Integrand size = 18, antiderivative size = 62 \[ \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {x^2}{15 a}-\frac {a x^4}{20}+\frac {1}{3} x^3 \text {arctanh}(a x)-\frac {1}{5} a^2 x^5 \text {arctanh}(a x)+\frac {\log \left (1-a^2 x^2\right )}{15 a^3} \] Output:
1/15*x^2/a-1/20*a*x^4+1/3*x^3*arctanh(a*x)-1/5*a^2*x^5*arctanh(a*x)+1/15*l n(-a^2*x^2+1)/a^3
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {x^2}{15 a}-\frac {a x^4}{20}+\frac {1}{3} x^3 \text {arctanh}(a x)-\frac {1}{5} a^2 x^5 \text {arctanh}(a x)+\frac {\log \left (1-a^2 x^2\right )}{15 a^3} \] Input:
Integrate[x^2*(1 - a^2*x^2)*ArcTanh[a*x],x]
Output:
x^2/(15*a) - (a*x^4)/20 + (x^3*ArcTanh[a*x])/3 - (a^2*x^5*ArcTanh[a*x])/5 + Log[1 - a^2*x^2]/(15*a^3)
Time = 0.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.60, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6576, 6452, 243, 49, 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.
\(\displaystyle \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx\) |
\(\Big \downarrow \) 6576 |
\(\displaystyle \int x^2 \text {arctanh}(a x)dx-a^2 \int x^4 \text {arctanh}(a x)dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle -a^2 \left (\frac {1}{5} x^5 \text {arctanh}(a x)-\frac {1}{5} a \int \frac {x^5}{1-a^2 x^2}dx\right )-\frac {1}{3} a \int \frac {x^3}{1-a^2 x^2}dx+\frac {1}{3} x^3 \text {arctanh}(a x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -a^2 \left (\frac {1}{5} x^5 \text {arctanh}(a x)-\frac {1}{10} a \int \frac {x^4}{1-a^2 x^2}dx^2\right )-\frac {1}{6} a \int \frac {x^2}{1-a^2 x^2}dx^2+\frac {1}{3} x^3 \text {arctanh}(a x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {1}{6} a \int \left (-\frac {1}{a^2}-\frac {1}{a^2 \left (a^2 x^2-1\right )}\right )dx^2-a^2 \left (\frac {1}{5} x^5 \text {arctanh}(a x)-\frac {1}{10} a \int \left (-\frac {x^2}{a^2}-\frac {1}{a^4 \left (a^2 x^2-1\right )}-\frac {1}{a^4}\right )dx^2\right )+\frac {1}{3} x^3 \text {arctanh}(a x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{6} a \left (-\frac {x^2}{a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^4}\right )-a^2 \left (\frac {1}{5} x^5 \text {arctanh}(a x)-\frac {1}{10} a \left (-\frac {x^2}{a^4}-\frac {x^4}{2 a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^6}\right )\right )+\frac {1}{3} x^3 \text {arctanh}(a x)\) |
Input:
Int[x^2*(1 - a^2*x^2)*ArcTanh[a*x],x]
Output:
(x^3*ArcTanh[a*x])/3 - (a*(-(x^2/a^2) - Log[1 - a^2*x^2]/a^4))/6 - a^2*((x ^5*ArcTanh[a*x])/5 - (a*(-(x^2/a^4) - x^4/(2*a^2) - Log[1 - a^2*x^2]/a^6)) /10)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_ .)*(x_)^2)^(q_.), x_Symbol] :> Simp[d Int[(f*x)^m*(d + e*x^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Simp[c^2*(d/f^2) Int[(f*x)^(m + 2)*(d + e*x ^2)^(q - 1)*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[q, 0] && IGtQ[p, 0] && (RationalQ[m] || (EqQ [p, 1] && IntegerQ[q]))
Time = 0.32 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98
method | result | size |
parallelrisch | \(-\frac {12 \,\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}+3 a^{4} x^{4}-20 a^{3} x^{3} \operatorname {arctanh}\left (a x \right )-4 a^{2} x^{2}-8 \ln \left (a x -1\right )-8 \,\operatorname {arctanh}\left (a x \right )}{60 a^{3}}\) | \(61\) |
parts | \(-\frac {a^{2} x^{5} \operatorname {arctanh}\left (a x \right )}{5}+\frac {x^{3} \operatorname {arctanh}\left (a x \right )}{3}-\frac {a \left (\frac {\frac {3}{2} a^{2} x^{4}-2 x^{2}}{2 a^{2}}-\frac {\ln \left (a^{2} x^{2}-1\right )}{a^{4}}\right )}{15}\) | \(61\) |
derivativedivides | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}}{5}+\frac {a^{3} x^{3} \operatorname {arctanh}\left (a x \right )}{3}-\frac {a^{4} x^{4}}{20}+\frac {a^{2} x^{2}}{15}+\frac {\ln \left (a x -1\right )}{15}+\frac {\ln \left (a x +1\right )}{15}}{a^{3}}\) | \(62\) |
default | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}}{5}+\frac {a^{3} x^{3} \operatorname {arctanh}\left (a x \right )}{3}-\frac {a^{4} x^{4}}{20}+\frac {a^{2} x^{2}}{15}+\frac {\ln \left (a x -1\right )}{15}+\frac {\ln \left (a x +1\right )}{15}}{a^{3}}\) | \(62\) |
risch | \(\left (-\frac {1}{10} a^{2} x^{5}+\frac {1}{6} x^{3}\right ) \ln \left (a x +1\right )+\frac {a^{2} x^{5} \ln \left (-a x +1\right )}{10}-\frac {a \,x^{4}}{20}-\frac {x^{3} \ln \left (-a x +1\right )}{6}+\frac {x^{2}}{15 a}+\frac {\ln \left (a^{2} x^{2}-1\right )}{15 a^{3}}-\frac {1}{45 a^{3}}\) | \(84\) |
