Integrand size = 62, antiderivative size = 28 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=5+\frac {\left (\frac {e^x}{5}+\frac {-1+e^3}{3 x}\right )^2}{x^2} \] Output:
5+(1/3*(exp(3)-1)/x+1/5*exp(x))^2/x^2
Time = 1.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {\left (-5+5 e^3+3 e^x x\right )^2}{225 x^4} \] Input:
Integrate[(-100 + 200*E^3 - 100*E^6 + E^(2*x)*(-18*x^2 + 18*x^3) + E^x*(90 *x - 30*x^2 + E^3*(-90*x + 30*x^2)))/(225*x^5),x]
Output:
(-5 + 5*E^3 + 3*E^x*x)^2/(225*x^4)
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 27, 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.
\(\displaystyle \int \frac {e^x \left (-30 x^2+e^3 \left (30 x^2-90 x\right )+90 x\right )+e^{2 x} \left (18 x^3-18 x^2\right )-100 e^6+200 e^3-100}{225 x^5} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{225} \int -\frac {2 \left (9 e^{2 x} \left (x^2-x^3\right )-15 e^x \left (-x^2+3 x-e^3 \left (3 x-x^2\right )\right )+50 \left (1-e^3\right )^2\right )}{x^5}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{225} \int \frac {9 e^{2 x} \left (x^2-x^3\right )-15 e^x \left (-x^2+3 x-e^3 \left (3 x-x^2\right )\right )+50 \left (1-e^3\right )^2}{x^5}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {2}{225} \int \left (-\frac {15 (-1+e) e^x \left (1+e+e^2\right ) (x-3)}{x^4}-\frac {9 e^{2 x} (x-1)}{x^3}+\frac {50 \left (-1+e^3\right )^2}{x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2}{225} \left (-\frac {25 \left (1-e^3\right )^2}{2 x^4}+\frac {15 (1-e) \left (1+e+e^2\right ) e^x}{x^3}-\frac {9 e^{2 x}}{2 x^2}\right )\) |
Input:
Int[(-100 + 200*E^3 - 100*E^6 + E^(2*x)*(-18*x^2 + 18*x^3) + E^x*(90*x - 3 0*x^2 + E^3*(-90*x + 30*x^2)))/(225*x^5),x]
Output:
(-2*((-25*(1 - E^3)^2)/(2*x^4) + (15*(1 - E)*E^x*(1 + E + E^2))/x^3 - (9*E ^(2*x))/(2*x^2)))/225
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 = 0.33 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29
method | result | size |
norman | \(\frac {\left (\frac {2 \,{\mathrm e}^{3}}{15}-\frac {2}{15}\right ) x \,{\mathrm e}^{x}+\frac {{\mathrm e}^{2 x} x^{2}}{25}+\frac {{\mathrm e}^{6}}{9}-\frac {2 \,{\mathrm e}^{3}}{9}+\frac {1}{9}}{x^{4}}\) | \(36\) |
risch | \(\frac {9 \,{\mathrm e}^{2 x} x^{2}+30 \,{\mathrm e}^{3+x} x -30 \,{\mathrm e}^{x} x +25 \,{\mathrm e}^{6}-50 \,{\mathrm e}^{3}+25}{225 x^{4}}\) | \(37\) |
parallelrisch | \(\frac {9 \,{\mathrm e}^{2 x} x^{2}+30 x \,{\mathrm e}^{3} {\mathrm e}^{x}-30 \,{\mathrm e}^{x} x +25 \,{\mathrm e}^{6}-50 \,{\mathrm e}^{3}+25}{225 x^{4}}\) | \(39\) |
parts | \(-\frac {-\frac {4 \,{\mathrm e}^{6}}{9}+\frac {8 \,{\mathrm e}^{3}}{9}-\frac {4}{9}}{4 x^{4}}+\frac {2 \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{2 x^{2}}-\frac {{\mathrm e}^{x}}{2 x}-\frac {\operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )}{15}-\frac {2 \,{\mathrm e}^{x}}{15 x^{3}}-\frac {2 \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{3 x^{3}}-\frac {{\mathrm e}^{x}}{6 x^{2}}-\frac {{\mathrm e}^{x}}{6 x}-\frac {\operatorname {expIntegral}_{1}\left (-x \right )}{6}\right )}{5}+\frac {{\mathrm e}^{2 x}}{25 x^{2}}\) | \(94\) |
