Integrand size = 32, antiderivative size = 95 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx=-\frac {143}{324} (3-2 x) \sqrt {2-x+3 x^2}+\frac {11}{27} (1+2 x)^2 \sqrt {2-x+3 x^2}+\frac {1}{6} (1+2 x)^3 \sqrt {2-x+3 x^2}+\frac {4147 \text {arcsinh}\left (\frac {1-6 x}{\sqrt {23}}\right )}{648 \sqrt {3}} \]
4147/1944*arcsinh(1/23*(1-6*x)*23^(1/2))*3^(1/2)-143/324*(3-2*x)*(3*x^2-x+ 2)^(1/2)+11/27*(1+2*x)^2*(3*x^2-x+2)^(1/2)+1/6*(1+2*x)^3*(3*x^2-x+2)^(1/2)
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx=\frac {6 \sqrt {2-x+3 x^2} \left (-243+1138 x+1176 x^2+432 x^3\right )+4147 \sqrt {3} \log \left (1-6 x+2 \sqrt {6-3 x+9 x^2}\right )}{1944} \]
(6*Sqrt[2 - x + 3*x^2]*(-243 + 1138*x + 1176*x^2 + 432*x^3) + 4147*Sqrt[3] *Log[1 - 6*x + 2*Sqrt[6 - 3*x + 9*x^2]])/1944
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2184, 27, 1236, 27, 1225, 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.
\(\displaystyle \int \frac {(2 x+1)^2 \left (4 x^2+3 x+1\right )}{\sqrt {3 x^2-x+2}} \, dx\) |
\(\Big \downarrow \) 2184 |
\(\displaystyle \frac {1}{48} \int -\frac {44 (1-4 x) (2 x+1)^2}{\sqrt {3 x^2-x+2}}dx+\frac {1}{6} \sqrt {3 x^2-x+2} (2 x+1)^3\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} (2 x+1)^3 \sqrt {3 x^2-x+2}-\frac {11}{12} \int \frac {(1-4 x) (2 x+1)^2}{\sqrt {3 x^2-x+2}}dx\) |
\(\Big \downarrow \) 1236 |
\(\displaystyle \frac {1}{6} (2 x+1)^3 \sqrt {3 x^2-x+2}-\frac {11}{12} \left (\frac {1}{9} \int \frac {13 (3-2 x) (2 x+1)}{\sqrt {3 x^2-x+2}}dx-\frac {4}{9} (2 x+1)^2 \sqrt {3 x^2-x+2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} (2 x+1)^3 \sqrt {3 x^2-x+2}-\frac {11}{12} \left (\frac {13}{9} \int \frac {(3-2 x) (2 x+1)}{\sqrt {3 x^2-x+2}}dx-\frac {4}{9} (2 x+1)^2 \sqrt {3 x^2-x+2}\right )\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {1}{6} (2 x+1)^3 \sqrt {3 x^2-x+2}-\frac {11}{12} \left (\frac {13}{9} \left (\frac {29}{6} \int \frac {1}{\sqrt {3 x^2-x+2}}dx+\frac {1}{3} \sqrt {3 x^2-x+2} (3-2 x)\right )-\frac {4}{9} (2 x+1)^2 \sqrt {3 x^2-x+2}\right )\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{6} (2 x+1)^3 \sqrt {3 x^2-x+2}-\frac {11}{12} \left (\frac {13}{9} \left (\frac {29 \int \frac {1}{\sqrt {\frac {1}{23} (6 x-1)^2+1}}d(6 x-1)}{6 \sqrt {69}}+\frac {1}{3} \sqrt {3 x^2-x+2} (3-2 x)\right )-\frac {4}{9} (2 x+1)^2 \sqrt {3 x^2-x+2}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{6} (2 x+1)^3 \sqrt {3 x^2-x+2}-\frac {11}{12} \left (\frac {13}{9} \left (\frac {29 \text {arcsinh}\left (\frac {6 x-1}{\sqrt {23}}\right )}{6 \sqrt {3}}+\frac {1}{3} \sqrt {3 x^2-x+2} (3-2 x)\right )-\frac {4}{9} (2 x+1)^2 \sqrt {3 x^2-x+2}\right )\) |
((1 + 2*x)^3*Sqrt[2 - x + 3*x^2])/6 - (11*((-4*(1 + 2*x)^2*Sqrt[2 - x + 3* x^2])/9 + (13*(((3 - 2*x)*Sqrt[2 - x + 3*x^2])/3 + (29*ArcSinh[(-1 + 6*x)/ Sqrt[23]])/(6*Sqrt[3])))/9))/12
3.3.41.3.1 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[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 = 0.70 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.47
method | result | size |
