Integrand size = 20, antiderivative size = 86 \[ \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {4 x^2}{105 a}-\frac {9 a x^4}{140}+\frac {a^3 x^6}{42}+\frac {1}{3} x^3 \text {arctanh}(a x)-\frac {2}{5} a^2 x^5 \text {arctanh}(a x)+\frac {1}{7} a^4 x^7 \text {arctanh}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{105 a^3} \] Output:
4/105*x^2/a-9/140*a*x^4+1/42*a^3*x^6+1/3*x^3*arctanh(a*x)-2/5*a^2*x^5*arct anh(a*x)+1/7*a^4*x^7*arctanh(a*x)+4/105*ln(-a^2*x^2+1)/a^3
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {4 x^2}{105 a}-\frac {9 a x^4}{140}+\frac {a^3 x^6}{42}+\frac {1}{3} x^3 \text {arctanh}(a x)-\frac {2}{5} a^2 x^5 \text {arctanh}(a x)+\frac {1}{7} a^4 x^7 \text {arctanh}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{105 a^3} \] Input:
Integrate[x^2*(1 - a^2*x^2)^2*ArcTanh[a*x],x]
Output:
(4*x^2)/(105*a) - (9*a*x^4)/140 + (a^3*x^6)/42 + (x^3*ArcTanh[a*x])/3 - (2 *a^2*x^5*ArcTanh[a*x])/5 + (a^4*x^7*ArcTanh[a*x])/7 + (4*Log[1 - a^2*x^2]) /(105*a^3)
Time = 0.39 (sec) , antiderivative size = 86, 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.
\(\displaystyle \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx\) |
\(\Big \downarrow \) 6574 |
\(\displaystyle \int \left (a^4 x^6 \text {arctanh}(a x)-2 a^2 x^4 \text {arctanh}(a x)+x^2 \text {arctanh}(a x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{7} a^4 x^7 \text {arctanh}(a x)+\frac {a^3 x^6}{42}-\frac {2}{5} a^2 x^5 \text {arctanh}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{105 a^3}+\frac {1}{3} x^3 \text {arctanh}(a x)-\frac {9 a x^4}{140}+\frac {4 x^2}{105 a}\) |
Input:
Int[x^2*(1 - a^2*x^2)^2*ArcTanh[a*x],x]
Output:
(4*x^2)/(105*a) - (9*a*x^4)/140 + (a^3*x^6)/42 + (x^3*ArcTanh[a*x])/3 - (2 *a^2*x^5*ArcTanh[a*x])/5 + (a^4*x^7*ArcTanh[a*x])/7 + (4*Log[1 - a^2*x^2]) /(105*a^3)
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 = 0.45 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(-\frac {-60 \,\operatorname {arctanh}\left (a x \right ) a^{7} x^{7}-10 a^{6} x^{6}+168 \,\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}+27 a^{4} x^{4}-140 a^{3} x^{3} \operatorname {arctanh}\left (a x \right )-16 a^{2} x^{2}-32 \ln \left (a x -1\right )-32 \,\operatorname {arctanh}\left (a x \right )}{420 a^{3}}\) | \(81\) |
parts | \(\frac {a^{4} x^{7} \operatorname {arctanh}\left (a x \right )}{7}-\frac {2 a^{2} x^{5} \operatorname {arctanh}\left (a x \right )}{5}+\frac {x^{3} \operatorname {arctanh}\left (a x \right )}{3}-\frac {a \left (-\frac {5 a^{4} x^{6}-\frac {27}{2} a^{2} x^{4}+8 x^{2}}{2 a^{2}}-\frac {4 \ln \left (a^{2} x^{2}-1\right )}{a^{4}}\right )}{105}\) | \(81\) |
derivativedivides | \(\frac {\frac {\operatorname {arctanh}\left (a x \right ) a^{7} x^{7}}{7}-\frac {2 \,\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}}{5}+\frac {a^{3} x^{3} \operatorname {arctanh}\left (a x \right )}{3}+\frac {a^{6} x^{6}}{42}-\frac {9 a^{4} x^{4}}{140}+\frac {4 a^{2} x^{2}}{105}+\frac {4 \ln \left (a x -1\right )}{105}+\frac {4 \ln \left (a x +1\right )}{105}}{a^{3}}\) | \(82\) |
default | \(\frac {\frac {\operatorname {arctanh}\left (a x \right ) a^{7} x^{7}}{7}-\frac {2 \,\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}}{5}+\frac {a^{3} x^{3} \operatorname {arctanh}\left (a x \right )}{3}+\frac {a^{6} x^{6}}{42}-\frac {9 a^{4} x^{4}}{140}+\frac {4 a^{2} x^{2}}{105}+\frac {4 \ln \left (a x -1\right )}{105}+\frac {4 \ln \left (a x +1\right )}{105}}{a^{3}}\) | \(82\) |
