Integrand size = 39, antiderivative size = 31 \[ \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (165+33 e^{e^5}+33 x\right )}{11 x^3} \, dx=\frac {x^2}{11}+\frac {x-3 \left (e^{\frac {5+e^{e^5}}{x}}+x\right )}{x} \] Output:
1/11*x^2+(-2*x-3*exp((exp(exp(5))+5)/x))/x
Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (165+33 e^{e^5}+33 x\right )}{11 x^3} \, dx=-\frac {3 e^{\frac {5+e^{e^5}}{x}}}{x}+\frac {x^2}{11} \] Input:
Integrate[(2*x^4 + E^((5 + E^E^5)/x)*(165 + 33*E^E^5 + 33*x))/(11*x^3),x]
Output:
(-3*E^((5 + E^E^5)/x))/x + x^2/11
Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {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.
\(\displaystyle \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (33 x+33 e^{e^5}+165\right )}{11 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{11} \int \frac {2 x^4+33 e^{\frac {5+e^{e^5}}{x}} \left (x+e^{e^5}+5\right )}{x^3}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{11} \int \left (2 x+\frac {33 e^{\frac {5+e^{e^5}}{x}} \left (x+e^{e^5}+5\right )}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{11} \left (x^2-\frac {33 e^{\frac {5+e^{e^5}}{x}}}{x}\right )\) |
Input:
Int[(2*x^4 + E^((5 + E^E^5)/x)*(165 + 33*E^E^5 + 33*x))/(11*x^3),x]
Output:
((-33*E^((5 + E^E^5)/x))/x + x^2)/11
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 = 0.55 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71
method | result | size |
risch | \(\frac {x^{2}}{11}-\frac {3 \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}}{x}\) | \(22\) |
parallelrisch | \(\frac {x^{3}-33 \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}}{11 x}\) | \(22\) |
norman | \(\frac {\frac {x^{4}}{11}-3 \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}} x}{x^{2}}\) | \(24\) |
parts | \(\frac {x^{2}}{11}-\frac {3 \left ({\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}} {\mathrm e}^{{\mathrm e}^{5}}+{\mathrm e}^{{\mathrm e}^{5}} \left (\frac {\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right ) {\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}}{x}-{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}\right )+\frac {5 \left ({\mathrm e}^{{\mathrm e}^{5}}+5\right ) {\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}}{x}\right )}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}\) | \(87\) |
derivativedivides | \(-\frac {-\frac {625 x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}-\frac {500 \,{\mathrm e}^{{\mathrm e}^{5}} x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}-\frac {150 \,{\mathrm e}^{2 \,{\mathrm e}^{5}} x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}-\frac {20 \,{\mathrm e}^{3 \,{\mathrm e}^{5}} x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}-\frac {{\mathrm e}^{4 \,{\mathrm e}^{5}} x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}+\frac {165 \left ({\mathrm e}^{{\mathrm e}^{5}}+5\right ) {\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}}{x}+33 \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}} {\mathrm e}^{{\mathrm e}^{5}}+33 \,{\mathrm e}^{{\mathrm e}^{5}} \left (\frac {\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right ) {\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}}{x}-{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}\right )}{11 \left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}\) | \(161\) |
