Integrand size = 29, antiderivative size = 77 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \sqrt {2+3 x^2}} \, dx=-\frac {\sqrt {2+3 x^2}}{22 (1+2 x)^2}+\frac {13 \sqrt {2+3 x^2}}{242 (1+2 x)}-\frac {103 \text {arctanh}\left (\frac {4-3 x}{\sqrt {11} \sqrt {2+3 x^2}}\right )}{121 \sqrt {11}} \]
-103/1331*arctanh(1/11*(4-3*x)*11^(1/2)/(3*x^2+2)^(1/2))*11^(1/2)-1/22*(3* x^2+2)^(1/2)/(1+2*x)^2+13/242*(3*x^2+2)^(1/2)/(1+2*x)
Time = 0.39 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {\frac {11 (1+13 x) \sqrt {2+3 x^2}}{(1+2 x)^2}+206 \sqrt {11} \text {arctanh}\left (\frac {\sqrt {3}+2 \sqrt {3} x-2 \sqrt {2+3 x^2}}{\sqrt {11}}\right )}{1331} \]
((11*(1 + 13*x)*Sqrt[2 + 3*x^2])/(1 + 2*x)^2 + 206*Sqrt[11]*ArcTanh[(Sqrt[ 3] + 2*Sqrt[3]*x - 2*Sqrt[2 + 3*x^2])/Sqrt[11]])/1331
Time = 0.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2182, 25, 679, 488, 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.
\(\displaystyle \int \frac {4 x^2+3 x+1}{(2 x+1)^3 \sqrt {3 x^2+2}} \, dx\) |
\(\Big \downarrow \) 2182 |
\(\displaystyle -\frac {1}{22} \int -\frac {41 x+14}{(2 x+1)^2 \sqrt {3 x^2+2}}dx-\frac {\sqrt {3 x^2+2}}{22 (2 x+1)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{22} \int \frac {41 x+14}{(2 x+1)^2 \sqrt {3 x^2+2}}dx-\frac {\sqrt {3 x^2+2}}{22 (2 x+1)^2}\) |
\(\Big \downarrow \) 679 |
\(\displaystyle \frac {1}{22} \left (\frac {206}{11} \int \frac {1}{(2 x+1) \sqrt {3 x^2+2}}dx+\frac {13 \sqrt {3 x^2+2}}{11 (2 x+1)}\right )-\frac {\sqrt {3 x^2+2}}{22 (2 x+1)^2}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {1}{22} \left (\frac {13 \sqrt {3 x^2+2}}{11 (2 x+1)}-\frac {206}{11} \int \frac {1}{11-\frac {(4-3 x)^2}{3 x^2+2}}d\frac {4-3 x}{\sqrt {3 x^2+2}}\right )-\frac {\sqrt {3 x^2+2}}{22 (2 x+1)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{22} \left (\frac {13 \sqrt {3 x^2+2}}{11 (2 x+1)}-\frac {206 \text {arctanh}\left (\frac {4-3 x}{\sqrt {11} \sqrt {3 x^2+2}}\right )}{11 \sqrt {11}}\right )-\frac {\sqrt {3 x^2+2}}{22 (2 x+1)^2}\) |
-1/22*Sqrt[2 + 3*x^2]/(1 + 2*x)^2 + ((13*Sqrt[2 + 3*x^2])/(11*(1 + 2*x)) - (206*ArcTanh[(4 - 3*x)/(Sqrt[11]*Sqrt[2 + 3*x^2])])/(11*Sqrt[11]))/22
3.2.23.3.1 Defintions of rubi rules used
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/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 )/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[e*R*(d + e*x)^(m + 1)*((a + b*x^2)^(p + 1)/((m + 1)*(b* d^2 + a*e^2))), x] + Simp[1/((m + 1)*(b*d^2 + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[(m + 1)*(b*d^2 + a*e^2)*Qx + b*d*R*(m + 1) - b *e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[m, -1]
Time = 0.49 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {39 x^{3}+3 x^{2}+26 x +2}{121 \left (1+2 x \right )^{2} \sqrt {3 x^{2}+2}}-\frac {103 \sqrt {11}\, \operatorname {arctanh}\left (\frac {2 \left (4-3 x \right ) \sqrt {11}}{11 \sqrt {12 \left (x +\frac {1}{2}\right )^{2}-12 x +5}}\right )}{1331}\) | \(65\) |
trager | \(\frac {\left (13 x +1\right ) \sqrt {3 x^{2}+2}}{121 \left (1+2 x \right )^{2}}-\frac {103 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right ) x +11 \sqrt {3 x^{2}+2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-11\right )}{1+2 x}\right )}{1331}\) | \(71\) |
