Integrand size = 85, antiderivative size = 23 \[ \int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+\left (-80 x+40 x^2-48 x^3-5 x^4\right ) \log \left (\frac {80-40 x+48 x^2+5 x^3}{5 x^2}\right )}{80 x-40 x^2+48 x^3+5 x^4} \, dx=(3-x) \log \left (\frac {43}{5}+\frac {(4-x)^2}{x^2}+x\right ) \] Output:
ln(43/5+1/x^2*(4-x)^2+x)*(3-x)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.10 (sec) , antiderivative size = 166, normalized size of antiderivative = 7.22 \[ \int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+\left (-80 x+40 x^2-48 x^3-5 x^4\right ) \log \left (\frac {80-40 x+48 x^2+5 x^3}{5 x^2}\right )}{80 x-40 x^2+48 x^3+5 x^4} \, dx=-6 \log (x)-x \log \left (\frac {48}{5}+\frac {16}{x^2}-\frac {8}{x}+x\right )-16 \text {RootSum}\left [80-40 \text {$\#$1}+48 \text {$\#$1}^2+5 \text {$\#$1}^3\&,\frac {15 \log (x-\text {$\#$1})-5 \log (x-\text {$\#$1}) \text {$\#$1}+3 \log (x-\text {$\#$1}) \text {$\#$1}^2}{-40+96 \text {$\#$1}+15 \text {$\#$1}^2}\&\right ]+\text {RootSum}\left [80-40 \text {$\#$1}+48 \text {$\#$1}^2+5 \text {$\#$1}^3\&,\frac {120 \log (x-\text {$\#$1})+208 \log (x-\text {$\#$1}) \text {$\#$1}+93 \log (x-\text {$\#$1}) \text {$\#$1}^2}{-40+96 \text {$\#$1}+15 \text {$\#$1}^2}\&\right ] \] Input:
Integrate[(-480 + 280*x - 40*x^2 + 15*x^3 - 5*x^4 + (-80*x + 40*x^2 - 48*x ^3 - 5*x^4)*Log[(80 - 40*x + 48*x^2 + 5*x^3)/(5*x^2)])/(80*x - 40*x^2 + 48 *x^3 + 5*x^4),x]
Output:
-6*Log[x] - x*Log[48/5 + 16/x^2 - 8/x + x] - 16*RootSum[80 - 40*#1 + 48*#1 ^2 + 5*#1^3 & , (15*Log[x - #1] - 5*Log[x - #1]*#1 + 3*Log[x - #1]*#1^2)/( -40 + 96*#1 + 15*#1^2) & ] + RootSum[80 - 40*#1 + 48*#1^2 + 5*#1^3 & , (12 0*Log[x - #1] + 208*Log[x - #1]*#1 + 93*Log[x - #1]*#1^2)/(-40 + 96*#1 + 1 5*#1^2) & ]
Time = 15.77 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {2026, 7293, 7239, 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.
