Integrand size = 52, antiderivative size = 29 \[ \int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+\left (-60+4 x+8 x^2\right ) \log \left ((3+x) \log ^2(5)\right )}{15+5 x} \, dx=\left (-2+e^2+x-\frac {x^2}{5}\right ) \left (x-4 \left (2+\log \left ((3+x) \log ^2(5)\right )\right )\right ) \] Output:
(x-4*ln((3+x)*ln(5)^2)-8)*(x-1/5*x^2-2+exp(2))
Leaf count is larger than twice the leaf count of optimal. \(86\) vs. \(2(29)=58\).
Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.97 \[ \int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+\left (-60+4 x+8 x^2\right ) \log \left ((3+x) \log ^2(5)\right )}{15+5 x} \, dx=\frac {1}{5} \left (-50 x+5 e^2 x+13 x^2-x^3+100 \log (3+x)-20 e^2 \log (3+x)+4 x^2 \log \left ((3+x) \log ^2(5)\right )-60 \log \left (3 \log ^2(5)+x \log ^2(5)\right )-20 x \log \left (3 \log ^2(5)+x \log ^2(5)\right )\right ) \] Input:
Integrate[(-110 + 8*x + 21*x^2 - 3*x^3 + E^2*(-5 + 5*x) + (-60 + 4*x + 8*x ^2)*Log[(3 + x)*Log[5]^2])/(15 + 5*x),x]
Output:
(-50*x + 5*E^2*x + 13*x^2 - x^3 + 100*Log[3 + x] - 20*E^2*Log[3 + x] + 4*x ^2*Log[(3 + x)*Log[5]^2] - 60*Log[3*Log[5]^2 + x*Log[5]^2] - 20*x*Log[3*Lo g[5]^2 + x*Log[5]^2])/5
Leaf count is larger than twice the leaf count of optimal. \(89\) vs. \(2(29)=58\).
Time = 0.46 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.07, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {7293, 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.
\(\displaystyle \int \frac {-3 x^3+21 x^2+\left (8 x^2+4 x-60\right ) \log \left ((x+3) \log ^2(5)\right )+8 x+e^2 (5 x-5)-110}{5 x+15} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-3 x^3+21 x^2+\left (8+5 e^2\right ) x-5 \left (22+e^2\right )}{5 (x+3)}+\frac {4}{5} (2 x-5) \log \left (x \log ^2(5)+3 \log ^2(5)\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {x^3}{5}+3 x^2-\frac {1}{5} \left (82-5 e^2\right ) x+\frac {22 x}{5}-\frac {1}{10} (5-2 x)^2+\frac {1}{5} (5-2 x)^2 \log \left (x \log ^2(5)+3 \log ^2(5)\right )+\frac {4}{5} \left (34-5 e^2\right ) \log (x+3)-\frac {121}{5} \log (x+3)\) |
Input:
Int[(-110 + 8*x + 21*x^2 - 3*x^3 + E^2*(-5 + 5*x) + (-60 + 4*x + 8*x^2)*Lo g[(3 + x)*Log[5]^2])/(15 + 5*x),x]
Output:
-1/10*(5 - 2*x)^2 + (22*x)/5 - ((82 - 5*E^2)*x)/5 + 3*x^2 - x^3/5 - (121*L og[3 + x])/5 + (4*(34 - 5*E^2)*Log[3 + x])/5 + ((5 - 2*x)^2*Log[3*Log[5]^2 + x*Log[5]^2])/5
Leaf count of result is larger than twice the leaf count of optimal. \(51\) vs. \(2(25)=50\).
