Integrand size = 70, antiderivative size = 31 \[ \int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )-32 x^2-e^3 x^2-x^3}{32 e^3 x^2+32 x^3} \, dx=\log \left (\frac {2 e^{\frac {1}{32} \left (\left (5+e^{3/x}\right )^2-x\right )}}{e^3+x}\right ) \]
Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )-32 x^2-e^3 x^2-x^3}{32 e^3 x^2+32 x^3} \, dx=\frac {1}{32} \left (10 e^{3/x}+e^{6/x}-x-32 \log \left (e^3+x\right )\right ) \]
Integrate[(E^(3/x)*(-30*E^3 - 30*x) + E^(6/x)*(-6*E^3 - 6*x) - 32*x^2 - E^ 3*x^2 - x^3)/(32*E^3*x^2 + 32*x^3),x]
Time = 0.53 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6, 2026, 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.
\(\displaystyle \int \frac {-x^3-e^3 x^2-32 x^2+e^{3/x} \left (-30 x-30 e^3\right )+e^{6/x} \left (-6 x-6 e^3\right )}{32 x^3+32 e^3 x^2} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x^3+\left (-32-e^3\right ) x^2+e^{3/x} \left (-30 x-30 e^3\right )+e^{6/x} \left (-6 x-6 e^3\right )}{32 x^3+32 e^3 x^2}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^3+\left (-32-e^3\right ) x^2+e^{3/x} \left (-30 x-30 e^3\right )+e^{6/x} \left (-6 x-6 e^3\right )}{x^2 \left (32 x+32 e^3\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {15 e^{3/x}}{16 x^2}-\frac {3 e^{6/x}}{16 x^2}+\frac {-x-e^3-32}{32 \left (x+e^3\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {x}{32}+\frac {5 e^{3/x}}{16}+\frac {e^{6/x}}{32}-\log \left (x+e^3\right )\) |
Int[(E^(3/x)*(-30*E^3 - 30*x) + E^(6/x)*(-6*E^3 - 6*x) - 32*x^2 - E^3*x^2 - x^3)/(32*E^3*x^2 + 32*x^3),x]
3.26.12.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.56 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90
method | result | size |
risch | \(-\frac {x}{32}-\ln \left ({\mathrm e}^{3}+x \right )+\frac {{\mathrm e}^{\frac {6}{x}}}{32}+\frac {5 \,{\mathrm e}^{\frac {3}{x}}}{16}\) | \(28\) |
parallelrisch | \(-\frac {x}{32}-\ln \left ({\mathrm e}^{3}+x \right )+\frac {{\mathrm e}^{\frac {6}{x}}}{32}+\frac {5 \,{\mathrm e}^{\frac {3}{x}}}{16}\) | \(30\) |
parts | \(-\frac {x}{32}-\ln \left ({\mathrm e}^{3}+x \right )+\frac {{\mathrm e}^{\frac {6}{x}}}{32}+\frac {5 \,{\mathrm e}^{\frac {3}{x}}}{16}\) | \(30\) |
norman | \(\frac {-\frac {x^{2}}{32}+\frac {5 x \,{\mathrm e}^{\frac {3}{x}}}{16}+\frac {x \,{\mathrm e}^{\frac {6}{x}}}{32}}{x}-\ln \left ({\mathrm e}^{3}+x \right )\) | \(39\) |
derivativedivides | \(-\frac {x}{32}-\frac {{\mathrm e}^{3} \ln \left (\frac {3}{x}\right )}{32}+\frac {{\mathrm e}^{3} \ln \left (\frac {3 \,{\mathrm e}^{3}}{x}+3\right )}{32}+\ln \left (\frac {3}{x}\right )-\ln \left (\frac {3 \,{\mathrm e}^{3}}{x}+3\right )+\frac {3 \,{\mathrm e}^{3} \left (\frac {\ln \left (\frac {3}{x}\right )}{3}-\frac {\ln \left (\frac {3 \,{\mathrm e}^{3}}{x}+3\right )}{3}\right )}{32}-\frac {15 \,{\mathrm e}^{-3} {\mathrm e}^{-3 \,{\mathrm e}^{-3}} \operatorname {Ei}_{1}\left (-\frac {3}{x}-3 \,{\mathrm e}^{-3}\right )}{16}-\frac {3 \,{\mathrm e}^{-3} {\mathrm e}^{-6 \,{\mathrm e}^{-3}} \operatorname {Ei}_{1}\left (-\frac {6}{x}-6 \,{\mathrm e}^{-3}\right )}{16}+\frac {5 \,{\mathrm e}^{3} \left ({\mathrm e}^{-3} {\mathrm e}^{\frac {3}{x}}+3 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{-3 \,{\mathrm e}^{-3}} \operatorname {Ei}_{1}\left (-\frac {3}{x}-3 \,{\mathrm e}^{-3}\right )\right )}{16}+\frac {{\mathrm e}^{3} \left (\frac {{\mathrm e}^{-3} {\mathrm e}^{\frac {6}{x}}}{2}+3 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{-6 \,{\mathrm e}^{-3}} \operatorname {Ei}_{1}\left (-\frac {6}{x}-6 \,{\mathrm e}^{-3}\right )\right )}{16}\) | \(219\) |
