Integrand size = 32, antiderivative size = 118 \[ \int (1+2 x)^2 \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {235 (1-6 x) \sqrt {2-x+3 x^2}}{1296}+\frac {1}{5} (1+2 x)^2 \left (2-x+3 x^2\right )^{3/2}+\frac {1}{9} (1+2 x)^3 \left (2-x+3 x^2\right )^{3/2}+\frac {1}{810} (25+306 x) \left (2-x+3 x^2\right )^{3/2}+\frac {5405 \text {arcsinh}\left (\frac {1-6 x}{\sqrt {23}}\right )}{2592 \sqrt {3}} \]
1/5*(1+2*x)^2*(3*x^2-x+2)^(3/2)+1/9*(1+2*x)^3*(3*x^2-x+2)^(3/2)+1/810*(25+ 306*x)*(3*x^2-x+2)^(3/2)+5405/7776*arcsinh(1/23*(1-6*x)*23^(1/2))*3^(1/2)+ 235/1296*(1-6*x)*(3*x^2-x+2)^(1/2)
Time = 0.34 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64 \[ \int (1+2 x)^2 \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {6 \sqrt {2-x+3 x^2} \left (5607+14638 x+22344 x^2+33552 x^3+35712 x^4+17280 x^5\right )+27025 \sqrt {3} \log \left (1-6 x+2 \sqrt {6-3 x+9 x^2}\right )}{38880} \]
(6*Sqrt[2 - x + 3*x^2]*(5607 + 14638*x + 22344*x^2 + 33552*x^3 + 35712*x^4 + 17280*x^5) + 27025*Sqrt[3]*Log[1 - 6*x + 2*Sqrt[6 - 3*x + 9*x^2]])/3888 0
Time = 0.34 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2184, 27, 1236, 27, 1225, 1087, 1090, 222}
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.
| \(\displaystyle \int (2 x+1)^2 \sqrt {3 x^2-x+2} \left (4 x^2+3 x+1\right ) \, dx\) |
| \(\Big \downarrow \) 2184 |
| \(\displaystyle \frac {1}{72} \int -12 (1-18 x) (2 x+1)^2 \sqrt {3 x^2-x+2}dx+\frac {1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} (2 x+1)^3 \left (3 x^2-x+2\right )^{3/2}-\frac {1}{6} \int (1-18 x) (2 x+1)^2 \sqrt {3 x^2-x+2}dx\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {1}{6} \left (\frac {6}{5} (2 x+1)^2 \left (3 x^2-x+2\right )^{3/2}-\frac {1}{15} \int 12 (11-17 x) (2 x+1) \sqrt {3 x^2-x+2}dx\right )+\frac {1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {6}{5} (2 x+1)^2 \left (3 x^2-x+2\right )^{3/2}-\frac {4}{5} \int (11-17 x) (2 x+1) \sqrt {3 x^2-x+2}dx\right )+\frac {1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {1}{6} \left (\frac {6}{5} (2 x+1)^2 \left (3 x^2-x+2\right )^{3/2}-\frac {4}{5} \left (\frac {1175}{72} \int \sqrt {3 x^2-x+2}dx-\frac {1}{108} (306 x+25) \left (3 x^2-x+2\right )^{3/2}\right )\right )+\frac {1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {1}{6} \left (\frac {6}{5} (2 x+1)^2 \left (3 x^2-x+2\right )^{3/2}-\frac {4}{5} \left (\frac {1175}{72} \left (\frac {23}{24} \int \frac {1}{\sqrt {3 x^2-x+2}}dx-\frac {1}{12} (1-6 x) \sqrt {3 x^2-x+2}\right )-\frac {1}{108} (306 x+25) \left (3 x^2-x+2\right )^{3/2}\right )\right )+\frac {1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3\) |
| \(\Big \downarrow \) 1090 |
| \(\displaystyle \frac {1}{6} \left (\frac {6}{5} (2 x+1)^2 \left (3 x^2-x+2\right )^{3/2}-\frac {4}{5} \left (\frac {1175}{72} \left (\frac {1}{24} \sqrt {\frac {23}{3}} \int \frac {1}{\sqrt {\frac {1}{23} (6 x-1)^2+1}}d(6 x-1)-\frac {1}{12} (1-6 x) \sqrt {3 x^2-x+2}\right )-\frac {1}{108} (306 x+25) \left (3 x^2-x+2\right )^{3/2}\right )\right )+\frac {1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{6} \left (\frac {6}{5} (2 x+1)^2 \left (3 x^2-x+2\right )^{3/2}-\frac {4}{5} \left (\frac {1175}{72} \left (\frac {23 \text {arcsinh}\left (\frac {6 x-1}{\sqrt {23}}\right )}{24 \sqrt {3}}-\frac {1}{12} (1-6 x) \sqrt {3 x^2-x+2}\right )-\frac {1}{108} (306 x+25) \left (3 x^2-x+2\right )^{3/2}\right )\right )+\frac {1}{9} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^3\) |
