Integrand size = 64, antiderivative size = 18 \[ \int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx=x^4 \left (2+\log \left (-1+e x^2-\log (9)\right )\right ) \] Output:
x^4*(ln(-2*ln(3)+x^2*exp(1)-1)+2)
Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39 \[ \int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx=2 \left (x^4+\frac {1}{2} x^4 \log \left (-1+e x^2-\log (9)\right )\right ) \] Input:
Integrate[(8*x^3 - 10*E*x^5 + 8*x^3*Log[9] + (4*x^3 - 4*E*x^5 + 4*x^3*Log[ 9])*Log[-1 + E*x^2 - Log[9]])/(1 - E*x^2 + Log[9]),x]
Output:
2*(x^4 + (x^4*Log[-1 + E*x^2 - Log[9]])/2)
Time = 0.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {6, 7276, 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.
\(\displaystyle \int \frac {-10 e x^5+8 x^3+8 x^3 \log (9)+\left (-4 e x^5+4 x^3+4 x^3 \log (9)\right ) \log \left (e x^2-1-\log (9)\right )}{-e x^2+1+\log (9)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-10 e x^5+x^3 (8+8 \log (9))+\left (-4 e x^5+4 x^3+4 x^3 \log (9)\right ) \log \left (e x^2-1-\log (9)\right )}{-e x^2+1+\log (9)}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (4 x^3 \log \left (e x^2-1-\log (9)\right )+\frac {2 x^3 \left (5 e x^2-4-4 \log (9)\right )}{e x^2-1-\log (9)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 x^4+x^4 \log \left (e x^2-1-\log (9)\right )\) |
Input:
Int[(8*x^3 - 10*E*x^5 + 8*x^3*Log[9] + (4*x^3 - 4*E*x^5 + 4*x^3*Log[9])*Lo g[-1 + E*x^2 - Log[9]])/(1 - E*x^2 + Log[9]),x]
Output:
2*x^4 + x^4*Log[-1 + E*x^2 - Log[9]]
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.41 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33
method | result | size |
norman | \(x^{4} \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right )+2 x^{4}\) | \(24\) |
risch | \(x^{4} \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right )+2 x^{4}\) | \(24\) |
parallelrisch | \(\left ({\mathrm e}^{2} \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right ) x^{4}+2 x^{4} {\mathrm e}^{2}-2-8 \ln \left (3\right )^{2}-8 \ln \left (3\right )\right ) {\mathrm e}^{-2}\) | \(48\) |
default | \(x^{4} \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right )-2 \,{\mathrm e} \left (\frac {{\mathrm e}^{-2} \left (\frac {x^{4} {\mathrm e}}{2}+2 x^{2} \ln \left (3\right )+x^{2}\right )}{2}+\frac {{\mathrm e}^{-2} \left (4 \ln \left (3\right )^{2}+4 \ln \left (3\right )+1\right ) {\mathrm e}^{-1} \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right )}{2}\right )+{\mathrm e}^{-1} \left (\frac {5 x^{4} {\mathrm e}}{2}+2 x^{2} \ln \left (3\right )+x^{2}\right )+\left ({\mathrm e}^{-1}\right )^{2} \left (4 \ln \left (3\right )^{2}+4 \ln \left (3\right )+1\right ) \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right )\) | \(136\) |
