Integrand size = 30, antiderivative size = 93 \[ \int (1+2 x) \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {19 (1-6 x) \sqrt {2-x+3 x^2}}{2592}+\frac {2}{15} (1+2 x)^2 \left (2-x+3 x^2\right )^{3/2}+\frac {(745+738 x) \left (2-x+3 x^2\right )^{3/2}}{1620}+\frac {437 \text {arcsinh}\left (\frac {1-6 x}{\sqrt {23}}\right )}{5184 \sqrt {3}} \]
2/15*(1+2*x)^2*(3*x^2-x+2)^(3/2)+1/1620*(745+738*x)*(3*x^2-x+2)^(3/2)+437/ 15552*arcsinh(1/23*(1-6*x)*23^(1/2))*3^(1/2)+19/2592*(1-6*x)*(3*x^2-x+2)^( 1/2)
Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.75 \[ \int (1+2 x) \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {6 \sqrt {2-x+3 x^2} \left (15471+17374 x+24072 x^2+31536 x^3+20736 x^4\right )+2185 \sqrt {3} \log \left (1-6 x+2 \sqrt {6-3 x+9 x^2}\right )}{77760} \]
(6*Sqrt[2 - x + 3*x^2]*(15471 + 17374*x + 24072*x^2 + 31536*x^3 + 20736*x^ 4) + 2185*Sqrt[3]*Log[1 - 6*x + 2*Sqrt[6 - 3*x + 9*x^2]])/77760
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2184, 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.
\(\displaystyle \int (2 x+1) \sqrt {3 x^2-x+2} \left (4 x^2+3 x+1\right ) \, dx\) |
\(\Big \downarrow \) 2184 |
\(\displaystyle \frac {1}{60} \int 4 (2 x+1) (41 x+2) \sqrt {3 x^2-x+2}dx+\frac {2}{15} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^2\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{15} \int (2 x+1) (41 x+2) \sqrt {3 x^2-x+2}dx+\frac {2}{15} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^2\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{108} (738 x+745) \left (3 x^2-x+2\right )^{3/2}-\frac {95}{72} \int \sqrt {3 x^2-x+2}dx\right )+\frac {2}{15} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^2\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{108} (738 x+745) \left (3 x^2-x+2\right )^{3/2}-\frac {95}{72} \left (\frac {23}{24} \int \frac {1}{\sqrt {3 x^2-x+2}}dx-\frac {1}{12} (1-6 x) \sqrt {3 x^2-x+2}\right )\right )+\frac {2}{15} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^2\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{108} (738 x+745) \left (3 x^2-x+2\right )^{3/2}-\frac {95}{72} \left (\frac {1}{24} \sqrt {\frac {23}{3}} \int \frac {1}{\sqrt {\frac {1}{23} (6 x-1)^2+1}}d(6 x-1)-\frac {1}{12} (1-6 x) \sqrt {3 x^2-x+2}\right )\right )+\frac {2}{15} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^2\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{15} \left (\frac {1}{108} (738 x+745) \left (3 x^2-x+2\right )^{3/2}-\frac {95}{72} \left (\frac {23 \text {arcsinh}\left (\frac {6 x-1}{\sqrt {23}}\right )}{24 \sqrt {3}}-\frac {1}{12} (1-6 x) \sqrt {3 x^2-x+2}\right )\right )+\frac {2}{15} \left (3 x^2-x+2\right )^{3/2} (2 x+1)^2\) |
(2*(1 + 2*x)^2*(2 - x + 3*x^2)^(3/2))/15 + (((745 + 738*x)*(2 - x + 3*x^2) ^(3/2))/108 - (95*(-1/12*((1 - 6*x)*Sqrt[2 - x + 3*x^2]) + (23*ArcSinh[(-1 + 6*x)/Sqrt[23]])/(24*Sqrt[3])))/72)/15
3.3.10.3.1 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[(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 = 50, normalized size of antiderivative = 0.54
method | result | size |
risch | \(\frac {\left (20736 x^{4}+31536 x^{3}+24072 x^{2}+17374 x +15471\right ) \sqrt {3 x^{2}-x +2}}{12960}-\frac {437 \sqrt {3}\, \operatorname {arcsinh}\left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{15552}\) | \(50\) |
trager | \(\left (\frac {8}{5} x^{4}+\frac {73}{30} x^{3}+\frac {1003}{540} x^{2}+\frac {8687}{6480} x +\frac {191}{160}\right ) \sqrt {3 x^{2}-x +2}-\frac {437 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {3 x^{2}-x +2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )\right )}{15552}\) | \(76\) |
default | \(-\frac {19 \left (-1+6 x \right ) \sqrt {3 x^{2}-x +2}}{2592}-\frac {437 \sqrt {3}\, \operatorname {arcsinh}\left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{15552}+\frac {961 \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}{1620}+\frac {8 x^{2} \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}{15}+\frac {89 x \left (3 x^{2}-x +2\right )^{\frac {3}{2}}}{90}\) | \(81\) |
