Integrand size = 63, antiderivative size = 28 \begin {dmath*} \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \end {dmath*}
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \begin {dmath*} \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \log ^2\left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right ) \end {dmath*}
Integrate[((-3 - 3*Log[x])*Log[(3 + 16*x*Log[-2 + E^5]*Log[x])/(4*x*Log[-2 + E^5]*Log[x])])/(6*x*Log[x] + 32*x^2*Log[-2 + E^5]*Log[x]^2),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.
\(\displaystyle \int \frac {(-3 \log (x)-3) \log \left (\frac {16 x \log \left (e^5-2\right ) \log (x)+3}{4 x \log \left (e^5-2\right ) \log (x)}\right )}{32 x^2 \log \left (e^5-2\right ) \log ^2(x)+6 x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 (-\log (x)-1) \log \left (\frac {16 x \log \left (e^5-2\right ) \log (x)+3}{4 x \log \left (e^5-2\right ) \log (x)}\right )}{2 x \log (x) \left (16 x \log \left (e^5-2\right ) \log (x)+3\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{2} \int -\frac {(\log (x)+1) \log \left (\frac {16 x \log \left (-2+e^5\right ) \log (x)+3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {3}{2} \int \frac {(\log (x)+1) \log \left (\frac {16 x \log \left (-2+e^5\right ) \log (x)+3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {3}{2} \int \left (\frac {\log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}+\frac {\log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{2} \left (\int \frac {\log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}dx+\int \frac {\log \left (4+\frac {3}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{x \log (x) \left (16 x \log \left (-2+e^5\right ) \log (x)+3\right )}dx\right )\) |
Int[((-3 - 3*Log[x])*Log[(3 + 16*x*Log[-2 + E^5]*Log[x])/(4*x*Log[-2 + E^5 ]*Log[x])])/(6*x*Log[x] + 32*x^2*Log[-2 + E^5]*Log[x]^2),x]
3.2.97.3.1 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. \(88\) vs. \(2(23)=46\).
Time = 4.44 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.18
method | result | size |
default | \(\frac {3 \ln \left (\ln \left ({\mathrm e}^{5}-2\right )\right ) \left (\frac {\ln \left (x \right )}{3}+\frac {\ln \left (\ln \left (x \right )\right )}{3}-\frac {\ln \left (16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3\right )}{3}\right )}{2}+3 \ln \left (2\right ) \left (\frac {\ln \left (x \right )}{3}+\frac {\ln \left (\ln \left (x \right )\right )}{3}-\frac {\ln \left (16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3\right )}{3}\right )+\frac {{\ln \left (\frac {16 x \ln \left (x \right ) \ln \left ({\mathrm e}^{5}-2\right )+3}{x \ln \left (x \right )}\right )}^{2}}{4}\) | \(89\) |
int((-3*ln(x)-3)*ln(1/4*(16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x)/ln(exp(5)-2))/ (32*x^2*ln(x)^2*ln(exp(5)-2)+6*x*ln(x)),x,method=_RETURNVERBOSE)
3/2*ln(ln(exp(5)-2))*(1/3*ln(x)+1/3*ln(ln(x))-1/3*ln(16*x*ln(x)*ln(exp(5)- 2)+3))+3*ln(2)*(1/3*ln(x)+1/3*ln(ln(x))-1/3*ln(16*x*ln(x)*ln(exp(5)-2)+3)) +1/4*ln((16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x))^2
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \begin {dmath*} \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {1}{4} \, \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )^{2} \end {dmath*}
integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log (exp(5)-2))/(32*x^2*log(x)^2*log(exp(5)-2)+6*x*log(x)),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \begin {dmath*} \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {\log {\left (\frac {4 x \log {\left (x \right )} \log {\left (-2 + e^{5} \right )} + \frac {3}{4}}{x \log {\left (x \right )} \log {\left (-2 + e^{5} \right )}} \right )}^{2}}{4} \end {dmath*}
integrate((-3*ln(x)-3)*ln(1/4*(16*x*ln(x)*ln(exp(5)-2)+3)/x/ln(x)/ln(exp(5 )-2))/(32*x**2*ln(x)**2*ln(exp(5)-2)+6*x*ln(x)),x)
\begin {dmath*} \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\int { -\frac {3 \, {\left (\log \left (x\right ) + 1\right )} \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )}{2 \, {\left (16 \, x^{2} \log \left (x\right )^{2} \log \left (e^{5} - 2\right ) + 3 \, x \log \left (x\right )\right )}} \,d x } \end {dmath*}
integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log (exp(5)-2))/(32*x^2*log(x)^2*log(exp(5)-2)+6*x*log(x)),x, algorithm=\
-3/2*integrate((log(x) + 1)*log(1/4*(16*x*log(x)*log(e^5 - 2) + 3)/(x*log( x)*log(e^5 - 2)))/(16*x^2*log(x)^2*log(e^5 - 2) + 3*x*log(x)), x)
\begin {dmath*} \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\int { -\frac {3 \, {\left (\log \left (x\right ) + 1\right )} \log \left (\frac {16 \, x \log \left (x\right ) \log \left (e^{5} - 2\right ) + 3}{4 \, x \log \left (x\right ) \log \left (e^{5} - 2\right )}\right )}{2 \, {\left (16 \, x^{2} \log \left (x\right )^{2} \log \left (e^{5} - 2\right ) + 3 \, x \log \left (x\right )\right )}} \,d x } \end {dmath*}
integrate((-3*log(x)-3)*log(1/4*(16*x*log(x)*log(exp(5)-2)+3)/x/log(x)/log (exp(5)-2))/(32*x^2*log(x)^2*log(exp(5)-2)+6*x*log(x)),x, algorithm=\
integrate(-3/2*(log(x) + 1)*log(1/4*(16*x*log(x)*log(e^5 - 2) + 3)/(x*log( x)*log(e^5 - 2)))/(16*x^2*log(x)^2*log(e^5 - 2) + 3*x*log(x)), x)
Time = 14.53 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \begin {dmath*} \int \frac {(-3-3 \log (x)) \log \left (\frac {3+16 x \log \left (-2+e^5\right ) \log (x)}{4 x \log \left (-2+e^5\right ) \log (x)}\right )}{6 x \log (x)+32 x^2 \log \left (-2+e^5\right ) \log ^2(x)} \, dx=\frac {{\ln \left (\frac {4\,x\,\ln \left ({\mathrm {e}}^5-2\right )\,\ln \left (x\right )+\frac {3}{4}}{x\,\ln \left ({\mathrm {e}}^5-2\right )\,\ln \left (x\right )}\right )}^2}{4} \end {dmath*}