parts | \(x^{4} \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right )-2 \,{\mathrm e} \left (\frac {{\mathrm e}^{-2} \left (\frac {x^{4} {\mathrm e}}{2}+2 x^{2} \ln \left (3\right )+x^{2}\right )}{2}+\frac {{\mathrm e}^{-2} \left (4 \ln \left (3\right )^{2}+4 \ln \left (3\right )+1\right ) {\mathrm e}^{-1} \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right )}{2}\right )+{\mathrm e}^{-1} \left (\frac {5 x^{4} {\mathrm e}}{2}+2 x^{2} \ln \left (3\right )+x^{2}\right )+\left ({\mathrm e}^{-1}\right )^{2} \left (4 \ln \left (3\right )^{2}+4 \ln \left (3\right )+1\right ) \ln \left (-2 \ln \left (3\right )+x^{2} {\mathrm e}-1\right )\) | \(136\) |
Input:
int(((8*x^3*ln(3)-4*x^5*exp(1)+4*x^3)*ln(-2*ln(3)+x^2*exp(1)-1)+16*x^3*ln( 3)-10*x^5*exp(1)+8*x^3)/(2*ln(3)-x^2*exp(1)+1),x,method=_RETURNVERBOSE)
Output:
x^4*ln(-2*ln(3)+x^2*exp(1)-1)+2*x^4
Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx=x^{4} \log \left (x^{2} e - 2 \, \log \left (3\right ) - 1\right ) + 2 \, x^{4} \] Input:
integrate(((8*x^3*log(3)-4*x^5*exp(1)+4*x^3)*log(-2*log(3)+x^2*exp(1)-1)+1 6*x^3*log(3)-10*x^5*exp(1)+8*x^3)/(2*log(3)-x^2*exp(1)+1),x, algorithm="fr icas")
Output:
x^4*log(x^2*e - 2*log(3) - 1) + 2*x^4
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx=x^{4} \log {\left (e x^{2} - 2 \log {\left (3 \right )} - 1 \right )} + 2 x^{4} \] Input:
integrate(((8*x**3*ln(3)-4*x**5*exp(1)+4*x**3)*ln(-2*ln(3)+x**2*exp(1)-1)+ 16*x**3*ln(3)-10*x**5*exp(1)+8*x**3)/(2*ln(3)-x**2*exp(1)+1),x)
Output:
x**4*log(E*x**2 - 2*log(3) - 1) + 2*x**4
Leaf count of result is larger than twice the leaf count of optimal. 467 vs. \(2 (19) = 38\).
Time = 0.13 (sec) , antiderivative size = 467, normalized size of antiderivative = 25.94 \[ \int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx =\text {Too large to display} \] Input:
integrate(((8*x^3*log(3)-4*x^5*exp(1)+4*x^3)*log(-2*log(3)+x^2*exp(1)-1)+1 6*x^3*log(3)-10*x^5*exp(1)+8*x^3)/(2*log(3)-x^2*exp(1)+1),x, algorithm="ma xima")
Output:
-4*x^2*e^(-1) + 2*(2*x^2*e + (2*log(3) + 1)*log(x^2*e - 2*log(3) - 1)^2 + 2*(2*log(3) + 1)*log(x^2*e - 2*log(3) - 1))*e^(-2)*log(3) + (2*(4*log(3)^2 + 4*log(3) + 1)*e^(-3)*log(x^2*e - 2*log(3) - 1) + (x^4*e + 2*x^2*(2*log( 3) + 1))*e^(-2))*e*log(x^2*e - 2*log(3) - 1) - 4*(2*log(3) + 1)*e^(-2)*log (x^2*e - 2*log(3) - 1) - 4*(x^2*e^(-1) + (2*log(3) + 1)*e^(-2)*log(x^2*e - 2*log(3) - 1))*log(3)*log(x^2*e - 2*log(3) - 1) + 5/2*(2*(4*log(3)^2 + 4* log(3) + 1)*e^(-3)*log(x^2*e - 2*log(3) - 1) + (x^4*e + 2*x^2*(2*log(3) + 1))*e^(-2))*e - 1/2*(x^4*e^2 + 6*x^2*(2*log(3) + 1)*e + 2*(4*log(3)^2 + 4* log(3) + 1)*log(x^2*e - 2*log(3) - 1)^2 + 6*(4*log(3)^2 + 4*log(3) + 1)*lo g(x^2*e - 2*log(3) - 1))*e^(-2) + (2*x^2*e + (2*log(3) + 1)*log(x^2*e - 2* log(3) - 1)^2 + 2*(2*log(3) + 1)*log(x^2*e - 2*log(3) - 1))*e^(-2) - 8*(x^ 2*e^(-1) + (2*log(3) + 1)*e^(-2)*log(x^2*e - 2*log(3) - 1))*log(3) - 2*(x^ 2*e^(-1) + (2*log(3) + 1)*e^(-2)*log(x^2*e - 2*log(3) - 1))*log(x^2*e - 2* log(3) - 1)
Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (19) = 38\).
