Integrand size = 140, antiderivative size = 33 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-x+\left (1+e^4+\log (3)\right ) \log \left (\frac {5 \log \left (4-\frac {3}{5+x}\right )}{x (2+x)}\right ) \] Output:
(1+exp(4)+ln(3))*ln(5/x/(2+x)*ln(4-3/(5+x)))-x
Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-x-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (x)-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (2+x)+\left (1+e^4+\log (3)\right ) \log \left (\log \left (\frac {17+4 x}{5+x}\right )\right ) \] Input:
Integrate[(6*x + 3*x^2 + E^4*(6*x + 3*x^2) + (6*x + 3*x^2)*Log[3] + (-170 - 414*x - 241*x^2 - 53*x^3 - 4*x^4 + E^4*(-170 - 244*x - 82*x^2 - 8*x^3) + (-170 - 244*x - 82*x^2 - 8*x^3)*Log[3])*Log[(17 + 4*x)/(5 + x)])/((170*x + 159*x^2 + 45*x^3 + 4*x^4)*Log[(17 + 4*x)/(5 + x)]),x]
Output:
-x - ((2 + 2*E^4 + Log[9])*Log[x])/2 - ((2 + 2*E^4 + Log[9])*Log[2 + x])/2 + (1 + E^4 + Log[3])*Log[Log[(17 + 4*x)/(5 + 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.
| \(\displaystyle \int \frac {3 x^2+e^4 \left (3 x^2+6 x\right )+\left (3 x^2+6 x\right ) \log (3)+\left (-4 x^4-53 x^3-241 x^2+e^4 \left (-8 x^3-82 x^2-244 x-170\right )+\left (-8 x^3-82 x^2-244 x-170\right ) \log (3)-414 x-170\right ) \log \left (\frac {4 x+17}{x+5}\right )+6 x}{\left (4 x^4+45 x^3+159 x^2+170 x\right ) \log \left (\frac {4 x+17}{x+5}\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {3 x^2+\left (3 x^2+6 x\right ) \left (e^4+\log (3)\right )+\left (-4 x^4-53 x^3-241 x^2+e^4 \left (-8 x^3-82 x^2-244 x-170\right )+\left (-8 x^3-82 x^2-244 x-170\right ) \log (3)-414 x-170\right ) \log \left (\frac {4 x+17}{x+5}\right )+6 x}{\left (4 x^4+45 x^3+159 x^2+170 x\right ) \log \left (\frac {4 x+17}{x+5}\right )}dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {3 x^2+\left (3 x^2+6 x\right ) \left (e^4+\log (3)\right )+\left (-4 x^4-53 x^3-241 x^2+e^4 \left (-8 x^3-82 x^2-244 x-170\right )+\left (-8 x^3-82 x^2-244 x-170\right ) \log (3)-414 x-170\right ) \log \left (\frac {4 x+17}{x+5}\right )+6 x}{x \left (4 x^3+45 x^2+159 x+170\right ) \log \left (\frac {4 x+17}{x+5}\right )}dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (\frac {3 x^2+\left (3 x^2+6 x\right ) \left (e^4+\log (3)\right )+\left (-4 x^4-53 x^3-241 x^2+e^4 \left (-8 x^3-82 x^2-244 x-170\right )+\left (-8 x^3-82 x^2-244 x-170\right ) \log (3)-414 x-170\right ) \log \left (\frac {4 x+17}{x+5}\right )+6 x}{27 x (x+2) \log \left (\frac {4 x+17}{x+5}\right )}+\frac {3 x^2+\left (3 x^2+6 x\right ) \left (e^4+\log (3)\right )+\left (-4 x^4-53 x^3-241 x^2+e^4 \left (-8 x^3-82 x^2-244 x-170\right )+\left (-8 x^3-82 x^2-244 x-170\right ) \log (3)-414 x-170\right ) \log \left (\frac {4 x+17}{x+5}\right )+6 x}{9 x (x+5) \log \left (\frac {4 x+17}{x+5}\right )}-\frac {16 \left (3 x^2+\left (3 x^2+6 x\right ) \left (e^4+\log (3)\right )+\left (-4 x^4-53 x^3-241 x^2+e^4 \left (-8 x^3-82 x^2-244 x-170\right )+\left (-8 x^3-82 x^2-244 x-170\right ) \log (3)-414 x-170\right ) \log \left (\frac {4 x+17}{x+5}\right )+6 x\right )}{27 x (4 x+17) \log \left (\frac {4 