Integrand size = 18, antiderivative size = 42 \[ \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx=-\frac {a}{12 x^3}+\frac {a^3}{4 x}-\frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{4 x^4} \] Output:
-1/12*a/x^3+1/4*a^3/x-1/4*(-a^2*x^2+1)^2*arctanh(a*x)/x^4
Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.69 \[ \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx=-\frac {a}{12 x^3}+\frac {a^3}{4 x}-\frac {\text {arctanh}(a x)}{4 x^4}+\frac {a^2 \text {arctanh}(a x)}{2 x^2}+\frac {1}{8} a^4 \log (1-a x)-\frac {1}{8} a^4 \log (1+a x) \] Input:
Integrate[((1 - a^2*x^2)*ArcTanh[a*x])/x^5,x]
Output:
-1/12*a/x^3 + a^3/(4*x) - ArcTanh[a*x]/(4*x^4) + (a^2*ArcTanh[a*x])/(2*x^2 ) + (a^4*Log[1 - a*x])/8 - (a^4*Log[1 + a*x])/8
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6570, 244, 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.
\(\displaystyle \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx\) |
\(\Big \downarrow \) 6570 |
\(\displaystyle \frac {1}{4} a \int \frac {1-a^2 x^2}{x^4}dx-\frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{4 x^4}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {1}{4} a \int \left (\frac {1}{x^4}-\frac {a^2}{x^2}\right )dx-\frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{4 x^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} a \left (\frac {a^2}{x}-\frac {1}{3 x^3}\right )-\frac {\left (1-a^2 x^2\right )^2 \text {arctanh}(a x)}{4 x^4}\) |
Input:
Int[((1 - a^2*x^2)*ArcTanh[a*x])/x^5,x]
Output:
(a*(-1/3*1/x^3 + a^2/x))/4 - ((1 - a^2*x^2)^2*ArcTanh[a*x])/(4*x^4)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e _.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^p/(d*(m + 1))), x] - Simp[b*c*(p/(m + 1)) Int[(f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d + e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]
Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(-\frac {3 a^{4} x^{4} \operatorname {arctanh}\left (a x \right )-3 a^{3} x^{3}-6 a^{2} x^{2} \operatorname {arctanh}\left (a x \right )+a x +3 \,\operatorname {arctanh}\left (a x \right )}{12 x^{4}}\) | \(48\) |
derivativedivides | \(a^{4} \left (-\frac {\operatorname {arctanh}\left (a x \right )}{4 a^{4} x^{4}}+\frac {\operatorname {arctanh}\left (a x \right )}{2 a^{2} x^{2}}+\frac {1}{4 a x}-\frac {1}{12 a^{3} x^{3}}+\frac {\ln \left (a x -1\right )}{8}-\frac {\ln \left (a x +1\right )}{8}\right )\) | \(62\) |
default | \(a^{4} \left (-\frac {\operatorname {arctanh}\left (a x \right )}{4 a^{4} x^{4}}+\frac {\operatorname {arctanh}\left (a x \right )}{2 a^{2} x^{2}}+\frac {1}{4 a x}-\frac {1}{12 a^{3} x^{3}}+\frac {\ln \left (a x -1\right )}{8}-\frac {\ln \left (a x +1\right )}{8}\right )\) | \(62\) |
parts | \(-\frac {\operatorname {arctanh}\left (a x \right )}{4 x^{4}}+\frac {a^{2} \operatorname {arctanh}\left (a x \right )}{2 x^{2}}-\frac {a \left (\frac {1}{3 x^{3}}-\frac {a^{2}}{x}+\frac {a^{3} \ln \left (a x +1\right )}{2}-\frac {a^{3} \ln \left (a x -1\right )}{2}\right )}{4}\) | \(62\) |
risch | \(\frac {\left (2 a^{2} x^{2}-1\right ) \ln \left (a x +1\right )}{8 x^{4}}+\frac {3 x^{4} \ln \left (-a x +1\right ) a^{4}-3 \ln \left (-a x -1\right ) a^{4} x^{4}+6 a^{3} x^{3}-6 x^{2} \ln \left (-a x +1\right ) a^{2}-2 a x +3 \ln \left (-a x +1\right )}{24 x^{4}}\) | \(95\) |
orering | \(\frac {\left (3 a^{4} x^{4}-6 a^{2} x^{2}+2\right ) \left (-a^{2} x^{2}+1\right ) \operatorname {arctanh}\left (a x \right )}{3 x^{4} \left (a^{2} x^{2}-1\right )}+\frac {\left (3 a^{2} x^{2}-1\right ) x^{2} \left (-\frac {2 a^{2} \operatorname {arctanh}\left (a x \right )}{x^{4}}+\frac {a}{x^{5}}-\frac {5 \left (-a^{2} x^{2}+1\right ) \operatorname {arctanh}\left (a x \right )}{x^{6}}\right )}{12}\) | \(102\) |
