Integrand size = 121, antiderivative size = 23 \[ \int \frac {5 x^3+2 x^4+e^5 \left (-2 x^2-x^3\right )+\left (15 x^2-3 e^5 x^2+6 x^3\right ) \log \left (\frac {4}{x}\right )+\left (15 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )+(5+2 x) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )} \, dx=x \left (5+x-\frac {e^5 x^2}{\left (x+\log \left (\frac {4}{x}\right )\right )^2}\right ) \]
Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {5 x^3+2 x^4+e^5 \left (-2 x^2-x^3\right )+\left (15 x^2-3 e^5 x^2+6 x^3\right ) \log \left (\frac {4}{x}\right )+\left (15 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )+(5+2 x) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )} \, dx=5 x+x^2-\frac {e^5 x^3}{\left (x+\log \left (\frac {4}{x}\right )\right )^2} \]
Integrate[(5*x^3 + 2*x^4 + E^5*(-2*x^2 - x^3) + (15*x^2 - 3*E^5*x^2 + 6*x^ 3)*Log[4/x] + (15*x + 6*x^2)*Log[4/x]^2 + (5 + 2*x)*Log[4/x]^3)/(x^3 + 3*x ^2*Log[4/x] + 3*x*Log[4/x]^2 + Log[4/x]^3),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.
\(\displaystyle \int \frac {2 x^4+5 x^3+\left (6 x^2+15 x\right ) \log ^2\left (\frac {4}{x}\right )+e^5 \left (-x^3-2 x^2\right )+\left (6 x^3-3 e^5 x^2+15 x^2\right ) \log \left (\frac {4}{x}\right )+(2 x+5) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 x^4+5 x^3+\left (6 x^2+15 x\right ) \log ^2\left (\frac {4}{x}\right )+e^5 \left (-x^3-2 x^2\right )+\left (6 x^3-3 e^5 x^2+15 x^2\right ) \log \left (\frac {4}{x}\right )+(2 x+5) \log ^3\left (\frac {4}{x}\right )}{\left (x+\log \left (\frac {4}{x}\right )\right )^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {3 e^5 x^2}{\left (x+\log \left (\frac {4}{x}\right )\right )^2}+\frac {2 e^5 (x-1) x^2}{\left (x+\log \left (\frac {4}{x}\right )\right )^3}+2 x+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 e^5 \int \frac {x^3}{\left (x+\log \left (\frac {4}{x}\right )\right )^3}dx-2 e^5 \int \frac {x^2}{\left (x+\log \left (\frac {4}{x}\right )\right )^3}dx-3 e^5 \int \frac {x^2}{\left (x+\log \left (\frac {4}{x}\right )\right )^2}dx+x^2+5 x\) |
Int[(5*x^3 + 2*x^4 + E^5*(-2*x^2 - x^3) + (15*x^2 - 3*E^5*x^2 + 6*x^3)*Log [4/x] + (15*x + 6*x^2)*Log[4/x]^2 + (5 + 2*x)*Log[4/x]^3)/(x^3 + 3*x^2*Log [4/x] + 3*x*Log[4/x]^2 + Log[4/x]^3),x]
3.28.67.3.1 Defintions of rubi rules used
Time = 0.89 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
method | result | size |
risch | \(x^{2}+5 x -\frac {{\mathrm e}^{5} x^{3}}{\left (\ln \left (\frac {4}{x}\right )+x \right )^{2}}\) | \(25\) |
norman | \(\frac {x^{4}+x^{2} \ln \left (\frac {4}{x}\right )^{2}+\left (5-{\mathrm e}^{5}\right ) x^{3}+5 x \ln \left (\frac {4}{x}\right )^{2}+10 x^{2} \ln \left (\frac {4}{x}\right )+2 x^{3} \ln \left (\frac {4}{x}\right )}{\left (\ln \left (\frac {4}{x}\right )+x \right )^{2}}\) | \(71\) |
