-81 x^2\right )}{\left (4-x^2\right )^2}-4300 x+4000\right )}{81 x^2-324} \, dx\) |
\(\Big \downarrow \) 281 |
\(\displaystyle -\frac {1}{81} \int e^{2 x-4} \left (4-x^2\right ) \left (-50 x^4+350 x^3+200 x^2-4300 x+\frac {81 e^{4-2 x}}{4-x^2}+4000\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{81} \int \left (50 e^{2 x-4} \left (x^6-7 x^5-8 x^4+114 x^3-64 x^2-344 x+320\right )+81\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{81} \left (-25 e^{2 x-4} x^6+250 e^{2 x-4} x^5-425 e^{2 x-4} x^4-2000 e^{2 x-4} x^3+4600 e^{2 x-4} x^2+4000 e^{2 x-4} x-81 x-10000 e^{2 x-4}\right )\) |
Int[(E^(-4 + 2*x)*(4 - x^2)^2*(4000 - 4300*x + 200*x^2 + 350*x^3 - 50*x^4 + (E^(4 - 2*x)*(324 - 81*x^2))/(4 - x^2)^2))/(-324 + 81*x^2),x]
(-10000*E^(-4 + 2*x) - 81*x + 4000*E^(-4 + 2*x)*x + 4600*E^(-4 + 2*x)*x^2 - 2000*E^(-4 + 2*x)*x^3 - 425*E^(-4 + 2*x)*x^4 + 250*E^(-4 + 2*x)*x^5 - 25 *E^(-4 + 2*x)*x^6)/81
3.15.18.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ Symbol] :> Simp[(b/d)^p Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] && !(IntegerQ[q] & & SimplerQ[a + b*x^n, c + d*x^n])
Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
default | \(-x +\frac {25 \left (-4 x^{2}+40 x -100\right ) \left (-x^{2}+4\right )^{2} {\mathrm e}^{2 x -4}}{324}\) | \(35\) |
parts | \(-x +\frac {25 \left (-4 x^{2}+40 x -100\right ) \left (-x^{2}+4\right )^{2} {\mathrm e}^{2 x -4}}{324}\) | \(35\) |
risch | \(-x +\left (-\frac {10000}{81}+\frac {4000}{81} x +\frac {4600}{81} x^{2}-\frac {2000}{81} x^{3}-\frac {425}{81} x^{4}+\frac {250}{81} x^{5}-\frac {25}{81} x^{6}\right ) {\mathrm e}^{2 x -4}\) | \(42\) |
parallelrisch | \(\frac {\left (-5000-\frac {648 x \,{\mathrm e}^{4-2 x}}{\left (-x^{2}+4\right )^{2}}-200 x^{2}+2000 x \right ) \left (-x^{2}+4\right )^{2} {\mathrm e}^{2 x -4}}{648}\) | \(52\) |
meijerg | \(-\frac {250 \,2^{2+4 \,{\mathrm e}^{-4}} {\mathrm e}^{2 x} \left (-x^{2}+4\right )^{2-2 \,{\mathrm e}^{-4}} x \operatorname {hypergeom}\left (\left [\frac {1}{2}, 1-2 \,{\mathrm e}^{-4}\right ], \left [\frac {3}{2}\right ], \frac {x^{2}}{4}\right ) {\mathrm e}^{-4}}{81}+\frac {2 \,2^{-3-4 \,{\mathrm e}^{4}+4 \,{\mathrm e}^{-4}} {\mathrm e}^{4} \left (-x^{2}+4\right )^{2 \,{\mathrm e}^{4}+2-2 \,{\mathrm e}^{-4}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{2}, -2 \,{\mathrm e}^{-4}+2 \,{\mathrm e}^{4}+1\right ], \left [\frac {5}{2}\right ], \frac {x^{2}}{4}\right ) {\mathrm e}^{-4}}{3 \left (x^{2}-4\right )^{2}}+\frac {10 \,2^{-2+4 \,{\mathrm e}^{-4}} {\mathrm e}^{2 x} \left (-x^{2}+4\right )^{2-2 \,{\mathrm