\(\displaystyle \int \frac {-5 x^4+15 x^3-40 x^2+\left (-5 x^4-48 x^3+40 x^2-80 x\right ) \log \left (\frac {5 x^3+48 x^2-40 x+80}{5 x^2}\right )+280 x-480}{5 x^4+48 x^3-40 x^2+80 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-5 x^4+15 x^3-40 x^2+\left (-5 x^4-48 x^3+40 x^2-80 x\right ) \log \left (\frac {5 x^3+48 x^2-40 x+80}{5 x^2}\right )+280 x-480}{x \left (5 x^3+48 x^2-40 x+80\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\log \left (\frac {16}{x^2}+x-\frac {8}{x}+\frac {48}{5}\right )-\frac {5 x^3}{5 x^3+48 x^2-40 x+80}+\frac {15 x^2}{5 x^3+48 x^2-40 x+80}-\frac {40 x}{5 x^3+48 x^2-40 x+80}+\frac {280}{5 x^3+48 x^2-40 x+80}-\frac {480}{\left (5 x^3+48 x^2-40 x+80\right ) x}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (-\log \left (\frac {16}{x^2}+x-\frac {8}{x}+\frac {48}{5}\right )-\frac {5 \left (x^4-3 x^3+8 x^2-56 x+96\right )}{x \left (5 x^3+48 x^2-40 x+80\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x \log \left (\frac {16}{x^2}+x-\frac {8}{x}+\frac {48}{5}\right )+3 \log \left (5 x^3+48 x^2-40 x+80\right )-6 \log (x)\) |
Input:
Int[(-480 + 280*x - 40*x^2 + 15*x^3 - 5*x^4 + (-80*x + 40*x^2 - 48*x^3 - 5 *x^4)*Log[(80 - 40*x + 48*x^2 + 5*x^3)/(5*x^2)])/(80*x - 40*x^2 + 48*x^3 + 5*x^4),x]
Output:
-6*Log[x] - x*Log[48/5 + 16/x^2 - 8/x + x] + 3*Log[80 - 40*x + 48*x^2 + 5* x^3]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(47\) vs. \(2(21)=42\).
Time = 0.76 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.09
method | result | size |
risch | \(-\ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right ) x -6 \ln \left (x \right )+3 \ln \left (5 x^{3}+48 x^{2}-40 x +80\right )\) | \(48\) |
norman | \(3 \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right )-\ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right ) x\) | \(49\) |
parallelrisch | \(3 \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right )-\ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right ) x\) | \(49\) |
default | \(-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (5 \textit {\_Z}^{3}+48 \textit {\_Z}^{2}-40 \textit {\_Z} +80\right )}{\sum }\frac {\left (-93 \textit {\_R}^{2}-208 \textit {\_R} -120\right ) \ln \left (x -\textit {\_R} \right )}{15 \textit {\_R}^{2}+96 \textit {\_R} -40}\right )-6 \ln \left (x \right )-x \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{x^{2}}\right )+16 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (5 \textit {\_Z}^{3}+48 \textit {\_Z}^{2}-40 \textit {\_Z} +80\right )}{\sum }\frac {\left (-3 \textit {\_R}^{2}+5 \textit {\_R} -15\right ) \ln \left (x -\textit {\_R} \right )}{15 \textit {\_R}^{2}+96 \textit {\_R} -40}\right )+x \ln \left (5\right )\) | \(133\) |
parts | \(-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (5 \textit {\_Z}^{3}+48 \textit {\_Z}^{2}-40 \textit {\_Z} +80\right )}{\sum }\frac {\left (-93 \textit {\_R}^{2}-208 \textit {\_R} -120\right ) \ln \left (x -\textit {\_R} \right )}{15 \textit {\_R}^{2}+96 \textit {\_R} -40}\right )-6 \ln \left (x \right )-x \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{x^{2}}\right )+16 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (5 \textit {\_Z}^{3}+48 \textit {\_Z}^{2}-40 \textit {\_Z} +80\right )}{\sum }\frac {\left (-3 \textit {\_R}^{2}+5 \textit {\_R} -15\right ) \ln \left (x -\textit {\_R} \right )}{15 \textit {\_R}^{2}+96 \textit {\_R} -40}\right )+x \ln \left (5\right )\) | \(133\) |
