Integrand size = 20, antiderivative size = 64 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^2} \, dx=\frac {a^3 x^2}{6}-\frac {\text {arctanh}(a x)}{x}-2 a^2 x \text {arctanh}(a x)+\frac {1}{3} a^4 x^3 \text {arctanh}(a x)+a \log (x)-\frac {4}{3} a \log \left (1-a^2 x^2\right ) \]
1/6*a^3*x^2-arctanh(a*x)/x-2*a^2*x*arctanh(a*x)+1/3*a^4*x^3*arctanh(a*x)+a *ln(x)-4/3*a*ln(-a^2*x^2+1)
Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^2} \, dx=\frac {a^3 x^2}{6}-\frac {\text {arctanh}(a x)}{x}-2 a^2 x \text {arctanh}(a x)+\frac {1}{3} a^4 x^3 \text {arctanh}(a x)+a \log (x)-\frac {4}{3} a \log \left (1-a^2 x^2\right ) \]
(a^3*x^2)/6 - ArcTanh[a*x]/x - 2*a^2*x*ArcTanh[a*x] + (a^4*x^3*ArcTanh[a*x ])/3 + a*Log[x] - (4*a*Log[1 - a^2*x^2])/3
Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6574, 2009}
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.
\(\displaystyle \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^2} \, dx\) |
\(\Big \downarrow \) 6574 |
\(\displaystyle \int \left (a^4 x^2 \text {arctanh}(a x)-2 a^2 \text {arctanh}(a x)+\frac {\text {arctanh}(a x)}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} a^4 x^3 \text {arctanh}(a x)+\frac {a^3 x^2}{6}-2 a^2 x \text {arctanh}(a x)-\frac {4}{3} a \log \left (1-a^2 x^2\right )-\frac {\text {arctanh}(a x)}{x}+a \log (x)\) |
(a^3*x^2)/6 - ArcTanh[a*x]/x - 2*a^2*x*ArcTanh[a*x] + (a^4*x^3*ArcTanh[a*x ])/3 + a*Log[x] - (4*a*Log[1 - a^2*x^2])/3
3.2.98.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_ .)*(x_)^2)^(q_), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0] && IGtQ[q, 1]
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(a \left (\frac {a^{3} x^{3} \operatorname {arctanh}\left (a x \right )}{3}-2 a x \,\operatorname {arctanh}\left (a x \right )-\frac {\operatorname {arctanh}\left (a x \right )}{a x}+\frac {a^{2} x^{2}}{6}+\ln \left (a x \right )-\frac {4 \ln \left (a x -1\right )}{3}-\frac {4 \ln \left (a x +1\right )}{3}\right )\) | \(64\) |
default | \(a \left (\frac {a^{3} x^{3} \operatorname {arctanh}\left (a x \right )}{3}-2 a x \,\operatorname {arctanh}\left (a x \right )-\frac {\operatorname {arctanh}\left (a x \right )}{a x}+\frac {a^{2} x^{2}}{6}+\ln \left (a x \right )-\frac {4 \ln \left (a x -1\right )}{3}-\frac {4 \ln \left (a x +1\right )}{3}\right )\) | \(64\) |
parts | \(\frac {a^{4} x^{3} \operatorname {arctanh}\left (a x \right )}{3}-2 a^{2} x \,\operatorname {arctanh}\left (a x \right )-\frac {\operatorname {arctanh}\left (a x \right )}{x}-\frac {a \left (-\frac {a^{2} x^{2}}{2}-3 \ln \left (x \right )+4 \ln \left (a x +1\right )+4 \ln \left (a x -1\right )\right )}{3}\) | \(65\) |
parallelrisch | \(\frac {2 a^{4} x^{4} \operatorname {arctanh}\left (a x \right )+a^{3} x^{3}-12 a^{2} x^{2} \operatorname {arctanh}\left (a x \right )+6 a \ln \left (x \right ) x -16 \ln \left (a x -1\right ) a x -16 a x \,\operatorname {arctanh}\left (a x \right )-6 \,\operatorname {arctanh}\left (a x \right )}{6 x}\) | \(68\) |
risch | \(\frac {\left (a^{4} x^{4}-6 a^{2} x^{2}-3\right ) \ln \left (a x +1\right )}{6 x}+\frac {-a^{4} x^{4} \ln \left (-a x +1\right )+a^{3} x^{3}+6 x^{2} \ln \left (-a x +1\right ) a^{2}+6 a \ln \left (x \right ) x -8 a \ln \left (a^{2} x^{2}-1\right ) x +3 \ln \left (-a x +1\right )}{6 x}\) | \(102\) |
