Integrand size = 51, antiderivative size = 31 \[ \int \frac {e^{-2 x} \left (-7 e^{5+2 x}+e^5 \left (2 x^2-6 x^3+2 x^4\right )+e^{5+2 x} \log \left (x^2\right )\right )}{x^2} \, dx=\frac {e^5 \left (5+2 x-e^{-2 x} (-2+x) x^2-\log \left (x^2\right )\right )}{x} \] Output:
(2*x-ln(x^2)-(-2+x)*x^2/exp(2*x)+5)/x*exp(5)
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-2 x} \left (-7 e^{5+2 x}+e^5 \left (2 x^2-6 x^3+2 x^4\right )+e^{5+2 x} \log \left (x^2\right )\right )}{x^2} \, dx=-\frac {e^5 \left (-5+e^{-2 x} (-2+x) x^2+\log \left (x^2\right )\right )}{x} \] Input:
Integrate[(-7*E^(5 + 2*x) + E^5*(2*x^2 - 6*x^3 + 2*x^4) + E^(5 + 2*x)*Log[ x^2])/(E^(2*x)*x^2),x]
Output:
-((E^5*(-5 + ((-2 + x)*x^2)/E^(2*x) + Log[x^2]))/x)
Time = 0.46 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {7239, 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.
\(\displaystyle \int \frac {e^{-2 x} \left (e^{2 x+5} \log \left (x^2\right )+e^5 \left (2 x^4-6 x^3+2 x^2\right )-7 e^{2 x+5}\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^5 \left (2 e^{-2 x} \left (x^2-3 x+1\right ) x^2+\log \left (x^2\right )-7\right )}{x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^5 \int -\frac {-2 e^{-2 x} \left (x^2-3 x+1\right ) x^2-\log \left (x^2\right )+7}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -e^5 \int \frac {-2 e^{-2 x} \left (x^2-3 x+1\right ) x^2-\log \left (x^2\right )+7}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -e^5 \int \left (\frac {7-\log \left (x^2\right )}{x^2}-2 e^{-2 x} \left (x^2-3 x+1\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -e^5 \left (e^{-2 x} x^2-\frac {7-\log \left (x^2\right )}{x}-2 e^{-2 x} x+\frac {2}{x}\right )\) |
Input:
Int[(-7*E^(5 + 2*x) + E^5*(2*x^2 - 6*x^3 + 2*x^4) + E^(5 + 2*x)*Log[x^2])/ (E^(2*x)*x^2),x]
Output:
-(E^5*(2/x - (2*x)/E^(2*x) + x^2/E^(2*x) - (7 - Log[x^2])/x))
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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48
method | result | size |
norman | \(\frac {\left (2 x^{2} {\mathrm e}^{5}-x^{3} {\mathrm e}^{5}+5 \,{\mathrm e}^{5} {\mathrm e}^{2 x}-{\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )\right ) {\mathrm e}^{-2 x}}{x}\) | \(46\) |
parallelrisch | \(-\frac {\left (2 x^{3} {\mathrm e}^{5}-4 x^{2} {\mathrm e}^{5}+2 \,{\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )-10 \,{\mathrm e}^{5} {\mathrm e}^{2 x}\right ) {\mathrm e}^{-2 x}}{2 x}\) | \(47\) |
default | \({\mathrm e}^{5} \left (-\frac {\ln \left (x^{2}\right )}{x}-\frac {2}{x}\right )+\frac {7 \,{\mathrm e}^{5}}{x}+2 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{-2 x} x^{2}}{2}+x \,{\mathrm e}^{-2 x}\right )\) | \(51\) |
parts | \({\mathrm e}^{5} \left (-\frac {\ln \left (x^{2}\right )}{x}-\frac {2}{x}\right )+\frac {7 \,{\mathrm e}^{5}}{x}+2 \,{\mathrm e}^{5} \left (-\frac {{\mathrm e}^{-2 x} x^{2}}{2}+x \,{\mathrm e}^{-2 x}\right )\) | \(51\) |
risch | \(\frac {i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{5}-2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{5}+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{5}-2 \,{\mathrm e}^{5-2 x} x^{3}+4 \,{\mathrm e}^{5-2 x} x^{2}+10 \,{\mathrm e}^{5}-4 \,{\mathrm e}^{5} \ln \left (x \right )}{2 x}\) | \(94\) |
