Integrand size = 77, antiderivative size = 27 \[ \int \frac {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{-x+x^2} \, dx=5 \left (\log (2-2 x)+\left (3-2 \left (e^{-1+3 x}+\log \left (x^2\right )\right )\right )^2\right ) \] Output:
5*(3-2*ln(x^2)-2*exp(-1+3*x))^2+5*ln(2-2*x)
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{-x+x^2} \, dx=\frac {5 \left (e^2 \log (1-x)+\left (-3 e+2 e^{3 x}+2 e \log \left (x^2\right )\right )^2\right )}{e^2} \] Input:
Integrate[(120 - 115*x + E^(-1 + 3*x)*(-80 + 260*x - 180*x^2) + E^(-2 + 6* x)*(-120*x + 120*x^2) + (-80 + 80*x + E^(-1 + 3*x)*(-120*x + 120*x^2))*Log [x^2])/(-x + x^2),x]
Output:
(5*(E^2*Log[1 - x] + (-3*E + 2*E^(3*x) + 2*E*Log[x^2])^2))/E^2
Time = 1.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2026, 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.
\(\displaystyle \int \frac {e^{3 x-1} \left (-180 x^2+260 x-80\right )+e^{6 x-2} \left (120 x^2-120 x\right )+\left (e^{3 x-1} \left (120 x^2-120 x\right )+80 x-80\right ) \log \left (x^2\right )-115 x+120}{x^2-x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{3 x-1} \left (-180 x^2+260 x-80\right )+e^{6 x-2} \left (120 x^2-120 x\right )+\left (e^{3 x-1} \left (120 x^2-120 x\right )+80 x-80\right ) \log \left (x^2\right )-115 x+120}{(x-1) x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {20 e^{3 x-1} \left (6 x \log \left (x^2\right )-9 x+4\right )}{x}+\frac {5 \left (16 x \log \left (x^2\right )-16 \log \left (x^2\right )-23 x+24\right )}{(x-1) x}+120 e^{6 x-2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 20 \log ^2\left (x^2\right )-\frac {20 e^{3 x-1} \left (3 x-2 x \log \left (x^2\right )\right )}{x}+20 e^{6 x-2}+5 \log (1-x)-120 \log (x)\) |
Input:
Int[(120 - 115*x + E^(-1 + 3*x)*(-80 + 260*x - 180*x^2) + E^(-2 + 6*x)*(-1 20*x + 120*x^2) + (-80 + 80*x + E^(-1 + 3*x)*(-120*x + 120*x^2))*Log[x^2]) /(-x + x^2),x]
Output:
20*E^(-2 + 6*x) + 5*Log[1 - x] - 120*Log[x] + 20*Log[x^2]^2 - (20*E^(-1 + 3*x)*(3*x - 2*x*Log[x^2]))/x
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])
Time = 10.57 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.93
method | result | size |
norman | \(-60 \ln \left (x^{2}\right )+20 \,{\mathrm e}^{6 x -2}+20 \ln \left (x^{2}\right )^{2}+40 \,{\mathrm e}^{-1+3 x} \ln \left (x^{2}\right )-60 \,{\mathrm e}^{-1+3 x}+5 \ln \left (-1+x \right )\) | \(52\) |
parallelrisch | \(-60 \ln \left (x^{2}\right )+20 \,{\mathrm e}^{6 x -2}+20 \ln \left (x^{2}\right )^{2}+40 \,{\mathrm e}^{-1+3 x} \ln \left (x^{2}\right )-60 \,{\mathrm e}^{-1+3 x}+5 \ln \left (-1+x \right )\) | \(52\) |
default | \(20 \left (-3+2 \ln \left (x^{2}\right )-4 \ln \left (x \right )\right ) {\mathrm e}^{-1+3 x}+80 \ln \left (x \right ) {\mathrm e}^{-1+3 x}+20 \,{\mathrm e}^{6 x -2}+20 \ln \left (x^{2}\right )^{2}+5 \ln \left (-1+x \right )-120 \ln \left (x \right )\) | \(60\) |