orering | \(\frac {\left (-3+x \right ) \left (\left (-5 x^{4}-48 x^{3}+40 x^{2}-80 x \right ) \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right )-5 x^{4}+15 x^{3}-40 x^{2}+280 x -480\right )}{5 x^{4}+48 x^{3}-40 x^{2}+80 x}-\frac {\left (5 x^{5}+373 x^{3}-640 x^{2}+1800 x -1440\right ) x \left (5 x^{3}+48 x^{2}-40 x +80\right ) \left (\frac {\left (-20 x^{3}-144 x^{2}+80 x -80\right ) \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right )+\frac {5 \left (-5 x^{4}-48 x^{3}+40 x^{2}-80 x \right ) \left (\frac {15 x^{2}+96 x -40}{5 x^{2}}-\frac {2 \left (5 x^{3}+48 x^{2}-40 x +80\right )}{5 x^{3}}\right ) x^{2}}{5 x^{3}+48 x^{2}-40 x +80}-20 x^{3}+45 x^{2}-80 x +280}{5 x^{4}+48 x^{3}-40 x^{2}+80 x}-\frac {\left (\left (-5 x^{4}-48 x^{3}+40 x^{2}-80 x \right ) \ln \left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right )-5 x^{4}+15 x^{3}-40 x^{2}+280 x -480\right ) \left (20 x^{3}+144 x^{2}-80 x +80\right )}{\left (5 x^{4}+48 x^{3}-40 x^{2}+80 x \right )^{2}}\right )}{5 \left (5 x^{7}+111 x^{6}-160 x^{5}+1120 x^{4}+1120 x^{3}-13504 x^{2}+5120 x -7680\right )}\) | \(414\) |
Input:
int(((-5*x^4-48*x^3+40*x^2-80*x)*ln(1/5*(5*x^3+48*x^2-40*x+80)/x^2)-5*x^4+ 15*x^3-40*x^2+280*x-480)/(5*x^4+48*x^3-40*x^2+80*x),x,method=_RETURNVERBOS E)
Output:
-ln(1/5*(5*x^3+48*x^2-40*x+80)/x^2)*x-6*ln(x)+3*ln(5*x^3+48*x^2-40*x+80)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+\left (-80 x+40 x^2-48 x^3-5 x^4\right ) \log \left (\frac {80-40 x+48 x^2+5 x^3}{5 x^2}\right )}{80 x-40 x^2+48 x^3+5 x^4} \, dx=-{\left (x - 3\right )} \log \left (\frac {5 \, x^{3} + 48 \, x^{2} - 40 \, x + 80}{5 \, x^{2}}\right ) \] Input:
integrate(((-5*x^4-48*x^3+40*x^2-80*x)*log(1/5*(5*x^3+48*x^2-40*x+80)/x^2) -5*x^4+15*x^3-40*x^2+280*x-480)/(5*x^4+48*x^3-40*x^2+80*x),x, algorithm="f ricas")
Output:
-(x - 3)*log(1/5*(5*x^3 + 48*x^2 - 40*x + 80)/x^2)
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (17) = 34\).
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+\left (-80 x+40 x^2-48 x^3-5 x^4\right ) \log \left (\frac {80-40 x+48 x^2+5 x^3}{5 x^2}\right )}{80 x-40 x^2+48 x^3+5 x^4} \, dx=- x \log {\left (\frac {x^{3} + \frac {48 x^{2}}{5} - 8 x + 16}{x^{2}} \right )} - 6 \log {\left (x \right )} + 3 \log {\left (5 x^{3} + 48 x^{2} - 40 x + 80 \right )} \] Input:
integrate(((-5*x**4-48*x**3+40*x**2-80*x)*ln(1/5*(5*x**3+48*x**2-40*x+80)/ x**2)-5*x**4+15*x**3-40*x**2+280*x-480)/(5*x**4+48*x**3-40*x**2+80*x),x)
Output:
-x*log((x**3 + 48*x**2/5 - 8*x + 16)/x**2) - 6*log(x) + 3*log(5*x**3 + 48* x**2 - 40*x + 80)
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+\left (-80 x+40 x^2-48 x^3-5 x^4\right ) \log \left (\frac {80-40 x+48 x^2+5 x^3}{5 x^2}\right )}{80 x-40 x^2+48 x^3+5 x^4} \, dx=x \log \left (5\right ) - {\left (x - 3\right )} \log \left (5 \, x^{3} + 48 \, x^{2} - 40 \, x + 80\right ) + 2 \, {\left (x - 3\right )} \log \left (x\right ) \] Input:
integrate(((-5*x^4-48*x^3+40*x^2-80*x)*log(1/5*(5*x^3+48*x^2-40*x+80)/x^2) -5*x^4+15*x^3-40*x^2+280*x-480)/(5*x^4+48*x^3-40*x^2+80*x),x, algorithm="m axima")
Output:
x*log(5) - (x - 3)*log(5*x^3 + 48*x^2 - 40*x + 80) + 2*(x - 3)*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (18) = 36\).