default | \(-\frac {x}{32}-\frac {{\mathrm e}^{3} \ln \left (\frac {3}{x}\right )}{32}+\frac {{\mathrm e}^{3} \ln \left (\frac {3 \,{\mathrm e}^{3}}{x}+3\right )}{32}+\ln \left (\frac {3}{x}\right )-\ln \left (\frac {3 \,{\mathrm e}^{3}}{x}+3\right )+\frac {3 \,{\mathrm e}^{3} \left (\frac {\ln \left (\frac {3}{x}\right )}{3}-\frac {\ln \left (\frac {3 \,{\mathrm e}^{3}}{x}+3\right )}{3}\right )}{32}-\frac {15 \,{\mathrm e}^{-3} {\mathrm e}^{-3 \,{\mathrm e}^{-3}} \operatorname {Ei}_{1}\left (-\frac {3}{x}-3 \,{\mathrm e}^{-3}\right )}{16}-\frac {3 \,{\mathrm e}^{-3} {\mathrm e}^{-6 \,{\mathrm e}^{-3}} \operatorname {Ei}_{1}\left (-\frac {6}{x}-6 \,{\mathrm e}^{-3}\right )}{16}+\frac {5 \,{\mathrm e}^{3} \left ({\mathrm e}^{-3} {\mathrm e}^{\frac {3}{x}}+3 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{-3 \,{\mathrm e}^{-3}} \operatorname {Ei}_{1}\left (-\frac {3}{x}-3 \,{\mathrm e}^{-3}\right )\right )}{16}+\frac {{\mathrm e}^{3} \left (\frac {{\mathrm e}^{-3} {\mathrm e}^{\frac {6}{x}}}{2}+3 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{-6 \,{\mathrm e}^{-3}} \operatorname {Ei}_{1}\left (-\frac {6}{x}-6 \,{\mathrm e}^{-3}\right )\right )}{16}\) | \(219\) |
int(((-6*exp(3)-6*x)*exp(3/x)^2+(-30*exp(3)-30*x)*exp(3/x)-x^2*exp(3)-x^3- 32*x^2)/(32*x^2*exp(3)+32*x^3),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )-32 x^2-e^3 x^2-x^3}{32 e^3 x^2+32 x^3} \, dx=-\frac {1}{32} \, x + \frac {1}{32} \, e^{\frac {6}{x}} + \frac {5}{16} \, e^{\frac {3}{x}} - \log \left (x + e^{3}\right ) \]
integrate(((-6*exp(3)-6*x)*exp(3/x)^2+(-30*exp(3)-30*x)*exp(3/x)-x^2*exp(3 )-x^3-32*x^2)/(32*x^2*exp(3)+32*x^3),x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )-32 x^2-e^3 x^2-x^3}{32 e^3 x^2+32 x^3} \, dx=- \frac {x}{32} + \frac {e^{\frac {6}{x}}}{32} + \frac {5 e^{\frac {3}{x}}}{16} - \log {\left (x + e^{3} \right )} \]
integrate(((-6*exp(3)-6*x)*exp(3/x)**2+(-30*exp(3)-30*x)*exp(3/x)-x**2*exp (3)-x**3-32*x**2)/(32*x**2*exp(3)+32*x**3),x)
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )-32 x^2-e^3 x^2-x^3}{32 e^3 x^2+32 x^3} \, dx=-\frac {1}{32} \, x + \frac {1}{32} \, e^{\frac {6}{x}} + \frac {5}{16} \, e^{\frac {3}{x}} - \log \left (x + e^{3}\right ) \]
integrate(((-6*exp(3)-6*x)*exp(3/x)^2+(-30*exp(3)-30*x)*exp(3/x)-x^2*exp(3 )-x^3-32*x^2)/(32*x^2*exp(3)+32*x^3),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )-32 x^2-e^3 x^2-x^3}{32 e^3 x^2+32 x^3} \, dx=\frac {1}{32} \, x {\left (\frac {e^{\frac {6}{x}}}{x} + \frac {10 \, e^{\frac {3}{x}}}{x} - \frac {32 \, \log \left (x\right )}{x} - \frac {32 \, \log \left (\frac {e^{3}}{x} + 1\right )}{x} - 1\right )} \]
integrate(((-6*exp(3)-6*x)*exp(3/x)^2+(-30*exp(3)-30*x)*exp(3/x)-x^2*exp(3 )-x^3-32*x^2)/(32*x^2*exp(3)+32*x^3),x, algorithm=\
Time = 10.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{3/x} \left (-30 e^3-30 x\right )+e^{6/x} \left (-6 e^3-6 x\right )-32 x^2-e^3 x^2-x^3}{32 e^3 x^2+32 x^3} \, dx=\frac {5\,{\mathrm {e}}^{3/x}}{16}-\frac {x}{32}+\frac {{\mathrm {e}}^{6/x}}{32}-\ln \left (x+{\mathrm {e}}^3\right ) \]