Integrand size = 145, antiderivative size = 28 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=-2+x-\frac {5 e^2}{5+x-\log \left (-2+x^2 \log \left (e^{2 x}\right )\right )} \] Output:
x-5/(5+x-ln(ln(exp(x)^2)*x^2-2))*exp(2)-2
Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=x+\frac {5 e^2}{-5-x+\log \left (2 \left (-1+x^3\right )\right )} \] Input:
Integrate[(-25 - 10*x - x^2 + 25*x^3 + 10*x^4 + x^5 + E^2*(-5 - 15*x^2 + 5 *x^3) + (10 + 2*x - 10*x^3 - 2*x^4)*Log[-2 + 2*x^3] + (-1 + x^3)*Log[-2 + 2*x^3]^2)/(-25 - 10*x - x^2 + 25*x^3 + 10*x^4 + x^5 + (10 + 2*x - 10*x^3 - 2*x^4)*Log[-2 + 2*x^3] + (-1 + x^3)*Log[-2 + 2*x^3]^2),x]
Output:
x + (5*E^2)/(-5 - x + Log[2*(-1 + x^3)])
Time = 0.93 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {7292, 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.
\(\displaystyle \int \frac {x^5+10 x^4+25 x^3+\left (x^3-1\right ) \log ^2\left (2 x^3-2\right )-x^2+\left (-2 x^4-10 x^3+2 x+10\right ) \log \left (2 x^3-2\right )+e^2 \left (5 x^3-15 x^2-5\right )-10 x-25}{x^5+10 x^4+25 x^3+\left (x^3-1\right ) \log ^2\left (2 x^3-2\right )-x^2+\left (-2 x^4-10 x^3+2 x+10\right ) \log \left (2 x^3-2\right )-10 x-25} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-x^5-10 x^4-25 x^3-\left (x^3-1\right ) \log ^2\left (2 x^3-2\right )+x^2-\left (-2 x^4-10 x^3+2 x+10\right ) \log \left (2 x^3-2\right )-e^2 \left (5 x^3-15 x^2-5\right )+10 x+25}{\left (1-x^3\right ) \left (-\log \left (2 x^3-2\right )+x+5\right )^2}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {5 e^2 \left (x^3-3 x^2-1\right )}{\left (x^3-1\right ) \left (-\log \left (2 x^3-2\right )+x+5\right )^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x-\frac {5 e^2}{-\log \left (2 x^3-2\right )+x+5}\) |
Input:
Int[(-25 - 10*x - x^2 + 25*x^3 + 10*x^4 + x^5 + E^2*(-5 - 15*x^2 + 5*x^3) + (10 + 2*x - 10*x^3 - 2*x^4)*Log[-2 + 2*x^3] + (-1 + x^3)*Log[-2 + 2*x^3] ^2)/(-25 - 10*x - x^2 + 25*x^3 + 10*x^4 + x^5 + (10 + 2*x - 10*x^3 - 2*x^4 )*Log[-2 + 2*x^3] + (-1 + x^3)*Log[-2 + 2*x^3]^2),x]
Output:
x - (5*E^2)/(5 + x - Log[-2 + 2*x^3])
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 = 2.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79
method | result | size |
risch | \(x -\frac {5 \,{\mathrm e}^{2}}{5+x -\ln \left (2 x^{3}-2\right )}\) | \(22\) |
parallelrisch | \(\frac {x^{2}-\ln \left (2 x^{3}-2\right ) x -5 \,{\mathrm e}^{2}+5 x}{5+x -\ln \left (2 x^{3}-2\right )}\) | \(39\) |
norman | \(\frac {x^{2}-25+5 \ln \left (2 x^{3}-2\right )-\ln \left (2 x^{3}-2\right ) x -5 \,{\mathrm e}^{2}}{5+x -\ln \left (2 x^{3}-2\right )}\) | \(47\) |
Input:
int(((x^3-1)*ln(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*ln(2*x^3-2)+(5*x^3-15*x^ 2-5)*exp(2)+x^5+10*x^4+25*x^3-x^2-10*x-25)/((x^3-1)*ln(2*x^3-2)^2+(-2*x^4- 10*x^3+2*x+10)*ln(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x,method=_RETURN VERBOSE)
Output:
x-5*exp(2)/(5+x-ln(2*x^3-2))
Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=\frac {x^{2} - x \log \left (2 \, x^{3} - 2\right ) + 5 \, x - 5 \, e^{2}}{x - \log \left (2 \, x^{3} - 2\right ) + 5} \] Input:
integrate(((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+(5*x ^3-15*x^2-5)*exp(2)+x^5+10*x^4+25*x^3-x^2-10*x-25)/((x^3-1)*log(2*x^3-2)^2 +(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x, alg orithm="fricas")
Output:
(x^2 - x*log(2*x^3 - 2) + 5*x - 5*e^2)/(x - log(2*x^3 - 2) + 5)
Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=x + \frac {5 e^{2}}{- x + \log {\left (2 x^{3} - 2 \right )} - 5} \] Input:
integrate(((x**3-1)*ln(2*x**3-2)**2+(-2*x**4-10*x**3+2*x+10)*ln(2*x**3-2)+ (5*x**3-15*x**2-5)*exp(2)+x**5+10*x**4+25*x**3-x**2-10*x-25)/((x**3-1)*ln( 2*x**3-2)**2+(-2*x**4-10*x**3+2*x+10)*ln(2*x**3-2)+x**5+10*x**4+25*x**3-x* *2-10*x-25),x)
Output:
x + 5*exp(2)/(-x + log(2*x**3 - 2) - 5)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (26) = 52\).
