Integrand size = 81, antiderivative size = 24 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {\log (4)+\log (\log (4 \log (2)))}{x \left (3+\frac {5+x}{e^3}\right )} \] Output:
1/x*(ln(ln(4*ln(2)))+2*ln(2))/((5+x)/exp(3)+3)
Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {e^3 \log (4 \log (\log (16)))}{x \left (5+3 e^3+x\right )} \] Input:
Integrate[((-3*E^6 + E^3*(-5 - 2*x))*Log[4] + (-3*E^6 + E^3*(-5 - 2*x))*Lo g[Log[4*Log[2]]])/(25*x^2 + 9*E^6*x^2 + 10*x^3 + x^4 + E^3*(30*x^2 + 6*x^3 )),x]
Output:
(E^3*Log[4*Log[Log[16]]])/(x*(5 + 3*E^3 + x))
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.099, Rules used = {6, 6, 27, 25, 2026, 2007, 205, 83}
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.
\(\displaystyle \int \frac {\left (e^3 (-2 x-5)-3 e^6\right ) \log (\log (4 \log (2)))+\left (e^3 (-2 x-5)-3 e^6\right ) \log (4)}{x^4+10 x^3+9 e^6 x^2+25 x^2+e^3 \left (6 x^3+30 x^2\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (e^3 (-2 x-5)-3 e^6\right ) \log (\log (4 \log (2)))+\left (e^3 (-2 x-5)-3 e^6\right ) \log (4)}{x^4+10 x^3+\left (25+9 e^6\right ) x^2+e^3 \left (6 x^3+30 x^2\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (e^3 (-2 x-5)-3 e^6\right ) (\log (4)+\log (\log (4 \log (2))))}{x^4+10 x^3+\left (25+9 e^6\right ) x^2+e^3 \left (6 x^3+30 x^2\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (4 \log (\log (16))) \int -\frac {e^3 (2 x+5)+3 e^6}{x^4+10 x^3+\left (25+9 e^6\right ) x^2+6 e^3 \left (x^3+5 x^2\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\log (4 \log (\log (16))) \int \frac {e^3 (2 x+5)+3 e^6}{x^4+10 x^3+\left (25+9 e^6\right ) x^2+6 e^3 \left (x^3+5 x^2\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle -\log (4 \log (\log (16))) \int \frac {e^3 (2 x+5)+3 e^6}{x^2 \left (x^2+2 \left (5+3 e^3\right ) x+\left (5+3 e^3\right )^2\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle -\log (4 \log (\log (16))) \int \frac {e^3 (2 x+5)+3 e^6}{x^2 \left (x+3 e^3+5\right )^2}dx\) |
\(\Big \downarrow \) 205 |
\(\displaystyle -\log (4 \log (\log (16))) \int \frac {2 e^3 x+e^3 \left (5+3 e^3\right )}{x^2 \left (x+3 e^3+5\right )^2}dx\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \frac {e^3 \log (4 \log (\log (16)))}{x \left (x+3 e^3+5\right )}\) |
Input:
Int[((-3*E^6 + E^3*(-5 - 2*x))*Log[4] + (-3*E^6 + E^3*(-5 - 2*x))*Log[Log[ 4*Log[2]]])/(25*x^2 + 9*E^6*x^2 + 10*x^3 + x^4 + E^3*(30*x^2 + 6*x^3)),x]
Output:
(E^3*Log[4*Log[Log[16]]])/(x*(5 + 3*E^3 + x))
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[(u_)^(m_.)*(v_)^(n_.)*(w_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m* ExpandToSum[v, x]^n*ExpandToSum[w, x]^p, x] /; FreeQ[{m, n, p}, x] && Linea rQ[{u, v, w}, x] && !LinearMatchQ[{u, v, w}, x]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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 = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
gosper | \(\frac {{\mathrm e}^{3} \left (\ln \left (\ln \left (4 \ln \left (2\right )\right )\right )+2 \ln \left (2\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{3} \ln \left (\ln \left (4 \ln \left (2\right )\right )\right )+2 \,{\mathrm e}^{3} \ln \left (2\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\) | \(30\) |
norman | \(\frac {2 \,{\mathrm e}^{3} \ln \left (2\right )+{\mathrm e}^{3} \ln \left (2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\) | \(33\) |
risch | \(\frac {2 \,{\mathrm e}^{3} \ln \left (2\right )+{\mathrm e}^{3} \ln \left (2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\) | \(35\) |
Input:
int(((-3*exp(3)^2+(-2*x-5)*exp(3))*ln(ln(4*ln(2)))+2*(-3*exp(3)^2+(-2*x-5) *exp(3))*ln(2))/(9*x^2*exp(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x ,method=_RETURNVERBOSE)
Output:
exp(3)*(ln(ln(4*ln(2)))+2*ln(2))/x/(5+3*exp(3)+x)
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {2 \, e^{3} \log \left (2\right ) + e^{3} \log \left (\log \left (4 \, \log \left (2\right )\right )\right )}{x^{2} + 3 \, x e^{3} + 5 \, x} \] Input:
integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2 +(-2*x-5)*exp(3))*log(2))/(9*x^2*exp(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3 +25*x^2),x, algorithm="fricas")
Output:
(2*e^3*log(2) + e^3*log(log(4*log(2))))/(x^2 + 3*x*e^3 + 5*x)
Time = 0.59 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=- \frac {- 2 e^{3} \log {\left (2 \right )} - e^{3} \log {\left (\log {\left (\log {\left (2 \right )} \right )} + 2 \log {\left (2 \right )} \right )}}{x^{2} + x \left (5 + 3 e^{3}\right )} \] Input:
integrate(((-3*exp(3)**2+(-2*x-5)*exp(3))*ln(ln(4*ln(2)))+2*(-3*exp(3)**2+ (-2*x-5)*exp(3))*ln(2))/(9*x**2*exp(3)**2+(6*x**3+30*x**2)*exp(3)+x**4+10* x**3+25*x**2),x)
Output:
-(-2*exp(3)*log(2) - exp(3)*log(log(log(2)) + 2*log(2)))/(x**2 + x*(5 + 3* exp(3)))
Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {2 \, e^{3} \log \left (2\right ) + e^{3} \log \left (\log \left (4 \, \log \left (2\right )\right )\right )}{x^{2} + x {\left (3 \, e^{3} + 5\right )}} \] Input:
integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2 +(-2*x-5)*exp(3))*log(2))/(9*x^2*exp(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3 +25*x^2),x, algorithm="maxima")
Output:
(2*e^3*log(2) + e^3*log(log(4*log(2))))/(x^2 + x*(3*e^3 + 5))
Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (25) = 50\).
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.29 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {4 \, e^{6} \log \left (2\right )^{2} + 4 \, e^{6} \log \left (2\right ) \log \left (\log \left (4 \, \log \left (2\right )\right )\right ) + e^{6} \log \left (\log \left (4 \, \log \left (2\right )\right )\right )^{2}}{2 \, {\left (3 \, x e^{6} + {\left (x^{2} + 5 \, x\right )} e^{3}\right )} \log \left (2\right ) + {\left (3 \, x e^{6} + {\left (x^{2} + 5 \, x\right )} e^{3}\right )} \log \left (\log \left (4 \, \log \left (2\right )\right )\right )} \] Input:
integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2 +(-2*x-5)*exp(3))*log(2))/(9*x^2*exp(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3 +25*x^2),x, algorithm="giac")
Output:
(4*e^6*log(2)^2 + 4*e^6*log(2)*log(log(4*log(2))) + e^6*log(log(4*log(2))) ^2)/(2*(3*x*e^6 + (x^2 + 5*x)*e^3)*log(2) + (3*x*e^6 + (x^2 + 5*x)*e^3)*lo g(log(4*log(2))))
Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {{\mathrm {e}}^3\,\ln \left (4\,\ln \left (\ln \left (16\right )\right )\right )}{x^2+\left (3\,{\mathrm {e}}^3+5\right )\,x} \] Input:
int(-(log(log(4*log(2)))*(3*exp(6) + exp(3)*(2*x + 5)) + 2*log(2)*(3*exp(6 ) + exp(3)*(2*x + 5)))/(exp(3)*(30*x^2 + 6*x^3) + 9*x^2*exp(6) + 25*x^2 + 10*x^3 + x^4),x)
Output:
(exp(3)*log(4*log(log(16))))/(x^2 + x*(3*exp(3) + 5))
Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx=\frac {e^{3} \left (\mathrm {log}\left (\mathrm {log}\left (4 \,\mathrm {log}\left (2\right )\right )\right )+2 \,\mathrm {log}\left (2\right )\right )}{x \left (3 e^{3}+x +5\right )} \] Input:
int(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2+(-2*x -5)*exp(3))*log(2))/(9*x^2*exp(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^ 2),x)
Output:
(e**3*(log(log(4*log(2))) + 2*log(2)))/(x*(3*e**3 + x + 5))