Integrand size = 20, antiderivative size = 42 \[ \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx=-\frac {x \text {arctanh}(a x)}{a^2}+\frac {\text {arctanh}(a x)^2}{2 a^3}-\frac {\log \left (1-a^2 x^2\right )}{2 a^3} \] Output:
-x*arctanh(a*x)/a^2+1/2*arctanh(a*x)^2/a^3-1/2*ln(-a^2*x^2+1)/a^3
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx=-\frac {x \text {arctanh}(a x)}{a^2}+\frac {\text {arctanh}(a x)^2}{2 a^3}-\frac {\log \left (1-a^2 x^2\right )}{2 a^3} \] Input:
Integrate[(x^2*ArcTanh[a*x])/(1 - a^2*x^2),x]
Output:
-((x*ArcTanh[a*x])/a^2) + ArcTanh[a*x]^2/(2*a^3) - Log[1 - a^2*x^2]/(2*a^3 )
Time = 0.40 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6542, 6436, 240, 6510}
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.
\(\displaystyle \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle \frac {\int \frac {\text {arctanh}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\int \text {arctanh}(a x)dx}{a^2}\) |
\(\Big \downarrow \) 6436 |
\(\displaystyle \frac {\int \frac {\text {arctanh}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {x \text {arctanh}(a x)-a \int \frac {x}{1-a^2 x^2}dx}{a^2}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {\int \frac {\text {arctanh}(a x)}{1-a^2 x^2}dx}{a^2}-\frac {\frac {\log \left (1-a^2 x^2\right )}{2 a}+x \text {arctanh}(a x)}{a^2}\) |
\(\Big \downarrow \) 6510 |
\(\displaystyle \frac {\text {arctanh}(a x)^2}{2 a^3}-\frac {\frac {\log \left (1-a^2 x^2\right )}{2 a}+x \text {arctanh}(a x)}{a^2}\) |
Input:
Int[(x^2*ArcTanh[a*x])/(1 - a^2*x^2),x]
Output:
ArcTanh[a*x]^2/(2*a^3) - (x*ArcTanh[a*x] + Log[1 - a^2*x^2]/(2*a))/a^2
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Simp[b*c*n*p Int[x^n*((a + b*ArcTanh[c*x^n]) ^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0] && (EqQ[n, 1] || EqQ[p, 1])
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb ol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b , c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( e_.)*(x_)^2), x_Symbol] :> Simp[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcTanh[c* x])^p, x], x] - Simp[d*(f^2/e) Int[(f*x)^(m - 2)*((a + b*ArcTanh[c*x])^p/ (d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(-\frac {2 a x \,\operatorname {arctanh}\left (a x \right )-\operatorname {arctanh}\left (a x \right )^{2}+2 \ln \left (a x -1\right )+2 \,\operatorname {arctanh}\left (a x \right )}{2 a^{3}}\) | \(37\) |
risch | \(\frac {\ln \left (a x +1\right )^{2}}{8 a^{3}}-\frac {\left (2 a x +\ln \left (-a x +1\right )\right ) \ln \left (a x +1\right )}{4 a^{3}}+\frac {\ln \left (-a x +1\right ) x}{2 a^{2}}+\frac {\ln \left (-a x +1\right )^{2}}{8 a^{3}}-\frac {\ln \left (a^{2} x^{2}-1\right )}{2 a^{3}}\) | \(80\) |
derivativedivides | \(\frac {-a x \,\operatorname {arctanh}\left (a x \right )-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{2}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {\ln \left (a x -1\right )^{2}}{8}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x -1\right )}{2}-\frac {\ln \left (a x +1\right )}{2}-\frac {\ln \left (a x +1\right )^{2}}{8}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}}{a^{3}}\) | \(114\) |
default | \(\frac {-a x \,\operatorname {arctanh}\left (a x \right )-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{2}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{2}-\frac {\ln \left (a x -1\right )^{2}}{8}+\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{4}-\frac {\ln \left (a x -1\right )}{2}-\frac {\ln \left (a x +1\right )}{2}-\frac {\ln \left (a x +1\right )^{2}}{8}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{4}}{a^{3}}\) | \(114\) |
