Integrand size = 80, antiderivative size = 27 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=\log \left ((-2+x)^2\right ) \log \left (\frac {5-\frac {x}{2}}{-16+(5-x) x}\right ) \]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.18 (sec) , antiderivative size = 220, normalized size of antiderivative = 8.15 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=-2 \log (8) \log (-10+x)+2 \log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log (-2+x)+2 \log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log (-2+x)-\log \left (\frac {5-i \sqrt {39}-2 x}{1-i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )-\log \left (\frac {5+i \sqrt {39}-2 x}{1+i \sqrt {39}}\right ) \log \left ((-2+x)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((-2+x)^2\right )+2 \log (-2+x) \log \left (-\frac {10-x}{2 \left (16-5 x+x^2\right )}\right )+2 \operatorname {PolyLog}\left (2,\frac {10-x}{8}\right )+2 \operatorname {PolyLog}\left (2,\frac {1}{8} (-2+x)\right ) \]
Integrate[((68 - 74*x + 22*x^2 - x^3)*Log[4 - 4*x + x^2] + (-320 + 132*x - 30*x^2 + 2*x^3)*Log[(-10 + x)/(32 - 10*x + 2*x^2)])/(320 - 292*x + 96*x^2 - 17*x^3 + x^4),x]
-2*Log[8]*Log[-10 + x] + 2*Log[(5 - I*Sqrt[39] - 2*x)/(1 - I*Sqrt[39])]*Lo g[-2 + x] + 2*Log[(5 + I*Sqrt[39] - 2*x)/(1 + I*Sqrt[39])]*Log[-2 + x] - L og[(5 - I*Sqrt[39] - 2*x)/(1 - I*Sqrt[39])]*Log[(-2 + x)^2] - Log[(5 + I*S qrt[39] - 2*x)/(1 + I*Sqrt[39])]*Log[(-2 + x)^2] + Log[(10 - x)/8]*Log[(-2 + x)^2] + 2*Log[-2 + x]*Log[-1/2*(10 - x)/(16 - 5*x + x^2)] + 2*PolyLog[2 , (10 - x)/8] + 2*PolyLog[2, (-2 + x)/8]
Result contains complex when optimal does not.
Time = 2.00 (sec) , antiderivative size = 204, normalized size of antiderivative = 7.56, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {2463, 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.
\(\displaystyle \int \frac {\left (-x^3+22 x^2-74 x+68\right ) \log \left (x^2-4 x+4\right )+\left (2 x^3-30 x^2+132 x-320\right ) \log \left (\frac {x-10}{2 x^2-10 x+32}\right )}{x^4-17 x^3+96 x^2-292 x+320} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (-x^3+22 x^2-74 x+68\right ) \log \left (x^2-4 x+4\right )+\left (2 x^3-30 x^2+132 x-320\right ) \log \left (\frac {x-10}{2 x^2-10 x+32}\right )}{528 (x-10)}-\frac {\left (-x^3+22 x^2-74 x+68\right ) \log \left (x^2-4 x+4\right )+\left (2 x^3-30 x^2+132 x-320\right ) \log \left (\frac {x-10}{2 x^2-10 x+32}\right )}{80 (x-2)}+\frac {(7 x-31) \left (\left (-x^3+22 x^2-74 x+68\right ) \log \left (x^2-4 x+4\right )+\left (2 x^3-30 x^2+132 x-320\right ) \log \left (\frac {x-10}{2 x^2-10 x+32}\right )\right )}{660 \left (x^2-5 x+16\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \log \left (-\frac {10-x}{2 \left (x^2-5 x+16\right )}\right ) \log (x-2)+2 \log \left (\frac {-2 x-i \sqrt {39}+5}{1-i \sqrt {39}}\right ) \log (x-2)+2 \log \left (\frac {-2 x+i \sqrt {39}+5}{1+i \sqrt {39}}\right ) \log (x-2)-2 \log \left (\frac {10-x}{8}\right ) \log (x-2)-\log \left (\frac {-2 x-i \sqrt {39}+5}{1-i \sqrt {39}}\right ) \log \left ((x-2)^2\right )-\log \left (\frac {-2 x+i \sqrt {39}+5}{1+i \sqrt {39}}\right ) \log \left ((x-2)^2\right )+\log \left (\frac {10-x}{8}\right ) \log \left ((x-2)^2\right )\) |
