Integrand size = 32, antiderivative size = 62 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2-x+3 x^2\right )^{3/2}} \, dx=-\frac {2 (101-77 x)}{299 \sqrt {2-x+3 x^2}}-\frac {2 \text {arctanh}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {2-x+3 x^2}}\right )}{13 \sqrt {13}} \]
-2/169*arctanh(1/26*(9-8*x)*13^(1/2)/(3*x^2-x+2)^(1/2))*13^(1/2)-2/299*(10 1-77*x)/(3*x^2-x+2)^(1/2)
Time = 0.37 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2-x+3 x^2\right )^{3/2}} \, dx=\frac {2 (-101+77 x)}{299 \sqrt {2-x+3 x^2}}+\frac {4 \text {arctanh}\left (\frac {\sqrt {3}+2 \sqrt {3} x-2 \sqrt {2-x+3 x^2}}{\sqrt {13}}\right )}{13 \sqrt {13}} \]
(2*(-101 + 77*x))/(299*Sqrt[2 - x + 3*x^2]) + (4*ArcTanh[(Sqrt[3] + 2*Sqrt [3]*x - 2*Sqrt[2 - x + 3*x^2])/Sqrt[13]])/(13*Sqrt[13])
Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2177, 27, 1154, 219}
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.
\(\displaystyle \int \frac {4 x^2+3 x+1}{(2 x+1) \left (3 x^2-x+2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2177 |
\(\displaystyle \frac {2}{23} \int \frac {23}{13 (2 x+1) \sqrt {3 x^2-x+2}}dx-\frac {2 (101-77 x)}{299 \sqrt {3 x^2-x+2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{13} \int \frac {1}{(2 x+1) \sqrt {3 x^2-x+2}}dx-\frac {2 (101-77 x)}{299 \sqrt {3 x^2-x+2}}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle -\frac {4}{13} \int \frac {1}{52-\frac {(9-8 x)^2}{3 x^2-x+2}}d\frac {9-8 x}{\sqrt {3 x^2-x+2}}-\frac {2 (101-77 x)}{299 \sqrt {3 x^2-x+2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 \text {arctanh}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {3 x^2-x+2}}\right )}{13 \sqrt {13}}-\frac {2 (101-77 x)}{299 \sqrt {3 x^2-x+2}}\) |
(-2*(101 - 77*x))/(299*Sqrt[2 - x + 3*x^2]) - (2*ArcTanh[(9 - 8*x)/(2*Sqrt [13]*Sqrt[2 - x + 3*x^2])])/(13*Sqrt[13])
3.3.49.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x + c* x^2, x], R = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 0], S = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*R - 2*a*S + (2*c*R - b*S)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^ m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Qx)/(d + e*x )^m - ((2*p + 3)*(2*c*R - b*S))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a* e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
Time = 0.80 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.82
method | result | size |
risch | \(\frac {-\frac {202}{299}+\frac {154 x}{299}}{\sqrt {3 x^{2}-x +2}}-\frac {2 \sqrt {13}\, \operatorname {arctanh}\left (\frac {2 \left (\frac {9}{2}-4 x \right ) \sqrt {13}}{13 \sqrt {12 \left (x +\frac {1}{2}\right )^{2}-16 x +5}}\right )}{169}\) | \(51\) |
trager | \(\frac {-\frac {202}{299}+\frac {154 x}{299}}{\sqrt {3 x^{2}-x +2}}-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right ) \ln \left (-\frac {8 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right ) x -9 \operatorname {RootOf}\left (\textit {\_Z}^{2}-13\right )-26 \sqrt {3 x^{2}-x +2}}{1+2 x}\right )}{169}\) | \(71\) |
