Integrand size = 56, antiderivative size = 24 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=2+\frac {\left (1+\log \left (2+e^5\right )\right )^2}{\left (-1+\frac {e^3}{x^2}\right )^2} \] Output:
(ln(exp(5)+2)+1)^2/(-1+exp(3-ln(x^2)))^2+2
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=-\frac {e^3 \left (e^3-2 x^2\right ) \left (1+\log \left (2+e^5\right )\right )^2}{\left (e^3-x^2\right )^2} \] Input:
Integrate[(E^3*(4 + 8*Log[2 + E^5] + 4*Log[2 + E^5]^2))/((E^9/x^5 - (3*E^6 )/x^3 + (3*E^3)/x - x)*x^2),x]
Output:
-((E^3*(E^3 - 2*x^2)*(1 + Log[2 + E^5])^2)/(E^3 - x^2)^2)
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12, 27, 2070, 242}
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.
\(\displaystyle \int \frac {e^3 \left (4+4 \log ^2\left (2+e^5\right )+8 \log \left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}-x+\frac {3 e^3}{x}\right ) x^2} \, dx\) |
\(\Big \downarrow \) 12 |
\(\displaystyle \int \frac {4 e^3 x^3 \left (1+\log \left (2+e^5\right )\right )^2}{-x^6+3 e^3 x^4-3 e^6 x^2+e^9}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 e^3 \left (1+\log \left (2+e^5\right )\right )^2 \int \frac {x^3}{-x^6+3 e^3 x^4-3 e^6 x^2+e^9}dx\) |
\(\Big \downarrow \) 2070 |
\(\displaystyle 4 e^3 \left (1+\log \left (2+e^5\right )\right )^2 \int \frac {x^3}{\left (e^3-x^2\right )^3}dx\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {x^4 \left (1+\log \left (2+e^5\right )\right )^2}{\left (e^3-x^2\right )^2}\) |
Input:
Int[(E^3*(4 + 8*Log[2 + E^5] + 4*Log[2 + E^5]^2))/((E^9/x^5 - (3*E^6)/x^3 + (3*E^3)/x - x)*x^2),x]
Output:
(x^4*(1 + Log[2 + E^5])^2)/(E^3 - x^2)^2
Int[(u_.)*((e_.)*(x_))^(m_.)*((d_.)*(x_)^(q_.) + (a_.)*(x_)^(r_.) + (b_.)*( x_)^(s_.) + (c_.)*(x_)^(t_.))^(p_.), x_Symbol] :> Simp[1/e^(p*r) Int[u*(e *x)^(m + p*r)*(a + b*x^(s - r) + c*x^(t - r) + d*x^(q - r))^p, x], x] /; Fr eeQ[{a, b, c, d, e, m, r, s, t, q}, x] && IntegerQ[p] && (IntegerQ[p*r] || GtQ[e, 0]) && PosQ[s - r] && PosQ[t - r] && PosQ[q - r]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x^2, 0], Expon[Px , x^2]], b = Rt[Coeff[Px, x^2, Expon[Px, x^2]], Expon[Px, x^2]]}, Int[u*(a + b*x^2)^(Expon[Px, x^2]*p), x] /; EqQ[Px, (a + b*x^2)^Expon[Px, x^2]]] /; IntegerQ[p] && PolyQ[Px, x^2] && GtQ[Expon[Px, x^2], 1] && NeQ[Coeff[Px, x^ 2, 0], 0]
Time = 0.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96
method | result | size |
risch | \(\frac {\left (4 \ln \left ({\mathrm e}^{5}+2\right )^{2}+8 \ln \left ({\mathrm e}^{5}+2\right )+4\right ) {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{3}}{4}+\frac {x^{2}}{2}\right )}{{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{4}}\) | \(47\) |
parallelrisch | \(\frac {4 \ln \left ({\mathrm e}^{5}+2\right )^{2}+8 \ln \left ({\mathrm e}^{5}+2\right )+4}{\frac {4 \,{\mathrm e}^{6}}{x^{4}}-8 \,{\mathrm e}^{3-\ln \left (x^{2}\right )}+4}\) | \(47\) |
norman | \(\frac {-{\mathrm e}^{6} \left (\ln \left ({\mathrm e}^{5}+2\right )^{2}+2 \ln \left ({\mathrm e}^{5}+2\right )+1\right ) x +\left (2 \,{\mathrm e}^{3} \ln \left ({\mathrm e}^{5}+2\right )^{2}+4 \,{\mathrm e}^{3} \ln \left ({\mathrm e}^{5}+2\right )+2 \,{\mathrm e}^{3}\right ) x^{3}}{x \left (-x^{2}+{\mathrm e}^{3}\right )^{2}}\) | \(68\) |
default | \(\frac {\left (4 \ln \left ({\mathrm e}^{5}+2\right )^{2}+8 \ln \left ({\mathrm e}^{5}+2\right )+4\right ) {\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \left (x \right )} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (-3 \,{\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \left (x \right )} \textit {\_Z}^{2}+\textit {\_Z}^{3}+3 \,{\mathrm e}^{6-2 \ln \left (x^{2}\right )+4 \ln \left (x \right )} \textit {\_Z} -{\mathrm e}^{9-3 \ln \left (x^{2}\right )+6 \ln \left (x \right )}\right )}{\sum }\frac {\textit {\_R} \ln \left (x^{2}-\textit {\_R} \right )}{-\textit {\_R}^{2}+2 \,{\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \left (x \right )} \textit {\_R} -{\mathrm e}^{6-2 \ln \left (x^{2}\right )+4 \ln \left (x \right )}}\right )}{6}\) | \(140\) |
Input:
int((4*ln(exp(5)+2)^2+8*ln(exp(5)+2)+4)*exp(3-ln(x^2))/(x*exp(3-ln(x^2))^3 -3*x*exp(3-ln(x^2))^2+3*x*exp(3-ln(x^2))-x),x,method=_RETURNVERBOSE)
Output:
(4*ln(exp(5)+2)^2+8*ln(exp(5)+2)+4)*exp(3)*(-1/4*exp(3)+1/2*x^2)/(exp(6)-2 *x^2*exp(3)+x^4)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (25) = 50\).
