Integrand size = 160, antiderivative size = 27 \[ \int \frac {3 x^3-x^4+\left (3 x+x^2\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (-6 x^3+6 x \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^3-x \log \left (\frac {16}{5}\right )+\left (-3 x^2+3 \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \, dx=\frac {x^2}{\log \left (\frac {x}{3}-\log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )} \]
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {3 x^3-x^4+\left (3 x+x^2\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (-6 x^3+6 x \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^3-x \log \left (\frac {16}{5}\right )+\left (-3 x^2+3 \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \, dx=\frac {x^2}{\log \left (\frac {1}{3} \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \]
Integrate[(3*x^3 - x^4 + (3*x + x^2)*Log[16/5] + (2*x^4 - 2*x^2*Log[16/5] + (-6*x^3 + 6*x*Log[16/5])*Log[(x^2 - Log[16/5])/x])*Log[(x - 3*Log[(x^2 - Log[16/5])/x])/3])/((x^3 - x*Log[16/5] + (-3*x^2 + 3*Log[16/5])*Log[(x^2 - Log[16/5])/x])*Log[(x - 3*Log[(x^2 - Log[16/5])/x])/3]^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.
\(\displaystyle \int \frac {-x^4+3 x^3+\left (x^2+3 x\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (6 x \log \left (\frac {16}{5}\right )-6 x^3\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^3+\left (3 \log \left (\frac {16}{5}\right )-3 x^2\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )-x \log \left (\frac {16}{5}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^4+3 x^3+\left (x^2+3 x\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (6 x \log \left (\frac {16}{5}\right )-6 x^3\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^2-\log \left (\frac {16}{5}\right )\right ) \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {2 x}{\log \left (\frac {1}{3} \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}-\frac {x \left (x^3-3 x^2-x \log \left (\frac {16}{5}\right )-3 \log \left (\frac {16}{5}\right )\right )}{\left (x^2-\log \left (\frac {16}{5}\right )\right ) \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {x^2}{\left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}dx+3 \int \frac {x}{\left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}dx-3 \log \left (\frac {16}{5}\right ) \int \frac {1}{\left (\sqrt {\log \left (\frac {16}{5}\right )}-x\right ) \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}dx+3 \log \left (\frac {16}{5}\right ) \int \frac {1}{\left (x+\sqrt {\log \left (\frac {16}{5}\right )}\right ) \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}dx+2 \int \frac {x}{\log \left (\frac {1}{3} \left (x-3 \log \left (x-\frac {\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}dx\) |
Int[(3*x^3 - x^4 + (3*x + x^2)*Log[16/5] + (2*x^4 - 2*x^2*Log[16/5] + (-6* x^3 + 6*x*Log[16/5])*Log[(x^2 - Log[16/5])/x])*Log[(x - 3*Log[(x^2 - Log[1 6/5])/x])/3])/((x^3 - x*Log[16/5] + (-3*x^2 + 3*Log[16/5])*Log[(x^2 - Log[ 16/5])/x])*Log[(x - 3*Log[(x^2 - Log[16/5])/x])/3]^2),x]
3.21.44.3.1 Defintions of rubi rules used
