Integrand size = 67, antiderivative size = 37 \[ \int \frac {100-100 x^2+e^{6-2 e^3-2 x} \left (50 x^2+50 x^3\right )+e^{3-e^3-x} \left (150 x+50 x^2-50 x^3-50 x^4\right )}{x^5} \, dx=-\frac {\left (5-5 \left (-e^{3-e^3-x}-\frac {1-x}{x}+x\right )\right )^2}{x^2} \] Output:
-(5-5*x+5*(1-x)/x+5*exp(-exp(3)+3-x))^2/x^2
Time = 2.57 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.57 \[ \int \frac {100-100 x^2+e^{6-2 e^3-2 x} \left (50 x^2+50 x^3\right )+e^{3-e^3-x} \left (150 x+50 x^2-50 x^3-50 x^4\right )}{x^5} \, dx=-50 \left (\frac {1}{2 x^4}-\frac {1}{x^2}+\frac {e^{6-2 e^3-2 x}}{2 x^2}-\frac {e^{3-e^3-x} \left (-x+x^3\right )}{x^4}\right ) \] Input:
Integrate[(100 - 100*x^2 + E^(6 - 2*E^3 - 2*x)*(50*x^2 + 50*x^3) + E^(3 - E^3 - x)*(150*x + 50*x^2 - 50*x^3 - 50*x^4))/x^5,x]
Output:
-50*(1/(2*x^4) - x^(-2) + E^(6 - 2*E^3 - 2*x)/(2*x^2) - (E^(3 - E^3 - x)*( -x + x^3))/x^4)
Time = 0.50 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.76, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {2010, 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.
\(\displaystyle \int \frac {-100 x^2+e^{-2 x-2 e^3+6} \left (50 x^3+50 x^2\right )+e^{-x-e^3+3} \left (-50 x^4-50 x^3+50 x^2+150 x\right )+100}{x^5} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {50 e^{-2 x-2 e^3+6} (x+1)}{x^3}-\frac {100 \left (x^2-1\right )}{x^5}-\frac {50 e^{-x-e^3+3} \left (x^3+x^2-x-3\right )}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {25}{x^4}-\frac {50 e^{-x-e^3+3}}{x^3}-\frac {25 e^{2 \left (3-e^3\right )-2 x}}{x^2}+\frac {50}{x^2}+\frac {50 e^{-x-e^3+3}}{x}\) |
Input:
Int[(100 - 100*x^2 + E^(6 - 2*E^3 - 2*x)*(50*x^2 + 50*x^3) + E^(3 - E^3 - x)*(150*x + 50*x^2 - 50*x^3 - 50*x^4))/x^5,x]
Output:
-25/x^4 - (50*E^(3 - E^3 - x))/x^3 + 50/x^2 - (25*E^(2*(3 - E^3) - 2*x))/x ^2 + (50*E^(3 - E^3 - x))/x
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.56 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30
method | result | size |
risch | \(\frac {50 x^{2}-25}{x^{4}}-\frac {25 \,{\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}}{x^{2}}+\frac {50 \left (x^{2}-1\right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}}{x^{3}}\) | \(48\) |
norman | \(\frac {-25+50 x^{2}-50 \,{\mathrm e}^{-{\mathrm e}^{3}+3-x} x +50 \,{\mathrm e}^{-{\mathrm e}^{3}+3-x} x^{3}-25 \,{\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x} x^{2}}{x^{4}}\) | \(57\) |
parallelrisch | \(\frac {-25+50 x^{2}-50 \,{\mathrm e}^{-{\mathrm e}^{3}+3-x} x +50 \,{\mathrm e}^{-{\mathrm e}^{3}+3-x} x^{3}-25 \,{\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x} x^{2}}{x^{4}}\) | \(57\) |