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.79 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {2 \, x^{2} e^{3} + {\left (2 \, x^{2} e^{3} - e^{6}\right )} \log \left (e^{5} + 2\right )^{2} + 2 \, {\left (2 \, x^{2} e^{3} - e^{6}\right )} \log \left (e^{5} + 2\right ) - e^{6}}{x^{4} - 2 \, x^{2} e^{3} + e^{6}} \] Input:
integrate((4*log(exp(5)+2)^2+8*log(exp(5)+2)+4)*exp(3-log(x^2))/(x*exp(3-l og(x^2))^3-3*x*exp(3-log(x^2))^2+3*x*exp(3-log(x^2))-x),x, algorithm="fric as")
Output:
(2*x^2*e^3 + (2*x^2*e^3 - e^6)*log(e^5 + 2)^2 + 2*(2*x^2*e^3 - e^6)*log(e^ 5 + 2) - e^6)/(x^4 - 2*x^2*e^3 + e^6)
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (20) = 40\).
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.42 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {\left (- 2 x^{2} + e^{3}\right ) \left (- 4 e^{3} \log {\left (2 + e^{5} \right )}^{2} - 8 e^{3} \log {\left (2 + e^{5} \right )} - 4 e^{3}\right )}{4 x^{4} - 8 x^{2} e^{3} + 4 e^{6}} \] Input:
integrate((4*ln(exp(5)+2)**2+8*ln(exp(5)+2)+4)*exp(3-ln(x**2))/(x*exp(3-ln (x**2))**3-3*x*exp(3-ln(x**2))**2+3*x*exp(3-ln(x**2))-x),x)
Output:
(-2*x**2 + exp(3))*(-4*exp(3)*log(2 + exp(5))**2 - 8*exp(3)*log(2 + exp(5) ) - 4*exp(3))/(4*x**4 - 8*x**2*exp(3) + 4*exp(6))
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {{\left (2 \, x^{2} - e^{3}\right )} {\left (\log \left (e^{5} + 2\right )^{2} + 2 \, \log \left (e^{5} + 2\right ) + 1\right )} e^{3}}{x^{4} - 2 \, x^{2} e^{3} + e^{6}} \] Input:
integrate((4*log(exp(5)+2)^2+8*log(exp(5)+2)+4)*exp(3-log(x^2))/(x*exp(3-l og(x^2))^3-3*x*exp(3-log(x^2))^2+3*x*exp(3-log(x^2))-x),x, algorithm="maxi ma")
Output:
(2*x^2 - e^3)*(log(e^5 + 2)^2 + 2*log(e^5 + 2) + 1)*e^3/(x^4 - 2*x^2*e^3 + e^6)
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {{\left (2 \, x^{2} - e^{3}\right )} {\left (\log \left (e^{5} + 2\right )^{2} + 2 \, \log \left (e^{5} + 2\right ) + 1\right )} e^{3}}{{\left (x^{2} - e^{3}\right )}^{2}} \] Input:
integrate((4*log(exp(5)+2)^2+8*log(exp(5)+2)+4)*exp(3-log(x^2))/(x*exp(3-l og(x^2))^3-3*x*exp(3-log(x^2))^2+3*x*exp(3-log(x^2))-x),x, algorithm="giac ")
Output:
(2*x^2 - e^3)*(log(e^5 + 2)^2 + 2*log(e^5 + 2) + 1)*e^3/(x^2 - e^3)^2
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {x^4\,\left (2\,\ln \left ({\mathrm {e}}^5+2\right )+{\ln \left ({\mathrm {e}}^5+2\right )}^2+1\right )}{x^4-2\,{\mathrm {e}}^3\,x^2+{\mathrm {e}}^6} \] Input:
int(-(exp(3 - log(x^2))*(8*log(exp(5) + 2) + 4*log(exp(5) + 2)^2 + 4))/(x - 3*x*exp(3 - log(x^2)) + 3*x*exp(6 - 2*log(x^2)) - x*exp(9 - 3*log(x^2))) ,x)
Output:
(x^4*(2*log(exp(5) + 2) + log(exp(5) + 2)^2 + 1))/(exp(6) - 2*x^2*exp(3) + x^4)
Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {x^{4} \left (\mathrm {log}\left (e^{5}+2\right )^{2}+2 \,\mathrm {log}\left (e^{5}+2\right )+1\right )}{e^{6}-2 e^{3} x^{2}+x^{4}} \] Input:
int((4*log(exp(5)+2)^2+8*log(exp(5)+2)+4)*exp(3-log(x^2))/(x*exp(3-log(x^2 ))^3-3*x*exp(3-log(x^2))^2+3*x*exp(3-log(x^2))-x),x)
Output:
(x**4*(log(e**5 + 2)**2 + 2*log(e**5 + 2) + 1))/(e**6 - 2*e**3*x**2 + x**4 )