meijerg | \(\frac {a \left (4 \ln \left (x \right )+4 \ln \left (i a \right )+\frac {2 \ln \left (1-\sqrt {a^{2} x^{2}}\right )-2 \ln \left (1+\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{4}+\frac {a \left (\frac {2 a^{2} x^{2}}{3}-\frac {2 a^{4} x^{4} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}\right )}{4}+\frac {a \left (\frac {2 a^{2} x^{2} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{2}\) | \(204\) |
a*(1/3*a^3*x^3*arctanh(a*x)-2*a*x*arctanh(a*x)-arctanh(a*x)/a/x+1/6*a^2*x^ 2+ln(a*x)-4/3*ln(a*x-1)-4/3*ln(a*x+1))
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^2} \, dx=\frac {a^{3} x^{3} - 8 \, a x \log \left (a^{2} x^{2} - 1\right ) + 6 \, a x \log \left (x\right ) + {\left (a^{4} x^{4} - 6 \, a^{2} x^{2} - 3\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{6 \, x} \]
1/6*(a^3*x^3 - 8*a*x*log(a^2*x^2 - 1) + 6*a*x*log(x) + (a^4*x^4 - 6*a^2*x^ 2 - 3)*log(-(a*x + 1)/(a*x - 1)))/x
Time = 0.43 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^2} \, dx=\begin {cases} \frac {a^{4} x^{3} \operatorname {atanh}{\left (a x \right )}}{3} + \frac {a^{3} x^{2}}{6} - 2 a^{2} x \operatorname {atanh}{\left (a x \right )} + a \log {\left (x \right )} - \frac {8 a \log {\left (x - \frac {1}{a} \right )}}{3} - \frac {8 a \operatorname {atanh}{\left (a x \right )}}{3} - \frac {\operatorname {atanh}{\left (a x \right )}}{x} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((a**4*x**3*atanh(a*x)/3 + a**3*x**2/6 - 2*a**2*x*atanh(a*x) + a* log(x) - 8*a*log(x - 1/a)/3 - 8*a*atanh(a*x)/3 - atanh(a*x)/x, Ne(a, 0)), (0, True))
Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^2} \, dx=\frac {1}{6} \, {\left (a^{2} x^{2} - 8 \, \log \left (a x + 1\right ) - 8 \, \log \left (a x - 1\right ) + 6 \, \log \left (x\right )\right )} a + \frac {1}{3} \, {\left (a^{4} x^{3} - 6 \, a^{2} x - \frac {3}{x}\right )} \operatorname {artanh}\left (a x\right ) \]
1/6*(a^2*x^2 - 8*log(a*x + 1) - 8*log(a*x - 1) + 6*log(x))*a + 1/3*(a^4*x^ 3 - 6*a^2*x - 3/x)*arctanh(a*x)
Leaf count of result is larger than twice the leaf count of optimal. 249 vs. \(2 (58) = 116\).
Time = 0.28 (sec) , antiderivative size = 249, normalized size of antiderivative = 3.89 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^2} \, dx=\frac {1}{3} \, {\left ({\left (\frac {3}{\frac {a x + 1}{a x - 1} + 1} - \frac {\frac {3 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - \frac {12 \, {\left (a x + 1\right )}}{a x - 1} + 5}{{\left (\frac {a x + 1}{a x - 1} - 1\right )}^{3}}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right ) + \frac {2 \, {\left (a x + 1\right )}}{{\left (a x - 1\right )} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{2}} - 8 \, \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right ) + 5 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right ) + 3 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} - 1 \right |}\right )\right )} a \]
1/3*((3/((a*x + 1)/(a*x - 1) + 1) - (3*(a*x + 1)^2/(a*x - 1)^2 - 12*(a*x + 1)/(a*x - 1) + 5)/((a*x + 1)/(a*x - 1) - 1)^3)*log(-(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x - 1) + 1)/((a* x + 1)*a/(a*x - 1) - a) - 1)) + 2*(a*x + 1)/((a*x - 1)*((a*x + 1)/(a*x - 1 ) - 1)^2) - 8*log(abs(-a*x - 1)/abs(a*x - 1)) + 5*log(abs(-(a*x + 1)/(a*x - 1) + 1)) + 3*log(abs(-(a*x + 1)/(a*x - 1) - 1)))*a
Time = 3.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^2} \, dx=a\,\ln \left (x\right )-\frac {4\,a\,\ln \left (a^2\,x^2-1\right )}{3}-\frac {\mathrm {atanh}\left (a\,x\right )}{x}+\frac {a^3\,x^2}{6}-2\,a^2\,x\,\mathrm {atanh}\left (a\,x\right )+\frac {a^4\,x^3\,\mathrm {atanh}\left (a\,x\right )}{3} \]