orering | \(-\frac {\left (4 x^{9}+20 x^{8}-2 x^{7}-38 x^{6}+37 x^{5}+89 x^{4}-49 x^{3}-63 x^{2}+60 x -36\right ) \left (\left (50 x^{3}+50 x^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}+\left (-50 x^{4}-50 x^{3}+50 x^{2}+150 x \right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-100 x^{2}+100\right )}{4 x^{4} \left (2 x^{6}-x^{5}-2 x^{4}-x^{3}+12 x^{2}-6 x -6\right ) \left (x^{3}+x^{2}-x +1\right )}-\frac {\left (6 x^{8}+10 x^{7}-9 x^{6}-12 x^{5}+36 x^{4}+18 x^{3}-51 x^{2}+30 x -12\right ) x^{2} \left (\frac {\left (150 x^{2}+100 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}-2 \left (50 x^{3}+50 x^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}+\left (-200 x^{3}-150 x^{2}+100 x +150\right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-\left (-50 x^{4}-50 x^{3}+50 x^{2}+150 x \right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-200 x}{x^{5}}-\frac {5 \left (\left (50 x^{3}+50 x^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}+\left (-50 x^{4}-50 x^{3}+50 x^{2}+150 x \right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-100 x^{2}+100\right )}{x^{6}}\right )}{4 \left (2 x^{6}-x^{5}-2 x^{4}-x^{3}+12 x^{2}-6 x -6\right ) \left (x^{3}+x^{2}-x +1\right )}-\frac {\left (2 x^{7}-3 x^{5}+3 x^{4}+4 x^{3}-6 x^{2}+3 x -1\right ) x^{3} \left (\frac {\left (300 x +100\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}-4 \left (150 x^{2}+100 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}+4 \left (50 x^{3}+50 x^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}+\left (-600 x^{2}-300 x +100\right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-2 \left (-200 x^{3}-150 x^{2}+100 x +150\right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}+\left (-50 x^{4}-50 x^{3}+50 x^{2}+150 x \right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-200}{x^{5}}-\frac {10 \left (\left (150 x^{2}+100 x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}-2 \left (50 x^{3}+50 x^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}+\left (-200 x^{3}-150 x^{2}+100 x +150\right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-\left (-50 x^{4}-50 x^{3}+50 x^{2}+150 x \right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-200 x \right )}{x^{6}}+\frac {30 \left (50 x^{3}+50 x^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{3}+6-2 x}+30 \left (-50 x^{4}-50 x^{3}+50 x^{2}+150 x \right ) {\mathrm e}^{-{\mathrm e}^{3}+3-x}-3000 x^{2}+3000}{x^{7}}\right )}{4 \left (2 x^{6}-x^{5}-2 x^{4}-x^{3}+12 x^{2}-6 x -6\right ) \left (x^{3}+x^{2}-x +1\right )}\) | \(837\) |
parts | \(\text {Expression too large to display}\) | \(1127\) |
derivativedivides | \(\text {Expression too large to display}\) | \(3528\) |
default | \(\text {Expression too large to display}\) | \(3528\) |
Input:
int(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x)*exp( -exp(3)+3-x)-100*x^2+100)/x^5,x,method=_RETURNVERBOSE)
Output:
(50*x^2-25)/x^4-25*exp(-2*exp(3)+6-2*x)/x^2+50*(x^2-1)/x^3*exp(-exp(3)+3-x )
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {100-100 x^2+e^{6-2 e^3-2 x} \left (50 x^2+50 x^3\right )+e^{3-e^3-x} \left (150 x+50 x^2-50 x^3-50 x^4\right )}{x^5} \, dx=-\frac {25 \, {\left (x^{2} e^{\left (-2 \, x - 2 \, e^{3} + 6\right )} - 2 \, x^{2} - 2 \, {\left (x^{3} - x\right )} e^{\left (-x - e^{3} + 3\right )} + 1\right )}}{x^{4}} \] Input:
integrate(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x )*exp(-exp(3)+3-x)-100*x^2+100)/x^5,x, algorithm="fricas")
Output:
-25*(x^2*e^(-2*x - 2*e^3 + 6) - 2*x^2 - 2*(x^3 - x)*e^(-x - e^3 + 3) + 1)/ x^4
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.30 \[ \int \frac {100-100 x^2+e^{6-2 e^3-2 x} \left (50 x^2+50 x^3\right )+e^{3-e^3-x} \left (150 x+50 x^2-50 x^3-50 x^4\right )}{x^5} \, dx=- \frac {25 - 50 x^{2}}{x^{4}} + \frac {- 25 x^{3} e^{- 2 x - 2 e^{3} + 6} + \left (50 x^{4} - 50 x^{2}\right ) e^{- x - e^{3} + 3}}{x^{5}} \] Input:
integrate(((50*x**3+50*x**2)*exp(-exp(3)+3-x)**2+(-50*x**4-50*x**3+50*x**2 +150*x)*exp(-exp(3)+3-x)-100*x**2+100)/x**5,x)
Output:
-(25 - 50*x**2)/x**4 + (-25*x**3*exp(-2*x - 2*exp(3) + 6) + (50*x**4 - 50* x**2)*exp(-x - exp(3) + 3))/x**5
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.38 \[ \int \frac {100-100 x^2+e^{6-2 e^3-2 x} \left (50 x^2+50 x^3\right )+e^{3-e^3-x} \left (150 x+50 x^2-50 x^3-50 x^4\right )}{x^5} \, dx=-50 \, {\rm Ei}\left (-x\right ) e^{\left (-e^{3} + 3\right )} - 100 \, e^{\left (-2 \, e^{3} + 6\right )} \Gamma \left (-1, 2 \, x\right ) + 50 \, e^{\left (-e^{3} + 3\right )} \Gamma \left (-1, x\right ) - 200 \, e^{\left (-2 \, e^{3} + 6\right )} \Gamma \left (-2, 2 \, x\right ) - 50 \, e^{\left (-e^{3} + 3\right )} \Gamma \left (-2, x\right ) - 150 \, e^{\left (-e^{3} + 3\right )} \Gamma \left (-3, x\right ) + \frac {50}{x^{2}} - \frac {25}{x^{4}} \] Input:
integrate(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x )*exp(-exp(3)+3-x)-100*x^2+100)/x^5,x, algorithm="maxima")
Output:
-50*Ei(-x)*e^(-e^3 + 3) - 100*e^(-2*e^3 + 6)*gamma(-1, 2*x) + 50*e^(-e^3 + 3)*gamma(-1, x) - 200*e^(-2*e^3 + 6)*gamma(-2, 2*x) - 50*e^(-e^3 + 3)*gam ma(-2, x) - 150*e^(-e^3 + 3)*gamma(-3, x) + 50/x^2 - 25/x^4
Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (29) = 58\).
Time = 0.18 (sec) , antiderivative size = 376, normalized size of antiderivative = 10.16 \[ \int \frac {100-100 x^2+e^{6-2 e^3-2 x} \left (50 x^2+50 x^3\right )+e^{3-e^3-x} \left (150 x+50 x^2-50 x^3-50 x^4\right )}{x^5} \, dx=\frac {25 \, {\left (2 \, {\left (x + e^{3} - 3\right )}^{3} e^{\left (-x - e^{3} + 3\right )} - 6 \, {\left (x + e^{3} - 3\right )}^{2} e^{\left (-x - e^{3} + 6\right )} + 18 \, {\left (x + e^{3} - 3\right )}^{2} e^{\left (-x - e^{3} + 3\right )} - {\left (x + e^{3} - 3\right )}^{2} e^{\left (-2 \, x - 2 \, e^{3} + 6\right )} + 2 \, {\left (x + e^{3} - 3\right )}^{2} - 4 \, {\left (x + e^{3} - 3\right )} e^{3} + 6 \, {\left (x + e^{3} - 3\right )} e^{\left (-x - e^{3} + 9\right )} - 36 \, {\left (x + e^{3} - 3\right )} e^{\left (-x - e^{3} + 6\right )} + 52 \, {\left (x + e^{3} - 3\right )} e^{\left (-x - e^{3} + 3\right )} + 2 \, {\left (x + e^{3} - 3\right )} e^{\left (-2 \, x - 2 \, e^{3} + 9\right )} - 6 \, {\left (x + e^{3} - 3\right )} e^{\left (-2 \, x - 2 \, e^{3} + 6\right )} + 12 \, x + 2 \, e^{6} - 2 \, e^{\left (-x - e^{3} + 12\right )} + 18 \, e^{\left (-x - e^{3} + 9\right )} - 52 \, e^{\left (-x - e^{3} + 6\right )} + 48 \, e^{\left (-x - e^{3} + 3\right )} - e^{\left (-2 \, x - 2 \, e^{3} + 12\right )} + 6 \, e^{\left (-2 \, x - 2 \, e^{3} + 9\right )} - 9 \, e^{\left (-2 \, x - 2 \, e^{3} + 6\right )} - 19\right )}}{{\left (x + e^{3} - 3\right )}^{4} - 4 \, {\left (x + e^{3} - 3\right )}^{3} e^{3} + 12 \, {\left (x + e^{3} - 3\right )}^{3} + 6 \, {\left (x + e^{3} - 3\right )}^{2} e^{6} - 36 \, {\left (x + e^{3} - 3\right )}^{2} e^{3} + 54 \, {\left (x + e^{3} - 3\right )}^{2} - 4 \, {\left (x + e^{3} - 3\right )} e^{9} + 36 \, {\left (x + e^{3} - 3\right )} e^{6} - 108 \, {\left (x + e^{3} - 3\right )} e^{3} + 108 \, x + e^{12} - 12 \, e^{9} + 54 \, e^{6} - 243} \] Input:
integrate(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x )*exp(-exp(3)+3-x)-100*x^2+100)/x^5,x, algorithm="giac")
Output:
25*(2*(x + e^3 - 3)^3*e^(-x - e^3 + 3) - 6*(x + e^3 - 3)^2*e^(-x - e^3 + 6 ) + 18*(x + e^3 - 3)^2*e^(-x - e^3 + 3) - (x + e^3 - 3)^2*e^(-2*x - 2*e^3 + 6) + 2*(x + e^3 - 3)^2 - 4*(x + e^3 - 3)*e^3 + 6*(x + e^3 - 3)*e^(-x - e ^3 + 9) - 36*(x + e^3 - 3)*e^(-x - e^3 + 6) + 52*(x + e^3 - 3)*e^(-x - e^3 + 3) + 2*(x + e^3 - 3)*e^(-2*x - 2*e^3 + 9) - 6*(x + e^3 - 3)*e^(-2*x - 2 *e^3 + 6) + 12*x + 2*e^6 - 2*e^(-x - e^3 + 12) + 18*e^(-x - e^3 + 9) - 52* e^(-x - e^3 + 6) + 48*e^(-x - e^3 + 3) - e^(-2*x - 2*e^3 + 12) + 6*e^(-2*x - 2*e^3 + 9) - 9*e^(-2*x - 2*e^3 + 6) - 19)/((x + e^3 - 3)^4 - 4*(x + e^3 - 3)^3*e^3 + 12*(x + e^3 - 3)^3 + 6*(x + e^3 - 3)^2*e^6 - 36*(x + e^3 - 3 )^2*e^3 + 54*(x + e^3 - 3)^2 - 4*(x + e^3 - 3)*e^9 + 36*(x + e^3 - 3)*e^6 - 108*(x + e^3 - 3)*e^3 + 108*x + e^12 - 12*e^9 + 54*e^6 - 243)
Time = 0.51 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.59 \[ \int \frac {100-100 x^2+e^{6-2 e^3-2 x} \left (50 x^2+50 x^3\right )+e^{3-e^3-x} \left (150 x+50 x^2-50 x^3-50 x^4\right )}{x^5} \, dx=\frac {50}{x^2}-\frac {25}{x^4}+\frac {50\,{\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x}-\frac {50\,{\mathrm {e}}^{-{\mathrm {e}}^3}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3}{x^3}-\frac {25\,{\mathrm {e}}^{-2\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^6}{x^2} \] Input:
int((exp(6 - 2*exp(3) - 2*x)*(50*x^2 + 50*x^3) + exp(3 - exp(3) - x)*(150* x + 50*x^2 - 50*x^3 - 50*x^4) - 100*x^2 + 100)/x^5,x)
Output:
50/x^2 - 25/x^4 + (50*exp(-exp(3))*exp(-x)*exp(3))/x - (50*exp(-exp(3))*ex p(-x)*exp(3))/x^3 - (25*exp(-2*exp(3))*exp(-2*x)*exp(6))/x^2
Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.27 \[ \int \frac {100-100 x^2+e^{6-2 e^3-2 x} \left (50 x^2+50 x^3\right )+e^{3-e^3-x} \left (150 x+50 x^2-50 x^3-50 x^4\right )}{x^5} \, dx=\frac {50 e^{2 e^{3}+2 x} x^{2}-25 e^{2 e^{3}+2 x}+50 e^{e^{3}+x} e^{3} x^{3}-50 e^{e^{3}+x} e^{3} x -25 e^{6} x^{2}}{e^{2 e^{3}+2 x} x^{4}} \] Input:
int(((50*x^3+50*x^2)*exp(-exp(3)+3-x)^2+(-50*x^4-50*x^3+50*x^2+150*x)*exp( -exp(3)+3-x)-100*x^2+100)/x^5,x)
Output:
(25*(2*e**(2*e**3 + 2*x)*x**2 - e**(2*e**3 + 2*x) + 2*e**(e**3 + x)*e**3*x **3 - 2*e**(e**3 + x)*e**3*x - e**6*x**2))/(e**(2*e**3 + 2*x)*x**4)