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+\left (-80 x+40 x^2-48 x^3-5 x^4\right ) \log \left (\frac {80-40 x+48 x^2+5 x^3}{5 x^2}\right )}{80 x-40 x^2+48 x^3+5 x^4} \, dx=-x \log \left (\frac {5 \, x^{3} + 48 \, x^{2} - 40 \, x + 80}{5 \, x^{2}}\right ) + 3 \, \log \left (5 \, x^{3} + 48 \, x^{2} - 40 \, x + 80\right ) - 6 \, \log \left (x\right ) \] Input:
integrate(((-5*x^4-48*x^3+40*x^2-80*x)*log(1/5*(5*x^3+48*x^2-40*x+80)/x^2) -5*x^4+15*x^3-40*x^2+280*x-480)/(5*x^4+48*x^3-40*x^2+80*x),x, algorithm="g iac")
Output:
-x*log(1/5*(5*x^3 + 48*x^2 - 40*x + 80)/x^2) + 3*log(5*x^3 + 48*x^2 - 40*x + 80) - 6*log(x)
Time = 3.79 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+\left (-80 x+40 x^2-48 x^3-5 x^4\right ) \log \left (\frac {80-40 x+48 x^2+5 x^3}{5 x^2}\right )}{80 x-40 x^2+48 x^3+5 x^4} \, dx=3\,\ln \left (x^3+\frac {48\,x^2}{5}-8\,x+16\right )-6\,\ln \left (x\right )-x\,\ln \left (\frac {x^3+\frac {48\,x^2}{5}-8\,x+16}{x^2}\right ) \] Input:
int(-(log(((48*x^2)/5 - 8*x + x^3 + 16)/x^2)*(80*x - 40*x^2 + 48*x^3 + 5*x ^4) - 280*x + 40*x^2 - 15*x^3 + 5*x^4 + 480)/(80*x - 40*x^2 + 48*x^3 + 5*x ^4),x)
Output:
3*log((48*x^2)/5 - 8*x + x^3 + 16) - 6*log(x) - x*log(((48*x^2)/5 - 8*x + x^3 + 16)/x^2)
Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 3.04 \[ \int \frac {-480+280 x-40 x^2+15 x^3-5 x^4+\left (-80 x+40 x^2-48 x^3-5 x^4\right ) \log \left (\frac {80-40 x+48 x^2+5 x^3}{5 x^2}\right )}{80 x-40 x^2+48 x^3+5 x^4} \, dx=\frac {214 \,\mathrm {log}\left (5 x^{3}+48 x^{2}-40 x +80\right )}{15}-\mathrm {log}\left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right ) x -\frac {169 \,\mathrm {log}\left (\frac {5 x^{3}+48 x^{2}-40 x +80}{5 x^{2}}\right )}{15}-\frac {428 \,\mathrm {log}\left (x \right )}{15} \] Input:
int(((-5*x^4-48*x^3+40*x^2-80*x)*log(1/5*(5*x^3+48*x^2-40*x+80)/x^2)-5*x^4 +15*x^3-40*x^2+280*x-480)/(5*x^4+48*x^3-40*x^2+80*x),x)
Output:
(214*log(5*x**3 + 48*x**2 - 40*x + 80) - 15*log((5*x**3 + 48*x**2 - 40*x + 80)/(5*x**2))*x - 169*log((5*x**3 + 48*x**2 - 40*x + 80)/(5*x**2)) - 428* log(x))/15