Integrand size = 60, antiderivative size = 21 \[ \int \frac {12+\log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )}{x \log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )} \, dx=\log (x)+\frac {4}{\log \left (\frac {(-16-e+\log (3))^2}{x^3}\right )} \] Output:
ln(x)+4/ln((ln(3)-16-exp(1))^2/x^3)
Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {12+\log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )}{x \log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )} \, dx=\log (x)+\frac {4}{\log \left (\frac {(16+e-\log (3))^2}{x^3}\right )} \] Input:
Integrate[(12 + Log[(256 + 32*E + E^2 + (-32 - 2*E)*Log[3] + Log[3]^2)/x^3 ]^2)/(x*Log[(256 + 32*E + E^2 + (-32 - 2*E)*Log[3] + Log[3]^2)/x^3]^2),x]
Output:
Log[x] + 4/Log[(16 + E - Log[3])^2/x^3]
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {3039, 244, 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.
\(\displaystyle \int \frac {\log ^2\left (\frac {256+32 e+e^2+\log ^2(3)+(-32-2 e) \log (3)}{x^3}\right )+12}{x \log ^2\left (\frac {256+32 e+e^2+\log ^2(3)+(-32-2 e) \log (3)}{x^3}\right )} \, dx\) |
\(\Big \downarrow \) 3039 |
\(\displaystyle -\frac {1}{3} \int \frac {\log ^2\left (\frac {(16+e-\log (3))^2}{x^3}\right )+12}{\log ^2\left (\frac {(16+e-\log (3))^2}{x^3}\right )}d\log \left (\frac {(16+e-\log (3))^2}{x^3}\right )\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {1}{3} \int \left (1+\frac {12}{\log ^2\left (\frac {(16+e-\log (3))^2}{x^3}\right )}\right )d\log \left (\frac {(16+e-\log (3))^2}{x^3}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {12}{\log \left (\frac {(16+e-\log (3))^2}{x^3}\right )}-\log \left (\frac {(16+e-\log (3))^2}{x^3}\right )\right )\) |
Input:
Int[(12 + Log[(256 + 32*E + E^2 + (-32 - 2*E)*Log[3] + Log[3]^2)/x^3]^2)/( x*Log[(256 + 32*E + E^2 + (-32 - 2*E)*Log[3] + Log[3]^2)/x^3]^2),x]
Output:
(12/Log[(16 + E - Log[3])^2/x^3] - Log[(16 + E - Log[3])^2/x^3])/3
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst [[3]] Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /; !FalseQ[lst]] /; NonsumQ[u]
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62
method | result | size |
risch | \(\frac {4}{\ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}+\ln \left (x \right )\) | \(34\) |
norman | \(\frac {4}{\ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}+\ln \left (x \right )\) | \(36\) |
parts | \(\frac {4}{\ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}+\ln \left (x \right )\) | \(36\) |
derivativedivides | \(-\frac {\ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}{3}+\frac {4}{\ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}\) | \(64\) |
default | \(-\frac {\ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}{3}+\frac {4}{\ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}\) | \(64\) |
parallelrisch | \(\frac {12-\ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )^{2}}{3 \ln \left (\frac {\ln \left (3\right )^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \left (3\right )+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}\) | \(67\) |
Input:
int((ln((ln(3)^2+(-2*exp(1)-32)*ln(3)+exp(1)^2+32*exp(1)+256)/x^3)^2+12)/x /ln((ln(3)^2+(-2*exp(1)-32)*ln(3)+exp(1)^2+32*exp(1)+256)/x^3)^2,x,method= _RETURNVERBOSE)
Output:
4/ln((ln(3)^2+(-2*exp(1)-32)*ln(3)+exp(2)+32*exp(1)+256)/x^3)+ln(x)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (22) = 44\).
