Integrand size = 97, antiderivative size = 35 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {4}{-4+x+\frac {3+x}{x}}+\left (2-\frac {1}{3} e^5 (2-\log (2))\right ) \log \left (x^2\right ) \] Output:
4/(x-4+(3+x)/x)+ln(x^2)*(2-1/3*(2-ln(2))*exp(5))
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2}{3} \left (\frac {6 x}{3-3 x+x^2}+\left (6+e^5 (-2+\log (2))\right ) \log (x)\right ) \] Input:
Integrate[(108 - 180*x + 180*x^2 - 84*x^3 + 12*x^4 + E^5*(-36 + 72*x - 60* x^2 + 24*x^3 - 4*x^4) + E^5*(18 - 36*x + 30*x^2 - 12*x^3 + 2*x^4)*Log[2])/ (27*x - 54*x^2 + 45*x^3 - 18*x^4 + 3*x^5),x]
Output:
(2*((6*x)/(3 - 3*x + x^2) + (6 + E^5*(-2 + Log[2]))*Log[x]))/3
Time = 0.48 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {2026, 2462, 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.
\(\displaystyle \int \frac {12 x^4-84 x^3+180 x^2+e^5 \left (-4 x^4+24 x^3-60 x^2+72 x-36\right )+e^5 \left (2 x^4-12 x^3+30 x^2-36 x+18\right ) \log (2)-180 x+108}{3 x^5-18 x^4+45 x^3-54 x^2+27 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {12 x^4-84 x^3+180 x^2+e^5 \left (-4 x^4+24 x^3-60 x^2+72 x-36\right )+e^5 \left (2 x^4-12 x^3+30 x^2-36 x+18\right ) \log (2)-180 x+108}{x \left (3 x^4-18 x^3+45 x^2-54 x+27\right )}dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (-\frac {12 (x-2)}{\left (x^2-3 x+3\right )^2}-\frac {4}{x^2-3 x+3}+\frac {2 \left (6-e^5 (2-\log (2))\right )}{3 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 x}{x^2-3 x+3}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x)\) |
Input:
Int[(108 - 180*x + 180*x^2 - 84*x^3 + 12*x^4 + E^5*(-36 + 72*x - 60*x^2 + 24*x^3 - 4*x^4) + E^5*(18 - 36*x + 30*x^2 - 12*x^3 + 2*x^4)*Log[2])/(27*x - 54*x^2 + 45*x^3 - 18*x^4 + 3*x^5),x]
Output:
(4*x)/(3 - 3*x + x^2) + (2*(6 - E^5*(2 - Log[2]))*Log[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_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {4 x}{x^{2}-3 x +3}+\frac {2 \left ({\mathrm e}^{5} \ln \left (2\right )-2 \,{\mathrm e}^{5}+6\right ) \ln \left (x \right )}{3}\) | \(30\) |
norman | \(\frac {4 x}{x^{2}-3 x +3}+\left (\frac {2 \,{\mathrm e}^{5} \ln \left (2\right )}{3}-\frac {4 \,{\mathrm e}^{5}}{3}+4\right ) \ln \left (x \right )\) | \(30\) |
risch | \(\frac {4 x}{x^{2}-3 x +3}+\frac {2 \,{\mathrm e}^{5} \ln \left (2\right ) \ln \left (x \right )}{3}-\frac {4 \,{\mathrm e}^{5} \ln \left (x \right )}{3}+4 \ln \left (x \right )\) | \(33\) |
parallelrisch | \(\frac {2 \ln \left (x \right ) {\mathrm e}^{5} \ln \left (2\right ) x^{2}-6 \ln \left (x \right ) \ln \left (2\right ) {\mathrm e}^{5} x -4 x^{2} {\mathrm e}^{5} \ln \left (x \right )+6 \,{\mathrm e}^{5} \ln \left (2\right ) \ln \left (x \right )+12 \,{\mathrm e}^{5} \ln \left (x \right ) x +12 x^{2} \ln \left (x \right )-12 \,{\mathrm e}^{5} \ln \left (x \right )-36 x \ln \left (x \right )+36 \ln \left (x \right )+12 x}{3 x^{2}-9 x +9}\) | \(83\) |
Input:
int(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*ln(2)+(-4*x^4+24*x^3-60*x^2+72*x -36)*exp(5)+12*x^4-84*x^3+180*x^2-180*x+108)/(3*x^5-18*x^4+45*x^3-54*x^2+2 7*x),x,method=_RETURNVERBOSE)
Output:
4*x/(x^2-3*x+3)+2/3*(exp(5)*ln(2)-2*exp(5)+6)*ln(x)
Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.54 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2 \, {\left ({\left ({\left (x^{2} - 3 \, x + 3\right )} e^{5} \log \left (2\right ) + 6 \, x^{2} - 2 \, {\left (x^{2} - 3 \, x + 3\right )} e^{5} - 18 \, x + 18\right )} \log \left (x\right ) + 6 \, x\right )}}{3 \, {\left (x^{2} - 3 \, x + 3\right )}} \] Input:
integrate(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*log(2)+(-4*x^4+24*x^3-60*x ^2+72*x-36)*exp(5)+12*x^4-84*x^3+180*x^2-180*x+108)/(3*x^5-18*x^4+45*x^3-5 4*x^2+27*x),x, algorithm="fricas")
Output:
2/3*(((x^2 - 3*x + 3)*e^5*log(2) + 6*x^2 - 2*(x^2 - 3*x + 3)*e^5 - 18*x + 18)*log(x) + 6*x)/(x^2 - 3*x + 3)
Time = 0.53 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {4 x}{x^{2} - 3 x + 3} + \frac {2 \left (- 2 e^{5} + 6 + e^{5} \log {\left (2 \right )}\right ) \log {\left (x \right )}}{3} \] Input:
integrate(((2*x**4-12*x**3+30*x**2-36*x+18)*exp(5)*ln(2)+(-4*x**4+24*x**3- 60*x**2+72*x-36)*exp(5)+12*x**4-84*x**3+180*x**2-180*x+108)/(3*x**5-18*x** 4+45*x**3-54*x**2+27*x),x)
Output:
4*x/(x**2 - 3*x + 3) + 2*(-2*exp(5) + 6 + exp(5)*log(2))*log(x)/3
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2}{3} \, {\left (e^{5} \log \left (2\right ) - 2 \, e^{5} + 6\right )} \log \left (x\right ) + \frac {4 \, x}{x^{2} - 3 \, x + 3} \] Input:
integrate(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*log(2)+(-4*x^4+24*x^3-60*x ^2+72*x-36)*exp(5)+12*x^4-84*x^3+180*x^2-180*x+108)/(3*x^5-18*x^4+45*x^3-5 4*x^2+27*x),x, algorithm="maxima")
Output:
2/3*(e^5*log(2) - 2*e^5 + 6)*log(x) + 4*x/(x^2 - 3*x + 3)
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2}{3} \, {\left (e^{5} \log \left (2\right ) - 2 \, e^{5} + 6\right )} \log \left ({\left | x \right |}\right ) + \frac {4 \, x}{x^{2} - 3 \, x + 3} \] Input:
integrate(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*log(2)+(-4*x^4+24*x^3-60*x ^2+72*x-36)*exp(5)+12*x^4-84*x^3+180*x^2-180*x+108)/(3*x^5-18*x^4+45*x^3-5 4*x^2+27*x),x, algorithm="giac")
Output:
2/3*(e^5*log(2) - 2*e^5 + 6)*log(abs(x)) + 4*x/(x^2 - 3*x + 3)
Time = 3.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\ln \left (x\right )\,\left (\frac {2\,{\mathrm {e}}^5\,\ln \left (2\right )}{3}-\frac {4\,{\mathrm {e}}^5}{3}+4\right )+\frac {12\,x}{3\,x^2-9\,x+9} \] Input:
int((180*x^2 - exp(5)*(60*x^2 - 72*x - 24*x^3 + 4*x^4 + 36) - 180*x - 84*x ^3 + 12*x^4 + exp(5)*log(2)*(30*x^2 - 36*x - 12*x^3 + 2*x^4 + 18) + 108)/( 27*x - 54*x^2 + 45*x^3 - 18*x^4 + 3*x^5),x)
Output:
log(x)*((2*exp(5)*log(2))/3 - (4*exp(5))/3 + 4) + (12*x)/(3*x^2 - 9*x + 9)
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.63 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right ) e^{5} x^{2}-6 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right ) e^{5} x +6 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right ) e^{5}-4 \,\mathrm {log}\left (x \right ) e^{5} x^{2}+12 \,\mathrm {log}\left (x \right ) e^{5} x -12 \,\mathrm {log}\left (x \right ) e^{5}+12 \,\mathrm {log}\left (x \right ) x^{2}-36 \,\mathrm {log}\left (x \right ) x +36 \,\mathrm {log}\left (x \right )+4 x^{2}+12}{3 x^{2}-9 x +9} \] Input:
int(((2*x^4-12*x^3+30*x^2-36*x+18)*exp(5)*log(2)+(-4*x^4+24*x^3-60*x^2+72* x-36)*exp(5)+12*x^4-84*x^3+180*x^2-180*x+108)/(3*x^5-18*x^4+45*x^3-54*x^2+ 27*x),x)
Output:
(2*(log(x)*log(2)*e**5*x**2 - 3*log(x)*log(2)*e**5*x + 3*log(x)*log(2)*e** 5 - 2*log(x)*e**5*x**2 + 6*log(x)*e**5*x - 6*log(x)*e**5 + 6*log(x)*x**2 - 18*log(x)*x + 18*log(x) + 2*x**2 + 6))/(3*(x**2 - 3*x + 3))