Integrand size = 76, antiderivative size = 25 \[ \int \frac {e^2 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx=\frac {3}{16} \left (\frac {1}{4}-x\right ) x \left (e^2 x-\log (\log (x))\right )^2 \]
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^2 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx=-\frac {3}{64} x (-1+4 x) \left (-e^2 x+\log (\log (x))\right )^2 \]
Integrate[(E^2*(-6*x + 24*x^2) + E^4*(9*x^2 - 48*x^3)*Log[x] + (6 - 24*x + E^2*(-12*x + 72*x^2)*Log[x])*Log[Log[x]] + (3 - 24*x)*Log[x]*Log[Log[x]]^ 2)/(64*Log[x]),x]
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.
\(\displaystyle \int \frac {e^2 \left (24 x^2-6 x\right )+\left (e^2 \left (72 x^2-12 x\right ) \log (x)-24 x+6\right ) \log (\log (x))+e^4 \left (9 x^2-48 x^3\right ) \log (x)+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{64} \int -\frac {3 \left (-\left ((1-8 x) \log (x) \log ^2(\log (x))\right )-2 \left (-4 x-2 e^2 \left (x-6 x^2\right ) \log (x)+1\right ) \log (\log (x))+2 e^2 \left (x-4 x^2\right )-e^4 \left (3 x^2-16 x^3\right ) \log (x)\right )}{\log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{64} \int \frac {-\left ((1-8 x) \log (x) \log ^2(\log (x))\right )-2 \left (-4 x-2 e^2 \left (x-6 x^2\right ) \log (x)+1\right ) \log (\log (x))+2 e^2 \left (x-4 x^2\right )-e^4 \left (3 x^2-16 x^3\right ) \log (x)}{\log (x)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle -\frac {3}{64} \int \frac {\left (e^2 x-\log (\log (x))\right ) \left (-8 x+\log (x) \left (e^2 x (16 x-3)+(1-8 x) \log (\log (x))\right )+2\right )}{\log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {3}{64} \int \left ((8 x-1) \log ^2(\log (x))-\frac {2 \left (12 e^2 \log (x) x^2-2 e^2 \log (x) x-4 x+1\right ) \log (\log (x))}{\log (x)}+\frac {e^2 x \left (16 e^2 \log (x) x^2-3 e^2 \log (x) x-8 x+2\right )}{\log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{64} \left (-\int \log ^2(\log (x))dx+8 \int x \log ^2(\log (x))dx-2 \int \frac {\log (\log (x))}{\log (x)}dx+8 \int \frac {x \log (\log (x))}{\log (x)}dx+4 e^4 x^4-e^4 x^3-8 e^2 x^3 \log (\log (x))+2 e^2 x^2 \log (\log (x))\right )\) |
Int[(E^2*(-6*x + 24*x^2) + E^4*(9*x^2 - 48*x^3)*Log[x] + (6 - 24*x + E^2*( -12*x + 72*x^2)*Log[x])*Log[Log[x]] + (3 - 24*x)*Log[x]*Log[Log[x]]^2)/(64 *Log[x]),x]
3.22.6.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.47 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80
method | result | size |
risch | \(\frac {\left (-12 x^{2}+3 x \right ) \ln \left (\ln \left (x \right )\right )^{2}}{64}+\frac {3 \,{\mathrm e}^{2} x^{2} \left (-1+4 x \right ) \ln \left (\ln \left (x \right )\right )}{32}-\frac {3 \,{\mathrm e}^{4} x^{3} \left (-1+4 x \right )}{64}\) | \(45\) |
parallelrisch | \(-\frac {3 x^{2} \ln \left (\ln \left (x \right )\right )^{2}}{16}+\frac {3 x \ln \left (\ln \left (x \right )\right )^{2}}{64}+\frac {3 \,{\mathrm e}^{2} \ln \left (\ln \left (x \right )\right ) x^{3}}{8}-\frac {3 \,{\mathrm e}^{2} \ln \left (\ln \left (x \right )\right ) x^{2}}{32}-\frac {3 x^{4} {\mathrm e}^{4}}{16}+\frac {3 x^{3} {\mathrm e}^{4}}{64}\) | \(58\) |
int(1/64*((-24*x+3)*ln(x)*ln(ln(x))^2+((72*x^2-12*x)*exp(2)*ln(x)-24*x+6)* ln(ln(x))+(-48*x^3+9*x^2)*exp(2)^2*ln(x)+(24*x^2-6*x)*exp(2))/ln(x),x,meth od=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (20) = 40\).
Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.00 \[ \int \frac {e^2 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx=\frac {3}{32} \, {\left (4 \, x^{3} - x^{2}\right )} e^{2} \log \left (\log \left (x\right )\right ) - \frac {3}{64} \, {\left (4 \, x^{2} - x\right )} \log \left (\log \left (x\right )\right )^{2} - \frac {3}{64} \, {\left (4 \, x^{4} - x^{3}\right )} e^{4} \]
integrate(1/64*((-24*x+3)*log(x)*log(log(x))^2+((72*x^2-12*x)*exp(2)*log(x )-24*x+6)*log(log(x))+(-48*x^3+9*x^2)*exp(2)^2*log(x)+(24*x^2-6*x)*exp(2)) /log(x),x, algorithm=\
3/32*(4*x^3 - x^2)*e^2*log(log(x)) - 3/64*(4*x^2 - x)*log(log(x))^2 - 3/64 *(4*x^4 - x^3)*e^4
Time = 0.23 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.52 \[ \int \frac {e^2 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx=- \frac {3 x^{4} e^{4}}{16} + \frac {3 x^{3} e^{4}}{64} + \left (- \frac {3 x^{2}}{16} + \frac {3 x}{64}\right ) \log {\left (\log {\left (x \right )} \right )}^{2} + \left (\frac {3 x^{3} e^{2}}{8} - \frac {3 x^{2} e^{2}}{32}\right ) \log {\left (\log {\left (x \right )} \right )} \]
integrate(1/64*((-24*x+3)*ln(x)*ln(ln(x))**2+((72*x**2-12*x)*exp(2)*ln(x)- 24*x+6)*ln(ln(x))+(-48*x**3+9*x**2)*exp(2)**2*ln(x)+(24*x**2-6*x)*exp(2))/ ln(x),x)
-3*x**4*exp(4)/16 + 3*x**3*exp(4)/64 + (-3*x**2/16 + 3*x/64)*log(log(x))** 2 + (3*x**3*exp(2)/8 - 3*x**2*exp(2)/32)*log(log(x))
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.48 \[ \int \frac {e^2 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx=-\frac {3}{16} \, x^{4} e^{4} + \frac {3}{64} \, x^{3} e^{4} - \frac {3}{64} \, {\left (4 \, x^{2} - x\right )} \log \left (\log \left (x\right )\right )^{2} + \frac {3}{8} \, {\left (x^{3} \log \left (\log \left (x\right )\right ) - {\rm Ei}\left (3 \, \log \left (x\right )\right )\right )} e^{2} - \frac {3}{32} \, {\left (x^{2} \log \left (\log \left (x\right )\right ) - {\rm Ei}\left (2 \, \log \left (x\right )\right )\right )} e^{2} + \frac {3}{8} \, {\rm Ei}\left (3 \, \log \left (x\right )\right ) e^{2} - \frac {3}{32} \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) e^{2} \]
integrate(1/64*((-24*x+3)*log(x)*log(log(x))^2+((72*x^2-12*x)*exp(2)*log(x )-24*x+6)*log(log(x))+(-48*x^3+9*x^2)*exp(2)^2*log(x)+(24*x^2-6*x)*exp(2)) /log(x),x, algorithm=\
-3/16*x^4*e^4 + 3/64*x^3*e^4 - 3/64*(4*x^2 - x)*log(log(x))^2 + 3/8*(x^3*l og(log(x)) - Ei(3*log(x)))*e^2 - 3/32*(x^2*log(log(x)) - Ei(2*log(x)))*e^2 + 3/8*Ei(3*log(x))*e^2 - 3/32*Ei(2*log(x))*e^2
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {e^2 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx=-\frac {3}{16} \, x^{4} e^{4} + \frac {3}{8} \, x^{3} e^{2} \log \left (\log \left (x\right )\right ) + \frac {3}{64} \, x^{3} e^{4} - \frac {3}{32} \, x^{2} e^{2} \log \left (\log \left (x\right )\right ) - \frac {3}{16} \, x^{2} \log \left (\log \left (x\right )\right )^{2} + \frac {3}{64} \, x \log \left (\log \left (x\right )\right )^{2} \]
integrate(1/64*((-24*x+3)*log(x)*log(log(x))^2+((72*x^2-12*x)*exp(2)*log(x )-24*x+6)*log(log(x))+(-48*x^3+9*x^2)*exp(2)^2*log(x)+(24*x^2-6*x)*exp(2)) /log(x),x, algorithm=\
-3/16*x^4*e^4 + 3/8*x^3*e^2*log(log(x)) + 3/64*x^3*e^4 - 3/32*x^2*e^2*log( log(x)) - 3/16*x^2*log(log(x))^2 + 3/64*x*log(log(x))^2
Time = 13.62 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^2 \left (-6 x+24 x^2\right )+e^4 \left (9 x^2-48 x^3\right ) \log (x)+\left (6-24 x+e^2 \left (-12 x+72 x^2\right ) \log (x)\right ) \log (\log (x))+(3-24 x) \log (x) \log ^2(\log (x))}{64 \log (x)} \, dx=-\frac {3\,x\,\left (4\,x-1\right )\,{\left (\ln \left (\ln \left (x\right )\right )-x\,{\mathrm {e}}^2\right )}^2}{64} \]