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 2.88 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\frac {x^{2}}{\ln \left (-\ln \left (-\frac {-x^{2}+\ln \left (\frac {16}{5}\right )}{x}\right )+\frac {x}{3}\right )}\) | \(28\) |
int((((6*x*ln(16/5)-6*x^3)*ln((-ln(16/5)+x^2)/x)-2*x^2*ln(16/5)+2*x^4)*ln( -ln((-ln(16/5)+x^2)/x)+1/3*x)+(x^2+3*x)*ln(16/5)-x^4+3*x^3)/((3*ln(16/5)-3 *x^2)*ln((-ln(16/5)+x^2)/x)-x*ln(16/5)+x^3)/ln(-ln((-ln(16/5)+x^2)/x)+1/3* x)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {3 x^3-x^4+\left (3 x+x^2\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (-6 x^3+6 x \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^3-x \log \left (\frac {16}{5}\right )+\left (-3 x^2+3 \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \, dx=\frac {x^{2}}{\log \left (\frac {1}{3} \, x - \log \left (\frac {x^{2} - \log \left (\frac {16}{5}\right )}{x}\right )\right )} \]
integrate((((6*x*log(16/5)-6*x^3)*log((-log(16/5)+x^2)/x)-2*x^2*log(16/5)+ 2*x^4)*log(-log((-log(16/5)+x^2)/x)+1/3*x)+(x^2+3*x)*log(16/5)-x^4+3*x^3)/ ((3*log(16/5)-3*x^2)*log((-log(16/5)+x^2)/x)-x*log(16/5)+x^3)/log(-log((-l og(16/5)+x^2)/x)+1/3*x)^2,x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {3 x^3-x^4+\left (3 x+x^2\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (-6 x^3+6 x \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^3-x \log \left (\frac {16}{5}\right )+\left (-3 x^2+3 \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \, dx=\frac {x^{2}}{\log {\left (\frac {x}{3} - \log {\left (\frac {x^{2} - \log {\left (\frac {16}{5} \right )}}{x} \right )} \right )}} \]
integrate((((6*x*ln(16/5)-6*x**3)*ln((-ln(16/5)+x**2)/x)-2*x**2*ln(16/5)+2 *x**4)*ln(-ln((-ln(16/5)+x**2)/x)+1/3*x)+(x**2+3*x)*ln(16/5)-x**4+3*x**3)/ ((3*ln(16/5)-3*x**2)*ln((-ln(16/5)+x**2)/x)-x*ln(16/5)+x**3)/ln(-ln((-ln(1 6/5)+x**2)/x)+1/3*x)**2,x)
Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {3 x^3-x^4+\left (3 x+x^2\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (-6 x^3+6 x \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^3-x \log \left (\frac {16}{5}\right )+\left (-3 x^2+3 \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \, dx=-\frac {x^{2}}{\log \left (3\right ) - \log \left (x - 3 \, \log \left (x^{2} + \log \left (5\right ) - 4 \, \log \left (2\right )\right ) + 3 \, \log \left (x\right )\right )} \]
integrate((((6*x*log(16/5)-6*x^3)*log((-log(16/5)+x^2)/x)-2*x^2*log(16/5)+ 2*x^4)*log(-log((-log(16/5)+x^2)/x)+1/3*x)+(x^2+3*x)*log(16/5)-x^4+3*x^3)/ ((3*log(16/5)-3*x^2)*log((-log(16/5)+x^2)/x)-x*log(16/5)+x^3)/log(-log((-l og(16/5)+x^2)/x)+1/3*x)^2,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 1559 vs. \(2 (23) = 46\).
Time = 1.07 (sec) , antiderivative size = 1559, normalized size of antiderivative = 57.74 \[ \int \frac {3 x^3-x^4+\left (3 x+x^2\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (-6 x^3+6 x \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^3-x \log \left (\frac {16}{5}\right )+\left (-3 x^2+3 \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \, dx=\text {Too large to display} \]
integrate((((6*x*log(16/5)-6*x^3)*log((-log(16/5)+x^2)/x)-2*x^2*log(16/5)+ 2*x^4)*log(-log((-log(16/5)+x^2)/x)+1/3*x)+(x^2+3*x)*log(16/5)-x^4+3*x^3)/ ((3*log(16/5)-3*x^2)*log((-log(16/5)+x^2)/x)-x*log(16/5)+x^3)/log(-log((-l og(16/5)+x^2)/x)+1/3*x)^2,x, algorithm=\