default | \(-\frac {-\frac {625 x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}-\frac {500 \,{\mathrm e}^{{\mathrm e}^{5}} x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}-\frac {150 \,{\mathrm e}^{2 \,{\mathrm e}^{5}} x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}-\frac {20 \,{\mathrm e}^{3 \,{\mathrm e}^{5}} x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}-\frac {{\mathrm e}^{4 \,{\mathrm e}^{5}} x^{2}}{\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}+\frac {165 \left ({\mathrm e}^{{\mathrm e}^{5}}+5\right ) {\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}}{x}+33 \,{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}} {\mathrm e}^{{\mathrm e}^{5}}+33 \,{\mathrm e}^{{\mathrm e}^{5}} \left (\frac {\left ({\mathrm e}^{{\mathrm e}^{5}}+5\right ) {\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}}{x}-{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{5}}+5}{x}}\right )}{11 \left ({\mathrm e}^{{\mathrm e}^{5}}+5\right )^{2}}\) | \(161\) |
Input:
int(1/11*((33*exp(exp(5))+33*x+165)*exp((exp(exp(5))+5)/x)+2*x^4)/x^3,x,me thod=_RETURNVERBOSE)
Output:
1/11*x^2-3/x*exp((exp(exp(5))+5)/x)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (165+33 e^{e^5}+33 x\right )}{11 x^3} \, dx=\frac {x^{3} - 33 \, e^{\left (\frac {e^{\left (e^{5}\right )} + 5}{x}\right )}}{11 \, x} \] Input:
integrate(1/11*((33*exp(exp(5))+33*x+165)*exp((exp(exp(5))+5)/x)+2*x^4)/x^ 3,x, algorithm="fricas")
Output:
1/11*(x^3 - 33*e^((e^(e^5) + 5)/x))/x
Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (165+33 e^{e^5}+33 x\right )}{11 x^3} \, dx=\frac {x^{2}}{11} - \frac {3 e^{\frac {5 + e^{e^{5}}}{x}}}{x} \] Input:
integrate(1/11*((33*exp(exp(5))+33*x+165)*exp((exp(exp(5))+5)/x)+2*x**4)/x **3,x)
Output:
x**2/11 - 3*exp((5 + exp(exp(5)))/x)/x
\[ \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (165+33 e^{e^5}+33 x\right )}{11 x^3} \, dx=\int { \frac {2 \, x^{4} + 33 \, {\left (x + e^{\left (e^{5}\right )} + 5\right )} e^{\left (\frac {e^{\left (e^{5}\right )} + 5}{x}\right )}}{11 \, x^{3}} \,d x } \] Input:
integrate(1/11*((33*exp(exp(5))+33*x+165)*exp((exp(exp(5))+5)/x)+2*x^4)/x^ 3,x, algorithm="maxima")
Output:
1/11*x^2 + 1/11*integrate(33*(x + e^(e^5) + 5)*e^(e^(e^5)/x + 5/x)/x^3, x)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (27) = 54\).
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77 \[ \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (165+33 e^{e^5}+33 x\right )}{11 x^3} \, dx=-\frac {x^{2} {\left (\frac {33 \, {\left (e^{\left (e^{5}\right )} + 5\right )}^{3} e^{\left (\frac {e^{\left (e^{5}\right )} + 5}{x}\right )}}{x^{3}} - e^{\left (3 \, e^{5}\right )} - 15 \, e^{\left (2 \, e^{5}\right )} - 75 \, e^{\left (e^{5}\right )} - 125\right )}}{11 \, {\left (e^{\left (e^{5}\right )} + 5\right )}^{3}} \] Input:
integrate(1/11*((33*exp(exp(5))+33*x+165)*exp((exp(exp(5))+5)/x)+2*x^4)/x^ 3,x, algorithm="giac")
Output:
-1/11*x^2*(33*(e^(e^5) + 5)^3*e^((e^(e^5) + 5)/x)/x^3 - e^(3*e^5) - 15*e^( 2*e^5) - 75*e^(e^5) - 125)/(e^(e^5) + 5)^3
Time = 1.66 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (165+33 e^{e^5}+33 x\right )}{11 x^3} \, dx=\frac {x^2}{11}-\frac {3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^5}}{x}}\,{\mathrm {e}}^{5/x}}{x} \] Input:
int(((exp((exp(exp(5)) + 5)/x)*(33*x + 33*exp(exp(5)) + 165))/11 + (2*x^4) /11)/x^3,x)
Output:
x^2/11 - (3*exp(exp(exp(5))/x)*exp(5/x))/x
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {2 x^4+e^{\frac {5+e^{e^5}}{x}} \left (165+33 e^{e^5}+33 x\right )}{11 x^3} \, dx=\frac {-33 e^{\frac {e^{e^{5}}+5}{x}}+x^{3}}{11 x} \] Input:
int(1/11*((33*exp(exp(5))+33*x+165)*exp((exp(exp(5))+5)/x)+2*x^4)/x^3,x)
Output:
( - 33*e**((e**(e**5) + 5)/x) + x**3)/(11*x)