default | \(-\frac {103 \sqrt {11}\, \operatorname {arctanh}\left (\frac {2 \left (4-3 x \right ) \sqrt {11}}{11 \sqrt {12 \left (x +\frac {1}{2}\right )^{2}-12 x +5}}\right )}{1331}+\frac {13 \sqrt {3 \left (x +\frac {1}{2}\right )^{2}-3 x +\frac {5}{4}}}{484 \left (x +\frac {1}{2}\right )}-\frac {\sqrt {3 \left (x +\frac {1}{2}\right )^{2}-3 x +\frac {5}{4}}}{88 \left (x +\frac {1}{2}\right )^{2}}\) | \(74\) |
1/121*(39*x^3+3*x^2+26*x+2)/(1+2*x)^2/(3*x^2+2)^(1/2)-103/1331*11^(1/2)*ar ctanh(2/11*(4-3*x)*11^(1/2)/(12*(x+1/2)^2-12*x+5)^(1/2))
Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.16 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {103 \, \sqrt {11} {\left (4 \, x^{2} + 4 \, x + 1\right )} \log \left (-\frac {\sqrt {11} \sqrt {3 \, x^{2} + 2} {\left (3 \, x - 4\right )} + 21 \, x^{2} - 12 \, x + 19}{4 \, x^{2} + 4 \, x + 1}\right ) + 22 \, \sqrt {3 \, x^{2} + 2} {\left (13 \, x + 1\right )}}{2662 \, {\left (4 \, x^{2} + 4 \, x + 1\right )}} \]
1/2662*(103*sqrt(11)*(4*x^2 + 4*x + 1)*log(-(sqrt(11)*sqrt(3*x^2 + 2)*(3*x - 4) + 21*x^2 - 12*x + 19)/(4*x^2 + 4*x + 1)) + 22*sqrt(3*x^2 + 2)*(13*x + 1))/(4*x^2 + 4*x + 1)
\[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \sqrt {2+3 x^2}} \, dx=\int \frac {4 x^{2} + 3 x + 1}{\left (2 x + 1\right )^{3} \sqrt {3 x^{2} + 2}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {103}{1331} \, \sqrt {11} \operatorname {arsinh}\left (\frac {\sqrt {6} x}{2 \, {\left | 2 \, x + 1 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 1 \right |}}\right ) - \frac {\sqrt {3 \, x^{2} + 2}}{22 \, {\left (4 \, x^{2} + 4 \, x + 1\right )}} + \frac {13 \, \sqrt {3 \, x^{2} + 2}}{242 \, {\left (2 \, x + 1\right )}} \]
103/1331*sqrt(11)*arcsinh(1/2*sqrt(6)*x/abs(2*x + 1) - 2/3*sqrt(6)/abs(2*x + 1)) - 1/22*sqrt(3*x^2 + 2)/(4*x^2 + 4*x + 1) + 13/242*sqrt(3*x^2 + 2)/( 2*x + 1)
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (62) = 124\).
Time = 0.29 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.34 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {103}{1331} \, \sqrt {11} \log \left (-\frac {{\left | -2 \, \sqrt {3} x - \sqrt {11} - \sqrt {3} + 2 \, \sqrt {3 \, x^{2} + 2} \right |}}{2 \, \sqrt {3} x - \sqrt {11} + \sqrt {3} - 2 \, \sqrt {3 \, x^{2} + 2}}\right ) + \frac {72 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{3} - 13 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{2} - 168 \, \sqrt {3} x + 104 \, \sqrt {3} + 168 \, \sqrt {3 \, x^{2} + 2}}{484 \, {\left ({\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{2} + \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )} - 2\right )}^{2}} \]
103/1331*sqrt(11)*log(-abs(-2*sqrt(3)*x - sqrt(11) - sqrt(3) + 2*sqrt(3*x^ 2 + 2))/(2*sqrt(3)*x - sqrt(11) + sqrt(3) - 2*sqrt(3*x^2 + 2))) + 1/484*(7 2*(sqrt(3)*x - sqrt(3*x^2 + 2))^3 - 13*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 2 ))^2 - 168*sqrt(3)*x + 104*sqrt(3) + 168*sqrt(3*x^2 + 2))/((sqrt(3)*x - sq rt(3*x^2 + 2))^2 + sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 2)) - 2)^2
Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {1+3 x+4 x^2}{(1+2 x)^3 \sqrt {2+3 x^2}} \, dx=\frac {103\,\sqrt {11}\,\ln \left (x+\frac {1}{2}\right )}{1331}-\frac {103\,\sqrt {11}\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {11}\,\sqrt {x^2+\frac {2}{3}}}{3}-\frac {4}{3}\right )}{1331}-\frac {\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{88\,\left (x^2+x+\frac {1}{4}\right )}+\frac {13\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{484\,\left (x+\frac {1}{2}\right )} \]