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.04 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=\frac {x^{2} - x {\left (\log \left (2\right ) - 5\right )} - x \log \left (x^{2} + x + 1\right ) - x \log \left (x - 1\right ) - 5 \, e^{2}}{x - \log \left (2\right ) - \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) + 5} \] Input:
integrate(((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+(5*x ^3-15*x^2-5)*exp(2)+x^5+10*x^4+25*x^3-x^2-10*x-25)/((x^3-1)*log(2*x^3-2)^2 +(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x, alg orithm="maxima")
Output:
(x^2 - x*(log(2) - 5) - x*log(x^2 + x + 1) - x*log(x - 1) - 5*e^2)/(x - lo g(2) - log(x^2 + x + 1) - log(x - 1) + 5)
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=\frac {x^{2} - x \log \left (2 \, x^{3} - 2\right ) + 5 \, x - 5 \, e^{2}}{x - \log \left (2 \, x^{3} - 2\right ) + 5} \] Input:
integrate(((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+(5*x ^3-15*x^2-5)*exp(2)+x^5+10*x^4+25*x^3-x^2-10*x-25)/((x^3-1)*log(2*x^3-2)^2 +(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x, alg orithm="giac")
Output:
(x^2 - x*log(2*x^3 - 2) + 5*x - 5*e^2)/(x - log(2*x^3 - 2) + 5)
Time = 2.52 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=x-\frac {5\,{\mathrm {e}}^2}{x-\ln \left (2\,x^3-2\right )+5} \] Input:
int((log(2*x^3 - 2)*(2*x - 10*x^3 - 2*x^4 + 10) - 10*x - exp(2)*(15*x^2 - 5*x^3 + 5) - x^2 + 25*x^3 + 10*x^4 + x^5 + log(2*x^3 - 2)^2*(x^3 - 1) - 25 )/(log(2*x^3 - 2)*(2*x - 10*x^3 - 2*x^4 + 10) - 10*x - x^2 + 25*x^3 + 10*x ^4 + x^5 + log(2*x^3 - 2)^2*(x^3 - 1) - 25),x)
Output:
x - (5*exp(2))/(x - log(2*x^3 - 2) + 5)
Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.89 \[ \int \frac {-25-10 x-x^2+25 x^3+10 x^4+x^5+e^2 \left (-5-15 x^2+5 x^3\right )+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )}{-25-10 x-x^2+25 x^3+10 x^4+x^5+\left (10+2 x-10 x^3-2 x^4\right ) \log \left (-2+2 x^3\right )+\left (-1+x^3\right ) \log ^2\left (-2+2 x^3\right )} \, dx=\frac {-\mathrm {log}\left (x^{2}+x +1\right ) \mathrm {log}\left (2 x^{3}-2\right )+\mathrm {log}\left (x^{2}+x +1\right ) x +5 \,\mathrm {log}\left (x^{2}+x +1\right )+\mathrm {log}\left (2 x^{3}-2\right )^{2}-\mathrm {log}\left (2 x^{3}-2\right ) \mathrm {log}\left (x -1\right )-10 \,\mathrm {log}\left (2 x^{3}-2\right )+\mathrm {log}\left (x -1\right ) x +5 \,\mathrm {log}\left (x -1\right )+5 e^{2}-x^{2}+25}{\mathrm {log}\left (2 x^{3}-2\right )-x -5} \] Input:
int(((x^3-1)*log(2*x^3-2)^2+(-2*x^4-10*x^3+2*x+10)*log(2*x^3-2)+(5*x^3-15* x^2-5)*exp(2)+x^5+10*x^4+25*x^3-x^2-10*x-25)/((x^3-1)*log(2*x^3-2)^2+(-2*x ^4-10*x^3+2*x+10)*log(2*x^3-2)+x^5+10*x^4+25*x^3-x^2-10*x-25),x)
Output:
( - log(x**2 + x + 1)*log(2*x**3 - 2) + log(x**2 + x + 1)*x + 5*log(x**2 + x + 1) + log(2*x**3 - 2)**2 - log(2*x**3 - 2)*log(x - 1) - 10*log(2*x**3 - 2) + log(x - 1)*x + 5*log(x - 1) + 5*e**2 - x**2 + 25)/(log(2*x**3 - 2) - x - 5)