x+17}{x+5}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{9} \left (1+e^4+\log (3)\right ) \int \frac {1}{\log \left (\frac {4 x+17}{x+5}\right )}dx+\frac {1}{3} \left (1+e^4+\log (3)\right ) \int \frac {x+2}{(x+5) \log \left (\frac {4 x+17}{x+5}\right )}dx-\frac {16}{9} \left (1+e^4+\log (3)\right ) \int \frac {x+2}{(4 x+17) \log \left (\frac {4 x+17}{x+5}\right )}dx-\frac {1}{54} x^2 \left (45+8 e^4+4 \log (9)\right )-\frac {1}{18} x^2 \left (33+8 e^4+4 \log (9)\right )+\frac {8}{27} x^2 \left (9+2 e^4+\log (9)\right )+\frac {32}{27} x \left (11+6 e^4+\log (729)\right )-\frac {1}{27} x \left (151+66 e^4+33 \log (9)\right )-\frac {1}{9} x \left (76+42 e^4+21 \log (9)\right )-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (x)-\frac {1}{2} \left (2+2 e^4+\log (9)\right ) \log (x+2)\) |
Input:
Int[(6*x + 3*x^2 + E^4*(6*x + 3*x^2) + (6*x + 3*x^2)*Log[3] + (-170 - 414* x - 241*x^2 - 53*x^3 - 4*x^4 + E^4*(-170 - 244*x - 82*x^2 - 8*x^3) + (-170 - 244*x - 82*x^2 - 8*x^3)*Log[3])*Log[(17 + 4*x)/(5 + x)])/((170*x + 159* x^2 + 45*x^3 + 4*x^4)*Log[(17 + 4*x)/(5 + x)]),x]
Output:
$Aborted
Time = 1.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52
| method | result | size |
| parts | \(\frac {\left (3+3 \,{\mathrm e}^{4}+3 \ln \left (3\right )\right ) \ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )}{3}-x -\left (1+{\mathrm e}^{4}+\ln \left (3\right )\right ) \ln \left (x \right )-\left (1+{\mathrm e}^{4}+\ln \left (3\right )\right ) \ln \left (2+x \right )\) | \(50\) |
| norman | \(-x +\left (1+{\mathrm e}^{4}+\ln \left (3\right )\right ) \ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right )+\left (-\ln \left (3\right )-{\mathrm e}^{4}-1\right ) \ln \left (x \right )+\left (-\ln \left (3\right )-{\mathrm e}^{4}-1\right ) \ln \left (2+x \right )\) | \(53\) |
| risch | \(-\ln \left (x^{2}+2 x \right ) \ln \left (3\right )-\ln \left (x^{2}+2 x \right ) {\mathrm e}^{4}+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right ) \ln \left (3\right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right ) {\mathrm e}^{4}-\ln \left (x^{2}+2 x \right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right )-x\) | \(84\) |
| parallelrisch | \(-\ln \left (3\right ) \ln \left (x \right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right ) \ln \left (3\right )-\ln \left (3\right ) \ln \left (2+x \right )-{\mathrm e}^{4} \ln \left (x \right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right ) {\mathrm e}^{4}-{\mathrm e}^{4} \ln \left (2+x \right )+\frac {37}{2}-\ln \left (x \right )+\ln \left (\ln \left (\frac {4 x +17}{5+x}\right )\right )-\ln \left (2+x \right )-x\) | \(89\) |
| derivativedivides | \(\ln \left (3\right ) \left (\ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )-\ln \left (1-\frac {3}{5+x}\right )-\ln \left (3-\frac {15}{5+x}\right )+2 \ln \left (-\frac {3}{5+x}\right )\right )+3 \left (\frac {{\mathrm e}^{4}}{3}+\frac {1}{3}\right ) \ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )+\left (-{\mathrm e}^{4}-1\right ) \ln \left (1-\frac {3}{5+x}\right )+\left (-{\mathrm e}^{4}-1\right ) \ln \left (3-\frac {15}{5+x}\right )+\left (2 \,{\mathrm e}^{4}+2\right ) \ln \left (-\frac {3}{5+x}\right )-5-x\) | \(123\) |