Time = 0.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 10.06 \[ \int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx={\left (4 \, x^{2} e + {\left (x^{2} e - 2 \, \log \left (3\right ) - 1\right )}^{2} \log \left (x^{2} e - 2 \, \log \left (3\right ) - 1\right ) + 4 \, {\left (x^{2} e - 2 \, \log \left (3\right ) - 1\right )} \log \left (3\right ) \log \left (x^{2} e - 2 \, \log \left (3\right ) - 1\right ) + 4 \, \log \left (3\right )^{2} \log \left (x^{2} e - 2 \, \log \left (3\right ) - 1\right ) + 2 \, {\left (x^{2} e - 2 \, \log \left (3\right ) - 1\right )}^{2} + 8 \, {\left (x^{2} e - 2 \, \log \left (3\right ) - 1\right )} \log \left (3\right ) + 2 \, {\left (x^{2} e - 2 \, \log \left (3\right ) - 1\right )} \log \left (x^{2} e - 2 \, \log \left (3\right ) - 1\right ) + 4 \, \log \left (3\right ) \log \left (x^{2} e - 2 \, \log \left (3\right ) - 1\right ) - 8 \, \log \left (3\right ) + \log \left (x^{2} e - 2 \, \log \left (3\right ) - 1\right ) - 4\right )} e^{\left (-2\right )} \] Input:
integrate(((8*x^3*log(3)-4*x^5*exp(1)+4*x^3)*log(-2*log(3)+x^2*exp(1)-1)+1 6*x^3*log(3)-10*x^5*exp(1)+8*x^3)/(2*log(3)-x^2*exp(1)+1),x, algorithm="gi ac")
Output:
(4*x^2*e + (x^2*e - 2*log(3) - 1)^2*log(x^2*e - 2*log(3) - 1) + 4*(x^2*e - 2*log(3) - 1)*log(3)*log(x^2*e - 2*log(3) - 1) + 4*log(3)^2*log(x^2*e - 2 *log(3) - 1) + 2*(x^2*e - 2*log(3) - 1)^2 + 8*(x^2*e - 2*log(3) - 1)*log(3 ) + 2*(x^2*e - 2*log(3) - 1)*log(x^2*e - 2*log(3) - 1) + 4*log(3)*log(x^2* e - 2*log(3) - 1) - 8*log(3) + log(x^2*e - 2*log(3) - 1) - 4)*e^(-2)
Time = 19.91 (sec) , antiderivative size = 156, normalized size of antiderivative = 8.67 \[ \int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx=x^4\,\ln \left (\mathrm {e}\,x^2-2\,\ln \left (3\right )-1\right )-4\,x^2\,{\mathrm {e}}^{-1}+2\,x^4-8\,x^2\,{\mathrm {e}}^{-1}\,\ln \left (3\right )-\ln \left (\mathrm {e}\,x^2-\ln \left (9\right )-1\right )\,{\mathrm {e}}^{-2}\,\left (4\,\ln \left (9\right )+4\right )+\ln \left (\mathrm {e}\,x^2-\ln \left (9\right )-1\right )\,{\mathrm {e}}^{-2}\,\left (10\,\ln \left (9\right )+5\,{\ln \left (9\right )}^2+5\right )-\ln \left (\mathrm {e}\,x^2-\ln \left (9\right )-1\right )\,{\mathrm {e}}^{-2}\,\left (8\,\ln \left (3\right )+8\,\ln \left (3\right )\,\ln \left (9\right )\right )+4\,x^2\,{\mathrm {e}}^{-1}\,\left (\ln \left (9\right )+1\right )-\ln \left (\mathrm {e}\,x^2-\ln \left (9\right )-1\right )\,{\mathrm {e}}^{-2}\,\left (2\,\ln \left (9\right )+{\ln \left (9\right )}^2+1\right ) \] Input:
int((log(x^2*exp(1) - 2*log(3) - 1)*(8*x^3*log(3) - 4*x^5*exp(1) + 4*x^3) - 10*x^5*exp(1) + 16*x^3*log(3) + 8*x^3)/(2*log(3) - x^2*exp(1) + 1),x)
Output:
x^4*log(x^2*exp(1) - 2*log(3) - 1) - 4*x^2*exp(-1) + 2*x^4 - 8*x^2*exp(-1) *log(3) - log(x^2*exp(1) - log(9) - 1)*exp(-2)*(4*log(9) + 4) + log(x^2*ex p(1) - log(9) - 1)*exp(-2)*(10*log(9) + 5*log(9)^2 + 5) - log(x^2*exp(1) - log(9) - 1)*exp(-2)*(8*log(3) + 8*log(3)*log(9)) + 4*x^2*exp(-1)*(log(9) + 1) - log(x^2*exp(1) - log(9) - 1)*exp(-2)*(2*log(9) + log(9)^2 + 1)
Time = 0.23 (sec) , antiderivative size = 129, normalized size of antiderivative = 7.17 \[ \int \frac {8 x^3-10 e x^5+8 x^3 \log (9)+\left (4 x^3-4 e x^5+4 x^3 \log (9)\right ) \log \left (-1+e x^2-\log (9)\right )}{1-e x^2+\log (9)} \, dx=\frac {-4 \,\mathrm {log}\left (-2 \,\mathrm {log}\left (3\right )+e \,x^{2}-1\right ) \mathrm {log}\left (3\right )^{2}-4 \,\mathrm {log}\left (-2 \,\mathrm {log}\left (3\right )+e \,x^{2}-1\right ) \mathrm {log}\left (3\right )+\mathrm {log}\left (-2 \,\mathrm {log}\left (3\right )+e \,x^{2}-1\right ) e^{2} x^{4}-\mathrm {log}\left (-2 \,\mathrm {log}\left (3\right )+e \,x^{2}-1\right )+4 \,\mathrm {log}\left (2 \,\mathrm {log}\left (3\right )-e \,x^{2}+1\right ) \mathrm {log}\left (3\right )^{2}+4 \,\mathrm {log}\left (2 \,\mathrm {log}\left (3\right )-e \,x^{2}+1\right ) \mathrm {log}\left (3\right )+\mathrm {log}\left (2 \,\mathrm {log}\left (3\right )-e \,x^{2}+1\right )+2 e^{2} x^{4}}{e^{2}} \] Input:
int(((8*x^3*log(3)-4*x^5*exp(1)+4*x^3)*log(-2*log(3)+x^2*exp(1)-1)+16*x^3* log(3)-10*x^5*exp(1)+8*x^3)/(2*log(3)-x^2*exp(1)+1),x)
Output:
( - 4*log( - 2*log(3) + e*x**2 - 1)*log(3)**2 - 4*log( - 2*log(3) + e*x**2 - 1)*log(3) + log( - 2*log(3) + e*x**2 - 1)*e**2*x**4 - log( - 2*log(3) + e*x**2 - 1) + 4*log(2*log(3) - e*x**2 + 1)*log(3)**2 + 4*log(2*log(3) - e *x**2 + 1)*log(3) + log(2*log(3) - e*x**2 + 1) + 2*e**2*x**4)/e**2