orering | \(-\frac {\left (12 x^{4}-82 x^{3}+142 x^{2}-53 x -4\right ) \left ({\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )-7 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (2 x^{4}-6 x^{3}+2 x^{2}\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x \left (4 x^{4}-30 x^{3}+62 x^{2}-37 x +4\right )}-\frac {x^{2} \left (8 x^{4}-40 x^{3}+14 x^{2}+71 x -32\right ) \left (\frac {\left (2 \,{\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )+\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{x}-14 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (8 x^{3}-18 x^{2}+4 x \right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{2}}-\frac {2 \left ({\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )-7 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (2 x^{4}-6 x^{3}+2 x^{2}\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{2}}-\frac {2 \left ({\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )-7 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (2 x^{4}-6 x^{3}+2 x^{2}\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{3}}\right )}{2 \left (4 x^{4}-30 x^{3}+62 x^{2}-37 x +4\right )}-\frac {x^{3} \left (4 x^{3}-22 x^{2}+29 x -8\right ) \left (\frac {\left (4 \,{\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )+\frac {8 \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{x}-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{x^{2}}-28 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (24 x^{2}-36 x +4\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{2}}-\frac {4 \left (2 \,{\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )+\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{x}-14 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (8 x^{3}-18 x^{2}+4 x \right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{2}}-\frac {4 \left (2 \,{\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )+\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{2 x}}{x}-14 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (8 x^{3}-18 x^{2}+4 x \right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{3}}+\frac {4 \left ({\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )-7 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (2 x^{4}-6 x^{3}+2 x^{2}\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{2}}+\frac {8 \left ({\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )-7 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (2 x^{4}-6 x^{3}+2 x^{2}\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{3}}+\frac {6 \left ({\mathrm e}^{5} {\mathrm e}^{2 x} \ln \left (x^{2}\right )-7 \,{\mathrm e}^{5} {\mathrm e}^{2 x}+\left (2 x^{4}-6 x^{3}+2 x^{2}\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-2 x}}{x^{4}}\right )}{2 \left (4 x^{4}-30 x^{3}+62 x^{2}-37 x +4\right )}\) | \(680\) |
Input:
int((exp(5)*exp(2*x)*ln(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)*exp(5)) /exp(2*x)/x^2,x,method=_RETURNVERBOSE)
Output:
(2*x^2*exp(5)-x^3*exp(5)+5*exp(5)*exp(2*x)-exp(5)*exp(2*x)*ln(x^2))/x/exp( 2*x)
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-2 x} \left (-7 e^{5+2 x}+e^5 \left (2 x^2-6 x^3+2 x^4\right )+e^{5+2 x} \log \left (x^2\right )\right )}{x^2} \, dx=-\frac {{\left ({\left (x^{3} - 2 \, x^{2}\right )} e^{10} + e^{\left (2 \, x + 10\right )} \log \left (x^{2}\right ) - 5 \, e^{\left (2 \, x + 10\right )}\right )} e^{\left (-2 \, x - 5\right )}}{x} \] Input:
integrate((exp(5)*exp(2*x)*log(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)* exp(5))/exp(2*x)/x^2,x, algorithm="fricas")
Output:
-((x^3 - 2*x^2)*e^10 + e^(2*x + 10)*log(x^2) - 5*e^(2*x + 10))*e^(-2*x - 5 )/x
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-2 x} \left (-7 e^{5+2 x}+e^5 \left (2 x^2-6 x^3+2 x^4\right )+e^{5+2 x} \log \left (x^2\right )\right )}{x^2} \, dx=\left (- x^{2} e^{5} + 2 x e^{5}\right ) e^{- 2 x} - \frac {e^{5} \log {\left (x^{2} \right )}}{x} + \frac {5 e^{5}}{x} \] Input:
integrate((exp(5)*exp(2*x)*ln(x**2)-7*exp(5)*exp(2*x)+(2*x**4-6*x**3+2*x** 2)*exp(5))/exp(2*x)/x**2,x)
Output:
(-x**2*exp(5) + 2*x*exp(5))*exp(-2*x) - exp(5)*log(x**2)/x + 5*exp(5)/x
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (27) = 54\).