Int[((68 - 74*x + 22*x^2 - x^3)*Log[4 - 4*x + x^2] + (-320 + 132*x - 30*x^ 2 + 2*x^3)*Log[(-10 + x)/(32 - 10*x + 2*x^2)])/(320 - 292*x + 96*x^2 - 17* x^3 + x^4),x]
2*Log[(5 - I*Sqrt[39] - 2*x)/(1 - I*Sqrt[39])]*Log[-2 + x] + 2*Log[(5 + I* Sqrt[39] - 2*x)/(1 + I*Sqrt[39])]*Log[-2 + x] - 2*Log[(10 - x)/8]*Log[-2 + x] - Log[(5 - I*Sqrt[39] - 2*x)/(1 - I*Sqrt[39])]*Log[(-2 + x)^2] - Log[( 5 + I*Sqrt[39] - 2*x)/(1 + I*Sqrt[39])]*Log[(-2 + x)^2] + Log[(10 - x)/8]* Log[(-2 + x)^2] + 2*Log[-2 + x]*Log[-1/2*(10 - x)/(16 - 5*x + x^2)]
3.28.91.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.45 (sec) , antiderivative size = 247, normalized size of antiderivative = 9.15
method | result | size |
default | \(-2 \ln \left (2\right ) \ln \left (-2+x \right )+2 \ln \left (-2+x \right ) \ln \left (\frac {x -10}{x^{2}-5 x +16}\right )-2 \left (\ln \left (-2+x \right )-\ln \left (-\frac {1}{4}+\frac {x}{8}\right )\right ) \ln \left (\frac {5}{4}-\frac {x}{8}\right )+2 \ln \left (-2+x \right ) \ln \left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \ln \left (-2+x \right ) \ln \left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )+2 \operatorname {dilog}\left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \operatorname {dilog}\left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )-\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}-5 \textit {\_Z} +16\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}-4 x +4\right )-2 \operatorname {dilog}\left (\frac {-2+x}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {-2+x}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )+\ln \left (x -10\right ) \ln \left (x^{2}-4 x +4\right )-2 \ln \left (x -10\right ) \ln \left (-\frac {1}{4}+\frac {x}{8}\right )\) | \(247\) |
parts | \(-2 \ln \left (2\right ) \ln \left (-2+x \right )+2 \ln \left (-2+x \right ) \ln \left (\frac {x -10}{x^{2}-5 x +16}\right )-2 \left (\ln \left (-2+x \right )-\ln \left (-\frac {1}{4}+\frac {x}{8}\right )\right ) \ln \left (\frac {5}{4}-\frac {x}{8}\right )+2 \ln \left (-2+x \right ) \ln \left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \ln \left (-2+x \right ) \ln \left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )+2 \operatorname {dilog}\left (\frac {i \sqrt {39}+5-2 x}{1+i \sqrt {39}}\right )+2 \operatorname {dilog}\left (\frac {i \sqrt {39}-5+2 x}{i \sqrt {39}-1}\right )-\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{2}-5 \textit {\_Z} +16\right )}{\sum }\left (\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}-4 x +4\right )-2 \operatorname {dilog}\left (\frac {-2+x}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {-2+x}{-2+\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )+\ln \left (x -10\right ) \ln \left (x^{2}-4 x +4\right )-2 \ln \left (x -10\right ) \ln \left (-\frac {1}{4}+\frac {x}{8}\right )\) | \(247\) |