| default | \(\ln \left (3\right ) \left (\ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )-\ln \left (1-\frac {3}{5+x}\right )-\ln \left (3-\frac {15}{5+x}\right )+2 \ln \left (-\frac {3}{5+x}\right )\right )+3 \left (\frac {{\mathrm e}^{4}}{3}+\frac {1}{3}\right ) \ln \left (\ln \left (4-\frac {3}{5+x}\right )\right )+\left (-{\mathrm e}^{4}-1\right ) \ln \left (1-\frac {3}{5+x}\right )+\left (-{\mathrm e}^{4}-1\right ) \ln \left (3-\frac {15}{5+x}\right )+\left (2 \,{\mathrm e}^{4}+2\right ) \ln \left (-\frac {3}{5+x}\right )-5-x\) | \(123\) |
Input:
int((((-8*x^3-82*x^2-244*x-170)*ln(3)+(-8*x^3-82*x^2-244*x-170)*exp(4)-4*x ^4-53*x^3-241*x^2-414*x-170)*ln((4*x+17)/(5+x))+(3*x^2+6*x)*ln(3)+(3*x^2+6 *x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/ln((4*x+17)/(5+x)),x,me thod=_RETURNVERBOSE)
Output:
1/3*(3+3*exp(4)+3*ln(3))*ln(ln(4-3/(5+x)))-x-(1+exp(4)+ln(3))*ln(x)-(1+exp (4)+ln(3))*ln(2+x)
Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.21 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-{\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (x^{2} + 2 \, x\right ) + {\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (\log \left (\frac {4 \, x + 17}{x + 5}\right )\right ) - x \] Input:
integrate((((-8*x^3-82*x^2-244*x-170)*log(3)+(-8*x^3-82*x^2-244*x-170)*exp (4)-4*x^4-53*x^3-241*x^2-414*x-170)*log((4*x+17)/(5+x))+(3*x^2+6*x)*log(3) +(3*x^2+6*x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/log((4*x+17)/( 5+x)),x, algorithm="fricas")
Output:
-(e^4 + log(3) + 1)*log(x^2 + 2*x) + (e^4 + log(3) + 1)*log(log((4*x + 17) /(x + 5))) - x
Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=- x - \left (1 + \log {\left (3 \right )} + e^{4}\right ) \log {\left (x^{2} + 2 x \right )} + \left (1 + \log {\left (3 \right )} + e^{4}\right ) \log {\left (\log {\left (\frac {4 x + 17}{x + 5} \right )} \right )} \] Input:
integrate((((-8*x**3-82*x**2-244*x-170)*ln(3)+(-8*x**3-82*x**2-244*x-170)* exp(4)-4*x**4-53*x**3-241*x**2-414*x-170)*ln((4*x+17)/(5+x))+(3*x**2+6*x)* ln(3)+(3*x**2+6*x)*exp(4)+3*x**2+6*x)/(4*x**4+45*x**3+159*x**2+170*x)/ln(( 4*x+17)/(5+x)),x)
Output:
-x - (1 + log(3) + exp(4))*log(x**2 + 2*x) + (1 + log(3) + exp(4))*log(log ((4*x + 17)/(x + 5)))
Time = 0.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=-{\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (x + 2\right ) - {\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (x\right ) + {\left (e^{4} + \log \left (3\right ) + 1\right )} \log \left (\log \left (4 \, x + 17\right ) - \log \left (x + 5\right )\right ) - x \] Input:
integrate((((-8*x^3-82*x^2-244*x-170)*log(3)+(-8*x^3-82*x^2-244*x-170)*exp (4)-4*x^4-53*x^3-241*x^2-414*x-170)*log((4*x+17)/(5+x))+(3*x^2+6*x)*log(3) +(3*x^2+6*x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/log((4*x+17)/( 5+x)),x, algorithm="maxima")
Output:
-(e^4 + log(3) + 1)*log(x + 2) - (e^4 + log(3) + 1)*log(x) + (e^4 + log(3) + 1)*log(log(4*x + 17) - log(x + 5)) - x
Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (32) = 64\).