risch | \(\left (\frac {1}{14} a^{4} x^{7}-\frac {1}{5} a^{2} x^{5}+\frac {1}{6} x^{3}\right ) \ln \left (a x +1\right )-\frac {a^{4} x^{7} \ln \left (-a x +1\right )}{14}+\frac {a^{3} x^{6}}{42}+\frac {a^{2} x^{5} \ln \left (-a x +1\right )}{5}-\frac {9 a \,x^{4}}{140}-\frac {x^{3} \ln \left (-a x +1\right )}{6}+\frac {4 x^{2}}{105 a}+\frac {4 \ln \left (a^{2} x^{2}-1\right )}{105 a^{3}}\) | \(110\) |
meijerg | \(\frac {\frac {x^{2} a^{2} \left (4 a^{4} x^{4}+6 a^{2} x^{2}+12\right )}{42}-\frac {2 x^{8} a^{8} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{7 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{7}}{4 a^{3}}+\frac {-\frac {a^{2} x^{2} \left (3 a^{2} x^{2}+6\right )}{15}+\frac {2 a^{6} x^{6} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{5}}{2 a^{3}}+\frac {\frac {2 a^{2} x^{2}}{3}-\frac {2 a^{4} x^{4} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}}{4 a^{3}}\) | \(249\) |
Input:
int(x^2*(-a^2*x^2+1)^2*arctanh(a*x),x,method=_RETURNVERBOSE)
Output:
-1/420*(-60*arctanh(a*x)*a^7*x^7-10*a^6*x^6+168*arctanh(a*x)*a^5*x^5+27*a^ 4*x^4-140*a^3*x^3*arctanh(a*x)-16*a^2*x^2-32*ln(a*x-1)-32*arctanh(a*x))/a^ 3
Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98 \[ \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {10 \, a^{6} x^{6} - 27 \, a^{4} x^{4} + 16 \, a^{2} x^{2} + 2 \, {\left (15 \, a^{7} x^{7} - 42 \, a^{5} x^{5} + 35 \, a^{3} x^{3}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 16 \, \log \left (a^{2} x^{2} - 1\right )}{420 \, a^{3}} \] Input:
integrate(x^2*(-a^2*x^2+1)^2*arctanh(a*x),x, algorithm="fricas")
Output:
1/420*(10*a^6*x^6 - 27*a^4*x^4 + 16*a^2*x^2 + 2*(15*a^7*x^7 - 42*a^5*x^5 + 35*a^3*x^3)*log(-(a*x + 1)/(a*x - 1)) + 16*log(a^2*x^2 - 1))/a^3
Time = 0.45 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05 \[ \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx=\begin {cases} \frac {a^{4} x^{7} \operatorname {atanh}{\left (a x \right )}}{7} + \frac {a^{3} x^{6}}{42} - \frac {2 a^{2} x^{5} \operatorname {atanh}{\left (a x \right )}}{5} - \frac {9 a x^{4}}{140} + \frac {x^{3} \operatorname {atanh}{\left (a x \right )}}{3} + \frac {4 x^{2}}{105 a} + \frac {8 \log {\left (x - \frac {1}{a} \right )}}{105 a^{3}} + \frac {8 \operatorname {atanh}{\left (a x \right )}}{105 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(x**2*(-a**2*x**2+1)**2*atanh(a*x),x)
Output:
Piecewise((a**4*x**7*atanh(a*x)/7 + a**3*x**6/42 - 2*a**2*x**5*atanh(a*x)/ 5 - 9*a*x**4/140 + x**3*atanh(a*x)/3 + 4*x**2/(105*a) + 8*log(x - 1/a)/(10 5*a**3) + 8*atanh(a*x)/(105*a**3), Ne(a, 0)), (0, True))
Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {1}{420} \, a {\left (\frac {10 \, a^{4} x^{6} - 27 \, a^{2} x^{4} + 16 \, x^{2}}{a^{2}} + \frac {16 \, \log \left (a x + 1\right )}{a^{4}} + \frac {16 \, \log \left (a x - 1\right )}{a^{4}}\right )} + \frac {1}{105} \, {\left (15 \, a^{4} x^{7} - 42 \, a^{2} x^{5} + 35 \, x^{3}\right )} \operatorname {artanh}\left (a x\right ) \] Input:
integrate(x^2*(-a^2*x^2+1)^2*arctanh(a*x),x, algorithm="maxima")
Output:
1/420*a*((10*a^4*x^6 - 27*a^2*x^4 + 16*x^2)/a^2 + 16*log(a*x + 1)/a^4 + 16 *log(a*x - 1)/a^4) + 1/105*(15*a^4*x^7 - 42*a^2*x^5 + 35*x^3)*arctanh(a*x)
Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (72) = 144\).
