Integrand size = 16, antiderivative size = 59 \[ \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx=-\frac {1}{8 x}+\frac {3}{20 x \left (3-2 x^2\right )^2}+\frac {1}{8 x \left (3-2 x^2\right )}+\frac {\text {arctanh}\left (\sqrt {\frac {2}{3}} x\right )}{4 \sqrt {6}} \] Output:
-1/8/x+3/20/x/(-2*x^2+3)^2+1/8/x/(-2*x^2+3)+1/24*arctanh(1/3*x*6^(1/2))*6^ (1/2)
Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.10 \[ \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx=\frac {1}{240} \left (-\frac {12 \left (12-25 x^2+10 x^4\right )}{x \left (3-2 x^2\right )^2}-5 \sqrt {6} \log \left (\sqrt {6}-2 x\right )+5 \sqrt {6} \log \left (\sqrt {6}+2 x\right )\right ) \] Input:
Integrate[9/(5*x^2*(3 - 2*x^2)^3),x]
Output:
((-12*(12 - 25*x^2 + 10*x^4))/(x*(3 - 2*x^2)^2) - 5*Sqrt[6]*Log[Sqrt[6] - 2*x] + 5*Sqrt[6]*Log[Sqrt[6] + 2*x])/240
Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.27, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {27, 253, 253, 264, 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.
\(\displaystyle \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {9}{5} \int \frac {1}{x^2 \left (3-2 x^2\right )^3}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {9}{5} \left (\frac {5}{12} \int \frac {1}{x^2 \left (3-2 x^2\right )^2}dx+\frac {1}{12 x \left (3-2 x^2\right )^2}\right )\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {9}{5} \left (\frac {5}{12} \left (\frac {1}{2} \int \frac {1}{x^2 \left (3-2 x^2\right )}dx+\frac {1}{6 x \left (3-2 x^2\right )}\right )+\frac {1}{12 x \left (3-2 x^2\right )^2}\right )\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {9}{5} \left (\frac {5}{12} \left (\frac {1}{2} \left (\frac {2}{3} \int \frac {1}{3-2 x^2}dx-\frac {1}{3 x}\right )+\frac {1}{6 x \left (3-2 x^2\right )}\right )+\frac {1}{12 x \left (3-2 x^2\right )^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {9}{5} \left (\frac {5}{12} \left (\frac {1}{2} \left (\frac {1}{3} \sqrt {\frac {2}{3}} \text {arctanh}\left (\sqrt {\frac {2}{3}} x\right )-\frac {1}{3 x}\right )+\frac {1}{6 x \left (3-2 x^2\right )}\right )+\frac {1}{12 x \left (3-2 x^2\right )^2}\right )\) |
Input:
Int[9/(5*x^2*(3 - 2*x^2)^3),x]
Output:
(9*(1/(12*x*(3 - 2*x^2)^2) + (5*(1/(6*x*(3 - 2*x^2)) + (-1/3*1/x + (Sqrt[2 /3]*ArcTanh[Sqrt[2/3]*x])/3)/2))/12))/5
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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66
method | result | size |
default | \(-\frac {1}{15 x}-\frac {8 \left (\frac {7}{16} x^{3}-\frac {27}{32} x \right )}{15 \left (2 x^{2}-3\right )^{2}}+\frac {\operatorname {arctanh}\left (\frac {x \sqrt {6}}{3}\right ) \sqrt {6}}{24}\) | \(39\) |
meijerg | \(\frac {i \sqrt {6}\, \left (\frac {i \sqrt {6}\, \left (\frac {20}{3} x^{4}-\frac {50}{3} x^{2}+8\right )}{4 x \left (-\frac {2 x^{2}}{3}+1\right )^{2}}-\frac {15 i \operatorname {arctanh}\left (\frac {x \sqrt {2}\, \sqrt {3}}{3}\right )}{2}\right )}{180}\) | \(51\) |
risch | \(\frac {-\frac {1}{2} x^{4}+\frac {5}{4} x^{2}-\frac {3}{5}}{\left (2 x^{2}-3\right )^{2} x}+\frac {\sqrt {6}\, \ln \left (2 x +\sqrt {6}\right )}{48}-\frac {\sqrt {6}\, \ln \left (2 x -\sqrt {6}\right )}{48}\) | \(56\) |
Input:
int(9/5/x^2/(-2*x^2+3)^3,x,method=_RETURNVERBOSE)
Output:
-1/15/x-8/15*(7/16*x^3-27/32*x)/(2*x^2-3)^2+1/24*arctanh(1/3*x*6^(1/2))*6^ (1/2)
Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.24 \[ \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx=-\frac {120 \, x^{4} - 5 \, \sqrt {6} {\left (4 \, x^{5} - 12 \, x^{3} + 9 \, x\right )} \log \left (\frac {2 \, x^{2} + 2 \, \sqrt {6} x + 3}{2 \, x^{2} - 3}\right ) - 300 \, x^{2} + 144}{240 \, {\left (4 \, x^{5} - 12 \, x^{3} + 9 \, x\right )}} \] Input:
integrate(9/5/x^2/(-2*x^2+3)^3,x, algorithm="fricas")
Output:
-1/240*(120*x^4 - 5*sqrt(6)*(4*x^5 - 12*x^3 + 9*x)*log((2*x^2 + 2*sqrt(6)* x + 3)/(2*x^2 - 3)) - 300*x^2 + 144)/(4*x^5 - 12*x^3 + 9*x)
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx=- \frac {9 \cdot \left (10 x^{4} - 25 x^{2} + 12\right )}{720 x^{5} - 2160 x^{3} + 1620 x} - \frac {\sqrt {6} \log {\left (x - \frac {\sqrt {6}}{2} \right )}}{48} + \frac {\sqrt {6} \log {\left (x + \frac {\sqrt {6}}{2} \right )}}{48} \] Input:
integrate(9/5/x**2/(-2*x**2+3)**3,x)
Output:
-9*(10*x**4 - 25*x**2 + 12)/(720*x**5 - 2160*x**3 + 1620*x) - sqrt(6)*log( x - sqrt(6)/2)/48 + sqrt(6)*log(x + sqrt(6)/2)/48
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.95 \[ \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx=-\frac {1}{48} \, \sqrt {6} \log \left (\frac {2 \, x - \sqrt {6}}{2 \, x + \sqrt {6}}\right ) - \frac {10 \, x^{4} - 25 \, x^{2} + 12}{20 \, {\left (4 \, x^{5} - 12 \, x^{3} + 9 \, x\right )}} \] Input:
integrate(9/5/x^2/(-2*x^2+3)^3,x, algorithm="maxima")
Output:
-1/48*sqrt(6)*log((2*x - sqrt(6))/(2*x + sqrt(6))) - 1/20*(10*x^4 - 25*x^2 + 12)/(4*x^5 - 12*x^3 + 9*x)
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx=-\frac {1}{48} \, \sqrt {6} \log \left (\frac {{\left | 4 \, x - 2 \, \sqrt {6} \right |}}{{\left | 4 \, x + 2 \, \sqrt {6} \right |}}\right ) - \frac {14 \, x^{3} - 27 \, x}{60 \, {\left (2 \, x^{2} - 3\right )}^{2}} - \frac {1}{15 \, x} \] Input:
integrate(9/5/x^2/(-2*x^2+3)^3,x, algorithm="giac")
Output:
-1/48*sqrt(6)*log(abs(4*x - 2*sqrt(6))/abs(4*x + 2*sqrt(6))) - 1/60*(14*x^ 3 - 27*x)/(2*x^2 - 3)^2 - 1/15/x
Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69 \[ \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx=\frac {\sqrt {6}\,\mathrm {atanh}\left (\frac {\sqrt {6}\,x}{3}\right )}{24}-\frac {\frac {x^4}{8}-\frac {5\,x^2}{16}+\frac {3}{20}}{x^5-3\,x^3+\frac {9\,x}{4}} \] Input:
int(-9/(5*x^2*(2*x^2 - 3)^3),x)
Output:
(6^(1/2)*atanh((6^(1/2)*x)/3))/24 - (x^4/8 - (5*x^2)/16 + 3/20)/((9*x)/4 - 3*x^3 + x^5)
Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.98 \[ \int \frac {9}{5 x^2 \left (3-2 x^2\right )^3} \, dx=\frac {-20 \sqrt {6}\, \mathrm {log}\left (-\sqrt {6}+2 x \right ) x^{5}+60 \sqrt {6}\, \mathrm {log}\left (-\sqrt {6}+2 x \right ) x^{3}-45 \sqrt {6}\, \mathrm {log}\left (-\sqrt {6}+2 x \right ) x +20 \sqrt {6}\, \mathrm {log}\left (\sqrt {6}+2 x \right ) x^{5}-60 \sqrt {6}\, \mathrm {log}\left (\sqrt {6}+2 x \right ) x^{3}+45 \sqrt {6}\, \mathrm {log}\left (\sqrt {6}+2 x \right ) x -120 x^{4}+300 x^{2}-144}{240 x \left (4 x^{4}-12 x^{2}+9\right )} \] Input:
int(9/5/x^2/(-2*x^2+3)^3,x)
Output:
( - 20*sqrt(6)*log( - sqrt(6) + 2*x)*x**5 + 60*sqrt(6)*log( - sqrt(6) + 2* x)*x**3 - 45*sqrt(6)*log( - sqrt(6) + 2*x)*x + 20*sqrt(6)*log(sqrt(6) + 2* x)*x**5 - 60*sqrt(6)*log(sqrt(6) + 2*x)*x**3 + 45*sqrt(6)*log(sqrt(6) + 2* x)*x - 120*x**4 + 300*x**2 - 144)/(240*x*(4*x**4 - 12*x**2 + 9))