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.23 \[ \int \frac {e^{-2 x} \left (-7 e^{5+2 x}+e^5 \left (2 x^2-6 x^3+2 x^4\right )+e^{5+2 x} \log \left (x^2\right )\right )}{x^2} \, dx=-{\left (\frac {\log \left (x^{2}\right )}{x} + \frac {2}{x}\right )} e^{5} - \frac {1}{2} \, {\left (2 \, x^{2} e^{5} + 2 \, x e^{5} + e^{5}\right )} e^{\left (-2 \, x\right )} + \frac {3}{2} \, {\left (2 \, x e^{5} + e^{5}\right )} e^{\left (-2 \, x\right )} + \frac {7 \, e^{5}}{x} - e^{\left (-2 \, x + 5\right )} \] Input:
integrate((exp(5)*exp(2*x)*log(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)* exp(5))/exp(2*x)/x^2,x, algorithm="maxima")
Output:
-(log(x^2)/x + 2/x)*e^5 - 1/2*(2*x^2*e^5 + 2*x*e^5 + e^5)*e^(-2*x) + 3/2*( 2*x*e^5 + e^5)*e^(-2*x) + 7*e^5/x - e^(-2*x + 5)
Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-2 x} \left (-7 e^{5+2 x}+e^5 \left (2 x^2-6 x^3+2 x^4\right )+e^{5+2 x} \log \left (x^2\right )\right )}{x^2} \, dx=-\frac {x^{3} e^{\left (-2 \, x + 5\right )} - 2 \, x^{2} e^{\left (-2 \, x + 5\right )} + e^{5} \log \left (x^{2}\right ) - 5 \, e^{5}}{x} \] Input:
integrate((exp(5)*exp(2*x)*log(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)* exp(5))/exp(2*x)/x^2,x, algorithm="giac")
Output:
-(x^3*e^(-2*x + 5) - 2*x^2*e^(-2*x + 5) + e^5*log(x^2) - 5*e^5)/x
Time = 3.97 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-2 x} \left (-7 e^{5+2 x}+e^5 \left (2 x^2-6 x^3+2 x^4\right )+e^{5+2 x} \log \left (x^2\right )\right )}{x^2} \, dx=2\,x\,{\mathrm {e}}^{5-2\,x}+\frac {5\,{\mathrm {e}}^5}{x}-x^2\,{\mathrm {e}}^{5-2\,x}-\frac {\ln \left (x^2\right )\,{\mathrm {e}}^5}{x} \] Input:
int((exp(-2*x)*(exp(5)*(2*x^2 - 6*x^3 + 2*x^4) - 7*exp(2*x)*exp(5) + log(x ^2)*exp(2*x)*exp(5)))/x^2,x)
Output:
2*x*exp(5 - 2*x) + (5*exp(5))/x - x^2*exp(5 - 2*x) - (log(x^2)*exp(5))/x
Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-2 x} \left (-7 e^{5+2 x}+e^5 \left (2 x^2-6 x^3+2 x^4\right )+e^{5+2 x} \log \left (x^2\right )\right )}{x^2} \, dx=\frac {e^{5} \left (-e^{2 x} \mathrm {log}\left (x^{2}\right )+5 e^{2 x}-x^{3}+2 x^{2}\right )}{e^{2 x} x} \] Input:
int((exp(5)*exp(2*x)*log(x^2)-7*exp(5)*exp(2*x)+(2*x^4-6*x^3+2*x^2)*exp(5) )/exp(2*x)/x^2,x)
Output:
(e**5*( - e**(2*x)*log(x**2) + 5*e**(2*x) - x**3 + 2*x**2))/(e**(2*x)*x)