Time = 0.17 (sec) , antiderivative size = 502, normalized size of antiderivative = 15.21 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx =\text {Too large to display} \] Input:
integrate((((-8*x^3-82*x^2-244*x-170)*log(3)+(-8*x^3-82*x^2-244*x-170)*exp (4)-4*x^4-53*x^3-241*x^2-414*x-170)*log((4*x+17)/(5+x))+(3*x^2+6*x)*log(3) +(3*x^2+6*x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/log((4*x+17)/( 5+x)),x, algorithm="giac")
Output:
-((4*x + 17)*e^4*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x + 5) + 51 )/(x + 5) - 4*e^4*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x + 5) + 5 1) + (4*x + 17)*log(3)*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x + 5 ) + 51)/(x + 5) - 4*log(3)*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17)/(x + 5) + 51) - 2*(4*x + 17)*e^4*log((4*x + 17)/(x + 5) - 4)/(x + 5) + 8*e^4 *log((4*x + 17)/(x + 5) - 4) - 2*(4*x + 17)*log(3)*log((4*x + 17)/(x + 5) - 4)/(x + 5) + 8*log(3)*log((4*x + 17)/(x + 5) - 4) - (4*x + 17)*e^4*log(l og((4*x + 17)/(x + 5)))/(x + 5) + 4*e^4*log(log((4*x + 17)/(x + 5))) - (4* x + 17)*log(3)*log(log((4*x + 17)/(x + 5)))/(x + 5) + 4*log(3)*log(log((4* x + 17)/(x + 5))) + (4*x + 17)*log(5*(4*x + 17)^2/(x + 5)^2 - 32*(4*x + 17 )/(x + 5) + 51)/(x + 5) - 2*(4*x + 17)*log((4*x + 17)/(x + 5) - 4)/(x + 5) - (4*x + 17)*log(log((4*x + 17)/(x + 5)))/(x + 5) - 4*log(5*(4*x + 17)^2/ (x + 5)^2 - 32*(4*x + 17)/(x + 5) + 51) + 8*log((4*x + 17)/(x + 5) - 4) + 4*log(log((4*x + 17)/(x + 5))) - 3)/((4*x + 17)/(x + 5) - 4)
Time = 0.40 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=\ln \left (\ln \left (\frac {4\,x+17}{x+5}\right )\right )\,\left ({\mathrm {e}}^4+\ln \left (3\right )+1\right )-x-\ln \left (x\,\left (x+2\right )\right )\,\left ({\mathrm {e}}^4+\ln \left (3\right )+1\right ) \] Input:
int((6*x + exp(4)*(6*x + 3*x^2) + log(3)*(6*x + 3*x^2) - log((4*x + 17)/(x + 5))*(414*x + exp(4)*(244*x + 82*x^2 + 8*x^3 + 170) + log(3)*(244*x + 82 *x^2 + 8*x^3 + 170) + 241*x^2 + 53*x^3 + 4*x^4 + 170) + 3*x^2)/(log((4*x + 17)/(x + 5))*(170*x + 159*x^2 + 45*x^3 + 4*x^4)),x)
Output:
log(log((4*x + 17)/(x + 5)))*(exp(4) + log(3) + 1) - x - log(x*(x + 2))*(e xp(4) + log(3) + 1)
Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.73 \[ \int \frac {6 x+3 x^2+e^4 \left (6 x+3 x^2\right )+\left (6 x+3 x^2\right ) \log (3)+\left (-170-414 x-241 x^2-53 x^3-4 x^4+e^4 \left (-170-244 x-82 x^2-8 x^3\right )+\left (-170-244 x-82 x^2-8 x^3\right ) \log (3)\right ) \log \left (\frac {17+4 x}{5+x}\right )}{\left (170 x+159 x^2+45 x^3+4 x^4\right ) \log \left (\frac {17+4 x}{5+x}\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\frac {4 x +17}{x +5}\right )\right ) \mathrm {log}\left (3\right )+\mathrm {log}\left (\mathrm {log}\left (\frac {4 x +17}{x +5}\right )\right ) e^{4}+\mathrm {log}\left (\mathrm {log}\left (\frac {4 x +17}{x +5}\right )\right )-\mathrm {log}\left (x +2\right ) \mathrm {log}\left (3\right )-\mathrm {log}\left (x +2\right ) e^{4}-\mathrm {log}\left (x +2\right )-\mathrm {log}\left (x \right ) \mathrm {log}\left (3\right )-\mathrm {log}\left (x \right ) e^{4}-\mathrm {log}\left (x \right )-x \] Input:
int((((-8*x^3-82*x^2-244*x-170)*log(3)+(-8*x^3-82*x^2-244*x-170)*exp(4)-4* x^4-53*x^3-241*x^2-414*x-170)*log((4*x+17)/(5+x))+(3*x^2+6*x)*log(3)+(3*x^ 2+6*x)*exp(4)+3*x^2+6*x)/(4*x^4+45*x^3+159*x^2+170*x)/log((4*x+17)/(5+x)), x)
Output:
log(log((4*x + 17)/(x + 5)))*log(3) + log(log((4*x + 17)/(x + 5)))*e**4 + log(log((4*x + 17)/(x + 5))) - log(x + 2)*log(3) - log(x + 2)*e**4 - log(x + 2) - log(x)*log(3) - log(x)*e**4 - log(x) - x