derivativedivides | \(\frac {\left (64+\frac {320-64 \,{\mathrm e}^{5}}{x}+\frac {128 \ln \left (\frac {4}{x}\right )}{x}+\frac {640 \ln \left (\frac {4}{x}\right )}{x^{2}}+\frac {64 \ln \left (\frac {4}{x}\right )^{2}}{x^{2}}+\frac {320 \ln \left (\frac {4}{x}\right )^{2}}{x^{3}}\right ) x^{2}}{4 \left (\frac {4 \ln \left (\frac {4}{x}\right )}{x}+4\right )^{2}}\) | \(82\) |
default | \(\frac {\left (64+\frac {320-64 \,{\mathrm e}^{5}}{x}+\frac {128 \ln \left (\frac {4}{x}\right )}{x}+\frac {640 \ln \left (\frac {4}{x}\right )}{x^{2}}+\frac {64 \ln \left (\frac {4}{x}\right )^{2}}{x^{2}}+\frac {320 \ln \left (\frac {4}{x}\right )^{2}}{x^{3}}\right ) x^{2}}{4 \left (\frac {4 \ln \left (\frac {4}{x}\right )}{x}+4\right )^{2}}\) | \(82\) |
parallelrisch | \(-\frac {x^{3} {\mathrm e}^{5}-x^{4}-2 x^{3} \ln \left (\frac {4}{x}\right )-x^{2} \ln \left (\frac {4}{x}\right )^{2}-5 x^{3}-10 x^{2} \ln \left (\frac {4}{x}\right )-5 x \ln \left (\frac {4}{x}\right )^{2}}{x^{2}+2 x \ln \left (\frac {4}{x}\right )+\ln \left (\frac {4}{x}\right )^{2}}\) | \(89\) |
int(((5+2*x)*ln(4/x)^3+(6*x^2+15*x)*ln(4/x)^2+(-3*x^2*exp(5)+6*x^3+15*x^2) *ln(4/x)+(-x^3-2*x^2)*exp(5)+2*x^4+5*x^3)/(ln(4/x)^3+3*x*ln(4/x)^2+3*x^2*l n(4/x)+x^3),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (22) = 44\).
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.17 \[ \int \frac {5 x^3+2 x^4+e^5 \left (-2 x^2-x^3\right )+\left (15 x^2-3 e^5 x^2+6 x^3\right ) \log \left (\frac {4}{x}\right )+\left (15 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )+(5+2 x) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )} \, dx=\frac {x^{4} - x^{3} e^{5} + 5 \, x^{3} + {\left (x^{2} + 5 \, x\right )} \log \left (\frac {4}{x}\right )^{2} + 2 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \left (\frac {4}{x}\right )}{x^{2} + 2 \, x \log \left (\frac {4}{x}\right ) + \log \left (\frac {4}{x}\right )^{2}} \]
integrate(((5+2*x)*log(4/x)^3+(6*x^2+15*x)*log(4/x)^2+(-3*x^2*exp(5)+6*x^3 +15*x^2)*log(4/x)+(-x^3-2*x^2)*exp(5)+2*x^4+5*x^3)/(log(4/x)^3+3*x*log(4/x )^2+3*x^2*log(4/x)+x^3),x, algorithm=\
(x^4 - x^3*e^5 + 5*x^3 + (x^2 + 5*x)*log(4/x)^2 + 2*(x^3 + 5*x^2)*log(4/x) )/(x^2 + 2*x*log(4/x) + log(4/x)^2)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {5 x^3+2 x^4+e^5 \left (-2 x^2-x^3\right )+\left (15 x^2-3 e^5 x^2+6 x^3\right ) \log \left (\frac {4}{x}\right )+\left (15 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )+(5+2 x) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )} \, dx=- \frac {x^{3} e^{5}}{x^{2} + 2 x \log {\left (\frac {4}{x} \right )} + \log {\left (\frac {4}{x} \right )}^{2}} + x^{2} + 5 x \]
integrate(((5+2*x)*ln(4/x)**3+(6*x**2+15*x)*ln(4/x)**2+(-3*x**2*exp(5)+6*x **3+15*x**2)*ln(4/x)+(-x**3-2*x**2)*exp(5)+2*x**4+5*x**3)/(ln(4/x)**3+3*x* ln(4/x)**2+3*x**2*ln(4/x)+x**3),x)
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (22) = 44\).
Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.57 \[ \int \frac {5 x^3+2 x^4+e^5 \left (-2 x^2-x^3\right )+\left (15 x^2-3 e^5 x^2+6 x^3\right ) \log \left (\frac {4}{x}\right )+\left (15 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )+(5+2 x) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )} \, dx=\frac {x^{4} - x^{3} {\left (e^{5} - 4 \, \log \left (2\right ) - 5\right )} + 4 \, {\left (\log \left (2\right )^{2} + 5 \, \log \left (2\right )\right )} x^{2} + 20 \, x \log \left (2\right )^{2} + {\left (x^{2} + 5 \, x\right )} \log \left (x\right )^{2} - 2 \, {\left (x^{3} + x^{2} {\left (2 \, \log \left (2\right ) + 5\right )} + 10 \, x \log \left (2\right )\right )} \log \left (x\right )}{x^{2} + 4 \, x \log \left (2\right ) + 4 \, \log \left (2\right )^{2} - 2 \, {\left (x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) + \log \left (x\right )^{2}} \]
integrate(((5+2*x)*log(4/x)^3+(6*x^2+15*x)*log(4/x)^2+(-3*x^2*exp(5)+6*x^3 +15*x^2)*log(4/x)+(-x^3-2*x^2)*exp(5)+2*x^4+5*x^3)/(log(4/x)^3+3*x*log(4/x )^2+3*x^2*log(4/x)+x^3),x, algorithm=\
(x^4 - x^3*(e^5 - 4*log(2) - 5) + 4*(log(2)^2 + 5*log(2))*x^2 + 20*x*log(2 )^2 + (x^2 + 5*x)*log(x)^2 - 2*(x^3 + x^2*(2*log(2) + 5) + 10*x*log(2))*lo g(x))/(x^2 + 4*x*log(2) + 4*log(2)^2 - 2*(x + 2*log(2))*log(x) + log(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 4.00 \[ \int \frac {5 x^3+2 x^4+e^5 \left (-2 x^2-x^3\right )+\left (15 x^2-3 e^5 x^2+6 x^3\right ) \log \left (\frac {4}{x}\right )+\left (15 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )+(5+2 x) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )} \, dx=-\frac {\frac {e^{5}}{x} - \frac {2 \, \log \left (\frac {4}{x}\right )}{x} - \frac {\log \left (\frac {4}{x}\right )^{2}}{x^{2}} - \frac {5}{x} - \frac {10 \, \log \left (\frac {4}{x}\right )}{x^{2}} - \frac {5 \, \log \left (\frac {4}{x}\right )^{2}}{x^{3}} - 1}{\frac {1}{x^{2}} + \frac {2 \, \log \left (\frac {4}{x}\right )}{x^{3}} + \frac {\log \left (\frac {4}{x}\right )^{2}}{x^{4}}} \]
integrate(((5+2*x)*log(4/x)^3+(6*x^2+15*x)*log(4/x)^2+(-3*x^2*exp(5)+6*x^3 +15*x^2)*log(4/x)+(-x^3-2*x^2)*exp(5)+2*x^4+5*x^3)/(log(4/x)^3+3*x*log(4/x )^2+3*x^2*log(4/x)+x^3),x, algorithm=\
-(e^5/x - 2*log(4/x)/x - log(4/x)^2/x^2 - 5/x - 10*log(4/x)/x^2 - 5*log(4/ x)^2/x^3 - 1)/(1/x^2 + 2*log(4/x)/x^3 + log(4/x)^2/x^4)
Time = 14.53 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.17 \[ \int \frac {5 x^3+2 x^4+e^5 \left (-2 x^2-x^3\right )+\left (15 x^2-3 e^5 x^2+6 x^3\right ) \log \left (\frac {4}{x}\right )+\left (15 x+6 x^2\right ) \log ^2\left (\frac {4}{x}\right )+(5+2 x) \log ^3\left (\frac {4}{x}\right )}{x^3+3 x^2 \log \left (\frac {4}{x}\right )+3 x \log ^2\left (\frac {4}{x}\right )+\log ^3\left (\frac {4}{x}\right )} \, dx=x\,\left (x+5\right )-\frac {x^3\,\left (x+5\right )+\ln \left (\frac {4}{x}\right )\,\left (2\,x^2\,\left (x+5\right )-x\,\left (2\,x^2+10\,x\right )\right )-x\,\left (5\,x^2-x^2\,{\mathrm {e}}^5+x^3\right )}{{\left (x+\ln \left (\frac {4}{x}\right )\right )}^2} \]
int((log(4/x)*(15*x^2 - 3*x^2*exp(5) + 6*x^3) - exp(5)*(2*x^2 + x^3) + log (4/x)^3*(2*x + 5) + log(4/x)^2*(15*x + 6*x^2) + 5*x^3 + 2*x^4)/(log(4/x)^3 + x^3 + 3*x*log(4/x)^2 + 3*x^2*log(4/x)),x)