risch | \(\frac {\left (432 x^{3}+1176 x^{2}+1138 x -243\right ) \sqrt {3 x^{2}-x +2}}{324}-\frac {4147 \sqrt {3}\, \operatorname {arcsinh}\left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{1944}\) | \(45\) |
trager | \(\left (\frac {4}{3} x^{3}+\frac {98}{27} x^{2}+\frac {569}{162} x -\frac {3}{4}\right ) \sqrt {3 x^{2}-x +2}-\frac {4147 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {3 x^{2}-x +2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )\right )}{1944}\) | \(71\) |
default | \(-\frac {4147 \sqrt {3}\, \operatorname {arcsinh}\left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{1944}-\frac {3 \sqrt {3 x^{2}-x +2}}{4}+\frac {4 x^{3} \sqrt {3 x^{2}-x +2}}{3}+\frac {98 x^{2} \sqrt {3 x^{2}-x +2}}{27}+\frac {569 x \sqrt {3 x^{2}-x +2}}{162}\) | \(79\) |
1/324*(432*x^3+1176*x^2+1138*x-243)*(3*x^2-x+2)^(1/2)-4147/1944*3^(1/2)*ar csinh(6/23*23^(1/2)*(x-1/6))
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.72 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx=\frac {1}{324} \, {\left (432 \, x^{3} + 1176 \, x^{2} + 1138 \, x - 243\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {4147}{3888} \, \sqrt {3} \log \left (4 \, \sqrt {3} \sqrt {3 \, x^{2} - x + 2} {\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) \]
1/324*(432*x^3 + 1176*x^2 + 1138*x - 243)*sqrt(3*x^2 - x + 2) + 4147/3888* 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 = 56, normalized size of antiderivative = 0.59 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx=\sqrt {3 x^{2} - x + 2} \cdot \left (\frac {4 x^{3}}{3} + \frac {98 x^{2}}{27} + \frac {569 x}{162} - \frac {3}{4}\right ) - \frac {4147 \sqrt {3} \operatorname {asinh}{\left (\frac {6 \sqrt {23} \left (x - \frac {1}{6}\right )}{23} \right )}}{1944} \]
sqrt(3*x**2 - x + 2)*(4*x**3/3 + 98*x**2/27 + 569*x/162 - 3/4) - 4147*sqrt (3)*asinh(6*sqrt(23)*(x - 1/6)/23)/1944
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.84 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx=\frac {4}{3} \, \sqrt {3 \, x^{2} - x + 2} x^{3} + \frac {98}{27} \, \sqrt {3 \, x^{2} - x + 2} x^{2} + \frac {569}{162} \, \sqrt {3 \, x^{2} - x + 2} x - \frac {4147}{1944} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (6 \, x - 1\right )}\right ) - \frac {3}{4} \, \sqrt {3 \, x^{2} - x + 2} \]
4/3*sqrt(3*x^2 - x + 2)*x^3 + 98/27*sqrt(3*x^2 - x + 2)*x^2 + 569/162*sqrt (3*x^2 - x + 2)*x - 4147/1944*sqrt(3)*arcsinh(1/23*sqrt(23)*(6*x - 1)) - 3 /4*sqrt(3*x^2 - x + 2)
Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.66 \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx=\frac {1}{324} \, {\left (2 \, {\left (12 \, {\left (18 \, x + 49\right )} x + 569\right )} x - 243\right )} \sqrt {3 \, x^{2} - x + 2} + \frac {4147}{1944} \, \sqrt {3} \log \left (-2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} - x + 2}\right )} + 1\right ) \]
1/324*(2*(12*(18*x + 49)*x + 569)*x - 243)*sqrt(3*x^2 - x + 2) + 4147/1944 *sqrt(3)*log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) + 1)
Timed out. \[ \int \frac {(1+2 x)^2 \left (1+3 x+4 x^2\right )}{\sqrt {2-x+3 x^2}} \, dx=\int \frac {{\left (2\,x+1\right )}^2\,\left (4\,x^2+3\,x+1\right )}{\sqrt {3\,x^2-x+2}} \,d x \]