parts | \(-\frac {x \,\operatorname {arctanh}\left (a x \right )}{a^{2}}+\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x +1\right )}{2 a^{3}}-\frac {\operatorname {arctanh}\left (a x \right ) \ln \left (a x -1\right )}{2 a^{3}}-\frac {a \left (\frac {\frac {\ln \left (a x -1\right )^{2}}{4}-\frac {\operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )}{2}-\frac {\ln \left (a x -1\right ) \ln \left (\frac {a x}{2}+\frac {1}{2}\right )}{2}}{a^{4}}-\frac {-\frac {\ln \left (a x +1\right )^{2}}{4}+\frac {\left (\ln \left (a x +1\right )-\ln \left (\frac {a x}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {a x}{2}+\frac {1}{2}\right )}{2}-\frac {\operatorname {dilog}\left (\frac {a x}{2}+\frac {1}{2}\right )}{2}}{a^{4}}+\frac {\ln \left (a^{2} x^{2}-1\right )}{a^{4}}\right )}{2}\) | \(149\) |
Input:
int(x^2*arctanh(a*x)/(-a^2*x^2+1),x,method=_RETURNVERBOSE)
Output:
-1/2*(2*a*x*arctanh(a*x)-arctanh(a*x)^2+2*ln(a*x-1)+2*arctanh(a*x))/a^3
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.33 \[ \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx=-\frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 4 \, \log \left (a^{2} x^{2} - 1\right )}{8 \, a^{3}} \] Input:
integrate(x^2*arctanh(a*x)/(-a^2*x^2+1),x, algorithm="fricas")
Output:
-1/8*(4*a*x*log(-(a*x + 1)/(a*x - 1)) - log(-(a*x + 1)/(a*x - 1))^2 + 4*lo g(a^2*x^2 - 1))/a^3
Time = 0.36 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx=\begin {cases} - \frac {x \operatorname {atanh}{\left (a x \right )}}{a^{2}} - \frac {\log {\left (x - \frac {1}{a} \right )}}{a^{3}} + \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{2 a^{3}} - \frac {\operatorname {atanh}{\left (a x \right )}}{a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(x**2*atanh(a*x)/(-a**2*x**2+1),x)
Output:
Piecewise((-x*atanh(a*x)/a**2 - log(x - 1/a)/a**3 + atanh(a*x)**2/(2*a**3) - atanh(a*x)/a**3, Ne(a, 0)), (0, True))
Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (38) = 76\).
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.02 \[ \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx=-\frac {1}{2} \, {\left (\frac {2 \, x}{a^{2}} - \frac {\log \left (a x + 1\right )}{a^{3}} + \frac {\log \left (a x - 1\right )}{a^{3}}\right )} \operatorname {artanh}\left (a x\right ) + \frac {2 \, {\left (\log \left (a x - 1\right ) - 2\right )} \log \left (a x + 1\right ) - \log \left (a x + 1\right )^{2} - \log \left (a x - 1\right )^{2} - 4 \, \log \left (a x - 1\right )}{8 \, a^{3}} \] Input:
integrate(x^2*arctanh(a*x)/(-a^2*x^2+1),x, algorithm="maxima")
Output:
-1/2*(2*x/a^2 - log(a*x + 1)/a^3 + log(a*x - 1)/a^3)*arctanh(a*x) + 1/8*(2 *(log(a*x - 1) - 2)*log(a*x + 1) - log(a*x + 1)^2 - log(a*x - 1)^2 - 4*log (a*x - 1))/a^3
\[ \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx=\int { -\frac {x^{2} \operatorname {artanh}\left (a x\right )}{a^{2} x^{2} - 1} \,d x } \] Input:
integrate(x^2*arctanh(a*x)/(-a^2*x^2+1),x, algorithm="giac")
Output:
integrate(-x^2*arctanh(a*x)/(a^2*x^2 - 1), x)
Time = 3.57 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.95 \[ \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx=\frac {{\ln \left (a\,x+1\right )}^2}{8\,a^3}-\ln \left (1-a\,x\right )\,\left (\frac {\ln \left (a\,x+1\right )}{4\,a^3}-\frac {x}{2\,a^2}\right )+\frac {{\ln \left (1-a\,x\right )}^2}{8\,a^3}-\frac {\ln \left (a^2\,x^2-1\right )}{2\,a^3}-\frac {x\,\ln \left (a\,x+1\right )}{2\,a^2} \] Input:
int(-(x^2*atanh(a*x))/(a^2*x^2 - 1),x)
Output:
log(a*x + 1)^2/(8*a^3) - log(1 - a*x)*(log(a*x + 1)/(4*a^3) - x/(2*a^2)) + log(1 - a*x)^2/(8*a^3) - log(a^2*x^2 - 1)/(2*a^3) - (x*log(a*x + 1))/(2*a ^2)
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \text {arctanh}(a x)}{1-a^2 x^2} \, dx=\frac {\mathit {atanh} \left (a x \right )^{2}-2 \mathit {atanh} \left (a x \right ) a x -2 \mathit {atanh} \left (a x \right )-2 \,\mathrm {log}\left (a^{2} x -a \right )}{2 a^{3}} \] Input:
int(x^2*atanh(a*x)/(-a^2*x^2+1),x)
Output:
(atanh(a*x)**2 - 2*atanh(a*x)*a*x - 2*atanh(a*x) - 2*log(a**2*x - a))/(2*a **3)