\(\int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log (\frac {3+8 x}{x})+3 e^{\log ^8(4)} \log (\log (2))}{(3 x+8 x^2) \log ^2(x) \log ^2(\frac {3+8 x}{x})+(6 x+16 x^2) \log (x) \log ^2(\frac {3+8 x}{x}) \log (\log (2))+(3 x+8 x^2) \log ^2(\frac {3+8 x}{x}) \log ^2(\log (2))} \, dx\) [3989]

Optimal result

Integrand size = 128, antiderivative size = 25 \[ \int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx=\frac {e^{\log ^8(4)}}{\log \left (8+\frac {3}{x}\right ) (\log (x)+\log (\log (2)))} \]

[Out]

exp(256*ln(2)^8)/ln(8+3/x)/(ln(x)+ln(ln(2)))

Rubi [F]

\[ \int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx=\int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx \]

[In]

Int[(3*E^Log[4]^8*Log[x] + E^Log[4]^8*(-3 - 8*x)*Log[(3 + 8*x)/x] + 3*E^Log[4]^8*Log[Log[2]])/((3*x + 8*x^2)*L
og[x]^2*Log[(3 + 8*x)/x]^2 + (6*x + 16*x^2)*Log[x]*Log[(3 + 8*x)/x]^2*Log[Log[2]] + (3*x + 8*x^2)*Log[(3 + 8*x
)/x]^2*Log[Log[2]]^2),x]

[Out]

-(E^Log[4]^8*Defer[Int][1/(x*Log[8 + 3/x]*Log[x*Log[2]]^2), x]) + E^Log[4]^8*Defer[Int][1/(x*Log[8 + 3/x]^2*Lo
g[x*Log[2]]), x] - 8*E^Log[4]^8*Defer[Int][1/((3 + 8*x)*Log[8 + 3/x]^2*Log[x*Log[2]]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\log ^8(4)} \left (-\left ((3+8 x) \log \left (8+\frac {3}{x}\right )\right )+3 \log (x \log (2))\right )}{x (3+8 x) \log ^2\left (8+\frac {3}{x}\right ) \log ^2(x \log (2))} \, dx \\ & = e^{\log ^8(4)} \int \frac {-\left ((3+8 x) \log \left (8+\frac {3}{x}\right )\right )+3 \log (x \log (2))}{x (3+8 x) \log ^2\left (8+\frac {3}{x}\right ) \log ^2(x \log (2))} \, dx \\ & = e^{\log ^8(4)} \int \left (-\frac {1}{x \log \left (8+\frac {3}{x}\right ) \log ^2(x \log (2))}+\frac {3}{x (3+8 x) \log ^2\left (8+\frac {3}{x}\right ) \log (x \log (2))}\right ) \, dx \\ & = -\left (e^{\log ^8(4)} \int \frac {1}{x \log \left (8+\frac {3}{x}\right ) \log ^2(x \log (2))} \, dx\right )+\left (3 e^{\log ^8(4)}\right ) \int \frac {1}{x (3+8 x) \log ^2\left (8+\frac {3}{x}\right ) \log (x \log (2))} \, dx \\ & = -\left (e^{\log ^8(4)} \int \frac {1}{x \log \left (8+\frac {3}{x}\right ) \log ^2(x \log (2))} \, dx\right )+\left (3 e^{\log ^8(4)}\right ) \int \left (\frac {1}{3 x \log ^2\left (8+\frac {3}{x}\right ) \log (x \log (2))}-\frac {8}{3 (3+8 x) \log ^2\left (8+\frac {3}{x}\right ) \log (x \log (2))}\right ) \, dx \\ & = -\left (e^{\log ^8(4)} \int \frac {1}{x \log \left (8+\frac {3}{x}\right ) \log ^2(x \log (2))} \, dx\right )+e^{\log ^8(4)} \int \frac {1}{x \log ^2\left (8+\frac {3}{x}\right ) \log (x \log (2))} \, dx-\left (8 e^{\log ^8(4)}\right ) \int \frac {1}{(3+8 x) \log ^2\left (8+\frac {3}{x}\right ) \log (x \log (2))} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx=\frac {e^{\log ^8(4)}}{\log \left (8+\frac {3}{x}\right ) \log (x \log (2))} \]

[In]

Integrate[(3*E^Log[4]^8*Log[x] + E^Log[4]^8*(-3 - 8*x)*Log[(3 + 8*x)/x] + 3*E^Log[4]^8*Log[Log[2]])/((3*x + 8*
x^2)*Log[x]^2*Log[(3 + 8*x)/x]^2 + (6*x + 16*x^2)*Log[x]*Log[(3 + 8*x)/x]^2*Log[Log[2]] + (3*x + 8*x^2)*Log[(3
+ 8*x)/x]^2*Log[Log[2]]^2),x]

