Integrand size = 45, antiderivative size = 26 \[ \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{45 x^2} \, dx=\frac {4}{5} \left (16-e^{\left .\frac {1}{2}\right /x}+\frac {1}{3} (-4+x)\right )^2 \] Output:
2/5*(44/3+1/3*x-exp(1/2/x))*(88/3+2/3*x-2*exp(1/2/x))
Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{45 x^2} \, dx=\frac {4}{45} \left (44-3 e^{\left .\frac {1}{2}\right /x}+x\right )^2 \] Input:
Integrate[(-36*E^x^(-1) + 352*x^2 + 8*x^3 + E^(1/(2*x))*(528 + 12*x - 24*x ^2))/(45*x^2),x]
Output:
(4*(44 - 3*E^(1/(2*x)) + x)^2)/45
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, 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.
\(\displaystyle \int \frac {8 x^3+352 x^2+e^{\left .\frac {1}{2}\right /x} \left (-24 x^2+12 x+528\right )-36 e^{\frac {1}{x}}}{45 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{45} \int -\frac {4 \left (-2 x^3-88 x^2+9 e^{\frac {1}{x}}-3 e^{\left .\frac {1}{2}\right /x} \left (-2 x^2+x+44\right )\right )}{x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{45} \int \frac {-2 x^3-88 x^2+9 e^{\frac {1}{x}}-3 e^{\left .\frac {1}{2}\right /x} \left (-2 x^2+x+44\right )}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {4}{45} \int \left (-2 (x+44)+\frac {3 e^{\left .\frac {1}{2}\right /x} \left (2 x^2-x-44\right )}{x^2}+\frac {9 e^{\frac {1}{x}}}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4}{45} \left (-(x+44)^2+6 e^{\left .\frac {1}{2}\right /x} (x+44)-9 e^{\frac {1}{x}}\right )\) |
Input:
Int[(-36*E^x^(-1) + 352*x^2 + 8*x^3 + E^(1/(2*x))*(528 + 12*x - 24*x^2))/( 45*x^2),x]
Output:
(-4*(-9*E^x^(-1) + 6*E^(1/(2*x))*(44 + x) - (44 + x)^2))/45
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 = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {4 x^{2}}{45}+\frac {4 \,{\mathrm e}^{\frac {1}{x}}}{5}+\frac {352 x}{45}+\frac {\left (-1056-24 x \right ) {\mathrm e}^{\frac {1}{2 x}}}{45}\) | \(29\) |
derivativedivides | \(\frac {4 x^{2}}{45}+\frac {4 \,{\mathrm e}^{\frac {1}{x}}}{5}+\frac {352 x}{45}-\frac {8 \,{\mathrm e}^{\frac {1}{2 x}} x}{15}-\frac {352 \,{\mathrm e}^{\frac {1}{2 x}}}{15}\) | \(37\) |
default | \(\frac {4 x^{2}}{45}+\frac {4 \,{\mathrm e}^{\frac {1}{x}}}{5}+\frac {352 x}{45}-\frac {8 \,{\mathrm e}^{\frac {1}{2 x}} x}{15}-\frac {352 \,{\mathrm e}^{\frac {1}{2 x}}}{15}\) | \(37\) |
parallelrisch | \(\frac {4 x^{2}}{45}+\frac {4 \,{\mathrm e}^{\frac {1}{x}}}{5}+\frac {352 x}{45}-\frac {8 \,{\mathrm e}^{\frac {1}{2 x}} x}{15}-\frac {352 \,{\mathrm e}^{\frac {1}{2 x}}}{15}\) | \(37\) |
