Integrand size = 57, antiderivative size = 27 \[ \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{400 x^2-200 x^3+25 x^4} \, dx=-\frac {2 \log (10)}{(-4+x) x}+\log ^2\left (2 e^{\frac {x^2}{25}}\right ) \] Output:
ln(2*exp(1/25*x^2))^2-2/x/(-4+x)*ln(10)
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{400 x^2-200 x^3+25 x^4} \, dx=-\frac {x^4}{625}-\frac {\log (10)}{2 (-4+x)}+\frac {\log (10)}{2 x}+\frac {2}{25} x^2 \log \left (2 e^{\frac {x^2}{25}}\right ) \] Input:
Integrate[((-200 + 100*x)*Log[10] + (64*x^3 - 32*x^4 + 4*x^5)*Log[2*E^(x^2 /25)])/(400*x^2 - 200*x^3 + 25*x^4),x]
Output:
-1/625*x^4 - Log[10]/(2*(-4 + x)) + Log[10]/(2*x) + (2*x^2*Log[2*E^(x^2/25 )])/25
Time = 0.69 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2026, 7277, 27, 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.
\(\displaystyle \int \frac {\left (4 x^5-32 x^4+64 x^3\right ) \log \left (2 e^{\frac {x^2}{25}}\right )+(100 x-200) \log (10)}{25 x^4-200 x^3+400 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (4 x^5-32 x^4+64 x^3\right ) \log \left (2 e^{\frac {x^2}{25}}\right )+(100 x-200) \log (10)}{x^2 \left (25 x^2-200 x+400\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 100 \int -\frac {25 (2-x) \log (10)-\left (x^5-8 x^4+16 x^3\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{625 (4-x)^2 x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{25} \int \frac {25 (2-x) \log (10)-\left (x^5-8 x^4+16 x^3\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{(4-x)^2 x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {4}{25} \int \left (-\frac {25 \log (10) (x-2)}{(x-4)^2 x^2}-x \log \left (2 e^{\frac {x^2}{25}}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4}{25} \left (\frac {x^4}{100}-\frac {1}{2} x^2 \log \left (2 e^{\frac {x^2}{25}}\right )-\frac {25 \log (10)}{2 (4-x) x}\right )\) |
Input:
Int[((-200 + 100*x)*Log[10] + (64*x^3 - 32*x^4 + 4*x^5)*Log[2*E^(x^2/25)]) /(400*x^2 - 200*x^3 + 25*x^4),x]
Output:
(-4*(x^4/100 - (25*Log[10])/(2*(4 - x)*x) - (x^2*Log[2*E^(x^2/25)])/2))/25
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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])
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.68 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
default | \(\ln \left (2 \,{\mathrm e}^{\frac {x^{2}}{25}}\right )^{2}+4 \ln \left (10\right ) \left (-\frac {1}{8 \left (x -4\right )}+\frac {1}{8 x}\right )\) | \(30\) |
parts | \(\ln \left (2 \,{\mathrm e}^{\frac {x^{2}}{25}}\right )^{2}+4 \ln \left (10\right ) \left (-\frac {1}{8 \left (x -4\right )}+\frac {1}{8 x}\right )\) | \(30\) |
parallelrisch | \(-\frac {-25 x^{2} \ln \left (2 \,{\mathrm e}^{\frac {x^{2}}{25}}\right )^{2}+100 x \ln \left (2 \,{\mathrm e}^{\frac {x^{2}}{25}}\right )^{2}+50 \ln \left (10\right )}{25 x \left (x -4\right )}\) | \(46\) |
risch | \(\frac {2 x^{2} \ln \left ({\mathrm e}^{\frac {x^{2}}{25}}\right )}{25}-\frac {-10000 x \ln \left (2\right )^{2}+800 x^{3} \ln \left (2\right )-200 x^{4} \ln \left (2\right )+2500 x^{2} \ln \left (2\right )^{2}+5000 \ln \left (5\right )+4 x^{6}+5000 \ln \left (2\right )-16 x^{5}}{2500 \left (x -4\right ) x}\) | \(73\) |
Input:
int(((4*x^5-32*x^4+64*x^3)*ln(2*exp(1/25*x^2))+(100*x-200)*ln(10))/(25*x^4 -200*x^3+400*x^2),x,method=_RETURNVERBOSE)
Output:
ln(2*exp(1/25*x^2))^2+4*ln(10)*(-1/8/(x-4)+1/8/x)
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{400 x^2-200 x^3+25 x^4} \, dx=\frac {x^{6} - 4 \, x^{5} + 50 \, {\left (x^{4} - 4 \, x^{3}\right )} \log \left (2\right ) - 1250 \, \log \left (10\right )}{625 \, {\left (x^{2} - 4 \, x\right )}} \] Input:
integrate(((4*x^5-32*x^4+64*x^3)*log(2*exp(1/25*x^2))+(100*x-200)*log(10)) /(25*x^4-200*x^3+400*x^2),x, algorithm="fricas")
Output:
1/625*(x^6 - 4*x^5 + 50*(x^4 - 4*x^3)*log(2) - 1250*log(10))/(x^2 - 4*x)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{400 x^2-200 x^3+25 x^4} \, dx=\frac {x^{4}}{625} + \frac {2 x^{2} \log {\left (2 \right )}}{25} - \frac {2 \log {\left (10 \right )}}{x^{2} - 4 x} \] Input:
integrate(((4*x**5-32*x**4+64*x**3)*ln(2*exp(1/25*x**2))+(100*x-200)*ln(10 ))/(25*x**4-200*x**3+400*x**2),x)
Output:
x**4/625 + 2*x**2*log(2)/25 - 2*log(10)/(x**2 - 4*x)
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (24) = 48\).