meijerg | \(\frac {-\frac {a^{2} x^{2} \left (3 a^{2} x^{2}+6\right )}{15}+\frac {2 a^{6} x^{6} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{5}}{4 a^{3}}+\frac {\frac {2 a^{2} x^{2}}{3}-\frac {2 a^{4} x^{4} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}}{4 a^{3}}\) | \(158\) |
Input:
int(x^2*(-a^2*x^2+1)*arctanh(a*x),x,method=_RETURNVERBOSE)
Output:
-1/60*(12*arctanh(a*x)*a^5*x^5+3*a^4*x^4-20*a^3*x^3*arctanh(a*x)-4*a^2*x^2 -8*ln(a*x-1)-8*arctanh(a*x))/a^3
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10 \[ \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {3 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 2 \, {\left (3 \, a^{5} x^{5} - 5 \, a^{3} x^{3}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 4 \, \log \left (a^{2} x^{2} - 1\right )}{60 \, a^{3}} \] Input:
integrate(x^2*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="fricas")
Output:
-1/60*(3*a^4*x^4 - 4*a^2*x^2 + 2*(3*a^5*x^5 - 5*a^3*x^3)*log(-(a*x + 1)/(a *x - 1)) - 4*log(a^2*x^2 - 1))/a^3
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02 \[ \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\begin {cases} - \frac {a^{2} x^{5} \operatorname {atanh}{\left (a x \right )}}{5} - \frac {a x^{4}}{20} + \frac {x^{3} \operatorname {atanh}{\left (a x \right )}}{3} + \frac {x^{2}}{15 a} + \frac {2 \log {\left (x - \frac {1}{a} \right )}}{15 a^{3}} + \frac {2 \operatorname {atanh}{\left (a x \right )}}{15 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(-a**2*x**2+1)*atanh(a*x),x)
Output:
Piecewise((-a**2*x**5*atanh(a*x)/5 - a*x**4/20 + x**3*atanh(a*x)/3 + x**2/ (15*a) + 2*log(x - 1/a)/(15*a**3) + 2*atanh(a*x)/(15*a**3), Ne(a, 0)), (0, True))
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.05 \[ \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {1}{60} \, a {\left (\frac {3 \, a^{2} x^{4} - 4 \, x^{2}}{a^{2}} - \frac {4 \, \log \left (a x + 1\right )}{a^{4}} - \frac {4 \, \log \left (a x - 1\right )}{a^{4}}\right )} - \frac {1}{15} \, {\left (3 \, a^{2} x^{5} - 5 \, x^{3}\right )} \operatorname {artanh}\left (a x\right ) \] Input:
integrate(x^2*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="maxima")
Output:
-1/60*a*((3*a^2*x^4 - 4*x^2)/a^2 - 4*log(a*x + 1)/a^4 - 4*log(a*x - 1)/a^4 ) - 1/15*(3*a^2*x^5 - 5*x^3)*arctanh(a*x)
Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (52) = 104\).
Time = 0.13 (sec) , antiderivative size = 268, normalized size of antiderivative = 4.32 \[ \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {2}{15} \, a {\left (\frac {\log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{4}} - \frac {\log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{4}} - \frac {\frac {{\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {4 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {a x + 1}{a x - 1}}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{4}} - \frac {{\left (\frac {15 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {5 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {5 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{5}}\right )} \] Input:
integrate(x^2*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="giac")
Output:
2/15*a*(log(abs(-a*x - 1)/abs(a*x - 1))/a^4 - log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^4 - ((a*x + 1)^3/(a*x - 1)^3 + 4*(a*x + 1)^2/(a*x - 1)^2 + (a*x + 1)/(a*x - 1))/(a^4*((a*x + 1)/(a*x - 1) - 1)^4) - (15*(a*x + 1)^3/(a*x - 1)^3 + 5*(a*x + 1)^2/(a*x - 1)^2 + 5*(a*x + 1)/(a*x - 1) - 1)*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^4*((a*x + 1)/(a*x - 1) - 1) ^5))
Time = 3.51 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {\frac {\ln \left (a^2\,x^2-1\right )}{15}+\frac {a^2\,x^2}{15}}{a^3}-\frac {a\,x^4}{20}+\frac {x^3\,\mathrm {atanh}\left (a\,x\right )}{3}-\frac {a^2\,x^5\,\mathrm {atanh}\left (a\,x\right )}{5} \] Input:
int(-x^2*atanh(a*x)*(a^2*x^2 - 1),x)
Output:
(log(a^2*x^2 - 1)/15 + (a^2*x^2)/15)/a^3 - (a*x^4)/20 + (x^3*atanh(a*x))/3 - (a^2*x^5*atanh(a*x))/5
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.03 \[ \int x^2 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {-12 \mathit {atanh} \left (a x \right ) a^{5} x^{5}+20 \mathit {atanh} \left (a x \right ) a^{3} x^{3}+8 \mathit {atanh} \left (a x \right )+8 \,\mathrm {log}\left (a^{2} x -a \right )-3 a^{4} x^{4}+4 a^{2} x^{2}}{60 a^{3}} \] Input:
int(x^2*(-a^2*x^2+1)*atanh(a*x),x)
Output:
( - 12*atanh(a*x)*a**5*x**5 + 20*atanh(a*x)*a**3*x**3 + 8*atanh(a*x) + 8*l og(a**2*x - a) - 3*a**4*x**4 + 4*a**2*x**2)/(60*a**3)