parts | \(20 \left (-3+2 \ln \left (x^{2}\right )-4 \ln \left (x \right )\right ) {\mathrm e}^{-1+3 x}+80 \ln \left (x \right ) {\mathrm e}^{-1+3 x}+20 \,{\mathrm e}^{6 x -2}+20 \ln \left (x^{2}\right )^{2}+5 \ln \left (-1+x \right )-120 \ln \left (x \right )\) | \(60\) |
risch | \(80 \ln \left (x \right )^{2}+80 \ln \left (x \right ) {\mathrm e}^{-1+3 x}-40 i \pi \ln \left (\left (-8 \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+16 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-8 \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+25 i\right ) x \right ) \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+80 i \pi \ln \left (\left (-8 \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+16 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-8 \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+25 i\right ) x \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-40 i \pi \ln \left (\left (-8 \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+16 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-8 \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+25 i\right ) x \right ) \operatorname {csgn}\left (i x^{2}\right )^{3}-120 \ln \left (\left (-8 \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+16 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-8 \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+25 i\right ) x \right )+5 \ln \left (1-x \right )+20 \,{\mathrm e}^{6 x -2}-20 i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{-1+3 x}+40 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{-1+3 x}-20 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{-1+3 x}-60 \,{\mathrm e}^{-1+3 x}\) | \(368\) |
Input:
int((((120*x^2-120*x)*exp(-1+3*x)+80*x-80)*ln(x^2)+(120*x^2-120*x)*exp(-1+ 3*x)^2+(-180*x^2+260*x-80)*exp(-1+3*x)-115*x+120)/(x^2-x),x,method=_RETURN VERBOSE)
Output:
-60*ln(x^2)+20*exp(-1+3*x)^2+20*ln(x^2)^2+40*exp(-1+3*x)*ln(x^2)-60*exp(-1 +3*x)+5*ln(-1+x)
Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{-x+x^2} \, dx=40 \, e^{\left (3 \, x - 1\right )} \log \left (x^{2}\right ) + 20 \, \log \left (x^{2}\right )^{2} + 20 \, e^{\left (6 \, x - 2\right )} - 60 \, e^{\left (3 \, x - 1\right )} + 5 \, \log \left (x - 1\right ) - 120 \, \log \left (x\right ) \] Input:
integrate((((120*x^2-120*x)*exp(-1+3*x)+80*x-80)*log(x^2)+(120*x^2-120*x)* exp(-1+3*x)^2+(-180*x^2+260*x-80)*exp(-1+3*x)-115*x+120)/(x^2-x),x, algori thm="fricas")
Output:
40*e^(3*x - 1)*log(x^2) + 20*log(x^2)^2 + 20*e^(6*x - 2) - 60*e^(3*x - 1) + 5*log(x - 1) - 120*log(x)
Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{-x+x^2} \, dx=\left (40 \log {\left (x^{2} \right )} - 60\right ) e^{3 x - 1} + 20 e^{6 x - 2} - 120 \log {\left (x \right )} + 20 \log {\left (x^{2} \right )}^{2} + 5 \log {\left (x - 1 \right )} \] Input:
integrate((((120*x**2-120*x)*exp(-1+3*x)+80*x-80)*ln(x**2)+(120*x**2-120*x )*exp(-1+3*x)**2+(-180*x**2+260*x-80)*exp(-1+3*x)-115*x+120)/(x**2-x),x)
Output:
(40*log(x**2) - 60)*exp(3*x - 1) + 20*exp(6*x - 2) - 120*log(x) + 20*log(x **2)**2 + 5*log(x - 1)
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{-x+x^2} \, dx=20 \, {\left (4 \, e^{2} \log \left (x\right )^{2} + {\left (4 \, e \log \left (x\right ) - 3 \, e\right )} e^{\left (3 \, x\right )} + e^{\left (6 \, x\right )}\right )} e^{\left (-2\right )} + 5 \, \log \left (x - 1\right ) - 120 \, \log \left (x\right ) \] Input:
integrate((((120*x^2-120*x)*exp(-1+3*x)+80*x-80)*log(x^2)+(120*x^2-120*x)* exp(-1+3*x)^2+(-180*x^2+260*x-80)*exp(-1+3*x)-115*x+120)/(x^2-x),x, algori thm="maxima")
Output:
20*(4*e^2*log(x)^2 + (4*e*log(x) - 3*e)*e^(3*x) + e^(6*x))*e^(-2) + 5*log( x - 1) - 120*log(x)
Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{-x+x^2} \, dx=5 \, {\left (4 \, e^{2} \log \left (x^{2}\right )^{2} + 8 \, e^{\left (3 \, x + 1\right )} \log \left (x^{2}\right ) + e^{2} \log \left (x - 1\right ) - 24 \, e^{2} \log \left (x\right ) + 4 \, e^{\left (6 \, x\right )} - 12 \, e^{\left (3 \, x + 1\right )}\right )} e^{\left (-2\right )} \] Input:
integrate((((120*x^2-120*x)*exp(-1+3*x)+80*x-80)*log(x^2)+(120*x^2-120*x)* exp(-1+3*x)^2+(-180*x^2+260*x-80)*exp(-1+3*x)-115*x+120)/(x^2-x),x, algori thm="giac")
Output:
5*(4*e^2*log(x^2)^2 + 8*e^(3*x + 1)*log(x^2) + e^2*log(x - 1) - 24*e^2*log (x) + 4*e^(6*x) - 12*e^(3*x + 1))*e^(-2)
Time = 3.82 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{-x+x^2} \, dx=5\,\ln \left (x-1\right )-60\,\ln \left (x^2\right )-60\,{\mathrm {e}}^{3\,x-1}+20\,{\mathrm {e}}^{6\,x-2}+40\,\ln \left (x^2\right )\,{\mathrm {e}}^{3\,x-1}+20\,{\ln \left (x^2\right )}^2 \] Input:
int((115*x + log(x^2)*(exp(3*x - 1)*(120*x - 120*x^2) - 80*x + 80) + exp(6 *x - 2)*(120*x - 120*x^2) + exp(3*x - 1)*(180*x^2 - 260*x + 80) - 120)/(x - x^2),x)
Output:
5*log(x - 1) - 60*log(x^2) - 60*exp(3*x - 1) + 20*exp(6*x - 2) + 40*log(x^ 2)*exp(3*x - 1) + 20*log(x^2)^2
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.19 \[ \int \frac {120-115 x+e^{-1+3 x} \left (-80+260 x-180 x^2\right )+e^{-2+6 x} \left (-120 x+120 x^2\right )+\left (-80+80 x+e^{-1+3 x} \left (-120 x+120 x^2\right )\right ) \log \left (x^2\right )}{-x+x^2} \, dx=\frac {20 e^{6 x}+40 e^{3 x} \mathrm {log}\left (x^{2}\right ) e -60 e^{3 x} e +20 \mathrm {log}\left (x^{2}\right )^{2} e^{2}+5 \,\mathrm {log}\left (x -1\right ) e^{2}-120 \,\mathrm {log}\left (x \right ) e^{2}}{e^{2}} \] Input:
int((((120*x^2-120*x)*exp(-1+3*x)+80*x-80)*log(x^2)+(120*x^2-120*x)*exp(-1 +3*x)^2+(-180*x^2+260*x-80)*exp(-1+3*x)-115*x+120)/(x^2-x),x)
Output:
(5*(4*e**(6*x) + 8*e**(3*x)*log(x**2)*e - 12*e**(3*x)*e + 4*log(x**2)**2*e **2 + log(x - 1)*e**2 - 24*log(x)*e**2))/e**2