((1 + 2*x)^3*(2 - x + 3*x^2)^(3/2))/9 + ((6*(1 + 2*x)^2*(2 - x + 3*x^2)^(3 /2))/5 - (4*(-1/108*((25 + 306*x)*(2 - x + 3*x^2)^(3/2)) + (1175*(-1/12*(( 1 - 6*x)*Sqrt[2 - x + 3*x^2]) + (23*ArcSinh[(-1 + 6*x)/Sqrt[23]])/(24*Sqrt [3])))/72))/5)/6
3.3.9.3.1 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]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1 )*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m *(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ {a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q - 1) - c *d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[ q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && Pol yQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && !(IGt Q[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Time = 0.67 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.47
| method | result | size |
| risch | \(\frac {\left (17280 x^{5}+35712 x^{4}+33552 x^{3}+22344 x^{2}+14638 x +5607\right ) \sqrt {3 x^{2}-x +2}}{6480}-\frac {5405 \sqrt {3}\, \operatorname {arcsinh}\left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{7776}\) | \(55\) |
| trager | \(\left (\frac {8}{3} x^{5}+\frac {248}{45} x^{4}+\frac {233}{45} x^{3}+\frac {931}{270} x^{2}+\frac {7319}{3240} x +\frac {623}{720}\right ) \sqrt {3 x^{2}-x +2}-\frac {5405 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {3 x^{2}-x +2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )\right )}{7776}\) | \(81\) |
| default | \(-\frac {235 \left (-1+6 x \right ) \sqrt {3 x^{2}-x +2}}{1296}-\frac {5405 \sqrt {3}\, \operatorname {arcsinh}\left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{7776}+\frac {277 \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}{810}+\frac {8 x^{3} \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}{9}+\frac {32 x^{2} \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}{15}+\frac {83 x \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}{45}\) | \(98\) |
1/6480*(17280*x^5+35712*x^4+33552*x^3+22344*x^2+14638*x+5607)*(3*x^2-x+2)^ (1/2)-5405/7776*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))
Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66 \[ \int (1+2 x)^2 \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {1}{6480} \, {\left (17280 \, x^{5} + 35712 \, x^{4} + 33552 \, x^{3} + 22344 \, x^{2} + 14638 \, x + 5607\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {5405}{15552} \, \sqrt {3} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} - x + 2} {\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) \]
1/6480*(17280*x^5 + 35712*x^4 + 33552*x^3 + 22344*x^2 + 14638*x + 5607)*sq rt(3*x^2 - x + 2) + 5405/15552*sqrt(3)*log(4*sqrt(3)*sqrt(3*x^2 - x + 2)*( 6*x - 1) - 72*x^2 + 24*x - 25)
