Integrand size = 66, antiderivative size = 19 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{5 x^2} \log \left (729 \left (-3+x+x^2\right )^2\right ) \]
Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{5 x^2} \log \left (729 \left (-3+x+x^2\right )^2\right ) \]
Integrate[(E^(5*x^2)*(2 + 4*x) + E^(5*x^2)*(-30*x + 10*x^2 + 10*x^3)*Log[6 561 - 4374*x - 3645*x^2 + 1458*x^3 + 729*x^4])/(-3 + x + x^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.
\(\displaystyle \int \frac {e^{5 x^2} (4 x+2)+e^{5 x^2} \left (10 x^3+10 x^2-30 x\right ) \log \left (729 x^4+1458 x^3-3645 x^2-4374 x+6561\right )}{x^2+x-3} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {4 e^{5 x^2} x}{x^2+x-3}+\frac {2 e^{5 x^2}}{x^2+x-3}+\frac {10 e^{5 x^2} x^2 \log \left (729 \left (x^2+x-3\right )^2\right )}{x^2+x-3}-\frac {30 e^{5 x^2} x \log \left (729 \left (x^2+x-3\right )^2\right )}{x^2+x-3}+\frac {10 e^{5 x^2} x^3 \log \left (729 \left (x^2+x-3\right )^2\right )}{x^2+x-3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{5 x^2} \log \left (729 \left (-x^2-x+3\right )^2\right )-\frac {30}{13} \left (13-\sqrt {13}\right ) \int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx \log \left (729 \left (-x^2-x+3\right )^2\right )+\frac {20}{13} \left (26-5 \sqrt {13}\right ) \int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx \log \left (729 \left (-x^2-x+3\right )^2\right )-\frac {10}{13} \left (13-7 \sqrt {13}\right ) \int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx \log \left (729 \left (-x^2-x+3\right )^2\right )-\frac {10}{13} \left (13+7 \sqrt {13}\right ) \int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx \log \left (729 \left (-x^2-x+3\right )^2\right )+\frac {20}{13} \left (26+5 \sqrt {13}\right ) \int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx \log \left (729 \left (-x^2-x+3\right )^2\right )-\frac {30}{13} \left (13+\sqrt {13}\right ) \int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx \log \left (729 \left (-x^2-x+3\right )^2\right )-\frac {4 \int \frac {e^{5 x^2}}{-2 x+\sqrt {13}-1}dx}{\sqrt {13}}+\frac {4}{13} \left (13-\sqrt {13}\right ) \int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx-4 \int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx+\frac {4}{13} \left (13+\sqrt {13}\right ) \int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx-\frac {4 \int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{\sqrt {13}}-4 \int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx+\frac {40}{13} \left (7-\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{-2 x+\sqrt {13}-1}dx+\frac {120}{13} \left (1-\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{-2 x+\sqrt {13}-1}dx-\frac {80}{13} \left (5-2 \sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{-2 x+\sqrt {13}-1}dx+\frac {120}{169} \left (13-\sqrt {13}\right )^2 \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{2 x-\sqrt {13}+1}dx+\frac {160}{13} \left (5-2 \sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{2 x-\sqrt {13}+1}dx-\frac {80}{13} \left (31-7 \sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{2 x-\sqrt {13}+1}dx-\frac {200}{13} \left (7-\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{2 x+\sqrt {13}+1}dx+\frac {360}{13} \left (1-\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{2 x+\sqrt {13}+1}dx-\frac {80}{13} \left (5-2 \sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{2 x+\sqrt {13}+1}dx+\frac {1440}{13} \int \frac {\int \frac {e^{5 x^2}}{2 x-\sqrt {13}+1}dx}{2 x+\sqrt {13}+1}dx+\frac {80}{13} \left (5+2 \sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{-2 x+\sqrt {13}-1}dx-\frac {40}{13} \left (7+\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{-2 