Integrand size = 75, antiderivative size = 24 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {x}{-1+x-\frac {x}{3 \log \left (5 e^{x+x^2}\right )}} \]
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3 \left (x+3 \log \left (5 e^{x+x^2}\right )\right )}{-3 x+9 (-1+x) \log \left (5 e^{x+x^2}\right )} \]
Integrate[(-3*x^2 - 6*x^3 - 9*Log[5*E^(x + x^2)]^2)/(x^2 + (6*x - 6*x^2)*L og[5*E^(x + x^2)] + (9 - 18*x + 9*x^2)*Log[5*E^(x + x^2)]^2),x]
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.
\(\displaystyle \int \frac {-6 x^3-3 x^2-9 \log ^2\left (5 e^{x^2+x}\right )}{x^2+\left (9 x^2-18 x+9\right ) \log ^2\left (5 e^{x^2+x}\right )+\left (6 x-6 x^2\right ) \log \left (5 e^{x^2+x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 \left (-2 x^3-x^2-3 \log ^2\left (5 e^{x^2+x}\right )\right )}{\left (-3 x \log \left (5 e^{x^2+x}\right )+3 \log \left (5 e^{x^2+x}\right )+x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int -\frac {2 x^3+x^2+3 \log ^2\left (5 e^{x^2+x}\right )}{\left (-3 \log \left (5 e^{x^2+x}\right ) x+x+3 \log \left (5 e^{x^2+x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -3 \int \frac {2 x^3+x^2+3 \log ^2\left (5 e^{x^2+x}\right )}{\left (-3 \log \left (5 e^{x^2+x}\right ) x+x+3 \log \left (5 e^{x^2+x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -3 \int \left (\frac {\left (6 x^3-9 x^2+4\right ) x^2}{3 (x-1)^2 \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}+\frac {2 x}{3 (x-1)^2 \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )}+\frac {1}{3 (x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \left (\frac {1}{3} \int \frac {1}{\left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\frac {1}{3} \int \frac {1}{(x-1)^2 \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\frac {2}{3} \int \frac {1}{(x-1) \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\int \frac {x^2}{\left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\frac {2}{3} \int \frac {1}{(x-1)^2 \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )}dx+\frac {2}{3} \int \frac {1}{(x-1) \left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )}dx+2 \int \frac {x^3}{\left (3 \log \left (5 e^{x^2+x}\right ) x-x-3 \log \left (5 e^{x^2+x}\right )\right )^2}dx+\frac {1}{3 (1-x)}\right )\) |
Int[(-3*x^2 - 6*x^3 - 9*Log[5*E^(x + x^2)]^2)/(x^2 + (6*x - 6*x^2)*Log[5*E ^(x + x^2)] + (9 - 18*x + 9*x^2)*Log[5*E^(x + x^2)]^2),x]
3.7.86.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(46\) vs. \(2(21)=42\).
Time = 0.14 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96
method | result | size |
parallelrisch | \(-\frac {-9 x -27 \ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )}{9 \left (3 \ln \left (5 \,{\mathrm e}^{x^{2}+x}\right ) x -3 \ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )-x \right )}\) | \(47\) |
risch | \(\frac {1}{-1+x}+\frac {2 i x^{2}}{\left (-1+x \right ) \left (6 i x \ln \left (5\right )+6 i x \ln \left ({\mathrm e}^{\left (1+x \right ) x}\right )-6 i \ln \left (5\right )-2 i x -6 i \ln \left ({\mathrm e}^{\left (1+x \right ) x}\right )\right )}\) | \(57\) |
default | \(-\frac {3 \left (-\frac {x}{9}-\frac {\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )}{3}\right )}{x^{3}+x \left (\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )-x^{2}-x \right )-\frac {x}{3}-\ln \left (5 \,{\mathrm e}^{x^{2}+x}\right )+x^{2}}\) | \(61\) |
int((-9*ln(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*ln(5*exp(x^2+x))^2 +(-6*x^2+6*x)*ln(5*exp(x^2+x))+x^2),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3 \, x^{2} + 4 \, x + 3 \, \log \left (5\right )}{3 \, x^{3} + 3 \, {\left (x - 1\right )} \log \left (5\right ) - 4 \, x} \]
integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x ^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+x))+x^2),x, algorithm=\
Time = 0.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=- \frac {- 3 x^{2} - 4 x - 3 \log {\left (5 \right )}}{3 x^{3} + x \left (-4 + 3 \log {\left (5 \right )}\right ) - 3 \log {\left (5 \right )}} \]
integrate((-9*ln(5*exp(x**2+x))**2-6*x**3-3*x**2)/((9*x**2-18*x+9)*ln(5*ex p(x**2+x))**2+(-6*x**2+6*x)*ln(5*exp(x**2+x))+x**2),x)
Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3 \, x^{2} + 4 \, x + 3 \, \log \left (5\right )}{3 \, x^{3} + x {\left (3 \, \log \left (5\right ) - 4\right )} - 3 \, \log \left (5\right )} \]
integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x ^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+x))+x^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3 \, x^{2} + 4 \, x + 3 \, \log \left (5\right )}{3 \, x^{3} + 3 \, x \log \left (5\right ) - 4 \, x - 3 \, \log \left (5\right )} \]
integrate((-9*log(5*exp(x^2+x))^2-6*x^3-3*x^2)/((9*x^2-18*x+9)*log(5*exp(x ^2+x))^2+(-6*x^2+6*x)*log(5*exp(x^2+x))+x^2),x, algorithm=\
Time = 8.63 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-3 x^2-6 x^3-9 \log ^2\left (5 e^{x+x^2}\right )}{x^2+\left (6 x-6 x^2\right ) \log \left (5 e^{x+x^2}\right )+\left (9-18 x+9 x^2\right ) \log ^2\left (5 e^{x+x^2}\right )} \, dx=\frac {3\,x^2+4\,x+\ln \left (125\right )}{3\,x^3+\left (\ln \left (125\right )-4\right )\,x-\ln \left (125\right )} \]