Time = 0.47 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.59 \[ \int (1+2 x)^2 \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\sqrt {3 x^{2} - x + 2} \cdot \left (\frac {8 x^{5}}{3} + \frac {248 x^{4}}{45} + \frac {233 x^{3}}{45} + \frac {931 x^{2}}{270} + \frac {7319 x}{3240} + \frac {623}{720}\right ) - \frac {5405 \sqrt {3} \operatorname {asinh}{\left (\frac {6 \sqrt {23} \left (x - \frac {1}{6}\right )}{23} \right )}}{7776} \]
sqrt(3*x**2 - x + 2)*(8*x**5/3 + 248*x**4/45 + 233*x**3/45 + 931*x**2/270 + 7319*x/3240 + 623/720) - 5405*sqrt(3)*asinh(6*sqrt(23)*(x - 1/6)/23)/777 6
Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92 \[ \int (1+2 x)^2 \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {8}{9} \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}} x^{3} + \frac {32}{15} \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}} x^{2} + \frac {83}{45} \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}} x + \frac {277}{810} \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}} - \frac {235}{216} \, \sqrt {3 \, x^{2} - x + 2} x - \frac {5405}{7776} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (6 \, x - 1\right )}\right ) + \frac {235}{1296} \, \sqrt {3 \, x^{2} - x + 2} \]
8/9*(3*x^2 - x + 2)^(3/2)*x^3 + 32/15*(3*x^2 - x + 2)^(3/2)*x^2 + 83/45*(3 *x^2 - x + 2)^(3/2)*x + 277/810*(3*x^2 - x + 2)^(3/2) - 235/216*sqrt(3*x^2 - x + 2)*x - 5405/7776*sqrt(3)*arcsinh(1/23*sqrt(23)*(6*x - 1)) + 235/129 6*sqrt(3*x^2 - x + 2)
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int (1+2 x)^2 \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {1}{6480} \, {\left (2 \, {\left (12 \, {\left (6 \, {\left (8 \, {\left (15 \, x + 31\right )} x + 233\right )} x + 931\right )} x + 7319\right )} x + 5607\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {5405}{7776} \, \sqrt {3} \log \left (-2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} - x + 2}\right )} + 1\right ) \]
1/6480*(2*(12*(6*(8*(15*x + 31)*x + 233)*x + 931)*x + 7319)*x + 5607)*sqrt (3*x^2 - x + 2) + 5405/7776*sqrt(3)*log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) + 1)
Time = 14.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.30 \[ \int (1+2 x)^2 \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {32\,x^2\,{\left (3\,x^2-x+2\right )}^{3/2}}{15}+\frac {8\,x^3\,{\left (3\,x^2-x+2\right )}^{3/2}}{9}-\frac {2783\,\sqrt {3}\,\ln \left (\sqrt {3\,x^2-x+2}+\frac {\sqrt {3}\,\left (3\,x-\frac {1}{2}\right )}{3}\right )}{3240}-\frac {121\,\left (\frac {x}{2}-\frac {1}{12}\right )\,\sqrt {3\,x^2-x+2}}{45}+\frac {277\,\sqrt {3\,x^2-x+2}\,\left (72\,x^2-6\,x+45\right )}{19440}+\frac {83\,x\,{\left (3\,x^2-x+2\right )}^{3/2}}{45}+\frac {6371\,\sqrt {3}\,\ln \left (2\,\sqrt {3\,x^2-x+2}+\frac {\sqrt {3}\,\left (6\,x-1\right )}{3}\right )}{38880} \]
(32*x^2*(3*x^2 - x + 2)^(3/2))/15 + (8*x^3*(3*x^2 - x + 2)^(3/2))/9 - (278 3*3^(1/2)*log((3*x^2 - x + 2)^(1/2) + (3^(1/2)*(3*x - 1/2))/3))/3240 - (12 1*(x/2 - 1/12)*(3*x^2 - x + 2)^(1/2))/45 + (277*(3*x^2 - x + 2)^(1/2)*(72* x^2 - 6*x + 45))/19440 + (83*x*(3*x^2 - x + 2)^(3/2))/45 + (6371*3^(1/2)*l og(2*(3*x^2 - x + 2)^(1/2) + (3^(1/2)*(6*x - 1))/3))/38880