Integrand size = 40, antiderivative size = 29 \[ \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{12 x^4} \, dx=\frac {1}{16} \left (1+e^{\frac {\frac {2}{x^2}+x}{x}}\right )^2+\frac {5}{6 x} \] Output:
4*(1/16*exp((2/x^2+x)/x)+1/16)*(1/4*exp((2/x^2+x)/x)+1/4)+5/6/x
Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{12 x^4} \, dx=\frac {1}{12} \left (\frac {3}{2} e^{1+\frac {2}{x^3}}+\frac {3}{4} e^{2+\frac {4}{x^3}}+\frac {10}{x}\right ) \] Input:
Integrate[(-9*E^((2 + x^3)/x^3) - 9*E^((2*(2 + x^3))/x^3) - 10*x^2)/(12*x^ 4),x]
Output:
((3*E^(1 + 2/x^3))/2 + (3*E^(2 + 4/x^3))/4 + 10/x)/12
Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {-9 e^{\frac {x^3+2}{x^3}}-9 e^{\frac {2 \left (x^3+2\right )}{x^3}}-10 x^2}{12 x^4} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{12} \int -\frac {10 x^2+9 e^{\frac {x^3+2}{x^3}}+9 e^{\frac {2 \left (x^3+2\right )}{x^3}}}{x^4}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{12} \int \frac {10 x^2+9 e^{\frac {x^3+2}{x^3}}+9 e^{\frac {2 \left (x^3+2\right )}{x^3}}}{x^4}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{12} \int \left (\frac {10}{x^2}+\frac {9 e^{1+\frac {2}{x^3}}}{x^4}+\frac {9 e^{2+\frac {4}{x^3}}}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{12} \left (\frac {3}{2} e^{\frac {2}{x^3}+1}+\frac {3}{4} e^{\frac {4}{x^3}+2}+\frac {10}{x}\right )\) |
Input:
Int[(-9*E^((2 + x^3)/x^3) - 9*E^((2*(2 + x^3))/x^3) - 10*x^2)/(12*x^4),x]
Output:
((3*E^(1 + 2/x^3))/2 + (3*E^(2 + 4/x^3))/4 + 10/x)/12
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 = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {5}{6 x}+\frac {{\mathrm e} \,{\mathrm e}^{\frac {2}{x^{3}}}}{8}+\frac {{\mathrm e}^{\frac {4}{x^{3}}} {\mathrm e}^{2}}{16}\) | \(29\) |
default | \(\frac {5}{6 x}+\frac {{\mathrm e} \,{\mathrm e}^{\frac {2}{x^{3}}}}{8}+\frac {{\mathrm e}^{\frac {4}{x^{3}}} {\mathrm e}^{2}}{16}\) | \(29\) |
risch | \(\frac {5}{6 x}+\frac {{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}}}{16}+\frac {{\mathrm e}^{\frac {x^{3}+2}{x^{3}}}}{8}\) | \(32\) |
parts | \(\frac {5}{6 x}+\frac {{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}}}{16}+\frac {{\mathrm e}^{\frac {x^{3}+2}{x^{3}}}}{8}\) | \(33\) |
norman | \(\frac {\frac {5 x^{2}}{6}+\frac {{\mathrm e}^{\frac {x^{3}+2}{x^{3}}} x^{3}}{8}+\frac {{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}} x^{3}}{16}}{x^{3}}\) | \(43\) |
orering | \(\frac {\left (4 x^{12}+81 x^{9}-378 x^{6}-1296 x^{3}-1296\right ) \left (-9 \,{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}}-9 \,{\mathrm e}^{\frac {x^{3}+2}{x^{3}}}-10 x^{2}\right )}{432 x^{3} \left (5 x^{6}+18 x^{3}+36\right )}+\frac {x^{5} \left (7 x^{9}+108 x^{6}-306 x^{3}-648\right ) \left (\frac {-18 \,{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}} \left (\frac {3}{x}-\frac {3 \left (x^{3}+2\right )}{x^{4}}\right )-9 \left (\frac {3}{x}-\frac {3 \left (x^{3}+2\right )}{x^{4}}\right ) {\mathrm e}^{\frac {x^{3}+2}{x^{3}}}-20 x}{12 x^{4}}-\frac {-9 \,{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}}-9 \,{\mathrm e}^{\frac {x^{3}+2}{x^{3}}}-10 x^{2}}{3 x^{5}}\right )}{360 x^{6}+1296 x^{3}+2592}+\frac {\left (x^{6}+9 x^{3}-36\right ) x^{9} \left (\frac {-36 \,{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}} {\left (\frac {3}{x}-\frac {3 \left (x^{3}+2\right )}{x^{4}}\right )}^{2}-18 \,{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}} \left (-\frac {12}{x^{2}}+\frac {12 x^{3}+24}{x^{5}}\right )-9 \left (-\frac {12}{x^{2}}+\frac {12 x^{3}+24}{x^{5}}\right ) {\mathrm e}^{\frac {x^{3}+2}{x^{3}}}-9 {\left (\frac {3}{x}-\frac {3 \left (x^{3}+2\right )}{x^{4}}\right )}^{2} {\mathrm e}^{\frac {x^{3}+2}{x^{3}}}-20}{12 x^{4}}-\frac {2 \left (-18 \,{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}} \left (\frac {3}{x}-\frac {3 \left (x^{3}+2\right )}{x^{4}}\right )-9 \left (\frac {3}{x}-\frac {3 \left (x^{3}+2\right )}{x^{4}}\right ) {\mathrm e}^{\frac {x^{3}+2}{x^{3}}}-20 x \right )}{3 x^{5}}+\frac {-15 \,{\mathrm e}^{\frac {2 x^{3}+4}{x^{3}}}-15 \,{\mathrm e}^{\frac {x^{3}+2}{x^{3}}}-\frac {50 x^{2}}{3}}{x^{6}}\right )}{360 x^{6}+1296 x^{3}+2592}\) | \(477\) |
Input:
int(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x,method=_R ETURNVERBOSE)
Output:
5/6/x+1/8*exp(1/x^3)^2*exp(1)+1/16*exp(1/x^3)^4*exp(1)^2
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{12 x^4} \, dx=\frac {3 \, x e^{\left (\frac {2 \, {\left (x^{3} + 2\right )}}{x^{3}}\right )} + 6 \, x e^{\left (\frac {x^{3} + 2}{x^{3}}\right )} + 40}{48 \, x} \] Input:
integrate(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x, al gorithm="fricas")
Output:
1/48*(3*x*e^(2*(x^3 + 2)/x^3) + 6*x*e^((x^3 + 2)/x^3) + 40)/x
Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{12 x^4} \, dx=\frac {e^{\frac {2 \left (x^{3} + 2\right )}{x^{3}}}}{16} + \frac {e^{\frac {x^{3} + 2}{x^{3}}}}{8} + \frac {5}{6 x} \] Input:
integrate(1/12*(-9*exp((x**3+2)/x**3)**2-9*exp((x**3+2)/x**3)-10*x**2)/x** 4,x)
Output:
exp(2*(x**3 + 2)/x**3)/16 + exp((x**3 + 2)/x**3)/8 + 5/(6*x)
Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{12 x^4} \, dx=\frac {5}{6 \, x} + \frac {1}{16} \, e^{\left (\frac {4}{x^{3}} + 2\right )} + \frac {1}{8} \, e^{\left (\frac {2}{x^{3}} + 1\right )} \] Input:
integrate(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x, al gorithm="maxima")
Output:
5/6/x + 1/16*e^(4/x^3 + 2) + 1/8*e^(2/x^3 + 1)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{12 x^4} \, dx=\frac {5}{6 \, x} + \frac {1}{16} \, e^{\left (\frac {4}{x^{3}} + 2\right )} + \frac {1}{8} \, e^{\left (\frac {2}{x^{3}} + 1\right )} \] Input:
integrate(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x, al gorithm="giac")
Output:
5/6/x + 1/16*e^(4/x^3 + 2) + 1/8*e^(2/x^3 + 1)
Time = 3.67 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{12 x^4} \, dx=\frac {\mathrm {e}\,{\mathrm {e}}^{\frac {2}{x^3}}}{8}+\frac {{\mathrm {e}}^2\,{\mathrm {e}}^{\frac {4}{x^3}}}{16}+\frac {5}{6\,x} \] Input:
int(-((3*exp((x^3 + 2)/x^3))/4 + (3*exp((2*(x^3 + 2))/x^3))/4 + (5*x^2)/6) /x^4,x)
Output:
(exp(1)*exp(2/x^3))/8 + (exp(2)*exp(4/x^3))/16 + 5/(6*x)
Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-9 e^{\frac {2+x^3}{x^3}}-9 e^{\frac {2 \left (2+x^3\right )}{x^3}}-10 x^2}{12 x^4} \, dx=\frac {3 e^{\frac {4}{x^{3}}} e^{2} x +6 e^{\frac {2}{x^{3}}} e x +40}{48 x} \] Input:
int(1/12*(-9*exp((x^3+2)/x^3)^2-9*exp((x^3+2)/x^3)-10*x^2)/x^4,x)
Output:
(3*e**(4/x**3)*e**2*x + 6*e**(2/x**3)*e*x + 40)/(48*x)