Integrand size = 72, antiderivative size = 26 \[ \int \frac {9-6 x-8 x^2+2 x^3+e \left (6 x-x^2\right )}{9 x-6 x^2-2 x^3+x^4+e \left (3 x^2-x^3\right )+\left (45-30 x+5 x^2\right ) \log (2)} \, dx=\log \left (\frac {1}{5} \left (x+\frac {(e-x) x^2}{3-x}\right )+\log (2)\right ) \] Output:
ln(ln(2)+1/5*x^2*(exp(1)-x)/(3-x)+1/5*x)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.12 (sec) , antiderivative size = 212, normalized size of antiderivative = 8.15 \[ \int \frac {9-6 x-8 x^2+2 x^3+e \left (6 x-x^2\right )}{9 x-6 x^2-2 x^3+x^4+e \left (3 x^2-x^3\right )+\left (45-30 x+5 x^2\right ) \log (2)} \, dx=-\log (-3+x)+\frac {\text {RootSum}\left [-27+9 e-30 \text {$\#$1}+6 e \text {$\#$1}-\log (32) \text {$\#$1}-10 \text {$\#$1}^2+e \text {$\#$1}^2-\text {$\#$1}^3\&,\frac {270 \log (-3+x-\text {$\#$1})-144 e \log (-3+x-\text {$\#$1})+18 e^2 \log (-3+x-\text {$\#$1})-e \log (32768) \log (-3+x-\text {$\#$1})+\log (35184372088832) \log (-3+x-\text {$\#$1})+180 \log (-3+x-\text {$\#$1}) \text {$\#$1}-78 e \log (-3+x-\text {$\#$1}) \text {$\#$1}+6 e^2 \log (-3+x-\text {$\#$1}) \text {$\#$1}+27 \log (-3+x-\text {$\#$1}) \text {$\#$1}^2-9 e \log (-3+x-\text {$\#$1}) \text {$\#$1}^2}{-30+6 e-\log (32)-20 \text {$\#$1}+2 e \text {$\#$1}-3 \text {$\#$1}^2}\&\right ]}{3 (-3+e)} \] Input:
Integrate[(9 - 6*x - 8*x^2 + 2*x^3 + E*(6*x - x^2))/(9*x - 6*x^2 - 2*x^3 + x^4 + E*(3*x^2 - x^3) + (45 - 30*x + 5*x^2)*Log[2]),x]
Output:
-Log[-3 + x] + RootSum[-27 + 9*E - 30*#1 + 6*E*#1 - Log[32]*#1 - 10*#1^2 + E*#1^2 - #1^3 & , (270*Log[-3 + x - #1] - 144*E*Log[-3 + x - #1] + 18*E^2 *Log[-3 + x - #1] - E*Log[32768]*Log[-3 + x - #1] + Log[35184372088832]*Lo g[-3 + x - #1] + 180*Log[-3 + x - #1]*#1 - 78*E*Log[-3 + x - #1]*#1 + 6*E^ 2*Log[-3 + x - #1]*#1 + 27*Log[-3 + x - #1]*#1^2 - 9*E*Log[-3 + x - #1]*#1 ^2)/(-30 + 6*E - Log[32] - 20*#1 + 2*E*#1 - 3*#1^2) & ]/(3*(-3 + E))
Time = 0.53 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {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.