e}^{-4}} x^{5} \operatorname {hypergeom}\left (\left [\frac {5}{2}, 1-2 \,{\mathrm e}^{-4}\right ], \left [\frac {7}{2}\right ], \frac {x^{2}}{4}\right ) {\mathrm e}^{-4}}{81}-\frac {350 \,2^{-4+4 \,{\mathrm e}^{-4}} {\mathrm e}^{2 x} \left (-x^{2}+4\right )^{2-2 \,{\mathrm e}^{-4}} x^{4} \operatorname {hypergeom}\left (\left [2, 1-2 \,{\mathrm e}^{-4}\right ], \left [3\right ], \frac {x^{2}}{4}\right ) {\mathrm e}^{-4}}{81}+\frac {2150 \,2^{-2+4 \,{\mathrm e}^{-4}} {\mathrm e}^{2 x} \left (-x^{2}+4\right )^{2-2 \,{\mathrm e}^{-4}} x^{2} \operatorname {hypergeom}\left (\left [1, 1-2 \,{\mathrm e}^{-4}\right ], \left [2\right ], \frac {x^{2}}{4}\right ) {\mathrm e}^{-4}}{81}-\frac {50 \,{\mathrm e}^{2 x} \left (-x^{2}+4\right )^{2-2 \,{\mathrm e}^{-4}} 2^{4 \,{\mathrm e}^{-4}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{2}, 1-2 \,{\mathrm e}^{-4}\right ], \left [\frac {5}{2}\right ], \frac {x^{2}}{4}\right ) {\mathrm e}^{-4}}{243}-\frac {2 \,2^{-4 \,{\mathrm e}^{4}-1+4 \,{\mathrm e}^{-4}} {\mathrm e}^{4} \left (-x^{2}+4\right )^{2 \,{\mathrm e}^{4}+2-2 \,{\mathrm e}^{-4}} x \operatorname {hypergeom}\left (\left [\frac {1}{2}, -2 \,{\mathrm e}^{-4}+2 \,{\mathrm e}^{4}+1\right ], \left [\frac {3}{2}\right ], \frac {x^{2}}{4}\right ) {\mathrm e}^{-4}}{\left (x^{2}-4\right )^{2}}\) | \(373\) |
int(((-81*x^2+324)*exp(-ln(-x^2+4)+2-x)^2-50*x^4+350*x^3+200*x^2-4300*x+40 00)/(81*x^2-324)/exp(-ln(-x^2+4)+2-x)^2,x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-4+2 x} \left (4-x^2\right )^2 \left (4000-4300 x+200 x^2+350 x^3-50 x^4+\frac {e^{4-2 x} \left (324-81 x^2\right )}{\left (4-x^2\right )^2}\right )}{-324+81 x^2} \, dx=-\frac {25}{81} \, {\left (x^{2} - 10 \, x + 25\right )} e^{\left (2 \, x + 2 \, \log \left (-x^{2} + 4\right ) - 4\right )} - x \]
integrate(((-81*x^2+324)*exp(-log(-x^2+4)+2-x)^2-50*x^4+350*x^3+200*x^2-43 00*x+4000)/(81*x^2-324)/exp(-log(-x^2+4)+2-x)^2,x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-4+2 x} \left (4-x^2\right )^2 \left (4000-4300 x+200 x^2+350 x^3-50 x^4+\frac {e^{4-2 x} \left (324-81 x^2\right )}{\left (4-x^2\right )^2}\right )}{-324+81 x^2} \, dx=- x + \frac {\left (- 25 x^{6} + 250 x^{5} - 425 x^{4} - 2000 x^{3} + 4600 x^{2} + 4000 x - 10000\right ) e^{2 x - 4}}{81} \]
integrate(((-81*x**2+324)*exp(-ln(-x**2+4)+2-x)**2-50*x**4+350*x**3+200*x* *2-4300*x+4000)/(81*x**2-324)/exp(-ln(-x**2+4)+2-x)**2,x)
-x + (-25*x**6 + 250*x**5 - 425*x**4 - 2000*x**3 + 4600*x**2 + 4000*x - 10 000)*exp(2*x - 4)/81
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (28) = 56\).