[Out]

E^Log[4]^8/(Log[8 + 3/x]*Log[x*Log[2]])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 660.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 5.00

\[\frac {2 \,{\mathrm e}^{256 \ln \left (2\right )^{8}}}{\left (\ln \left (x \right )+\ln \left (\ln \left (2\right )\right )\right ) \left (6 \ln \left (2\right )-2 \ln \left (x \right )+2 \ln \left (x +\frac {3}{8}\right )-i \pi \,\operatorname {csgn}\left (i \left (x +\frac {3}{8}\right )\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x +\frac {3}{8}\right )}{x}\right )+i \pi \,\operatorname {csgn}\left (i \left (x +\frac {3}{8}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x +\frac {3}{8}\right )}{x}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (\frac {i \left (x +\frac {3}{8}\right )}{x}\right )^{2}-i \pi \operatorname {csgn}\left (\frac {i \left (x +\frac {3}{8}\right )}{x}\right )^{3}\right )}\]

[In]

int((3*exp(256*ln(2)^8)*ln(ln(2))+3*exp(256*ln(2)^8)*ln(x)+(-8*x-3)*exp(256*ln(2)^8)*ln((8*x+3)/x))/((8*x^2+3*
x)*ln((8*x+3)/x)^2*ln(ln(2))^2+(16*x^2+6*x)*ln((8*x+3)/x)^2*ln(x)*ln(ln(2))+(8*x^2+3*x)*ln((8*x+3)/x)^2*ln(x)^
2),x)

[Out]

2*exp(256*ln(2)^8)/(ln(x)+ln(ln(2)))/(6*ln(2)-2*ln(x)+2*ln(x+3/8)-I*Pi*csgn(I*(x+3/8))*csgn(I/x)*csgn(I/x*(x+3
/8))+I*Pi*csgn(I*(x+3/8))*csgn(I/x*(x+3/8))^2+I*Pi*csgn(I/x)*csgn(I/x*(x+3/8))^2-I*Pi*csgn(I/x*(x+3/8))^3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx=\frac {e^{\left (256 \, \log \left (2\right )^{8}\right )}}{\log \left (x\right ) \log \left (\frac {8 \, x + 3}{x}\right ) + \log \left (\frac {8 \, x + 3}{x}\right ) \log \left (\log \left (2\right )\right )} \]

[In]

integrate((3*exp(256*log(2)^8)*log(log(2))+3*exp(256*log(2)^8)*log(x)+(-8*x-3)*exp(256*log(2)^8)*log((8*x+3)/x
))/((8*x^2+3*x)*log((8*x+3)/x)^2*log(log(2))^2+(16*x^2+6*x)*log((8*x+3)/x)^2*log(x)*log(log(2))+(8*x^2+3*x)*lo
g((8*x+3)/x)^2*log(x)^2),x, algorithm="fricas")

[Out]

e^(256*log(2)^8)/(log(x)*log((8*x + 3)/x) + log((8*x + 3)/x)*log(log(2)))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx=\frac {e^{256 \log {\left (2 \right )}^{8}}}{\left (\log {\left (x \right )} + \log {\left (\log {\left (2 \right )} \right )}\right ) \log {\left (\frac {8 x + 3}{x} \right )}} \]

[In]

integrate((3*exp(256*ln(2)**8)*ln(ln(2))+3*exp(256*ln(2)**8)*ln(x)+(-8*x-3)*exp(256*ln(2)**8)*ln((8*x+3)/x))/(
(8*x**2+3*x)*ln((8*x+3)/x)**2*ln(ln(2))**2+(16*x**2+6*x)*ln((8*x+3)/x)**2*ln(x)*ln(ln(2))+(8*x**2+3*x)*ln((8*x
+3)/x)**2*ln(x)**2),x)

[Out]

exp(256*log(2)**8)/((log(x) + log(log(2)))*log((8*x + 3)/x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx=\frac {e^{\left (256 \, \log \left (2\right )^{8}\right )}}{{\left (\log \left (x\right ) + \log \left (\log \left (2\right )\right )\right )} \log \left (8 \, x + 3\right ) - \log \left (x\right )^{2} - \log \left (x\right ) \log \left (\log \left (2\right )\right )} \]

[In]

integrate((3*exp(256*log(2)^8)*log(log(2))+3*exp(256*log(2)^8)*log(x)+(-8*x-3)*exp(256*log(2)^8)*log((8*x+3)/x
))/((8*x^2+3*x)*log((8*x+3)/x)^2*log(log(2))^2+(16*x^2+6*x)*log((8*x+3)/x)^2*log(x)*log(log(2))+(8*x^2+3*x)*lo
g((8*x+3)/x)^2*log(x)^2),x, algorithm="maxima")

