Integrand size = 20, antiderivative size = 83 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^6} \, dx=-\frac {a}{20 x^4}+\frac {7 a^3}{30 x^2}-\frac {\text {arctanh}(a x)}{5 x^5}+\frac {2 a^2 \text {arctanh}(a x)}{3 x^3}-\frac {a^4 \text {arctanh}(a x)}{x}+\frac {8}{15} a^5 \log (x)-\frac {4}{15} a^5 \log \left (1-a^2 x^2\right ) \]
-1/20*a/x^4+7/30*a^3/x^2-1/5*arctanh(a*x)/x^5+2/3*a^2*arctanh(a*x)/x^3-a^4 *arctanh(a*x)/x+8/15*a^5*ln(x)-4/15*a^5*ln(-a^2*x^2+1)
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^6} \, dx=-\frac {a}{20 x^4}+\frac {7 a^3}{30 x^2}-\frac {\text {arctanh}(a x)}{5 x^5}+\frac {2 a^2 \text {arctanh}(a x)}{3 x^3}-\frac {a^4 \text {arctanh}(a x)}{x}+\frac {8}{15} a^5 \log (x)-\frac {4}{15} a^5 \log \left (1-a^2 x^2\right ) \]
-1/20*a/x^4 + (7*a^3)/(30*x^2) - ArcTanh[a*x]/(5*x^5) + (2*a^2*ArcTanh[a*x ])/(3*x^3) - (a^4*ArcTanh[a*x])/x + (8*a^5*Log[x])/15 - (4*a^5*Log[1 - a^2 *x^2])/15
Time = 0.34 (sec) , antiderivative size = 83, 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.
\(\displaystyle \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^6} \, dx\) |
\(\Big \downarrow \) 6574 |
\(\displaystyle \int \left (\frac {a^4 \text {arctanh}(a x)}{x^2}-\frac {2 a^2 \text {arctanh}(a x)}{x^4}+\frac {\text {arctanh}(a x)}{x^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {8}{15} a^5 \log (x)-\frac {a^4 \text {arctanh}(a x)}{x}+\frac {7 a^3}{30 x^2}+\frac {2 a^2 \text {arctanh}(a x)}{3 x^3}-\frac {4}{15} a^5 \log \left (1-a^2 x^2\right )-\frac {\text {arctanh}(a x)}{5 x^5}-\frac {a}{20 x^4}\) |
-1/20*a/x^4 + (7*a^3)/(30*x^2) - ArcTanh[a*x]/(5*x^5) + (2*a^2*ArcTanh[a*x ])/(3*x^3) - (a^4*ArcTanh[a*x])/x + (8*a^5*Log[x])/15 - (4*a^5*Log[1 - a^2 *x^2])/15
3.3.2.3.1 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 = 80, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(a^{5} \left (-\frac {\operatorname {arctanh}\left (a x \right )}{5 a^{5} x^{5}}-\frac {\operatorname {arctanh}\left (a x \right )}{a x}+\frac {2 \,\operatorname {arctanh}\left (a x \right )}{3 a^{3} x^{3}}-\frac {1}{20 a^{4} x^{4}}+\frac {7}{30 a^{2} x^{2}}+\frac {8 \ln \left (a x \right )}{15}-\frac {4 \ln \left (a x -1\right )}{15}-\frac {4 \ln \left (a x +1\right )}{15}\right )\) | \(80\) |
default | \(a^{5} \left (-\frac {\operatorname {arctanh}\left (a x \right )}{5 a^{5} x^{5}}-\frac {\operatorname {arctanh}\left (a x \right )}{a x}+\frac {2 \,\operatorname {arctanh}\left (a x \right )}{3 a^{3} x^{3}}-\frac {1}{20 a^{4} x^{4}}+\frac {7}{30 a^{2} x^{2}}+\frac {8 \ln \left (a x \right )}{15}-\frac {4 \ln \left (a x -1\right )}{15}-\frac {4 \ln \left (a x +1\right )}{15}\right )\) | \(80\) |