\(\displaystyle \int \frac {2 x^3-8 x^2+e \left (6 x-x^2\right )-6 x+9}{x^4-2 x^3-6 x^2+\left (5 x^2-30 x+45\right ) \log (2)+e \left (3 x^2-x^3\right )+9 x} \, dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (\frac {3 x^2+2 (1-e) x-3+\log (32)}{x^3+(1-e) x^2-x (3-\log (32))-15 \log (2)}+\frac {1}{3-x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log \left (-x^3-(1-e) x^2+x (3-\log (32))+15 \log (2)\right )-\log (3-x)\) |
Input:
Int[(9 - 6*x - 8*x^2 + 2*x^3 + E*(6*x - x^2))/(9*x - 6*x^2 - 2*x^3 + x^4 + E*(3*x^2 - x^3) + (45 - 30*x + 5*x^2)*Log[2]),x]
Output:
-Log[3 - x] + Log[-((1 - E)*x^2) - x^3 + 15*Log[2] + x*(3 - Log[32])]
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 = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35
method | result | size |
default | \(\ln \left (-x^{2} {\mathrm e}+x^{3}+5 x \ln \left (2\right )+x^{2}-15 \ln \left (2\right )-3 x \right )-\ln \left (-3+x \right )\) | \(35\) |
parallelrisch | \(\ln \left (-x^{2} {\mathrm e}+x^{3}+5 x \ln \left (2\right )+x^{2}-15 \ln \left (2\right )-3 x \right )-\ln \left (-3+x \right )\) | \(35\) |
risch | \(-\ln \left (-x +3\right )+\ln \left (x^{3}+\left (1-{\mathrm e}\right ) x^{2}+\left (5 \ln \left (2\right )-3\right ) x -15 \ln \left (2\right )\right )\) | \(37\) |
norman | \(-\ln \left (-3+x \right )+\ln \left (x^{2} {\mathrm e}-x^{3}-5 x \ln \left (2\right )-x^{2}+15 \ln \left (2\right )+3 x \right )\) | \(38\) |
Input:
int(((-x^2+6*x)*exp(1)+2*x^3-8*x^2-6*x+9)/((5*x^2-30*x+45)*ln(2)+(-x^3+3*x ^2)*exp(1)+x^4-2*x^3-6*x^2+9*x),x,method=_RETURNVERBOSE)
Output:
ln(-x^2*exp(1)+x^3+5*x*ln(2)+x^2-15*ln(2)-3*x)-ln(-3+x)
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {9-6 x-8 x^2+2 x^3+e \left (6 x-x^2\right )}{9 x-6 x^2-2 x^3+x^4+e \left (3 x^2-x^3\right )+\left (45-30 x+5 x^2\right ) \log (2)} \, dx=\log \left (x^{3} - x^{2} e + x^{2} + 5 \, {\left (x - 3\right )} \log \left (2\right ) - 3 \, x\right ) - \log \left (x - 3\right ) \] Input:
integrate(((-x^2+6*x)*exp(1)+2*x^3-8*x^2-6*x+9)/((5*x^2-30*x+45)*log(2)+(- x^3+3*x^2)*exp(1)+x^4-2*x^3-6*x^2+9*x),x, algorithm="fricas")
Output:
log(x^3 - x^2*e + x^2 + 5*(x - 3)*log(2) - 3*x) - log(x - 3)
Time = 5.83 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {9-6 x-8 x^2+2 x^3+e \left (6 x-x^2\right )}{9 x-6 x^2-2 x^3+x^4+e \left (3 x^2-x^3\right )+\left (45-30 x+5 x^2\right ) \log (2)} \, dx=- \log {\left (x - 3 \right )} + \log {\left (x^{3} + x^{2} \cdot \left (1 - e\right ) + x \left (-3 + 5 \log {\left (2 \right )}\right ) - 15 \log {\left (2 \right )} \right )} \] Input:
integrate(((-x**2+6*x)*exp(1)+2*x**3-8*x**2-6*x+9)/((5*x**2-30*x+45)*ln(2) +(-x**3+3*x**2)*exp(1)+x**4-2*x**3-6*x**2+9*x),x)
Output:
-log(x - 3) + log(x**3 + x**2*(1 - E) + x*(-3 + 5*log(2)) - 15*log(2))
Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {9-6 x-8 x^2+2 x^3+e \left (6 x-x^2\right )}{9 x-6 x^2-2 x^3+x^4+e \left (3 x^2-x^3\right )+\left (45-30 x+5 x^2\right ) \log (2)} \, dx=\log \left (x^{3} - x^{2} {\left (e - 1\right )} + x {\left (5 \, \log \left (2\right ) - 3\right )} - 15 \, \log \left (2\right )\right ) - \log \left (x - 3\right ) \] Input:
integrate(((-x^2+6*x)*exp(1)+2*x^3-8*x^2-6*x+9)/((5*x^2-30*x+45)*log(2)+(- x^3+3*x^2)*exp(1)+x^4-2*x^3-6*x^2+9*x),x, algorithm="maxima")