x+\sqrt {13}-1}dx-\frac {120}{13} \left (1+\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{-2 x+\sqrt {13}-1}dx-\frac {240}{13} \left (7+\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{2 x-\sqrt {13}+1}dx+\frac {240}{13} \left (1+\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{2 x-\sqrt {13}+1}dx+\frac {1440}{13} \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{2 x-\sqrt {13}+1}dx-\frac {80}{13} \left (31+7 \sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{2 x+\sqrt {13}+1}dx+\frac {240}{13} \left (5+2 \sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{2 x+\sqrt {13}+1}dx+\frac {200}{13} \left (7+\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{2 x+\sqrt {13}+1}dx-\frac {120}{13} \left (1+\sqrt {13}\right ) \int \frac {\int \frac {e^{5 x^2}}{2 x+\sqrt {13}+1}dx}{2 x+\sqrt {13}+1}dx\) |
Int[(E^(5*x^2)*(2 + 4*x) + E^(5*x^2)*(-30*x + 10*x^2 + 10*x^3)*Log[6561 - 4374*x - 3645*x^2 + 1458*x^3 + 729*x^4])/(-3 + x + x^2),x]
3.13.73.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.34 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53
method | result | size |
norman | \({\mathrm e}^{5 x^{2}} \ln \left (729 x^{4}+1458 x^{3}-3645 x^{2}-4374 x +6561\right )\) | \(29\) |
parallelrisch | \({\mathrm e}^{5 x^{2}} \ln \left (729 x^{4}+1458 x^{3}-3645 x^{2}-4374 x +6561\right )\) | \(29\) |
default | \(\left (6 \ln \left (3\right )+\ln \left (\left (x^{2}+x -3\right )^{2}\right )-2 \ln \left (x^{2}+x -3\right )\right ) {\mathrm e}^{5 x^{2}}+2 \,{\mathrm e}^{5 x^{2}} \ln \left (x^{2}+x -3\right )\) | \(47\) |
risch | \(2 \,{\mathrm e}^{5 x^{2}} \ln \left (x^{2}+x -3\right )+\frac {\left (-i \pi {\operatorname {csgn}\left (i \left (x^{2}+x -3\right )\right )}^{2} \operatorname {csgn}\left (i \left (x^{2}+x -3\right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \left (x^{2}+x -3\right )\right ) {\operatorname {csgn}\left (i \left (x^{2}+x -3\right )^{2}\right )}^{2}-i \pi {\operatorname {csgn}\left (i \left (x^{2}+x -3\right )^{2}\right )}^{3}+12 \ln \left (3\right )\right ) {\mathrm e}^{5 x^{2}}}{2}\) | \(104\) |
int(((10*x^3+10*x^2-30*x)*exp(5*x^2)*ln(729*x^4+1458*x^3-3645*x^2-4374*x+6 561)+(4*x+2)*exp(5*x^2))/(x^2+x-3),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{\left (5 \, x^{2}\right )} \log \left (729 \, x^{4} + 1458 \, x^{3} - 3645 \, x^{2} - 4374 \, x + 6561\right ) \]
integrate(((10*x^3+10*x^2-30*x)*exp(5*x^2)*log(729*x^4+1458*x^3-3645*x^2-4 374*x+6561)+(4*x+2)*exp(5*x^2))/(x^2+x-3),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{5 x^{2}} \log {\left (729 x^{4} + 1458 x^{3} - 3645 x^{2} - 4374 x + 6561 \right )} \]
integrate(((10*x**3+10*x**2-30*x)*exp(5*x**2)*ln(729*x**4+1458*x**3-3645*x **2-4374*x+6561)+(4*x+2)*exp(5*x**2))/(x**2+x-3),x)
Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=6 \, e^{\left (5 \, x^{2}\right )} \log \left (3\right ) + 2 \, e^{\left (5 \, x^{2}\right )} \log \left (x^{2} + x - 3\right ) \]
integrate(((10*x^3+10*x^2-30*x)*exp(5*x^2)*log(729*x^4+1458*x^3-3645*x^2-4 374*x+6561)+(4*x+2)*exp(5*x^2))/(x^2+x-3),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx=e^{\left (5 \, x^{2}\right )} \log \left (729 \, x^{4} + 1458 \, x^{3} - 3645 \, x^{2} - 4374 \, x + 6561\right ) \]
integrate(((10*x^3+10*x^2-30*x)*exp(5*x^2)*log(729*x^4+1458*x^3-3645*x^2-4 374*x+6561)+(4*x+2)*exp(5*x^2))/(x^2+x-3),x, algorithm=\
Time = 9.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {e^{5 x^2} (2+4 x)+e^{5 x^2} \left (-30 x+10 x^2+10 x^3\right ) \log \left (6561-4374 x-3645 x^2+1458 x^3+729 x^4\right )}{-3+x+x^2} \, dx={\mathrm {e}}^{5\,x^2}\,\left (\ln \left (729\right )+\ln \left (x^4+2\,x^3-5\,x^2-6\,x+9\right )\right ) \]