Time = 0.13 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.71 \[ \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {4}{105} \, a {\left (\frac {2 \, \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{4}} - \frac {2 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{4}} - \frac {\frac {2 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} - \frac {11 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} - \frac {22 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {11 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {2 \, {\left (a x + 1\right )}}{a x - 1}}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{6}} + \frac {2 \, {\left (\frac {70 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {35 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {21 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - \frac {7 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{7}}\right )} \] Input:
integrate(x^2*(-a^2*x^2+1)^2*arctanh(a*x),x, algorithm="giac")
Output:
4/105*a*(2*log(abs(-a*x - 1)/abs(a*x - 1))/a^4 - 2*log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^4 - (2*(a*x + 1)^5/(a*x - 1)^5 - 11*(a*x + 1)^4/(a*x - 1)^4 - 22*(a*x + 1)^3/(a*x - 1)^3 - 11*(a*x + 1)^2/(a*x - 1)^2 + 2*(a*x + 1)/(a *x - 1))/(a^4*((a*x + 1)/(a*x - 1) - 1)^6) + 2*(70*(a*x + 1)^4/(a*x - 1)^4 + 35*(a*x + 1)^3/(a*x - 1)^3 + 21*(a*x + 1)^2/(a*x - 1)^2 - 7*(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)^7))
Time = 3.74 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.83 \[ \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {x^3\,\mathrm {atanh}\left (a\,x\right )}{3}-\frac {9\,a\,x^4}{140}+\frac {4\,\ln \left (a^2\,x^2-1\right )}{105\,a^3}+\frac {4\,x^2}{105\,a}+\frac {a^3\,x^6}{42}-\frac {2\,a^2\,x^5\,\mathrm {atanh}\left (a\,x\right )}{5}+\frac {a^4\,x^7\,\mathrm {atanh}\left (a\,x\right )}{7} \] Input:
int(x^2*atanh(a*x)*(a^2*x^2 - 1)^2,x)
Output:
(x^3*atanh(a*x))/3 - (9*a*x^4)/140 + (4*log(a^2*x^2 - 1))/(105*a^3) + (4*x ^2)/(105*a) + (a^3*x^6)/42 - (2*a^2*x^5*atanh(a*x))/5 + (a^4*x^7*atanh(a*x ))/7
Time = 0.18 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98 \[ \int x^2 \left (1-a^2 x^2\right )^2 \text {arctanh}(a x) \, dx=\frac {60 \mathit {atanh} \left (a x \right ) a^{7} x^{7}-168 \mathit {atanh} \left (a x \right ) a^{5} x^{5}+140 \mathit {atanh} \left (a x \right ) a^{3} x^{3}+32 \mathit {atanh} \left (a x \right )+32 \,\mathrm {log}\left (a^{2} x -a \right )+10 a^{6} x^{6}-27 a^{4} x^{4}+16 a^{2} x^{2}}{420 a^{3}} \] Input:
int(x^2*(-a^2*x^2+1)^2*atanh(a*x),x)
Output:
(60*atanh(a*x)*a**7*x**7 - 168*atanh(a*x)*a**5*x**5 + 140*atanh(a*x)*a**3* x**3 + 32*atanh(a*x) + 32*log(a**2*x - a) + 10*a**6*x**6 - 27*a**4*x**4 + 16*a**2*x**2)/(420*a**3)