Time = 0.56 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79
method | result | size |
risch | \(\left (\frac {4}{5} x^{2}-4 x \right ) \ln \left (\left (3+x \right ) \ln \left (5\right )^{2}\right )-\frac {x^{3}}{5}+{\mathrm e}^{2} x +\frac {13 x^{2}}{5}-10 x -4 \ln \left (3+x \right ) {\mathrm e}^{2}+8 \ln \left (3+x \right )\) | \(52\) |
norman | \(\left (-10+{\mathrm e}^{2}\right ) x +\left (8-4 \,{\mathrm e}^{2}\right ) \ln \left (\left (3+x \right ) \ln \left (5\right )^{2}\right )+\frac {13 x^{2}}{5}-\frac {x^{3}}{5}-4 \ln \left (\left (3+x \right ) \ln \left (5\right )^{2}\right ) x +\frac {4 \ln \left (\left (3+x \right ) \ln \left (5\right )^{2}\right ) x^{2}}{5}\) | \(60\) |
parallelrisch | \(-\frac {x^{3}}{5}+\frac {4 \ln \left (\left (3+x \right ) \ln \left (5\right )^{2}\right ) x^{2}}{5}+{\mathrm e}^{2} x -4 \ln \left (\left (3+x \right ) \ln \left (5\right )^{2}\right ) {\mathrm e}^{2}+\frac {183}{5}+\frac {13 x^{2}}{5}-4 \ln \left (\left (3+x \right ) \ln \left (5\right )^{2}\right ) x -6 \,{\mathrm e}^{2}-10 x +8 \ln \left (\left (3+x \right ) \ln \left (5\right )^{2}\right )\) | \(74\) |
parts | \(-4 \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right ) x -\frac {96 \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )}{5}-10 x +\frac {114}{5}+\frac {4 \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right ) x^{2}}{5}+\frac {13 x^{2}}{5}-\frac {x^{3}}{5}+{\mathrm e}^{2} x +\frac {\left (136-20 \,{\mathrm e}^{2}\right ) \ln \left (3+x \right )}{5}\) | \(84\) |
derivativedivides | \(\frac {-20 \ln \left (5\right )^{2} {\mathrm e}^{2} \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )+5 \,{\mathrm e}^{2} \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )-44 \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right ) \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )-155 x \ln \left (5\right )^{2}-465 \ln \left (5\right )^{2}+136 \ln \left (5\right )^{2} \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )+\frac {4 \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )^{2} \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )-2 \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )^{2}}{\ln \left (5\right )^{2}}+\frac {24 \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )^{2}}{\ln \left (5\right )^{2}}-\frac {\left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )^{3}}{\ln \left (5\right )^{4}}}{5 \ln \left (5\right )^{2}}\) | \(206\) |
default | \(\frac {-20 \ln \left (5\right )^{2} {\mathrm e}^{2} \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )+5 \,{\mathrm e}^{2} \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )-44 \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right ) \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )-155 x \ln \left (5\right )^{2}-465 \ln \left (5\right )^{2}+136 \ln \left (5\right )^{2} \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )+\frac {4 \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )^{2} \ln \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )-2 \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )^{2}}{\ln \left (5\right )^{2}}+\frac {24 \left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )^{2}}{\ln \left (5\right )^{2}}-\frac {\left (x \ln \left (5\right )^{2}+3 \ln \left (5\right )^{2}\right )^{3}}{\ln \left (5\right )^{4}}}{5 \ln \left (5\right )^{2}}\) | \(206\) |
Input:
int(((8*x^2+4*x-60)*ln((3+x)*ln(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+8*x-110) /(5*x+15),x,method=_RETURNVERBOSE)
Output:
(4/5*x^2-4*x)*ln((3+x)*ln(5)^2)-1/5*x^3+exp(2)*x+13/5*x^2-10*x-4*ln(3+x)*e xp(2)+8*ln(3+x)
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+\left (-60+4 x+8 x^2\right ) \log \left ((3+x) \log ^2(5)\right )}{15+5 x} \, dx=-\frac {1}{5} \, x^{3} + \frac {13}{5} \, x^{2} + x e^{2} + \frac {4}{5} \, {\left (x^{2} - 5 \, x - 5 \, e^{2} + 10\right )} \log \left ({\left (x + 3\right )} \log \left (5\right )^{2}\right ) - 10 \, x \] Input:
integrate(((8*x^2+4*x-60)*log((3+x)*log(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+ 8*x-110)/(5*x+15),x, algorithm="fricas")
Output:
-1/5*x^3 + 13/5*x^2 + x*e^2 + 4/5*(x^2 - 5*x - 5*e^2 + 10)*log((x + 3)*log (5)^2) - 10*x
Time = 0.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+\left (-60+4 x+8 x^2\right ) \log \left ((3+x) \log ^2(5)\right )}{15+5 x} \, dx=- \frac {x^{3}}{5} + \frac {13 x^{2}}{5} - x \left (10 - e^{2}\right ) + \left (\frac {4 x^{2}}{5} - 4 x\right ) \log {\left (\left (x + 3\right ) \log {\left (5 \right )}^{2} \right )} - 4 \left (-2 + e^{2}\right ) \log {\left (x + 3 \right )} \] Input:
integrate(((8*x**2+4*x-60)*ln((3+x)*ln(5)**2)+(5*x-5)*exp(2)-3*x**3+21*x** 2+8*x-110)/(5*x+15),x)
Output:
-x**3/5 + 13*x**2/5 - x*(10 - exp(2)) + (4*x**2/5 - 4*x)*log((x + 3)*log(5 )**2) - 4*(-2 + exp(2))*log(x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (28) = 56\).