Time = 0.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 4.06 \[ \int \frac {e^{-4+2 x} \left (4-x^2\right )^2 \left (4000-4300 x+200 x^2+350 x^3-50 x^4+\frac {e^{4-2 x} \left (324-81 x^2\right )}{\left (4-x^2\right )^2}\right )}{-324+81 x^2} \, dx=-\frac {25}{81} \, {\left (x^{6} - 10 \, x^{5} + 17 \, x^{4} + 80 \, x^{3} - 184 \, x^{2} - 160 \, x + 400\right )} e^{\left (2 \, x - 4\right )} - x + \frac {9 \, x^{3} - 28 \, x}{2 \, {\left (x^{4} - 8 \, x^{2} + 16\right )}} - \frac {3 \, {\left (5 \, x^{3} - 12 \, x\right )}}{2 \, {\left (x^{4} - 8 \, x^{2} + 16\right )}} + \frac {3 \, x^{3} - 20 \, x}{2 \, {\left (x^{4} - 8 \, x^{2} + 16\right )}} + \frac {3 \, {\left (x^{3} + 4 \, x\right )}}{2 \, {\left (x^{4} - 8 \, x^{2} + 16\right )}} \]
integrate(((-81*x^2+324)*exp(-log(-x^2+4)+2-x)^2-50*x^4+350*x^3+200*x^2-43 00*x+4000)/(81*x^2-324)/exp(-log(-x^2+4)+2-x)^2,x, algorithm=\
-25/81*(x^6 - 10*x^5 + 17*x^4 + 80*x^3 - 184*x^2 - 160*x + 400)*e^(2*x - 4 ) - x + 1/2*(9*x^3 - 28*x)/(x^4 - 8*x^2 + 16) - 3/2*(5*x^3 - 12*x)/(x^4 - 8*x^2 + 16) + 1/2*(3*x^3 - 20*x)/(x^4 - 8*x^2 + 16) + 3/2*(x^3 + 4*x)/(x^4 - 8*x^2 + 16)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (28) = 56\).
Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.12 \[ \int \frac {e^{-4+2 x} \left (4-x^2\right )^2 \left (4000-4300 x+200 x^2+350 x^3-50 x^4+\frac {e^{4-2 x} \left (324-81 x^2\right )}{\left (4-x^2\right )^2}\right )}{-324+81 x^2} \, dx=-\frac {1}{81} \, {\left (25 \, x^{6} e^{\left (2 \, x\right )} - 250 \, x^{5} e^{\left (2 \, x\right )} + 425 \, x^{4} e^{\left (2 \, x\right )} + 2000 \, x^{3} e^{\left (2 \, x\right )} - 4600 \, x^{2} e^{\left (2 \, x\right )} + 81 \, x e^{4} - 4000 \, x e^{\left (2 \, x\right )} + 10000 \, e^{\left (2 \, x\right )}\right )} e^{\left (-4\right )} \]
integrate(((-81*x^2+324)*exp(-log(-x^2+4)+2-x)^2-50*x^4+350*x^3+200*x^2-43 00*x+4000)/(81*x^2-324)/exp(-log(-x^2+4)+2-x)^2,x, algorithm=\
-1/81*(25*x^6*e^(2*x) - 250*x^5*e^(2*x) + 425*x^4*e^(2*x) + 2000*x^3*e^(2* x) - 4600*x^2*e^(2*x) + 81*x*e^4 - 4000*x*e^(2*x) + 10000*e^(2*x))*e^(-4)
Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.38 \[ \int \frac {e^{-4+2 x} \left (4-x^2\right )^2 \left (4000-4300 x+200 x^2+350 x^3-50 x^4+\frac {e^{4-2 x} \left (324-81 x^2\right )}{\left (4-x^2\right )^2}\right )}{-324+81 x^2} \, dx=\frac {4000\,x\,{\mathrm {e}}^{2\,x-4}}{81}-\frac {10000\,{\mathrm {e}}^{2\,x-4}}{81}-x+\frac {4600\,x^2\,{\mathrm {e}}^{2\,x-4}}{81}-\frac {2000\,x^3\,{\mathrm {e}}^{2\,x-4}}{81}-\frac {425\,x^4\,{\mathrm {e}}^{2\,x-4}}{81}+\frac {250\,x^5\,{\mathrm {e}}^{2\,x-4}}{81}-\frac {25\,x^6\,{\mathrm {e}}^{2\,x-4}}{81} \]
int(-(exp(2*x + 2*log(4 - x^2) - 4)*(4300*x - 200*x^2 - 350*x^3 + 50*x^4 + exp(4 - 2*log(4 - x^2) - 2*x)*(81*x^2 - 324) - 4000))/(81*x^2 - 324),x)