risch | \(\text {Expression too large to display}\) | \(50326\) |
int(((-x^3+22*x^2-74*x+68)*ln(x^2-4*x+4)+(2*x^3-30*x^2+132*x-320)*ln((x-10 )/(2*x^2-10*x+32)))/(x^4-17*x^3+96*x^2-292*x+320),x,method=_RETURNVERBOSE)
-2*ln(2)*ln(-2+x)+2*ln(-2+x)*ln((x-10)/(x^2-5*x+16))-2*(ln(-2+x)-ln(-1/4+1 /8*x))*ln(5/4-1/8*x)+2*ln(-2+x)*ln((I*39^(1/2)+5-2*x)/(1+I*39^(1/2)))+2*ln (-2+x)*ln((I*39^(1/2)-5+2*x)/(I*39^(1/2)-1))+2*dilog((I*39^(1/2)+5-2*x)/(1 +I*39^(1/2)))+2*dilog((I*39^(1/2)-5+2*x)/(I*39^(1/2)-1))-Sum(ln(x-_alpha)* ln(x^2-4*x+4)-2*dilog((-2+x)/(-2+_alpha))-2*ln(x-_alpha)*ln((-2+x)/(-2+_al pha)),_alpha=RootOf(_Z^2-5*_Z+16))+ln(x-10)*ln(x^2-4*x+4)-2*ln(x-10)*ln(-1 /4+1/8*x)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=\log \left (x^{2} - 4 \, x + 4\right ) \log \left (\frac {x - 10}{2 \, {\left (x^{2} - 5 \, x + 16\right )}}\right ) \]
integrate(((-x^3+22*x^2-74*x+68)*log(x^2-4*x+4)+(2*x^3-30*x^2+132*x-320)*l og((x-10)/(2*x^2-10*x+32)))/(x^4-17*x^3+96*x^2-292*x+320),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=\log {\left (\frac {x - 10}{2 x^{2} - 10 x + 32} \right )} \log {\left (x^{2} - 4 x + 4 \right )} \]
integrate(((-x**3+22*x**2-74*x+68)*ln(x**2-4*x+4)+(2*x**3-30*x**2+132*x-32 0)*ln((x-10)/(2*x**2-10*x+32)))/(x**4-17*x**3+96*x**2-292*x+320),x)
Time = 0.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=-2 \, {\left (\log \left (2\right ) - \log \left (x - 10\right )\right )} \log \left (x - 2\right ) - 2 \, \log \left (x^{2} - 5 \, x + 16\right ) \log \left (x - 2\right ) \]
integrate(((-x^3+22*x^2-74*x+68)*log(x^2-4*x+4)+(2*x^3-30*x^2+132*x-320)*l og((x-10)/(2*x^2-10*x+32)))/(x^4-17*x^3+96*x^2-292*x+320),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (22) = 44\).
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.22 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=-{\left (\log \left (x^{2} - 5 \, x + 16\right ) - \log \left (x - 10\right )\right )} \log \left (x^{2} - 4 \, x + 4\right ) - 2 \, \log \left (2 \, x^{2} - 10 \, x + 32\right ) \log \left (x - 2\right ) + 2 \, \log \left (x^{2} - 5 \, x + 16\right ) \log \left (x - 2\right ) \]
integrate(((-x^3+22*x^2-74*x+68)*log(x^2-4*x+4)+(2*x^3-30*x^2+132*x-320)*l og((x-10)/(2*x^2-10*x+32)))/(x^4-17*x^3+96*x^2-292*x+320),x, algorithm=\
-(log(x^2 - 5*x + 16) - log(x - 10))*log(x^2 - 4*x + 4) - 2*log(2*x^2 - 10 *x + 32)*log(x - 2) + 2*log(x^2 - 5*x + 16)*log(x - 2)
Time = 10.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {\left (68-74 x+22 x^2-x^3\right ) \log \left (4-4 x+x^2\right )+\left (-320+132 x-30 x^2+2 x^3\right ) \log \left (\frac {-10+x}{32-10 x+2 x^2}\right )}{320-292 x+96 x^2-17 x^3+x^4} \, dx=\ln \left (x^2-4\,x+4\right )\,\left (\ln \left (x-10\right )-\ln \left (2\,x^2-10\,x+32\right )\right ) \]