Output:
log(x^3 - x^2*(e - 1) + x*(5*log(2) - 3) - 15*log(2)) - log(x - 3)
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {9-6 x-8 x^2+2 x^3+e \left (6 x-x^2\right )}{9 x-6 x^2-2 x^3+x^4+e \left (3 x^2-x^3\right )+\left (45-30 x+5 x^2\right ) \log (2)} \, dx=\log \left ({\left | x^{3} - x^{2} e + x^{2} + 5 \, x \log \left (2\right ) - 3 \, x - 15 \, \log \left (2\right ) \right |}\right ) - \log \left ({\left | x - 3 \right |}\right ) \] Input:
integrate(((-x^2+6*x)*exp(1)+2*x^3-8*x^2-6*x+9)/((5*x^2-30*x+45)*log(2)+(- x^3+3*x^2)*exp(1)+x^4-2*x^3-6*x^2+9*x),x, algorithm="giac")
Output:
log(abs(x^3 - x^2*e + x^2 + 5*x*log(2) - 3*x - 15*log(2))) - log(abs(x - 3 ))
Time = 17.18 (sec) , antiderivative size = 9651, normalized size of antiderivative = 371.19 \[ \int \frac {9-6 x-8 x^2+2 x^3+e \left (6 x-x^2\right )}{9 x-6 x^2-2 x^3+x^4+e \left (3 x^2-x^3\right )+\left (45-30 x+5 x^2\right ) \log (2)} \, dx=\text {Too large to display} \] Input:
int((exp(1)*(6*x - x^2) - 6*x - 8*x^2 + 2*x^3 + 9)/(9*x + log(2)*(5*x^2 - 30*x + 45) + exp(1)*(3*x^2 - x^3) - 6*x^2 - 2*x^3 + x^4),x)
Output:
symsum(log(837*exp(1) - 216*exp(2) + 27*exp(3) + 13140*log(2) + 72*log(32) - 3780*exp(1)*log(2) - 45*exp(3)*log(2) - 36*exp(1)*log(32) - 9*exp(2)*lo g(32) + 1440*log(2)*log(32) - x*(1908*exp(1) - 702*exp(2) + 162*exp(3) - 1 8*exp(4) + 11640*log(2) - 168*log(32) - 4320*exp(1)*log(2) + 270*exp(2)*lo g(2) - 30*exp(3)*log(2) - 72*exp(1)*log(32) + 18*exp(2)*log(32) + 6*exp(3) *log(32) + 1320*log(2)*log(32) - 24*exp(1)*log(32)^2 - 7200*log(2)^2 + 24* log(32)^2 + 120*exp(1)*log(2)*log(32) - 2322) - 10800*log(2)^2 + 36*log(32 )^2 - root(54000*z^4*exp(1)*log(2)^3*log(32) - 33000*z^4*exp(2)*log(2)^2*l og(32) - 18600*z^4*exp(1)*log(2)^2*log(32) - 11040*z^4*exp(1)*log(2)*log(3 2)^2 + 3900*z^4*exp(2)*log(2)*log(32)^2 + 3300*z^4*exp(3)*log(2)^2*log(32) + 880*z^4*exp(1)*log(2)*log(32)^3 - 180*z^4*exp(3)*log(2)*log(32)^2 - 20* z^4*exp(2)*log(2)*log(32)^3 + 84060*z^4*exp(1)*log(2)*log(32) - 25020*z^4* exp(2)*log(2)*log(32) + 7470*z^4*exp(3)*log(2)*log(32) - 540*z^4*exp(4)*lo g(2)*log(32) - 15200*z^4*exp(1)*log(2)^2*log(32)^2 + 100*z^4*exp(2)*log(2) ^2*log(32)^2 - 45200*z^4*log(2)^2*log(32)^2 - 400*z^4*log(2)^2*log(32)^3 + 756000*z^4*log(2)^3*log(32) + 250800*z^4*log(2)^2*log(32) + 21360*z^4*log (2)*log(32)^2 - 1760*z^4*log(2)*log(32)^3 + 80*z^4*log(2)*log(32)^4 + 3960 00*z^4*exp(1)*log(2)^3 + 63000*z^4*exp(2)*log(2)^3 + 43200*z^4*exp(1)*log( 2)^2 - 27000*z^4*exp(3)*log(2)^2 - 12000*z^4*exp(3)*log(2)^3 + 2925*z^4*ex p(4)*log(2)^2 - 1800*z^4*exp(2)*log(2)^2 - 288*z^4*exp(1)*log(32)^2 + 9...
Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {9-6 x-8 x^2+2 x^3+e \left (6 x-x^2\right )}{9 x-6 x^2-2 x^3+x^4+e \left (3 x^2-x^3\right )+\left (45-30 x+5 x^2\right ) \log (2)} \, dx=\mathrm {log}\left (5 \,\mathrm {log}\left (2\right ) x -15 \,\mathrm {log}\left (2\right )-e \,x^{2}+x^{3}+x^{2}-3 x \right )-\mathrm {log}\left (x -3\right ) \] Input:
int(((-x^2+6*x)*exp(1)+2*x^3-8*x^2-6*x+9)/((5*x^2-30*x+45)*log(2)+(-x^3+3* x^2)*exp(1)+x^4-2*x^3-6*x^2+9*x),x)
Output:
log(5*log(2)*x - 15*log(2) - e*x**2 + x**3 + x**2 - 3*x) - log(x - 3)