parts | \(-\frac {\operatorname {arctanh}\left (a x \right )}{5 x^{5}}+\frac {2 a^{2} \operatorname {arctanh}\left (a x \right )}{3 x^{3}}-\frac {a^{4} \operatorname {arctanh}\left (a x \right )}{x}-\frac {a \left (\frac {3}{4 x^{4}}-\frac {7 a^{2}}{2 x^{2}}-8 a^{4} \ln \left (x \right )+4 a^{4} \ln \left (a x +1\right )+4 a^{4} \ln \left (a x -1\right )\right )}{15}\) | \(81\) |
parallelrisch | \(\frac {32 \ln \left (x \right ) a^{5} x^{5}-32 \ln \left (a x -1\right ) x^{5} a^{5}-32 \,\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}+14 a^{5} x^{5}-60 a^{4} x^{4} \operatorname {arctanh}\left (a x \right )+14 a^{3} x^{3}+40 a^{2} x^{2} \operatorname {arctanh}\left (a x \right )-3 a x -12 \,\operatorname {arctanh}\left (a x \right )}{60 x^{5}}\) | \(93\) |
risch | \(-\frac {\left (15 a^{4} x^{4}-10 a^{2} x^{2}+3\right ) \ln \left (a x +1\right )}{30 x^{5}}+\frac {32 \ln \left (x \right ) a^{5} x^{5}-16 \ln \left (a^{2} x^{2}-1\right ) a^{5} x^{5}+30 a^{4} x^{4} \ln \left (-a x +1\right )+14 a^{3} x^{3}-20 x^{2} \ln \left (-a x +1\right ) a^{2}-3 a x +6 \ln \left (-a x +1\right )}{60 x^{5}}\) | \(116\) |
meijerg | \(\frac {a^{5} \left (-\frac {1}{a^{4} x^{4}}-\frac {2}{3 a^{2} x^{2}}-\frac {4}{25}+\frac {4 \ln \left (x \right )}{5}+\frac {4 \ln \left (i a \right )}{5}+\frac {\frac {4}{25} a^{4} x^{4}+\frac {4}{15} a^{2} x^{2}+\frac {4}{5}}{a^{4} x^{4}}+\frac {\frac {2 \ln \left (1-\sqrt {a^{2} x^{2}}\right )}{5}-\frac {2 \ln \left (1+\sqrt {a^{2} x^{2}}\right )}{5}}{a^{4} x^{4} \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{5}\right )}{4}+\frac {a^{5} \left (4 \ln \left (x \right )+4 \ln \left (i a \right )+\frac {2 \ln \left (1-\sqrt {a^{2} x^{2}}\right )-2 \ln \left (1+\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )\right )}{4}+\frac {a^{5} \left (\frac {2}{a^{2} x^{2}}+\frac {4}{9}-\frac {4 \ln \left (x \right )}{3}-\frac {4 \ln \left (i a \right )}{3}-\frac {2 \left (10 a^{2} x^{2}+30\right )}{45 a^{2} x^{2}}-\frac {2 \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 a^{2} x^{2} \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}\right )}{2}\) | \(294\) |
a^5*(-1/5*arctanh(a*x)/a^5/x^5-arctanh(a*x)/a/x+2/3*arctanh(a*x)/a^3/x^3-1 /20/a^4/x^4+7/30/a^2/x^2+8/15*ln(a*x)-4/15*ln(a*x-1)-4/15*ln(a*x+1))
Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^6} \, dx=-\frac {16 \, a^{5} x^{5} \log \left (a^{2} x^{2} - 1\right ) - 32 \, a^{5} x^{5} \log \left (x\right ) - 14 \, a^{3} x^{3} + 3 \, a x + 2 \, {\left (15 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 3\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{60 \, x^{5}} \]
-1/60*(16*a^5*x^5*log(a^2*x^2 - 1) - 32*a^5*x^5*log(x) - 14*a^3*x^3 + 3*a* x + 2*(15*a^4*x^4 - 10*a^2*x^2 + 3)*log(-(a*x + 1)/(a*x - 1)))/x^5
