Integrand size = 29, antiderivative size = 197 \[ \int \frac {2+3 x}{\sqrt [3]{4+3 x^2} \left (-12+52 x+9 x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\frac {10 \sqrt [3]{2}}{\sqrt {3} 7^{2/3}}-\frac {\sqrt [3]{2} \sqrt {3} x}{7^{2/3}}+\frac {\sqrt [3]{4+3 x^2}}{\sqrt {3}}}{\sqrt [3]{4+3 x^2}}\right )}{14 \sqrt [3]{14}}+\frac {\log \left (-10 \sqrt [3]{14}+3 \sqrt [3]{14} x+14 \sqrt [3]{4+3 x^2}\right )}{14 \sqrt [3]{14}}-\frac {\log \left (100\ 14^{2/3}-60\ 14^{2/3} x+9\ 14^{2/3} x^2+\left (140 \sqrt [3]{14}-42 \sqrt [3]{14} x\right ) \sqrt [3]{4+3 x^2}+196 \left (4+3 x^2\right )^{2/3}\right )}{28 \sqrt [3]{14}} \]
-1/196*3^(1/2)*arctan((10/21*2^(1/3)*3^(1/2)*7^(1/3)-1/7*2^(1/3)*3^(1/2)*x *7^(1/3)+1/3*(3*x^2+4)^(1/3)*3^(1/2))/(3*x^2+4)^(1/3))*14^(2/3)+1/196*ln(- 10*14^(1/3)+3*14^(1/3)*x+14*(3*x^2+4)^(1/3))*14^(2/3)-1/392*ln(100*14^(2/3 )-60*14^(2/3)*x+9*14^(2/3)*x^2+(140*14^(1/3)-42*14^(1/3)*x)*(3*x^2+4)^(1/3 )+196*(3*x^2+4)^(2/3))*14^(2/3)
Time = 0.27 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.80 \[ \int \frac {2+3 x}{\sqrt [3]{4+3 x^2} \left (-12+52 x+9 x^2\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {10 \sqrt [3]{14}-3 \sqrt [3]{14} x+7 \sqrt [3]{4+3 x^2}}{7 \sqrt {3} \sqrt [3]{4+3 x^2}}\right )-2 \log \left (-10 \sqrt [3]{14}+3 \sqrt [3]{14} x+14 \sqrt [3]{4+3 x^2}\right )+\log \left (100\ 14^{2/3}-60\ 14^{2/3} x+9\ 14^{2/3} x^2+196 \left (4+3 x^2\right )^{2/3}+14 (10-3 x) \sqrt [3]{56+42 x^2}\right )}{28 \sqrt [3]{14}} \]
-1/28*(2*Sqrt[3]*ArcTan[(10*14^(1/3) - 3*14^(1/3)*x + 7*(4 + 3*x^2)^(1/3)) /(7*Sqrt[3]*(4 + 3*x^2)^(1/3))] - 2*Log[-10*14^(1/3) + 3*14^(1/3)*x + 14*( 4 + 3*x^2)^(1/3)] + Log[100*14^(2/3) - 60*14^(2/3)*x + 9*14^(2/3)*x^2 + 19 6*(4 + 3*x^2)^(2/3) + 14*(10 - 3*x)*(56 + 42*x^2)^(1/3)])/14^(1/3)
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.