1/12960*(20736*x^4+31536*x^3+24072*x^2+17374*x+15471)*(3*x^2-x+2)^(1/2)-43 7/15552*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))
Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int (1+2 x) \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {1}{12960} \, {\left (20736 \, x^{4} + 31536 \, x^{3} + 24072 \, x^{2} + 17374 \, x + 15471\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {437}{31104} \, \sqrt {3} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} - x + 2} {\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) \]
1/12960*(20736*x^4 + 31536*x^3 + 24072*x^2 + 17374*x + 15471)*sqrt(3*x^2 - x + 2) + 437/31104*sqrt(3)*log(4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25)
Time = 0.49 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68 \[ \int (1+2 x) \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\sqrt {3 x^{2} - x + 2} \cdot \left (\frac {8 x^{4}}{5} + \frac {73 x^{3}}{30} + \frac {1003 x^{2}}{540} + \frac {8687 x}{6480} + \frac {191}{160}\right ) - \frac {437 \sqrt {3} \operatorname {asinh}{\left (\frac {6 \sqrt {23} \left (x - \frac {1}{6}\right )}{23} \right )}}{15552} \]
sqrt(3*x**2 - x + 2)*(8*x**4/5 + 73*x**3/30 + 1003*x**2/540 + 8687*x/6480 + 191/160) - 437*sqrt(3)*asinh(6*sqrt(23)*(x - 1/6)/23)/15552
Time = 0.28 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99 \[ \int (1+2 x) \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {8}{15} \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}} x^{2} + \frac {89}{90} \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}} x + \frac {961}{1620} \, {\left (3 \, x^{2} - x + 2\right )}^{\frac {3}{2}} - \frac {19}{432} \, \sqrt {3 \, x^{2} - x + 2} x - \frac {437}{15552} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (6 \, x - 1\right )}\right ) + \frac {19}{2592} \, \sqrt {3 \, x^{2} - x + 2} \]
8/15*(3*x^2 - x + 2)^(3/2)*x^2 + 89/90*(3*x^2 - x + 2)^(3/2)*x + 961/1620* (3*x^2 - x + 2)^(3/2) - 19/432*sqrt(3*x^2 - x + 2)*x - 437/15552*sqrt(3)*a rcsinh(1/23*sqrt(23)*(6*x - 1)) + 19/2592*sqrt(3*x^2 - x + 2)
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.73 \[ \int (1+2 x) \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {1}{12960} \, {\left (2 \, {\left (12 \, {\left (18 \, {\left (48 \, x + 73\right )} x + 1003\right )} x + 8687\right )} x + 15471\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {437}{15552} \, \sqrt {3} \log \left (-2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} - x + 2}\right )} + 1\right ) \]
1/12960*(2*(12*(18*(48*x + 73)*x + 1003)*x + 8687)*x + 15471)*sqrt(3*x^2 - x + 2) + 437/15552*sqrt(3)*log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2 )) + 1)
Time = 13.87 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.46 \[ \int (1+2 x) \sqrt {2-x+3 x^2} \left (1+3 x+4 x^2\right ) \, dx=\frac {8\,x^2\,{\left (3\,x^2-x+2\right )}^{3/2}}{15}-\frac {253\,\sqrt {3}\,\ln \left (\sqrt {3\,x^2-x+2}+\frac {\sqrt {3}\,\left (3\,x-\frac {1}{2}\right )}{3}\right )}{810}-\frac {44\,\left (\frac {x}{2}-\frac {1}{12}\right )\,\sqrt {3\,x^2-x+2}}{45}+\frac {961\,\sqrt {3\,x^2-x+2}\,\left (72\,x^2-6\,x+45\right )}{38880}+\frac {89\,x\,{\left (3\,x^2-x+2\right )}^{3/2}}{90}+\frac {22103\,\sqrt {3}\,\ln \left (2\,\sqrt {3\,x^2-x+2}+\frac {\sqrt {3}\,\left (6\,x-1\right )}{3}\right )}{77760} \]
(8*x^2*(3*x^2 - x + 2)^(3/2))/15 - (253*3^(1/2)*log((3*x^2 - x + 2)^(1/2) + (3^(1/2)*(3*x - 1/2))/3))/810 - (44*(x/2 - 1/12)*(3*x^2 - x + 2)^(1/2))/ 45 + (961*(3*x^2 - x + 2)^(1/2)*(72*x^2 - 6*x + 45))/38880 + (89*x*(3*x^2 - x + 2)^(3/2))/90 + (22103*3^(1/2)*log(2*(3*x^2 - x + 2)^(1/2) + (3^(1/2) *(6*x - 1))/3))/77760