[Out]

e^(256*log(2)^8)/((log(x) + log(log(2)))*log(8*x + 3) - log(x)^2 - log(x)*log(log(2)))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx=\frac {e^{\left (256 \, \log \left (2\right )^{8}\right )}}{\log \left (8 \, x + 3\right ) \log \left (x\right ) - \log \left (x\right )^{2} + \log \left (8 \, x + 3\right ) \log \left (\log \left (2\right )\right ) - \log \left (x\right ) \log \left (\log \left (2\right )\right )} \]

[In]

integrate((3*exp(256*log(2)^8)*log(log(2))+3*exp(256*log(2)^8)*log(x)+(-8*x-3)*exp(256*log(2)^8)*log((8*x+3)/x
))/((8*x^2+3*x)*log((8*x+3)/x)^2*log(log(2))^2+(16*x^2+6*x)*log((8*x+3)/x)^2*log(x)*log(log(2))+(8*x^2+3*x)*lo
g((8*x+3)/x)^2*log(x)^2),x, algorithm="giac")

[Out]

e^(256*log(2)^8)/(log(8*x + 3)*log(x) - log(x)^2 + log(8*x + 3)*log(log(2)) - log(x)*log(log(2)))

Mupad [B] (verification not implemented)

Time = 10.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 7.24 \[ \int \frac {3 e^{\log ^8(4)} \log (x)+e^{\log ^8(4)} (-3-8 x) \log \left (\frac {3+8 x}{x}\right )+3 e^{\log ^8(4)} \log (\log (2))}{\left (3 x+8 x^2\right ) \log ^2(x) \log ^2\left (\frac {3+8 x}{x}\right )+\left (6 x+16 x^2\right ) \log (x) \log ^2\left (\frac {3+8 x}{x}\right ) \log (\log (2))+\left (3 x+8 x^2\right ) \log ^2\left (\frac {3+8 x}{x}\right ) \log ^2(\log (2))} \, dx=\frac {\frac {4\,x\,\left ({\mathrm {e}}^{256\,{\ln \left (2\right )}^8}-{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}\,\ln \left (\ln \left (2\right )\right )\right )}{3}-\frac {4\,x\,{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}\,\ln \left (x\right )}{3}}{\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )}+\frac {4\,x\,{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}}{3}+\frac {{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}+\frac {8\,x\,{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}}{3}-\frac {4\,x\,{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}\,\ln \left (\ln \left (2\right )\right )}{3}-\frac {4\,x\,{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}\,\ln \left (x\right )}{3}}{{\ln \left (x\right )}^2+2\,\ln \left (\ln \left (2\right )\right )\,\ln \left (x\right )+{\ln \left (\ln \left (2\right )\right )}^2}+\frac {\frac {{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}}{\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )}-\frac {{\mathrm {e}}^{256\,{\ln \left (2\right )}^8}\,\ln \left (\frac {8\,x+3}{x}\right )\,\left (8\,x+3\right )}{3\,{\left (\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )\right )}^2}}{\ln \left (\frac {8\,x+3}{x}\right )} \]

[In]

int((3*exp(256*log(2)^8)*log(log(2)) + 3*exp(256*log(2)^8)*log(x) - exp(256*log(2)^8)*log((8*x + 3)/x)*(8*x +
3))/(log(log(2))^2*log((8*x + 3)/x)^2*(3*x + 8*x^2) + log((8*x + 3)/x)^2*log(x)^2*(3*x + 8*x^2) + log(log(2))*
log((8*x + 3)/x)^2*log(x)*(6*x + 16*x^2)),x)

[Out]

((4*x*(exp(256*log(2)^8) - exp(256*log(2)^8)*log(log(2))))/3 - (4*x*exp(256*log(2)^8)*log(x))/3)/(log(log(2))
+ log(x)) + (4*x*exp(256*log(2)^8))/3 + (exp(256*log(2)^8) + (8*x*exp(256*log(2)^8))/3 - (4*x*exp(256*log(2)^8
)*log(log(2)))/3 - (4*x*exp(256*log(2)^8)*log(x))/3)/(2*log(log(2))*log(x) + log(log(2))^2 + log(x)^2) + (exp(
256*log(2)^8)/(log(log(2)) + log(x)) - (exp(256*log(2)^8)*log((8*x + 3)/x)*(8*x + 3))/(3*(log(log(2)) + log(x)
)^2))/log((8*x + 3)/x)