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

Optimal. Leaf size=25 \[ \frac {e^{\log ^8(4)}}{\log \left (8+\frac {3}{x}\right ) (\log (x)+\log (\log (2)))} \]

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Rubi [F]  time = 1.13, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 5.03, size = 24, normalized size = 0.96 \begin {gather*} \frac {e^{\log ^8(4)}}{\log \left (8+\frac {3}{x}\right ) \log (x \log (2))} \end {gather*}

Antiderivative was successfully verified.

[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]])

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fricas [A]  time = 0.67, size = 38, normalized size = 1.52 \begin {gather*} \frac {e^{\left (256 \, \log \relax (2)^{8}\right )}}{\log \relax (x) \log \left (\frac {8 \, x + 3}{x}\right ) + \log \left (\frac {8 \, x + 3}{x}\right ) \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [A]  time = 0.23, size = 43, normalized size = 1.72 \begin {gather*} \frac {e^{\left (256 \, \log \relax (2)^{8}\right )}}{\log \left (8 \, x + 3\right ) \log \relax (x) - \log \relax (x)^{2} + \log \left (8 \, x + 3\right ) \log \left (\log \relax (2)\right ) - \log \relax (x) \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [C]  time = 0.20, size = 123, normalized size = 4.92

method	result	size
risch	\(\frac {2 i {\mathrm e}^{256 \ln \relax (2)^{8}}}{\left (\ln \relax (x )+\ln \left (\ln \relax (2)\right )\right ) \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (x +\frac {3}{8}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {3}{8}\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {3}{8}\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x +\frac {3}{8}\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {3}{8}\right )}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left (x +\frac {3}{8}\right )}{x}\right )^{3}+6 i \ln \relax (2)-2 i \ln \relax (x )+2 i \ln \left (x +\frac {3}{8}\right )\right )}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[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,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 37, normalized size = 1.48 \begin {gather*} \frac {e^{\left (256 \, \log \relax (2)^{8}\right )}}{{\left (\log \relax (x) + \log \left (\log \relax (2)\right )\right )} \log \left (8 \, x + 3\right ) - \log \relax (x)^{2} - \log \relax (x) \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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mupad [B]  time = 3.50, size = 181, normalized size = 7.24 \begin {gather*} \frac {\frac {4\,x\,\left ({\mathrm {e}}^{256\,{\ln \relax (2)}^8}-{\mathrm {e}}^{256\,{\ln \relax (2)}^8}\,\ln \left (\ln \relax (2)\right )\right )}{3}-\frac {4\,x\,{\mathrm {e}}^{256\,{\ln \relax (2)}^8}\,\ln \relax (x)}{3}}{\ln \left (\ln \relax (2)\right )+\ln \relax (x)}+\frac {4\,x\,{\mathrm {e}}^{256\,{\ln \relax (2)}^8}}{3}+\frac {{\mathrm {e}}^{256\,{\ln \relax (2)}^8}+\frac {8\,x\,{\mathrm {e}}^{256\,{\ln \relax (2)}^8}}{3}-\frac {4\,x\,{\mathrm {e}}^{256\,{\ln \relax (2)}^8}\,\ln \left (\ln \relax (2)\right )}{3}-\frac {4\,x\,{\mathrm {e}}^{256\,{\ln \relax (2)}^8}\,\ln \relax (x)}{3}}{{\ln \relax (x)}^2+2\,\ln \left (\ln \relax (2)\right )\,\ln \relax (x)+{\ln \left (\ln \relax (2)\right )}^2}+\frac {\frac {{\mathrm {e}}^{256\,{\ln \relax (2)}^8}}{\ln \left (\ln \relax (2)\right )+\ln \relax (x)}-\frac {{\mathrm {e}}^{256\,{\ln \relax (2)}^8}\,\ln \left (\frac {8\,x+3}{x}\right )\,\left (8\,x+3\right )}{3\,{\left (\ln \left (\ln \relax (2)\right )+\ln \relax (x)\right )}^2}}{\ln \left (\frac {8\,x+3}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.30, size = 24, normalized size = 0.96 \begin {gather*} \frac {e^{256 \log {\relax (2 )}^{8}}}{\left (\log {\relax (x )} + \log {\left (\log {\relax (2 )} \right )}\right ) \log {\left (\frac {8 x + 3}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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