Time = 0.04 (sec) , antiderivative size = 170, normalized size of antiderivative = 6.30 \[ \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{400 x^2-200 x^3+25 x^4} \, dx=-\frac {1}{625} \, x^{4} + \frac {1}{4} \, {\left (\frac {4 \, {\left (x - 2\right )}}{x^{2} - 4 \, x} + \log \left (x - 4\right ) - \log \left (x\right )\right )} \log \left (10\right ) - \frac {1}{4} \, {\left (\frac {4}{x - 4} + \log \left (x - 4\right ) - \log \left (x\right )\right )} \log \left (10\right ) - \frac {64}{625} \, {\left (3 \, x^{2} - 80\right )} \log \left (x - 4\right ) + \frac {256}{625} \, {\left (x^{2} - 32\right )} \log \left (x - 4\right ) - \frac {64}{625} \, {\left (x^{2} - 48\right )} \log \left (x - 4\right ) + \frac {2}{25} \, {\left (x^{2} + 16 \, x - \frac {128}{x - 4} + 96 \, \log \left (x - 4\right )\right )} \log \left (2 \, e^{\left (\frac {1}{25} \, x^{2}\right )}\right ) - \frac {32}{25} \, {\left (x - \frac {16}{x - 4} + 8 \, \log \left (x - 4\right )\right )} \log \left (2 \, e^{\left (\frac {1}{25} \, x^{2}\right )}\right ) - \frac {64}{25} \, {\left (\frac {4}{x - 4} - \log \left (x - 4\right )\right )} \log \left (2 \, e^{\left (\frac {1}{25} \, x^{2}\right )}\right ) \] Input:
integrate(((4*x^5-32*x^4+64*x^3)*log(2*exp(1/25*x^2))+(100*x-200)*log(10)) /(25*x^4-200*x^3+400*x^2),x, algorithm="maxima")
Output:
-1/625*x^4 + 1/4*(4*(x - 2)/(x^2 - 4*x) + log(x - 4) - log(x))*log(10) - 1 /4*(4/(x - 4) + log(x - 4) - log(x))*log(10) - 64/625*(3*x^2 - 80)*log(x - 4) + 256/625*(x^2 - 32)*log(x - 4) - 64/625*(x^2 - 48)*log(x - 4) + 2/25* (x^2 + 16*x - 128/(x - 4) + 96*log(x - 4))*log(2*e^(1/25*x^2)) - 32/25*(x - 16/(x - 4) + 8*log(x - 4))*log(2*e^(1/25*x^2)) - 64/25*(4/(x - 4) - log( x - 4))*log(2*e^(1/25*x^2))
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{400 x^2-200 x^3+25 x^4} \, dx=\frac {1}{625} \, x^{4} + \frac {2}{25} \, x^{2} \log \left (2\right ) - \frac {2 \, \log \left (10\right )}{x^{2} - 4 \, x} \] Input:
integrate(((4*x^5-32*x^4+64*x^3)*log(2*exp(1/25*x^2))+(100*x-200)*log(10)) /(25*x^4-200*x^3+400*x^2),x, algorithm="giac")
Output:
1/625*x^4 + 2/25*x^2*log(2) - 2*log(10)/(x^2 - 4*x)
Time = 2.57 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{400 x^2-200 x^3+25 x^4} \, dx=\frac {2\,x^2\,\ln \left (2\right )}{25}+\frac {x^4}{625}-\frac {2\,\ln \left (10\right )}{x\,\left (x-4\right )} \] Input:
int((log(10)*(100*x - 200) + log(2*exp(x^2/25))*(64*x^3 - 32*x^4 + 4*x^5)) /(400*x^2 - 200*x^3 + 25*x^4),x)
Output:
(2*x^2*log(2))/25 + x^4/625 - (2*log(10))/(x*(x - 4))
Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {(-200+100 x) \log (10)+\left (64 x^3-32 x^4+4 x^5\right ) \log \left (2 e^{\frac {x^2}{25}}\right )}{400 x^2-200 x^3+25 x^4} \, dx=\frac {\mathrm {log}\left (2 e^{\frac {x^{2}}{25}}\right )^{2} x^{2}-4 \mathrm {log}\left (2 e^{\frac {x^{2}}{25}}\right )^{2} x -2 \,\mathrm {log}\left (10\right )}{x \left (x -4\right )} \] Input:
int(((4*x^5-32*x^4+64*x^3)*log(2*exp(1/25*x^2))+(100*x-200)*log(10))/(25*x ^4-200*x^3+400*x^2),x)
Output:
(log(2*e**(x**2/25))**2*x**2 - 4*log(2*e**(x**2/25))**2*x - 2*log(10))/(x* (x - 4))