Integrand size = 19, antiderivative size = 71 \[ \int \frac {1}{\sqrt {-1+x^2} \left (-4+3 x^2\right )^2} \, dx=-\frac {3 x \sqrt {-1+x^2}}{8 \left (-4+3 x^2\right )}+\frac {5}{32} \log \left (2-x^2+x \sqrt {-1+x^2}\right )-\frac {5}{32} \log \left (2-3 x^2+3 x \sqrt {-1+x^2}\right ) \]
-3*x*(x^2-1)^(1/2)/(24*x^2-32)+5/32*ln(2-x^2+x*(x^2-1)^(1/2))-5/32*ln(2-3* x^2+3*x*(x^2-1)^(1/2))
Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {-1+x^2} \left (-4+3 x^2\right )^2} \, dx=-\frac {3 x \sqrt {-1+x^2}}{8 \left (-4+3 x^2\right )}+\frac {5}{32} \log \left (2-x^2+x \sqrt {-1+x^2}\right )-\frac {5}{32} \log \left (2-3 x^2+3 x \sqrt {-1+x^2}\right ) \]
(-3*x*Sqrt[-1 + x^2])/(8*(-4 + 3*x^2)) + (5*Log[2 - x^2 + x*Sqrt[-1 + x^2] ])/32 - (5*Log[2 - 3*x^2 + 3*x*Sqrt[-1 + x^2]])/32
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.61, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {296, 25, 291, 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.
\(\displaystyle \int \frac {1}{\sqrt {x^2-1} \left (3 x^2-4\right )^2} \, dx\) |
\(\Big \downarrow \) 296 |
\(\displaystyle \frac {3 x \sqrt {x^2-1}}{8 \left (4-3 x^2\right )}-\frac {5}{8} \int -\frac {1}{\left (4-3 x^2\right ) \sqrt {x^2-1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {5}{8} \int \frac {1}{\left (4-3 x^2\right ) \sqrt {x^2-1}}dx+\frac {3 \sqrt {x^2-1} x}{8 \left (4-3 x^2\right )}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {5}{8} \int \frac {1}{4-\frac {x^2}{x^2-1}}d\frac {x}{\sqrt {x^2-1}}+\frac {3 \sqrt {x^2-1} x}{8 \left (4-3 x^2\right )}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5}{16} \text {arctanh}\left (\frac {x}{2 \sqrt {x^2-1}}\right )+\frac {3 \sqrt {x^2-1} x}{8 \left (4-3 x^2\right )}\) |
3.10.32.3.1 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/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[(b*c + 2*(p + 1)*(b*c - a*d))/(2*a*(p + 1)*(b*c - a*d)) Int[ (a + b*x^2)^(p + 1)*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, q}, x] && N eQ[b*c - a*d, 0] && EqQ[2*(p + q + 2) + 1, 0] && (LtQ[p, -1] || !LtQ[q, -1 ]) && NeQ[p, -1]
Time = 1.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.73
method | result | size |
trager | \(-\frac {3 x \sqrt {x^{2}-1}}{8 \left (3 x^{2}-4\right )}-\frac {5 \ln \left (-\frac {4 x \sqrt {x^{2}-1}-5 x^{2}+4}{3 x^{2}-4}\right )}{32}\) | \(52\) |
pseudoelliptic | \(\frac {\left (-15 x^{2}+20\right ) \ln \left (\frac {2 \sqrt {x^{2}-1}-x}{x}\right )+15 \ln \left (\frac {2 \sqrt {x^{2}-1}+x}{x}\right ) x^{2}-12 x \sqrt {x^{2}-1}-20 \ln \left (\frac {2 \sqrt {x^{2}-1}+x}{x}\right )}{96 x^{2}-128}\) | \(87\) |
risch | \(-\frac {3 x \sqrt {x^{2}-1}}{8 \left (3 x^{2}-4\right )}+\frac {5 \,\operatorname {arctanh}\left (\frac {3 \left (\frac {2}{3}+\frac {4 \sqrt {3}\, \left (x -\frac {2 \sqrt {3}}{3}\right )}{3}\right ) \sqrt {3}}{2 \sqrt {9 \left (x -\frac {2 \sqrt {3}}{3}\right )^{2}+12 \sqrt {3}\, \left (x -\frac {2 \sqrt {3}}{3}\right )+3}}\right )}{32}-\frac {5 \,\operatorname {arctanh}\left (\frac {3 \left (\frac {2}{3}-\frac {4 \sqrt {3}\, \left (x +\frac {2 \sqrt {3}}{3}\right )}{3}\right ) \sqrt {3}}{2 \sqrt {9 \left (x +\frac {2 \sqrt {3}}{3}\right )^{2}-12 \sqrt {3}\, \left (x +\frac {2 \sqrt {3}}{3}\right )+3}}\right )}{32}\) | \(119\) |