meijerg | \(-\frac {i a^{4} \left (-\frac {i}{3 x^{3} a^{3}}-\frac {i}{x a}+\frac {4 i \left (-\frac {3 a^{4} x^{4}}{8}+\frac {3}{8}\right ) \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 x^{3} a^{3} \sqrt {a^{2} x^{2}}}\right )}{4}-\frac {i a^{4} \left (\frac {2 i}{x a}+\frac {2 i \left (-a x +1\right ) \left (a x +1\right ) \operatorname {arctanh}\left (a x \right )}{x^{2} a^{2}}\right )}{4}\) | \(124\) |
Input:
int((-a^2*x^2+1)*arctanh(a*x)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/12*(3*a^4*x^4*arctanh(a*x)-3*a^3*x^3-6*a^2*x^2*arctanh(a*x)+a*x+3*arcta nh(a*x))/x^4
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.24 \[ \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx=\frac {6 \, a^{3} x^{3} - 2 \, a x - 3 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{24 \, x^{4}} \] Input:
integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="fricas")
Output:
1/24*(6*a^3*x^3 - 2*a*x - 3*(a^4*x^4 - 2*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))/x^4
Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10 \[ \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx=- \frac {a^{4} \operatorname {atanh}{\left (a x \right )}}{4} + \frac {a^{3}}{4 x} + \frac {a^{2} \operatorname {atanh}{\left (a x \right )}}{2 x^{2}} - \frac {a}{12 x^{3}} - \frac {\operatorname {atanh}{\left (a x \right )}}{4 x^{4}} \] Input:
integrate((-a**2*x**2+1)*atanh(a*x)/x**5,x)
Output:
-a**4*atanh(a*x)/4 + a**3/(4*x) + a**2*atanh(a*x)/(2*x**2) - a/(12*x**3) - atanh(a*x)/(4*x**4)
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.45 \[ \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx=-\frac {1}{24} \, {\left (3 \, a^{3} \log \left (a x + 1\right ) - 3 \, a^{3} \log \left (a x - 1\right ) - \frac {2 \, {\left (3 \, a^{2} x^{2} - 1\right )}}{x^{3}}\right )} a + \frac {{\left (2 \, a^{2} x^{2} - 1\right )} \operatorname {artanh}\left (a x\right )}{4 \, x^{4}} \] Input:
integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="maxima")
Output:
-1/24*(3*a^3*log(a*x + 1) - 3*a^3*log(a*x - 1) - 2*(3*a^2*x^2 - 1)/x^3)*a + 1/4*(2*a^2*x^2 - 1)*arctanh(a*x)/x^4
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (35) = 70\).
Time = 0.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 3.81 \[ \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx=-\frac {1}{3} \, a {\left (\frac {a^{3} {\left (\frac {3 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )}}{{\left (\frac {a x + 1}{a x - 1} + 1\right )}^{3}} + \frac {6 \, {\left (a x + 1\right )}^{2} a^{3} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{{\left (a x - 1\right )}^{2} {\left (\frac {a x + 1}{a x - 1} + 1\right )}^{4}}\right )} \] Input:
integrate((-a^2*x^2+1)*arctanh(a*x)/x^5,x, algorithm="giac")
Output:
-1/3*a*(a^3*(3*(a*x + 1)/(a*x - 1) + 1)/((a*x + 1)/(a*x - 1) + 1)^3 + 6*(a *x + 1)^2*a^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))/((a* x - 1)^2*((a*x + 1)/(a*x - 1) + 1)^4))
Time = 3.71 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.45 \[ \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx=\frac {a^3}{4\,x}-\frac {\mathrm {atanh}\left (a\,x\right )}{4\,x^4}-\frac {a}{12\,x^3}+\frac {a^5\,\mathrm {atan}\left (\frac {a^2\,x}{\sqrt {-a^2}}\right )}{4\,\sqrt {-a^2}}+\frac {a^2\,\mathrm {atanh}\left (a\,x\right )}{2\,x^2} \] Input:
int(-(atanh(a*x)*(a^2*x^2 - 1))/x^5,x)
Output:
a^3/(4*x) - atanh(a*x)/(4*x^4) - a/(12*x^3) + (a^5*atan((a^2*x)/(-a^2)^(1/ 2)))/(4*(-a^2)^(1/2)) + (a^2*atanh(a*x))/(2*x^2)
Time = 0.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.14 \[ \int \frac {\left (1-a^2 x^2\right ) \text {arctanh}(a x)}{x^5} \, dx=\frac {-3 \mathit {atanh} \left (a x \right ) a^{4} x^{4}+6 \mathit {atanh} \left (a x \right ) a^{2} x^{2}-3 \mathit {atanh} \left (a x \right )+3 a^{3} x^{3}-a x}{12 x^{4}} \] Input:
int((-a^2*x^2+1)*atanh(a*x)/x^5,x)
Output:
( - 3*atanh(a*x)*a**4*x**4 + 6*atanh(a*x)*a**2*x**2 - 3*atanh(a*x) + 3*a** 3*x**3 - a*x)/(12*x**4)