Integrand size = 18, antiderivative size = 63 \[ \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {x}{12 a^3}+\frac {x^3}{36 a}-\frac {a x^5}{30}-\frac {\text {arctanh}(a x)}{12 a^4}+\frac {1}{4} x^4 \text {arctanh}(a x)-\frac {1}{6} a^2 x^6 \text {arctanh}(a x) \] Output:
1/12*x/a^3+1/36*x^3/a-1/30*a*x^5-1/12*arctanh(a*x)/a^4+1/4*x^4*arctanh(a*x )-1/6*a^2*x^6*arctanh(a*x)
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25 \[ \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {x}{12 a^3}+\frac {x^3}{36 a}-\frac {a x^5}{30}+\frac {1}{4} x^4 \text {arctanh}(a x)-\frac {1}{6} a^2 x^6 \text {arctanh}(a x)+\frac {\log (1-a x)}{24 a^4}-\frac {\log (1+a x)}{24 a^4} \] Input:
Integrate[x^3*(1 - a^2*x^2)*ArcTanh[a*x],x]
Output:
x/(12*a^3) + x^3/(36*a) - (a*x^5)/30 + (x^4*ArcTanh[a*x])/4 - (a^2*x^6*Arc Tanh[a*x])/6 + Log[1 - a*x]/(24*a^4) - Log[1 + a*x]/(24*a^4)
Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.57, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6576, 6452, 254, 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.
\(\displaystyle \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx\) |
\(\Big \downarrow \) 6576 |
\(\displaystyle \int x^3 \text {arctanh}(a x)dx-a^2 \int x^5 \text {arctanh}(a x)dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle -a^2 \left (\frac {1}{6} x^6 \text {arctanh}(a x)-\frac {1}{6} a \int \frac {x^6}{1-a^2 x^2}dx\right )-\frac {1}{4} a \int \frac {x^4}{1-a^2 x^2}dx+\frac {1}{4} x^4 \text {arctanh}(a x)\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {1}{4} a \int \left (-\frac {x^2}{a^2}+\frac {1}{a^4 \left (1-a^2 x^2\right )}-\frac {1}{a^4}\right )dx-a^2 \left (\frac {1}{6} x^6 \text {arctanh}(a x)-\frac {1}{6} a \int \left (-\frac {x^4}{a^2}-\frac {x^2}{a^4}+\frac {1}{a^6 \left (1-a^2 x^2\right )}-\frac {1}{a^6}\right )dx\right )+\frac {1}{4} x^4 \text {arctanh}(a x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{4} a \left (\frac {\text {arctanh}(a x)}{a^5}-\frac {x}{a^4}-\frac {x^3}{3 a^2}\right )-a^2 \left (\frac {1}{6} x^6 \text {arctanh}(a x)-\frac {1}{6} a \left (\frac {\text {arctanh}(a x)}{a^7}-\frac {x}{a^6}-\frac {x^3}{3 a^4}-\frac {x^5}{5 a^2}\right )\right )+\frac {1}{4} x^4 \text {arctanh}(a x)\) |
Input:
Int[x^3*(1 - a^2*x^2)*ArcTanh[a*x],x]
Output:
(x^4*ArcTanh[a*x])/4 - (a*(-(x/a^4) - x^3/(3*a^2) + ArcTanh[a*x]/a^5))/4 - a^2*((x^6*ArcTanh[a*x])/6 - (a*(-(x/a^6) - x^3/(3*a^4) - x^5/(5*a^2) + Ar cTanh[a*x]/a^7))/6)
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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 = 0.36 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(-\frac {30 \,\operatorname {arctanh}\left (a x \right ) a^{6} x^{6}+6 a^{5} x^{5}-45 a^{4} x^{4} \operatorname {arctanh}\left (a x \right )-5 a^{3} x^{3}-15 a x +15 \,\operatorname {arctanh}\left (a x \right )}{180 a^{4}}\) | \(57\) |
derivativedivides | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right ) a^{6} x^{6}}{6}+\frac {a^{4} x^{4} \operatorname {arctanh}\left (a x \right )}{4}-\frac {a^{5} x^{5}}{30}+\frac {a^{3} x^{3}}{36}+\frac {a x}{12}+\frac {\ln \left (a x -1\right )}{24}-\frac {\ln \left (a x +1\right )}{24}}{a^{4}}\) | \(66\) |
default | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right ) a^{6} x^{6}}{6}+\frac {a^{4} x^{4} \operatorname {arctanh}\left (a x \right )}{4}-\frac {a^{5} x^{5}}{30}+\frac {a^{3} x^{3}}{36}+\frac {a x}{12}+\frac {\ln \left (a x -1\right )}{24}-\frac {\ln \left (a x +1\right )}{24}}{a^{4}}\) | \(66\) |
parts | \(-\frac {a^{2} x^{6} \operatorname {arctanh}\left (a x \right )}{6}+\frac {x^{4} \operatorname {arctanh}\left (a x \right )}{4}-\frac {a \left (\frac {\frac {2}{5} a^{4} x^{5}-\frac {1}{3} a^{2} x^{3}-x}{a^{4}}+\frac {\ln \left (a x +1\right )}{2 a^{5}}-\frac {\ln \left (a x -1\right )}{2 a^{5}}\right )}{12}\) | \(73\) |
risch | \(\left (-\frac {1}{12} a^{2} x^{6}+\frac {1}{8} x^{4}\right ) \ln \left (a x +1\right )+\frac {a^{2} x^{6} \ln \left (-a x +1\right )}{12}-\frac {a \,x^{5}}{30}-\frac {x^{4} \ln \left (-a x +1\right )}{8}+\frac {x^{3}}{36 a}+\frac {x}{12 a^{3}}+\frac {\ln \left (-a x +1\right )}{24 a^{4}}-\frac {\ln \left (a x +1\right )}{24 a^{4}}\) | \(93\) |
orering | \(\frac {\left (15 a^{6} x^{6}-22 a^{4} x^{4}-15 a^{2} x^{2}+15\right ) \left (-a^{2} x^{2}+1\right ) \operatorname {arctanh}\left (a x \right )}{45 a^{4} \left (a^{2} x^{2}-1\right )}-\frac {\left (6 a^{4} x^{4}-5 a^{2} x^{2}-15\right ) \left (3 x^{2} \left (-a^{2} x^{2}+1\right ) \operatorname {arctanh}\left (a x \right )-2 x^{4} a^{2} \operatorname {arctanh}\left (a x \right )+a \,x^{3}\right )}{180 x^{2} a^{4}}\) | \(121\) |
meijerg | \(-\frac {i \left (-\frac {2 i x a \left (21 a^{4} x^{4}+35 a^{2} x^{2}+105\right )}{315}-\frac {i x a \left (-7 a^{6} x^{6}+7\right ) \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{21 \sqrt {a^{2} x^{2}}}\right )}{4 a^{4}}-\frac {i \left (\frac {i x a \left (5 a^{2} x^{2}+15\right )}{15}+\frac {i x a \left (-5 a^{4} x^{4}+5\right ) \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{10 \sqrt {a^{2} x^{2}}}\right )}{4 a^{4}}\) | \(160\) |
Input:
int(x^3*(-a^2*x^2+1)*arctanh(a*x),x,method=_RETURNVERBOSE)
Output:
-1/180*(30*arctanh(a*x)*a^6*x^6+6*a^5*x^5-45*a^4*x^4*arctanh(a*x)-5*a^3*x^ 3-15*a*x+15*arctanh(a*x))/a^4
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97 \[ \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {12 \, a^{5} x^{5} - 10 \, a^{3} x^{3} - 30 \, a x + 15 \, {\left (2 \, a^{6} x^{6} - 3 \, a^{4} x^{4} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{360 \, a^{4}} \] Input:
integrate(x^3*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="fricas")
Output:
-1/360*(12*a^5*x^5 - 10*a^3*x^3 - 30*a*x + 15*(2*a^6*x^6 - 3*a^4*x^4 + 1)* log(-(a*x + 1)/(a*x - 1)))/a^4
Time = 0.39 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\begin {cases} - \frac {a^{2} x^{6} \operatorname {atanh}{\left (a x \right )}}{6} - \frac {a x^{5}}{30} + \frac {x^{4} \operatorname {atanh}{\left (a x \right )}}{4} + \frac {x^{3}}{36 a} + \frac {x}{12 a^{3}} - \frac {\operatorname {atanh}{\left (a x \right )}}{12 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(-a**2*x**2+1)*atanh(a*x),x)