default | \(\frac {-\frac {5}{69}+\frac {10 x}{23}}{\sqrt {3 x^{2}-x +2}}-\frac {2}{3 \sqrt {3 x^{2}-x +2}}+\frac {1}{13 \sqrt {3 \left (x +\frac {1}{2}\right )^{2}-4 x +\frac {5}{4}}}+\frac {-\frac {4}{299}+\frac {24 x}{299}}{\sqrt {3 \left (x +\frac {1}{2}\right )^{2}-4 x +\frac {5}{4}}}-\frac {2 \sqrt {13}\, \operatorname {arctanh}\left (\frac {2 \left (\frac {9}{2}-4 x \right ) \sqrt {13}}{13 \sqrt {12 \left (x +\frac {1}{2}\right )^{2}-16 x +5}}\right )}{169}\) | \(102\) |
2/299*(-101+77*x)/(3*x^2-x+2)^(1/2)-2/169*13^(1/2)*arctanh(2/13*(9/2-4*x)* 13^(1/2)/(12*(x+1/2)^2-16*x+5)^(1/2))
Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.55 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2-x+3 x^2\right )^{3/2}} \, dx=\frac {23 \, \sqrt {13} {\left (3 \, x^{2} - x + 2\right )} \log \left (-\frac {4 \, \sqrt {13} \sqrt {3 \, x^{2} - x + 2} {\left (8 \, x - 9\right )} + 220 \, x^{2} - 196 \, x + 185}{4 \, x^{2} + 4 \, x + 1}\right ) + 26 \, \sqrt {3 \, x^{2} - x + 2} {\left (77 \, x - 101\right )}}{3887 \, {\left (3 \, x^{2} - x + 2\right )}} \]
1/3887*(23*sqrt(13)*(3*x^2 - x + 2)*log(-(4*sqrt(13)*sqrt(3*x^2 - x + 2)*( 8*x - 9) + 220*x^2 - 196*x + 185)/(4*x^2 + 4*x + 1)) + 26*sqrt(3*x^2 - x + 2)*(77*x - 101))/(3*x^2 - x + 2)
\[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2-x+3 x^2\right )^{3/2}} \, dx=\int \frac {4 x^{2} + 3 x + 1}{\left (2 x + 1\right ) \left (3 x^{2} - x + 2\right )^{\frac {3}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2-x+3 x^2\right )^{3/2}} \, dx=\frac {2}{169} \, \sqrt {13} \operatorname {arsinh}\left (\frac {8 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 1 \right |}} - \frac {9 \, \sqrt {23}}{23 \, {\left | 2 \, x + 1 \right |}}\right ) + \frac {154 \, x}{299 \, \sqrt {3 \, x^{2} - x + 2}} - \frac {202}{299 \, \sqrt {3 \, x^{2} - x + 2}} \]
2/169*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9/23*sqrt(23)/abs(2* x + 1)) + 154/299*x/sqrt(3*x^2 - x + 2) - 202/299/sqrt(3*x^2 - x + 2)
Time = 0.31 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.47 \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2-x+3 x^2\right )^{3/2}} \, dx=\frac {2}{169} \, \sqrt {13} \log \left (-\frac {{\left | -4 \, \sqrt {3} x - 2 \, \sqrt {13} - 2 \, \sqrt {3} + 4 \, \sqrt {3 \, x^{2} - x + 2} \right |}}{2 \, {\left (2 \, \sqrt {3} x - \sqrt {13} + \sqrt {3} - 2 \, \sqrt {3 \, x^{2} - x + 2}\right )}}\right ) + \frac {2 \, {\left (77 \, x - 101\right )}}{299 \, \sqrt {3 \, x^{2} - x + 2}} \]
2/169*sqrt(13)*log(-1/2*abs(-4*sqrt(3)*x - 2*sqrt(13) - 2*sqrt(3) + 4*sqrt (3*x^2 - x + 2))/(2*sqrt(3)*x - sqrt(13) + sqrt(3) - 2*sqrt(3*x^2 - x + 2) )) + 2/299*(77*x - 101)/sqrt(3*x^2 - x + 2)
Timed out. \[ \int \frac {1+3 x+4 x^2}{(1+2 x) \left (2-x+3 x^2\right )^{3/2}} \, dx=\int \frac {4\,x^2+3\,x+1}{\left (2\,x+1\right )\,{\left (3\,x^2-x+2\right )}^{3/2}} \,d x \]