default | \(\frac {1}{9 x^{4}}-\frac {2 \,{\mathrm e}^{3}}{9 x^{4}}+\frac {{\mathrm e}^{6}}{9 x^{4}}-\frac {2 \,{\mathrm e}^{x}}{15 x^{3}}+\frac {{\mathrm e}^{2 x}}{25 x^{2}}-\frac {2 \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{3 x^{3}}-\frac {{\mathrm e}^{x}}{6 x^{2}}-\frac {{\mathrm e}^{x}}{6 x}-\frac {\operatorname {expIntegral}_{1}\left (-x \right )}{6}\right )}{5}+\frac {2 \,{\mathrm e}^{3} \left (-\frac {{\mathrm e}^{x}}{2 x^{2}}-\frac {{\mathrm e}^{x}}{2 x}-\frac {\operatorname {expIntegral}_{1}\left (-x \right )}{2}\right )}{15}\) | \(98\) |
orering | \(-\frac {\left (2 x^{4}-16 x^{3}-27 x^{2}+24 x +36\right ) \left (\left (18 x^{3}-18 x^{2}\right ) {\mathrm e}^{2 x}+\left (\left (30 x^{2}-90 x \right ) {\mathrm e}^{3}-30 x^{2}+90 x \right ) {\mathrm e}^{x}-100 \,{\mathrm e}^{6}+200 \,{\mathrm e}^{3}-100\right )}{900 x^{4} \left (x^{3}-6 x -6\right ) \left (-1+2 x \right )}+\frac {3 x^{2} \left (x^{3}-x^{2}-6 x -4\right ) \left (\frac {\left (54 x^{2}-36 x \right ) {\mathrm e}^{2 x}+2 \left (18 x^{3}-18 x^{2}\right ) {\mathrm e}^{2 x}+\left (\left (60 x -90\right ) {\mathrm e}^{3}-60 x +90\right ) {\mathrm e}^{x}+\left (\left (30 x^{2}-90 x \right ) {\mathrm e}^{3}-30 x^{2}+90 x \right ) {\mathrm e}^{x}}{225 x^{5}}-\frac {\left (18 x^{3}-18 x^{2}\right ) {\mathrm e}^{2 x}+\left (\left (30 x^{2}-90 x \right ) {\mathrm e}^{3}-30 x^{2}+90 x \right ) {\mathrm e}^{x}-100 \,{\mathrm e}^{6}+200 \,{\mathrm e}^{3}-100}{45 x^{6}}\right )}{4 \left (x^{3}-6 x -6\right ) \left (-1+2 x \right )}-\frac {x^{3} \left (x^{2}+2 x +1\right ) \left (\frac {\left (108 x -36\right ) {\mathrm e}^{2 x}+4 \left (54 x^{2}-36 x \right ) {\mathrm e}^{2 x}+4 \left (18 x^{3}-18 x^{2}\right ) {\mathrm e}^{2 x}+\left (60 \,{\mathrm e}^{3}-60\right ) {\mathrm e}^{x}+2 \left (\left (60 x -90\right ) {\mathrm e}^{3}-60 x +90\right ) {\mathrm e}^{x}+\left (\left (30 x^{2}-90 x \right ) {\mathrm e}^{3}-30 x^{2}+90 x \right ) {\mathrm e}^{x}}{225 x^{5}}-\frac {2 \left (\left (54 x^{2}-36 x \right ) {\mathrm e}^{2 x}+2 \left (18 x^{3}-18 x^{2}\right ) {\mathrm e}^{2 x}+\left (\left (60 x -90\right ) {\mathrm e}^{3}-60 x +90\right ) {\mathrm e}^{x}+\left (\left (30 x^{2}-90 x \right ) {\mathrm e}^{3}-30 x^{2}+90 x \right ) {\mathrm e}^{x}\right )}{45 x^{6}}+\frac {\frac {2 \left (18 x^{3}-18 x^{2}\right ) {\mathrm e}^{2 x}}{15}+\frac {2 \left (\left (30 x^{2}-90 x \right ) {\mathrm e}^{3}-30 x^{2}+90 x \right ) {\mathrm e}^{x}}{15}-\frac {40 \,{\mathrm e}^{6}}{3}+\frac {80 \,{\mathrm e}^{3}}{3}-\frac {40}{3}}{x^{7}}\right )}{4 \left (x^{3}-6 x -6\right ) \left (-1+2 x \right )}\) | \(529\) |
Input:
int(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90*x)*exp (x)-100*exp(3)^2+200*exp(3)-100)/x^5,x,method=_RETURNVERBOSE)
Output:
((2/15*exp(3)-2/15)*x*exp(x)+1/25*exp(x)^2*x^2+1/9*exp(3)^2-2/9*exp(3)+1/9 )/x^4