Output:
Piecewise((-a**2*x**6*atanh(a*x)/6 - a*x**5/30 + x**4*atanh(a*x)/4 + x**3/ (36*a) + x/(12*a**3) - atanh(a*x)/(12*a**4), Ne(a, 0)), (0, True))
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.14 \[ \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {1}{360} \, a {\left (\frac {2 \, {\left (6 \, a^{4} x^{5} - 5 \, a^{2} x^{3} - 15 \, x\right )}}{a^{4}} + \frac {15 \, \log \left (a x + 1\right )}{a^{5}} - \frac {15 \, \log \left (a x - 1\right )}{a^{5}}\right )} - \frac {1}{12} \, {\left (2 \, a^{2} x^{6} - 3 \, x^{4}\right )} \operatorname {artanh}\left (a x\right ) \] Input:
integrate(x^3*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="maxima")
Output:
-1/360*a*(2*(6*a^4*x^5 - 5*a^2*x^3 - 15*x)/a^4 + 15*log(a*x + 1)/a^5 - 15* log(a*x - 1)/a^5) - 1/12*(2*a^2*x^6 - 3*x^4)*arctanh(a*x)
Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (51) = 102\).
Time = 0.13 (sec) , antiderivative size = 227, normalized size of antiderivative = 3.60 \[ \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {1}{45} \, a {\left (\frac {\frac {45 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {25 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {35 \, {\left (a x + 1\right )}}{a x - 1} - 7}{a^{5} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{5}} + \frac {30 \, {\left (\frac {3 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {2 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {3 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{5} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{6}}\right )} \] Input:
integrate(x^3*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="giac")
Output:
-1/45*a*((45*(a*x + 1)^3/(a*x - 1)^3 - 25*(a*x + 1)^2/(a*x - 1)^2 + 35*(a* x + 1)/(a*x - 1) - 7)/(a^5*((a*x + 1)/(a*x - 1) - 1)^5) + 30*(3*(a*x + 1)^ 4/(a*x - 1)^4 + 2*(a*x + 1)^3/(a*x - 1)^3 + 3*(a*x + 1)^2/(a*x - 1)^2)*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^5*((a*x + 1)/(a*x - 1) - 1)^6))
Time = 3.66 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81 \[ \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {\frac {a\,x}{12}-\frac {\mathrm {atanh}\left (a\,x\right )}{12}+\frac {a^3\,x^3}{36}}{a^4}-\frac {a\,x^5}{30}+\frac {x^4\,\mathrm {atanh}\left (a\,x\right )}{4}-\frac {a^2\,x^6\,\mathrm {atanh}\left (a\,x\right )}{6} \] Input:
int(-x^3*atanh(a*x)*(a^2*x^2 - 1),x)
Output:
((a*x)/12 - atanh(a*x)/12 + (a^3*x^3)/36)/a^4 - (a*x^5)/30 + (x^4*atanh(a* x))/4 - (a^2*x^6*atanh(a*x))/6
Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int x^3 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {-30 \mathit {atanh} \left (a x \right ) a^{6} x^{6}+45 \mathit {atanh} \left (a x \right ) a^{4} x^{4}-15 \mathit {atanh} \left (a x \right )-6 a^{5} x^{5}+5 a^{3} x^{3}+15 a x}{180 a^{4}} \] Input:
int(x^3*(-a^2*x^2+1)*atanh(a*x),x)
Output:
( - 30*atanh(a*x)*a**6*x**6 + 45*atanh(a*x)*a**4*x**4 - 15*atanh(a*x) - 6* a**5*x**5 + 5*a**3*x**3 + 15*a*x)/(180*a**4)