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 3.24 \[ \int \frac {12+\log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )}{x \log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )} \, dx=-\frac {\log \left (-\frac {2 \, {\left (e + 16\right )} \log \left (3\right ) - \log \left (3\right )^{2} - e^{2} - 32 \, e - 256}{x^{3}}\right )^{2} - 12}{3 \, \log \left (-\frac {2 \, {\left (e + 16\right )} \log \left (3\right ) - \log \left (3\right )^{2} - e^{2} - 32 \, e - 256}{x^{3}}\right )} \] Input:
integrate((log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3 )^2+12)/x/log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3) ^2,x, algorithm="fricas")
Output:
-1/3*(log(-(2*(e + 16)*log(3) - log(3)^2 - e^2 - 32*e - 256)/x^3)^2 - 12)/ log(-(2*(e + 16)*log(3) - log(3)^2 - e^2 - 32*e - 256)/x^3)
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {12+\log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )}{x \log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )} \, dx=\log {\left (x \right )} + \frac {4}{\log {\left (\frac {\left (-32 - 2 e\right ) \log {\left (3 \right )} + \log {\left (3 \right )}^{2} + e^{2} + 32 e + 256}{x^{3}} \right )}} \] Input:
integrate((ln((ln(3)**2+(-2*exp(1)-32)*ln(3)+exp(1)**2+32*exp(1)+256)/x**3 )**2+12)/x/ln((ln(3)**2+(-2*exp(1)-32)*ln(3)+exp(1)**2+32*exp(1)+256)/x**3 )**2,x)
Output:
log(x) + 4/log(((-32 - 2*E)*log(3) + log(3)**2 + exp(2) + 32*E + 256)/x**3 )
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {12+\log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )}{x \log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )} \, dx=-\frac {4}{3 \, \log \left (x\right ) - 2 \, \log \left (-e + \log \left (3\right ) - 16\right )} + \log \left (x\right ) \] Input:
integrate((log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3 )^2+12)/x/log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3) ^2,x, algorithm="maxima")
Output:
-4/(3*log(x) - 2*log(-e + log(3) - 16)) + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (22) = 44\).
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.29 \[ \int \frac {12+\log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )}{x \log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )} \, dx=\frac {\log \left (x^{3}\right ) \log \left (x\right ) - \log \left (-2 \, e \log \left (3\right ) + \log \left (3\right )^{2} + e^{2} + 32 \, e - 32 \, \log \left (3\right ) + 256\right ) \log \left (x\right ) - 4}{\log \left (x^{3}\right ) - \log \left (-2 \, e \log \left (3\right ) + \log \left (3\right )^{2} + e^{2} + 32 \, e - 32 \, \log \left (3\right ) + 256\right )} \] Input:
integrate((log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3 )^2+12)/x/log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3) ^2,x, algorithm="giac")
Output:
(log(x^3)*log(x) - log(-2*e*log(3) + log(3)^2 + e^2 + 32*e - 32*log(3) + 2 56)*log(x) - 4)/(log(x^3) - log(-2*e*log(3) + log(3)^2 + e^2 + 32*e - 32*l og(3) + 256))
Time = 2.66 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.86 \[ \int \frac {12+\log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )}{x \log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )} \, dx=\frac {4}{\ln \left (\frac {1}{x^3}\right )+\ln \left (32\,\mathrm {e}+{\mathrm {e}}^2-32\,\ln \left (3\right )-2\,\mathrm {e}\,\ln \left (3\right )+{\ln \left (3\right )}^2+256\right )}-\frac {\ln \left (\frac {1}{x^3}\right )}{3} \] Input:
int((log((32*exp(1) + exp(2) + log(3)^2 - log(3)*(2*exp(1) + 32) + 256)/x^ 3)^2 + 12)/(x*log((32*exp(1) + exp(2) + log(3)^2 - log(3)*(2*exp(1) + 32) + 256)/x^3)^2),x)
Output:
4/(log(1/x^3) + log(32*exp(1) + exp(2) - 32*log(3) - 2*exp(1)*log(3) + log (3)^2 + 256)) - log(1/x^3)/3
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.86 \[ \int \frac {12+\log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )}{x \log ^2\left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )} \, dx=\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (3\right )^{2}-2 \,\mathrm {log}\left (3\right ) e -32 \,\mathrm {log}\left (3\right )+e^{2}+32 e +256}{x^{3}}\right ) \mathrm {log}\left (x \right )+4}{\mathrm {log}\left (\frac {\mathrm {log}\left (3\right )^{2}-2 \,\mathrm {log}\left (3\right ) e -32 \,\mathrm {log}\left (3\right )+e^{2}+32 e +256}{x^{3}}\right )} \] Input:
int((log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3)^2+12 )/x/log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3)^2,x)
Output:
(log((log(3)**2 - 2*log(3)*e - 32*log(3) + e**2 + 32*e + 256)/x**3)*log(x) + 4)/log((log(3)**2 - 2*log(3)*e - 32*log(3) + e**2 + 32*e + 256)/x**3)