Time = 0.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.06 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^6} \, dx=\begin {cases} \frac {8 a^{5} \log {\left (x \right )}}{15} - \frac {8 a^{5} \log {\left (x - \frac {1}{a} \right )}}{15} - \frac {8 a^{5} \operatorname {atanh}{\left (a x \right )}}{15} - \frac {a^{4} \operatorname {atanh}{\left (a x \right )}}{x} + \frac {7 a^{3}}{30 x^{2}} + \frac {2 a^{2} \operatorname {atanh}{\left (a x \right )}}{3 x^{3}} - \frac {a}{20 x^{4}} - \frac {\operatorname {atanh}{\left (a x \right )}}{5 x^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((8*a**5*log(x)/15 - 8*a**5*log(x - 1/a)/15 - 8*a**5*atanh(a*x)/1 5 - a**4*atanh(a*x)/x + 7*a**3/(30*x**2) + 2*a**2*atanh(a*x)/(3*x**3) - a/ (20*x**4) - atanh(a*x)/(5*x**5), Ne(a, 0)), (0, True))
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^6} \, dx=-\frac {1}{60} \, {\left (16 \, a^{4} \log \left (a^{2} x^{2} - 1\right ) - 16 \, a^{4} \log \left (x^{2}\right ) - \frac {14 \, a^{2} x^{2} - 3}{x^{4}}\right )} a - \frac {{\left (15 \, a^{4} x^{4} - 10 \, a^{2} x^{2} + 3\right )} \operatorname {artanh}\left (a x\right )}{15 \, x^{5}} \]
-1/60*(16*a^4*log(a^2*x^2 - 1) - 16*a^4*log(x^2) - (14*a^2*x^2 - 3)/x^4)*a - 1/15*(15*a^4*x^4 - 10*a^2*x^2 + 3)*arctanh(a*x)/x^5
Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (71) = 142\).
Time = 0.28 (sec) , antiderivative size = 265, normalized size of antiderivative = 3.19 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^6} \, dx=-\frac {4}{15} \, {\left (2 \, a^{4} \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right ) - 2 \, a^{4} \log \left ({\left | -\frac {a x + 1}{a x - 1} - 1 \right |}\right ) + \frac {\frac {2 \, {\left (a x + 1\right )}^{3} a^{4}}{{\left (a x - 1\right )}^{3}} + \frac {7 \, {\left (a x + 1\right )}^{2} a^{4}}{{\left (a x - 1\right )}^{2}} + \frac {2 \, {\left (a x + 1\right )} a^{4}}{a x - 1}}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{4}} - \frac {2 \, {\left (\frac {10 \, {\left (a x + 1\right )}^{2} a^{4}}{{\left (a x - 1\right )}^{2}} + \frac {5 \, {\left (a x + 1\right )} a^{4}}{a x - 1} + a^{4}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{5}}\right )} a \]
-4/15*(2*a^4*log(abs(-a*x - 1)/abs(a*x - 1)) - 2*a^4*log(abs(-(a*x + 1)/(a *x - 1) - 1)) + (2*(a*x + 1)^3*a^4/(a*x - 1)^3 + 7*(a*x + 1)^2*a^4/(a*x - 1)^2 + 2*(a*x + 1)*a^4/(a*x - 1))/((a*x + 1)/(a*x - 1) + 1)^4 - 2*(10*(a*x + 1)^2*a^4/(a*x - 1)^2 + 5*(a*x + 1)*a^4/(a*x - 1) + a^4)*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))/((a*x + 1)/(a*x - 1) + 1)^5)*a
Time = 3.43 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{x^6} \, dx=\frac {8\,a^5\,\ln \left (x\right )}{15}-\frac {a}{20\,x^4}-\frac {\mathrm {atanh}\left (a\,x\right )}{5\,x^5}-\frac {4\,a^5\,\ln \left (a^2\,x^2-1\right )}{15}+\frac {7\,a^3}{30\,x^2}+\frac {2\,a^2\,\mathrm {atanh}\left (a\,x\right )}{3\,x^3}-\frac {a^4\,\mathrm {atanh}\left (a\,x\right )}{x} \]