default | \(-\frac {\sqrt {\left (x -\frac {2 \sqrt {3}}{3}\right )^{2}+\frac {4 \sqrt {3}\, \left (x -\frac {2 \sqrt {3}}{3}\right )}{3}+\frac {1}{3}}}{16 \left (x -\frac {2 \sqrt {3}}{3}\right )}+\frac {5 \,\operatorname {arctanh}\left (\frac {3 \left (\frac {2}{3}+\frac {4 \sqrt {3}\, \left (x -\frac {2 \sqrt {3}}{3}\right )}{3}\right ) \sqrt {3}}{2 \sqrt {9 \left (x -\frac {2 \sqrt {3}}{3}\right )^{2}+12 \sqrt {3}\, \left (x -\frac {2 \sqrt {3}}{3}\right )+3}}\right )}{32}-\frac {\sqrt {\left (x +\frac {2 \sqrt {3}}{3}\right )^{2}-\frac {4 \sqrt {3}\, \left (x +\frac {2 \sqrt {3}}{3}\right )}{3}+\frac {1}{3}}}{16 \left (x +\frac {2 \sqrt {3}}{3}\right )}-\frac {5 \,\operatorname {arctanh}\left (\frac {3 \left (\frac {2}{3}-\frac {4 \sqrt {3}\, \left (x +\frac {2 \sqrt {3}}{3}\right )}{3}\right ) \sqrt {3}}{2 \sqrt {9 \left (x +\frac {2 \sqrt {3}}{3}\right )^{2}-12 \sqrt {3}\, \left (x +\frac {2 \sqrt {3}}{3}\right )+3}}\right )}{32}\) | \(172\) |
Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\sqrt {-1+x^2} \left (-4+3 x^2\right )^2} \, dx=-\frac {12 \, x^{2} + 5 \, {\left (3 \, x^{2} - 4\right )} \log \left (3 \, x^{2} - 3 \, \sqrt {x^{2} - 1} x - 2\right ) - 5 \, {\left (3 \, x^{2} - 4\right )} \log \left (x^{2} - \sqrt {x^{2} - 1} x - 2\right ) + 12 \, \sqrt {x^{2} - 1} x - 16}{32 \, {\left (3 \, x^{2} - 4\right )}} \]
-1/32*(12*x^2 + 5*(3*x^2 - 4)*log(3*x^2 - 3*sqrt(x^2 - 1)*x - 2) - 5*(3*x^ 2 - 4)*log(x^2 - sqrt(x^2 - 1)*x - 2) + 12*sqrt(x^2 - 1)*x - 16)/(3*x^2 - 4)
\[ \int \frac {1}{\sqrt {-1+x^2} \left (-4+3 x^2\right )^2} \, dx=\int \frac {1}{\sqrt {\left (x - 1\right ) \left (x + 1\right )} \left (3 x^{2} - 4\right )^{2}}\, dx \]
\[ \int \frac {1}{\sqrt {-1+x^2} \left (-4+3 x^2\right )^2} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 4\right )}^{2} \sqrt {x^{2} - 1}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\sqrt {-1+x^2} \left (-4+3 x^2\right )^2} \, dx=\frac {5 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{2} - 3}{4 \, {\left (3 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{4} - 10 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 3\right )}} - \frac {5}{32} \, \log \left ({\left | 3 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{2} - 1 \right |}\right ) + \frac {5}{32} \, \log \left ({\left | {\left (x - \sqrt {x^{2} - 1}\right )}^{2} - 3 \right |}\right ) \]
1/4*(5*(x - sqrt(x^2 - 1))^2 - 3)/(3*(x - sqrt(x^2 - 1))^4 - 10*(x - sqrt( x^2 - 1))^2 + 3) - 5/32*log(abs(3*(x - sqrt(x^2 - 1))^2 - 1)) + 5/32*log(a bs((x - sqrt(x^2 - 1))^2 - 3))
Timed out. \[ \int \frac {1}{\sqrt {-1+x^2} \left (-4+3 x^2\right )^2} \, dx=\int \frac {1}{\sqrt {x^2-1}\,{\left (3\,x^2-4\right )}^2} \,d x \]