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.76 \[ \int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+\left (-60+4 x+8 x^2\right ) \log \left ((3+x) \log ^2(5)\right )}{15+5 x} \, dx=-\frac {1}{5} \, x^{3} + \frac {13}{5} \, x^{2} + {\left (x - 3 \, \log \left (x + 3\right )\right )} e^{2} + \frac {4}{5} \, {\left (x^{2} - 6 \, x + 18 \, \log \left (x + 3\right )\right )} \log \left (x \log \left (5\right )^{2} + 3 \, \log \left (5\right )^{2}\right ) + \frac {4}{5} \, {\left (x - 3 \, \log \left (x + 3\right )\right )} \log \left (x \log \left (5\right )^{2} + 3 \, \log \left (5\right )^{2}\right ) - e^{2} \log \left (x + 3\right ) - 12 \, \log \left (x + 3\right )^{2} - 24 \, \log \left (x + 3\right ) \log \left (\log \left (5\right )\right ) - 10 \, x + 8 \, \log \left (x + 3\right ) \] Input:
integrate(((8*x^2+4*x-60)*log((3+x)*log(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+ 8*x-110)/(5*x+15),x, algorithm="maxima")
Output:
-1/5*x^3 + 13/5*x^2 + (x - 3*log(x + 3))*e^2 + 4/5*(x^2 - 6*x + 18*log(x + 3))*log(x*log(5)^2 + 3*log(5)^2) + 4/5*(x - 3*log(x + 3))*log(x*log(5)^2 + 3*log(5)^2) - e^2*log(x + 3) - 12*log(x + 3)^2 - 24*log(x + 3)*log(log(5 )) - 10*x + 8*log(x + 3)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (28) = 56\).
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.34 \[ \int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+\left (-60+4 x+8 x^2\right ) \log \left ((3+x) \log ^2(5)\right )}{15+5 x} \, dx=-\frac {1}{5} \, x^{3} + \frac {4}{5} \, x^{2} \log \left (x \log \left (5\right )^{2} + 3 \, \log \left (5\right )^{2}\right ) + \frac {13}{5} \, x^{2} + x e^{2} - 4 \, x \log \left (x \log \left (5\right )^{2} + 3 \, \log \left (5\right )^{2}\right ) - 4 \, e^{2} \log \left (x + 3\right ) - 10 \, x + 8 \, \log \left (x + 3\right ) \] Input:
integrate(((8*x^2+4*x-60)*log((3+x)*log(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+ 8*x-110)/(5*x+15),x, algorithm="giac")
Output:
-1/5*x^3 + 4/5*x^2*log(x*log(5)^2 + 3*log(5)^2) + 13/5*x^2 + x*e^2 - 4*x*l og(x*log(5)^2 + 3*log(5)^2) - 4*e^2*log(x + 3) - 10*x + 8*log(x + 3)
Time = 3.57 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.00 \[ \int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+\left (-60+4 x+8 x^2\right ) \log \left ((3+x) \log ^2(5)\right )}{15+5 x} \, dx=8\,\ln \left (x+3\right )-10\,x-4\,\ln \left (x+3\right )\,{\mathrm {e}}^2+x\,{\mathrm {e}}^2-4\,x\,\ln \left ({\ln \left (5\right )}^2\,\left (x+3\right )\right )+\frac {4\,x^2\,\ln \left ({\ln \left (5\right )}^2\,\left (x+3\right )\right )}{5}+\frac {13\,x^2}{5}-\frac {x^3}{5} \] Input:
int((8*x + log(log(5)^2*(x + 3))*(4*x + 8*x^2 - 60) + 21*x^2 - 3*x^3 + exp (2)*(5*x - 5) - 110)/(5*x + 15),x)
Output:
8*log(x + 3) - 10*x - 4*log(x + 3)*exp(2) + x*exp(2) - 4*x*log(log(5)^2*(x + 3)) + (4*x^2*log(log(5)^2*(x + 3)))/5 + (13*x^2)/5 - x^3/5
Time = 0.15 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.97 \[ \int \frac {-110+8 x+21 x^2-3 x^3+e^2 (-5+5 x)+\left (-60+4 x+8 x^2\right ) \log \left ((3+x) \log ^2(5)\right )}{15+5 x} \, dx=\frac {4 \,\mathrm {log}\left (x \mathrm {log}\left (5\right )^{2}+3 \mathrm {log}\left (5\right )^{2}\right ) x^{2}}{5}-4 \,\mathrm {log}\left (x \mathrm {log}\left (5\right )^{2}+3 \mathrm {log}\left (5\right )^{2}\right ) x -\frac {96 \,\mathrm {log}\left (x \mathrm {log}\left (5\right )^{2}+3 \mathrm {log}\left (5\right )^{2}\right )}{5}-4 \,\mathrm {log}\left (x +3\right ) e^{2}+\frac {136 \,\mathrm {log}\left (x +3\right )}{5}+e^{2} x -\frac {x^{3}}{5}+\frac {13 x^{2}}{5}-10 x \] Input:
int(((8*x^2+4*x-60)*log((3+x)*log(5)^2)+(5*x-5)*exp(2)-3*x^3+21*x^2+8*x-11 0)/(5*x+15),x)
Output:
(4*log(log(5)**2*x + 3*log(5)**2)*x**2 - 20*log(log(5)**2*x + 3*log(5)**2) *x - 96*log(log(5)**2*x + 3*log(5)**2) - 20*log(x + 3)*e**2 + 136*log(x + 3) + 5*e**2*x - x**3 + 13*x**2 - 50*x)/5