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {9 \, x^{2} e^{\left (2 \, x\right )} + 30 \, {\left (x e^{3} - x\right )} e^{x} + 25 \, e^{6} - 50 \, e^{3} + 25}{225 \, x^{4}} \] Input:
integrate(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90* x)*exp(x)-100*exp(3)^2+200*exp(3)-100)/x^5,x, algorithm="fricas")
Output:
1/225*(9*x^2*e^(2*x) + 30*(x*e^3 - x)*e^x + 25*e^6 - 50*e^3 + 25)/x^4
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=- \frac {- \frac {4 e^{6}}{9} - \frac {4}{9} + \frac {8 e^{3}}{9}}{4 x^{4}} + \frac {15 x^{3} e^{2 x} + \left (- 50 x^{2} + 50 x^{2} e^{3}\right ) e^{x}}{375 x^{5}} \] Input:
integrate(1/225*((18*x**3-18*x**2)*exp(x)**2+((30*x**2-90*x)*exp(3)-30*x** 2+90*x)*exp(x)-100*exp(3)**2+200*exp(3)-100)/x**5,x)
Output:
-(-4*exp(6)/9 - 4/9 + 8*exp(3)/9)/(4*x**4) + (15*x**3*exp(2*x) + (-50*x**2 + 50*x**2*exp(3))*exp(x))/(375*x**5)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=-\frac {2}{15} \, e^{3} \Gamma \left (-2, -x\right ) - \frac {2}{5} \, e^{3} \Gamma \left (-3, -x\right ) + \frac {e^{6}}{9 \, x^{4}} - \frac {2 \, e^{3}}{9 \, x^{4}} + \frac {1}{9 \, x^{4}} + \frac {4}{25} \, \Gamma \left (-1, -2 \, x\right ) + \frac {2}{15} \, \Gamma \left (-2, -x\right ) + \frac {8}{25} \, \Gamma \left (-2, -2 \, x\right ) + \frac {2}{5} \, \Gamma \left (-3, -x\right ) \] Input:
integrate(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90* x)*exp(x)-100*exp(3)^2+200*exp(3)-100)/x^5,x, algorithm="maxima")
Output:
-2/15*e^3*gamma(-2, -x) - 2/5*e^3*gamma(-3, -x) + 1/9*e^6/x^4 - 2/9*e^3/x^ 4 + 1/9/x^4 + 4/25*gamma(-1, -2*x) + 2/15*gamma(-2, -x) + 8/25*gamma(-2, - 2*x) + 2/5*gamma(-3, -x)
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {9 \, x^{2} e^{\left (2 \, x\right )} + 30 \, x e^{\left (x + 3\right )} - 30 \, x e^{x} + 25 \, e^{6} - 50 \, e^{3} + 25}{225 \, x^{4}} \] Input:
integrate(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90* x)*exp(x)-100*exp(3)^2+200*exp(3)-100)/x^5,x, algorithm="giac")
Output:
1/225*(9*x^2*e^(2*x) + 30*x*e^(x + 3) - 30*x*e^x + 25*e^6 - 50*e^3 + 25)/x ^4
Time = 2.81 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {25\,{\left ({\mathrm {e}}^3-1\right )}^2+9\,x^2\,{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^x\,\left (30\,{\mathrm {e}}^3-30\right )}{225\,x^4} \] Input:
int(-((4*exp(6))/9 - (8*exp(3))/9 + (exp(2*x)*(18*x^2 - 18*x^3))/225 + (ex p(x)*(exp(3)*(90*x - 30*x^2) - 90*x + 30*x^2))/225 + 4/9)/x^5,x)
Output:
(25*(exp(3) - 1)^2 + 9*x^2*exp(2*x) + x*exp(x)*(30*exp(3) - 30))/(225*x^4)
Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {-100+200 e^3-100 e^6+e^{2 x} \left (-18 x^2+18 x^3\right )+e^x \left (90 x-30 x^2+e^3 \left (-90 x+30 x^2\right )\right )}{225 x^5} \, dx=\frac {9 e^{2 x} x^{2}+30 e^{x} e^{3} x -30 e^{x} x +25 e^{6}-50 e^{3}+25}{225 x^{4}} \] Input:
int(1/225*((18*x^3-18*x^2)*exp(x)^2+((30*x^2-90*x)*exp(3)-30*x^2+90*x)*exp (x)-100*exp(3)^2+200*exp(3)-100)/x^5,x)
Output:
(9*e**(2*x)*x**2 + 30*e**x*e**3*x - 30*e**x*x + 25*e**6 - 50*e**3 + 25)/(2 25*x**4)