parts | \(\frac {4 x^{2}}{45}+\frac {4 \,{\mathrm e}^{\frac {1}{x}}}{5}+\frac {352 x}{45}-\frac {8 \,{\mathrm e}^{\frac {1}{2 x}} x}{15}-\frac {352 \,{\mathrm e}^{\frac {1}{2 x}}}{15}\) | \(37\) |
norman | \(\frac {\frac {352 x^{2}}{45}+\frac {4 x^{3}}{45}-\frac {352 \,{\mathrm e}^{\frac {1}{2 x}} x}{15}-\frac {8 \,{\mathrm e}^{\frac {1}{2 x}} x^{2}}{15}+\frac {4 x \,{\mathrm e}^{\frac {1}{x}}}{5}}{x}\) | \(47\) |
orering | \(\frac {\left (768 x^{7}+17568 x^{6}+25456 x^{5}+672 x^{4}-376648 x^{3}-8707956 x^{2}-4275481 x -704748\right ) \left (-36 \,{\mathrm e}^{\frac {1}{x}}+\left (-24 x^{2}+12 x +528\right ) {\mathrm e}^{\frac {1}{2 x}}+8 x^{3}+352 x^{2}\right )}{720 \left (2 x^{2}+x +44\right ) \left (24 x^{4}-533 x^{2}-265 x -44\right ) x^{2}}-\frac {\left (384 x^{6}+17376 x^{5}+12816 x^{4}+751872 x^{3}+751882 x^{2}+8597955 x +2114244\right ) x^{2} \left (\frac {\frac {36 \,{\mathrm e}^{\frac {1}{x}}}{x^{2}}+\left (-48 x +12\right ) {\mathrm e}^{\frac {1}{2 x}}-\frac {\left (-24 x^{2}+12 x +528\right ) {\mathrm e}^{\frac {1}{2 x}}}{2 x^{2}}+24 x^{2}+704 x}{45 x^{2}}-\frac {2 \left (-36 \,{\mathrm e}^{\frac {1}{x}}+\left (-24 x^{2}+12 x +528\right ) {\mathrm e}^{\frac {1}{2 x}}+8 x^{3}+352 x^{2}\right )}{45 x^{3}}\right )}{16 \left (2 x^{2}+x +44\right ) \left (24 x^{4}-533 x^{2}-265 x -44\right )}-\frac {\left (64 x^{6}+96 x^{5}+4272 x^{4}+93984 x^{2}+46993 x +704748\right ) x^{4} \left (\frac {-\frac {36 \,{\mathrm e}^{\frac {1}{x}}}{x^{4}}-\frac {72 \,{\mathrm e}^{\frac {1}{x}}}{x^{3}}-48 \,{\mathrm e}^{\frac {1}{2 x}}-\frac {\left (-48 x +12\right ) {\mathrm e}^{\frac {1}{2 x}}}{x^{2}}+\frac {\left (-24 x^{2}+12 x +528\right ) {\mathrm e}^{\frac {1}{2 x}}}{x^{3}}+\frac {\left (-24 x^{2}+12 x +528\right ) {\mathrm e}^{\frac {1}{2 x}}}{4 x^{4}}+48 x +704}{45 x^{2}}-\frac {4 \left (\frac {36 \,{\mathrm e}^{\frac {1}{x}}}{x^{2}}+\left (-48 x +12\right ) {\mathrm e}^{\frac {1}{2 x}}-\frac {\left (-24 x^{2}+12 x +528\right ) {\mathrm e}^{\frac {1}{2 x}}}{2 x^{2}}+24 x^{2}+704 x \right )}{45 x^{3}}+\frac {-\frac {24 \,{\mathrm e}^{\frac {1}{x}}}{5}+\frac {2 \left (-24 x^{2}+12 x +528\right ) {\mathrm e}^{\frac {1}{2 x}}}{15}+\frac {16 x^{3}}{15}+\frac {704 x^{2}}{15}}{x^{4}}\right )}{8 \left (2 x^{2}+x +44\right ) \left (24 x^{4}-533 x^{2}-265 x -44\right )}\) | \(535\) |
Input:
int(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x^2)/x^ 2,x,method=_RETURNVERBOSE)
Output:
4/45*x^2+4/5*exp(1/x)+352/45*x+1/45*(-1056-24*x)*exp(1/2/x)
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{45 x^2} \, dx=\frac {4}{45} \, x^{2} - \frac {8}{15} \, {\left (x + 44\right )} e^{\left (\frac {1}{2 \, x}\right )} + \frac {352}{45} \, x + \frac {4}{5} \, e^{\frac {1}{x}} \] Input:
integrate(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x ^2)/x^2,x, algorithm="fricas")
Output:
4/45*x^2 - 8/15*(x + 44)*e^(1/2/x) + 352/45*x + 4/5*e^(1/x)
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{45 x^2} \, dx=\frac {4 x^{2}}{45} + \frac {352 x}{45} + \frac {\left (- 40 x - 1760\right ) e^{\frac {1}{2 x}}}{75} + \frac {4 e^{\frac {1}{x}}}{5} \] Input:
integrate(1/45*(-36*exp(1/2/x)**2+(-24*x**2+12*x+528)*exp(1/2/x)+8*x**3+35 2*x**2)/x**2,x)
Output:
4*x**2/45 + 352*x/45 + (-40*x - 1760)*exp(1/(2*x))/75 + 4*exp(1/x)/5
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{45 x^2} \, dx=\frac {4}{45} \, x^{2} + \frac {352}{45} \, x - \frac {4}{15} \, {\rm Ei}\left (\frac {1}{2 \, x}\right ) - \frac {352}{15} \, e^{\left (\frac {1}{2 \, x}\right )} + \frac {4}{5} \, e^{\frac {1}{x}} + \frac {4}{15} \, \Gamma \left (-1, -\frac {1}{2 \, x}\right ) \] Input:
integrate(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x ^2)/x^2,x, algorithm="maxima")
Output:
4/45*x^2 + 352/45*x - 4/15*Ei(1/2/x) - 352/15*e^(1/2/x) + 4/5*e^(1/x) + 4/ 15*gamma(-1, -1/2/x)
Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (15) = 30\).
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{45 x^2} \, dx=\frac {4}{45} \, x^{2} - \frac {8}{15} \, x e^{\left (\frac {1}{2 \, x}\right )} + \frac {352}{45} \, x - \frac {352}{15} \, e^{\left (\frac {1}{2 \, x}\right )} + \frac {4}{5} \, e^{\frac {1}{x}} \] Input:
integrate(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x ^2)/x^2,x, algorithm="giac")
Output:
4/45*x^2 - 8/15*x*e^(1/2/x) + 352/45*x - 352/15*e^(1/2/x) + 4/5*e^(1/x)
Time = 2.68 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{45 x^2} \, dx=\frac {4\,{\mathrm {e}}^{1/x}}{5}-\frac {352\,{\mathrm {e}}^{\frac {1}{2\,x}}}{15}-x\,\left (\frac {8\,{\mathrm {e}}^{\frac {1}{2\,x}}}{15}-\frac {352}{45}\right )+\frac {4\,x^2}{45} \] Input:
int(((exp(1/(2*x))*(12*x - 24*x^2 + 528))/45 - (4*exp(1/x))/5 + (352*x^2)/ 45 + (8*x^3)/45)/x^2,x)
Output:
(4*exp(1/x))/5 - (352*exp(1/(2*x)))/15 - x*((8*exp(1/(2*x)))/15 - 352/45) + (4*x^2)/45
Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {-36 e^{\frac {1}{x}}+352 x^2+8 x^3+e^{\left .\frac {1}{2}\right /x} \left (528+12 x-24 x^2\right )}{45 x^2} \, dx=-\frac {8 e^{\frac {1}{2 x}} x}{15}-\frac {352 e^{\frac {1}{2 x}}}{15}+\frac {4 e^{\frac {1}{x}}}{5}+\frac {4 x^{2}}{45}+\frac {352 x}{45} \] Input:
int(1/45*(-36*exp(1/2/x)^2+(-24*x^2+12*x+528)*exp(1/2/x)+8*x^3+352*x^2)/x^ 2,x)
Output:
(4*( - 6*e**(1/(2*x))*x - 264*e**(1/(2*x)) + 9*e**(1/x) + x**2 + 88*x))/45