-(x^8 - 3*x^7*log(x^2 - log(16/5)) + 3*x^7*log(x) - 3*x^7 + x^6*log(5) - x ^6*log(16/5) - 4*x^6*log(2) + 9*x^6*log(x^2 - log(16/5)) - 3*x^5*log(5)*lo g(x^2 - log(16/5)) + 3*x^5*log(16/5)*log(x^2 - log(16/5)) + 12*x^5*log(2)* log(x^2 - log(16/5)) - 9*x^6*log(x) + 3*x^5*log(5)*log(x) - 3*x^5*log(16/5 )*log(x) - 12*x^5*log(2)*log(x) + 3*x^5*log(5) + 3*x^5*log(16/5) - x^4*log (5)*log(16/5) - 12*x^5*log(2) + 4*x^4*log(16/5)*log(2) - 9*x^4*log(5)*log( x^2 - log(16/5)) - 9*x^4*log(16/5)*log(x^2 - log(16/5)) + 3*x^3*log(5)*log (16/5)*log(x^2 - log(16/5)) + 36*x^4*log(2)*log(x^2 - log(16/5)) - 12*x^3* log(16/5)*log(2)*log(x^2 - log(16/5)) + 9*x^4*log(5)*log(x) + 9*x^4*log(16 /5)*log(x) - 3*x^3*log(5)*log(16/5)*log(x) - 36*x^4*log(2)*log(x) + 12*x^3 *log(16/5)*log(2)*log(x) - 3*x^3*log(5)*log(16/5) + 12*x^3*log(16/5)*log(2 ) + 9*x^2*log(5)*log(16/5)*log(x^2 - log(16/5)) - 36*x^2*log(16/5)*log(2)* log(x^2 - log(16/5)) - 9*x^2*log(5)*log(16/5)*log(x) + 36*x^2*log(16/5)*lo g(2)*log(x))/(x^6*log(3) - x^6*log(x - 3*log(x^2 - log(16/5)) + 3*log(x)) - 3*x^5*log(3)*log((x^2 + log(5) - 4*log(2))/x) + 3*x^5*log(x - 3*log(x^2 - log(16/5)) + 3*log(x))*log((x^2 + log(5) - 4*log(2))/x) - 3*x^5*log(3) + x^4*log(5)*log(3) - x^4*log(16/5)*log(3) - 4*x^4*log(3)*log(2) + 3*x^5*lo g(x - 3*log(x^2 - log(16/5)) + 3*log(x)) - x^4*log(5)*log(x - 3*log(x^2 - log(16/5)) + 3*log(x)) + x^4*log(16/5)*log(x - 3*log(x^2 - log(16/5)) + 3* log(x)) + 4*x^4*log(2)*log(x - 3*log(x^2 - log(16/5)) + 3*log(x)) + 9*x...
Timed out. \[ \int \frac {3 x^3-x^4+\left (3 x+x^2\right ) \log \left (\frac {16}{5}\right )+\left (2 x^4-2 x^2 \log \left (\frac {16}{5}\right )+\left (-6 x^3+6 x \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log \left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )}{\left (x^3-x \log \left (\frac {16}{5}\right )+\left (-3 x^2+3 \log \left (\frac {16}{5}\right )\right ) \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right ) \log ^2\left (\frac {1}{3} \left (x-3 \log \left (\frac {x^2-\log \left (\frac {16}{5}\right )}{x}\right )\right )\right )} \, dx=\int \frac {\ln \left (\frac {x}{3}-\ln \left (-\frac {\ln \left (\frac {16}{5}\right )-x^2}{x}\right )\right )\,\left (2\,x^4-2\,x^2\,\ln \left (\frac {16}{5}\right )+\ln \left (-\frac {\ln \left (\frac {16}{5}\right )-x^2}{x}\right )\,\left (6\,x\,\ln \left (\frac {16}{5}\right )-6\,x^3\right )\right )+3\,x^3-x^4+\ln \left (\frac {16}{5}\right )\,\left (x^2+3\,x\right )}{{\ln \left (\frac {x}{3}-\ln \left (-\frac {\ln \left (\frac {16}{5}\right )-x^2}{x}\right )\right )}^2\,\left (\ln \left (-\frac {\ln \left (\frac {16}{5}\right )-x^2}{x}\right )\,\left (3\,\ln \left (\frac {16}{5}\right )-3\,x^2\right )-x\,\ln \left (\frac {16}{5}\right )+x^3\right )} \,d x \]
int((log(x/3 - log(-(log(16/5) - x^2)/x))*(2*x^4 - 2*x^2*log(16/5) + log(- (log(16/5) - x^2)/x)*(6*x*log(16/5) - 6*x^3)) + 3*x^3 - x^4 + log(16/5)*(3 *x + x^2))/(log(x/3 - log(-(log(16/5) - x^2)/x))^2*(log(-(log(16/5) - x^2) /x)*(3*log(16/5) - 3*x^2) - x*log(16/5) + x^3)),x)