Integrand size = 74, antiderivative size = 29 \[ \int \frac {1}{8} \left (-16 e^4+\left (16 e^4-e^8 x\right ) \log (3)+e^8 x \log ^2(3)+\left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2\left (\frac {3}{x}\right )\right ) \, dx=\frac {\left (x+x \left (3+\frac {1}{4} e^4 x \left (\log (3)+\log \left (\frac {3}{x}\right )\right )\right )\right )^2}{x^2} \]
Leaf count is larger than twice the leaf count of optimal. \(76\) vs. \(2(29)=58\).
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.62 \[ \int \frac {1}{8} \left (-16 e^4+\left (16 e^4-e^8 x\right ) \log (3)+e^8 x \log ^2(3)+\left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2\left (\frac {3}{x}\right )\right ) \, dx=\frac {1}{32} e^4 x \left (64 \log (3)+e^4 x \log (3)-e^4 x \log (9)+e^4 x \log ^2(9)+\left (64+e^4 x (1+\log (81))\right ) \log \left (\frac {3}{x}\right )+e^4 x \log \left (\frac {1}{x}\right ) \left (-1+\log (9)+2 \log \left (\frac {3}{x}\right )\right )\right ) \]
Integrate[(-16*E^4 + (16*E^4 - E^8*x)*Log[3] + E^8*x*Log[3]^2 + (16*E^4 - E^8*x + 2*E^8*x*Log[3])*Log[3/x] + E^8*x*Log[3/x]^2)/8,x]
(E^4*x*(64*Log[3] + E^4*x*Log[3] - E^4*x*Log[9] + E^4*x*Log[9]^2 + (64 + E ^4*x*(1 + Log[81]))*Log[3/x] + E^4*x*Log[x^(-1)]*(-1 + Log[9] + 2*Log[3/x] )))/32
Leaf count is larger than twice the leaf count of optimal. \(141\) vs. \(2(29)=58\).
Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 4.86, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {27, 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.
\(\displaystyle \int \frac {1}{8} \left (e^8 x \log ^2\left (\frac {3}{x}\right )+e^8 x \log ^2(3)+\left (-e^8 x+2 e^8 x \log (3)+16 e^4\right ) \log \left (\frac {3}{x}\right )+\left (16 e^4-e^8 x\right ) \log (3)-16 e^4\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int \left (e^8 x \log ^2\left (\frac {3}{x}\right )+\left (2 e^8 \log (3) x-e^8 x+16 e^4\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2(3)+e^4 \left (16-e^4 x\right ) \log (3)-16 e^4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{8} \left (\frac {e^8 x^2}{4}+\frac {1}{2} e^8 x^2 \log ^2\left (\frac {3}{x}\right )+\frac {1}{2} e^8 x^2 \log ^2(3)+\frac {1}{2} e^8 x^2 \log \left (\frac {3}{x}\right )-\frac {1}{4} e^8 x^2 (1-\log (9))-\frac {\left (16-e^4 x (1-\log (9))\right )^2 \log \left (\frac {3}{x}\right )}{2 (1-\log (9))}-\frac {128 \log (x)}{1-\log (9)}-\frac {1}{2} \left (16-e^4 x\right )^2 \log (3)\right )\) |
Int[(-16*E^4 + (16*E^4 - E^8*x)*Log[3] + E^8*x*Log[3]^2 + (16*E^4 - E^8*x + 2*E^8*x*Log[3])*Log[3/x] + E^8*x*Log[3/x]^2)/8,x]
((E^8*x^2)/4 - ((16 - E^4*x)^2*Log[3])/2 + (E^8*x^2*Log[3]^2)/2 - (E^8*x^2 *(1 - Log[9]))/4 + (E^8*x^2*Log[3/x])/2 - ((16 - E^4*x*(1 - Log[9]))^2*Log [3/x])/(2*(1 - Log[9])) + (E^8*x^2*Log[3/x]^2)/2 - (128*Log[x])/(1 - Log[9 ]))/8
3.14.17.3.1 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]]
Leaf count of result is larger than twice the leaf count of optimal. \(60\) vs. \(2(29)=58\).
Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.10
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{8} \ln \left (\frac {3}{x}\right )^{2}}{16}+2 \,{\mathrm e}^{4} x \ln \left (3\right )+2 \,{\mathrm e}^{4} \ln \left (\frac {3}{x}\right ) x +\frac {{\mathrm e}^{8} \ln \left (3\right )^{2} x^{2}}{16}+\frac {{\mathrm e}^{8} \ln \left (3\right ) x^{2} \ln \left (\frac {3}{x}\right )}{8}\) | \(61\) |
norman | \(\frac {x^{2} {\mathrm e}^{8} \ln \left (\frac {3}{x}\right )^{2}}{16}+2 \,{\mathrm e}^{4} x \ln \left (3\right )+2 \,{\mathrm e}^{4} \ln \left (\frac {3}{x}\right ) x +\frac {{\mathrm e}^{8} \ln \left (3\right )^{2} x^{2}}{16}+\frac {{\mathrm e}^{8} \ln \left (3\right ) x^{2} \ln \left (\frac {3}{x}\right )}{8}\) | \(67\) |
parallelrisch | \(\frac {x^{2} {\mathrm e}^{8} \ln \left (\frac {3}{x}\right )^{2}}{16}+2 \,{\mathrm e}^{4} x \ln \left (3\right )+2 \,{\mathrm e}^{4} \ln \left (\frac {3}{x}\right ) x +\frac {{\mathrm e}^{8} \ln \left (3\right )^{2} x^{2}}{16}+\frac {{\mathrm e}^{8} \ln \left (3\right ) x^{2} \ln \left (\frac {3}{x}\right )}{8}\) | \(67\) |
default | \(-\frac {\ln \left (3\right ) {\mathrm e}^{4} \left (\frac {x^{2} {\mathrm e}^{4}}{2}-16 x \right )}{8}-\frac {9 \,{\mathrm e}^{8} \ln \left (3\right ) \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{4}+\frac {9 \,{\mathrm e}^{8} \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{8}-6 \,{\mathrm e}^{4} \left (-\frac {x \ln \left (\frac {3}{x}\right )}{3}-\frac {x}{3}\right )+\frac {{\mathrm e}^{8} \ln \left (3\right )^{2} x^{2}}{16}+\frac {{\mathrm e}^{8} \left (\frac {x^{2} \ln \left (\frac {3}{x}\right )^{2}}{2}+\frac {x^{2} \ln \left (\frac {3}{x}\right )}{2}+\frac {x^{2}}{4}\right )}{8}-2 x \,{\mathrm e}^{4}\) | \(138\) |
derivativedivides | \(-\frac {9 \,{\mathrm e}^{8} \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )^{2}}{18}-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{8}-\frac {9 \,{\mathrm e}^{8} \ln \left (3\right ) \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{4}+\frac {{\mathrm e}^{8} \ln \left (3\right )^{2} x^{2}}{16}+\frac {9 \,{\mathrm e}^{8} \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{8}-6 \,{\mathrm e}^{4} \left (-\frac {x \ln \left (\frac {3}{x}\right )}{3}-\frac {x}{3}\right )-\frac {{\mathrm e}^{8} \ln \left (3\right ) x^{2}}{16}+2 \,{\mathrm e}^{4} x \ln \left (3\right )-2 x \,{\mathrm e}^{4}\) | \(139\) |