| \(\displaystyle \int \frac {3 x+2}{\sqrt [3]{3 x^2+4} \left (9 x^2+52 x-12\right )} \, dx\) |
| \(\Big \downarrow \) 1376 |
| \(\displaystyle \int \frac {3 x+2}{\sqrt [3]{3 x^2+4} \left (9 x^2+52 x-12\right )}dx\) |
3.25.32.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))*((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Unintegrable[(g + h*x)*(a + c*x^2)^p*(d + e*x + f*x^2)^q, x] /; FreeQ[{a, c, d, e, f, g, h, p, q}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 13.46 (sec) , antiderivative size = 1411, normalized size of antiderivative = 7.16
1/2*RootOf(RootOf(_Z^3-196)^2+98*_Z*RootOf(_Z^3-196)+9604*_Z^2)*ln((684110 00939778*RootOf(RootOf(_Z^3-196)^2+98*_Z*RootOf(_Z^3-196)+9604*_Z^2)*RootO f(_Z^3-196)^3*x^3+4149815744330613*RootOf(RootOf(_Z^3-196)^2+98*_Z*RootOf( _Z^3-196)+9604*_Z^2)^2*RootOf(_Z^3-196)^2*x^3+1216195572262720*RootOf(Root Of(_Z^3-196)^2+98*_Z*RootOf(_Z^3-196)+9604*_Z^2)*RootOf(_Z^3-196)^3*x^2+73 774502121433120*RootOf(RootOf(_Z^3-196)^2+98*_Z*RootOf(_Z^3-196)+9604*_Z^2 )^2*RootOf(_Z^3-196)^2*x^2-1605300200763000*(3*x^2+4)^(2/3)*RootOf(RootOf( _Z^3-196)^2+98*_Z*RootOf(_Z^3-196)+9604*_Z^2)*RootOf(_Z^3-196)^2*x+2280366 697992600*RootOf(RootOf(_Z^3-196)^2+98*_Z*RootOf(_Z^3-196)+9604*_Z^2)*Root Of(_Z^3-196)^3*x+138327191477687100*RootOf(RootOf(_Z^3-196)^2+98*_Z*RootOf (_Z^3-196)+9604*_Z^2)^2*RootOf(_Z^3-196)^2*x+5351000669210000*(3*x^2+4)^(2 /3)*RootOf(RootOf(_Z^3-196)^2+98*_Z*RootOf(_Z^3-196)+9604*_Z^2)*RootOf(_Z^ 3-196)^2-49141842880500*(3*x^2+4)^(1/3)*RootOf(_Z^3-196)^2*x^2-15109141162 947363*(3*x^2+4)^(1/3)*RootOf(_Z^3-196)*RootOf(RootOf(_Z^3-196)^2+98*_Z*Ro otOf(_Z^3-196)+9604*_Z^2)*x^2+327612285870000*(3*x^2+4)^(1/3)*RootOf(_Z^3- 196)^2*x+100727607752982420*(3*x^2+4)^(1/3)*RootOf(_Z^3-196)*RootOf(RootOf (_Z^3-196)^2+98*_Z*RootOf(_Z^3-196)+9604*_Z^2)*x-174517859540250*RootOf(_Z ^3-196)*x^3-10586264653904625*RootOf(RootOf(_Z^3-196)^2+98*_Z*RootOf(_Z^3- 196)+9604*_Z^2)*x^3-546020476450000*(3*x^2+4)^(1/3)*RootOf(_Z^3-196)^2-167 879346254970700*(3*x^2+4)^(1/3)*RootOf(_Z^3-196)*RootOf(RootOf(_Z^3-196...
Exception generated. \[ \int \frac {2+3 x}{\sqrt [3]{4+3 x^2} \left (-12+52 x+9 x^2\right )} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (residue poly has multiple non-linear fac tors)
\[ \int \frac {2+3 x}{\sqrt [3]{4+3 x^2} \left (-12+52 x+9 x^2\right )} \, dx=\int \frac {3 x + 2}{\left (x + 6\right ) \left (9 x - 2\right ) \sqrt [3]{3 x^{2} + 4}}\, dx \]
\[ \int \frac {2+3 x}{\sqrt [3]{4+3 x^2} \left (-12+52 x+9 x^2\right )} \, dx=\int { \frac {3 \, x + 2}{{\left (9 \, x^{2} + 52 \, x - 12\right )} {\left (3 \, x^{2} + 4\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {2+3 x}{\sqrt [3]{4+3 x^2} \left (-12+52 x+9 x^2\right )} \, dx=\int { \frac {3 \, x + 2}{{\left (9 \, x^{2} + 52 \, x - 12\right )} {\left (3 \, x^{2} + 4\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {2+3 x}{\sqrt [3]{4+3 x^2} \left (-12+52 x+9 x^2\right )} \, dx=\int \frac {3\,x+2}{{\left (3\,x^2+4\right )}^{1/3}\,\left (9\,x^2+52\,x-12\right )} \,d x \]