parts | \(-\frac {9 \,{\mathrm e}^{8} \ln \left (3\right ) \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{4}+\frac {9 \,{\mathrm e}^{8} \left (-\frac {x^{2} \ln \left (\frac {3}{x}\right )}{18}-\frac {x^{2}}{36}\right )}{8}-6 \,{\mathrm e}^{4} \left (-\frac {x \ln \left (\frac {3}{x}\right )}{3}-\frac {x}{3}\right )+2 \,{\mathrm e}^{4} x \ln \left (3\right )-\frac {{\mathrm e}^{8} \ln \left (3\right ) x^{2}}{16}+\frac {{\mathrm e}^{8} \ln \left (3\right )^{2} x^{2}}{16}+\frac {{\mathrm e}^{8} \left (\frac {x^{2} \ln \left (\frac {3}{x}\right )^{2}}{2}+\frac {x^{2} \ln \left (\frac {3}{x}\right )}{2}+\frac {x^{2}}{4}\right )}{8}-2 x \,{\mathrm e}^{4}\) | \(139\) |
int(1/8*x*exp(4)^2*ln(3/x)^2+1/8*(2*x*exp(4)^2*ln(3)-x*exp(4)^2+16*exp(4)) *ln(3/x)+1/8*x*exp(4)^2*ln(3)^2+1/8*(-x*exp(4)^2+16*exp(4))*ln(3)-2*exp(4) ,x,method=_RETURNVERBOSE)
1/16*x^2*exp(8)*ln(3/x)^2+2*exp(4)*x*ln(3)+2*exp(4)*ln(3/x)*x+1/16*exp(8)* ln(3)^2*x^2+1/8*exp(8)*ln(3)*x^2*ln(3/x)
Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {1}{8} \left (-16 e^4+\left (16 e^4-e^8 x\right ) \log (3)+e^8 x \log ^2(3)+\left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2\left (\frac {3}{x}\right )\right ) \, dx=\frac {1}{16} \, x^{2} e^{8} \log \left (3\right )^{2} + \frac {1}{16} \, x^{2} e^{8} \log \left (\frac {3}{x}\right )^{2} + 2 \, x e^{4} \log \left (3\right ) + \frac {1}{8} \, {\left (x^{2} e^{8} \log \left (3\right ) + 16 \, x e^{4}\right )} \log \left (\frac {3}{x}\right ) \]
integrate(1/8*x*exp(4)^2*log(3/x)^2+1/8*(2*x*exp(4)^2*log(3)-x*exp(4)^2+16 *exp(4))*log(3/x)+1/8*x*exp(4)^2*log(3)^2+1/8*(-x*exp(4)^2+16*exp(4))*log( 3)-2*exp(4),x, algorithm=\
1/16*x^2*e^8*log(3)^2 + 1/16*x^2*e^8*log(3/x)^2 + 2*x*e^4*log(3) + 1/8*(x^ 2*e^8*log(3) + 16*x*e^4)*log(3/x)
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (26) = 52\).
Time = 0.10 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.10 \[ \int \frac {1}{8} \left (-16 e^4+\left (16 e^4-e^8 x\right ) \log (3)+e^8 x \log ^2(3)+\left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2\left (\frac {3}{x}\right )\right ) \, dx=\frac {x^{2} e^{8} \log {\left (\frac {3}{x} \right )}^{2}}{16} + \frac {x^{2} e^{8} \log {\left (3 \right )}^{2}}{16} + 2 x e^{4} \log {\left (3 \right )} + \left (\frac {x^{2} e^{8} \log {\left (3 \right )}}{8} + 2 x e^{4}\right ) \log {\left (\frac {3}{x} \right )} \]
integrate(1/8*x*exp(4)**2*ln(3/x)**2+1/8*(2*x*exp(4)**2*ln(3)-x*exp(4)**2+ 16*exp(4))*ln(3/x)+1/8*x*exp(4)**2*ln(3)**2+1/8*(-x*exp(4)**2+16*exp(4))*l n(3)-2*exp(4),x)
x**2*exp(8)*log(3/x)**2/16 + x**2*exp(8)*log(3)**2/16 + 2*x*exp(4)*log(3) + (x**2*exp(8)*log(3)/8 + 2*x*exp(4))*log(3/x)
Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (28) = 56\).
Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 3.72 \[ \int \frac {1}{8} \left (-16 e^4+\left (16 e^4-e^8 x\right ) \log (3)+e^8 x \log ^2(3)+\left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2\left (\frac {3}{x}\right )\right ) \, dx=\frac {1}{16} \, x^{2} e^{8} \log \left (3\right )^{2} + \frac {1}{16} \, x^{2} e^{8} \log \left (\frac {3}{x}\right )^{2} + \frac {1}{32} \, {\left (2 \, e^{8} \log \left (3\right ) - e^{8}\right )} x^{2} + \frac {1}{32} \, {\left (2 \, x^{2} \log \left (\frac {3}{x}\right ) + x^{2}\right )} e^{8} - \frac {1}{16} \, {\left (x^{2} e^{8} - 32 \, x e^{4}\right )} \log \left (3\right ) + \frac {1}{16} \, {\left (2 \, x^{2} e^{8} \log \left (3\right ) - x^{2} e^{8} + 32 \, x e^{4}\right )} \log \left (\frac {3}{x}\right ) \]
integrate(1/8*x*exp(4)^2*log(3/x)^2+1/8*(2*x*exp(4)^2*log(3)-x*exp(4)^2+16 *exp(4))*log(3/x)+1/8*x*exp(4)^2*log(3)^2+1/8*(-x*exp(4)^2+16*exp(4))*log( 3)-2*exp(4),x, algorithm=\
1/16*x^2*e^8*log(3)^2 + 1/16*x^2*e^8*log(3/x)^2 + 1/32*(2*e^8*log(3) - e^8 )*x^2 + 1/32*(2*x^2*log(3/x) + x^2)*e^8 - 1/16*(x^2*e^8 - 32*x*e^4)*log(3) + 1/16*(2*x^2*e^8*log(3) - x^2*e^8 + 32*x*e^4)*log(3/x)
Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (28) = 56\).
Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 4.07 \[ \int \frac {1}{8} \left (-16 e^4+\left (16 e^4-e^8 x\right ) \log (3)+e^8 x \log ^2(3)+\left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2\left (\frac {3}{x}\right )\right ) \, dx=\frac {1}{16} \, x^{2} e^{8} \log \left (3\right )^{2} + \frac {1}{32} \, {\left (2 \, e^{8} \log \left (3\right ) + \frac {64 \, e^{4}}{x} - e^{8}\right )} x^{2} + \frac {1}{32} \, {\left (2 \, x^{2} \log \left (\frac {3}{x}\right )^{2} + 2 \, x^{2} \log \left (\frac {3}{x}\right ) + x^{2}\right )} e^{8} - 2 \, x e^{4} - \frac {1}{16} \, {\left (x^{2} e^{8} - 32 \, x e^{4}\right )} \log \left (3\right ) + \frac {1}{16} \, {\left (2 \, x^{2} e^{8} \log \left (3\right ) - x^{2} e^{8} + 32 \, x e^{4}\right )} \log \left (\frac {3}{x}\right ) \]
integrate(1/8*x*exp(4)^2*log(3/x)^2+1/8*(2*x*exp(4)^2*log(3)-x*exp(4)^2+16 *exp(4))*log(3/x)+1/8*x*exp(4)^2*log(3)^2+1/8*(-x*exp(4)^2+16*exp(4))*log( 3)-2*exp(4),x, algorithm=\
1/16*x^2*e^8*log(3)^2 + 1/32*(2*e^8*log(3) + 64*e^4/x - e^8)*x^2 + 1/32*(2 *x^2*log(3/x)^2 + 2*x^2*log(3/x) + x^2)*e^8 - 2*x*e^4 - 1/16*(x^2*e^8 - 32 *x*e^4)*log(3) + 1/16*(2*x^2*e^8*log(3) - x^2*e^8 + 32*x*e^4)*log(3/x)
Time = 10.65 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {1}{8} \left (-16 e^4+\left (16 e^4-e^8 x\right ) \log (3)+e^8 x \log ^2(3)+\left (16 e^4-e^8 x+2 e^8 x \log (3)\right ) \log \left (\frac {3}{x}\right )+e^8 x \log ^2\left (\frac {3}{x}\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^4\,\left (\ln \left (\frac {1}{x}\right )+2\,\ln \left (3\right )\right )\,\left (x\,\ln \left (\frac {1}{x}\right )\,{\mathrm {e}}^4+2\,x\,{\mathrm {e}}^4\,\ln \left (3\right )+32\right )}{16} \]