Integrand size = 18, antiderivative size = 72 \[ \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {x^2}{35 a^3}+\frac {x^4}{70 a}-\frac {a x^6}{42}+\frac {1}{5} x^5 \text {arctanh}(a x)-\frac {1}{7} a^2 x^7 \text {arctanh}(a x)+\frac {\log \left (1-a^2 x^2\right )}{35 a^5} \] Output:
1/35*x^2/a^3+1/70*x^4/a-1/42*a*x^6+1/5*x^5*arctanh(a*x)-1/7*a^2*x^7*arctan h(a*x)+1/35*ln(-a^2*x^2+1)/a^5
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {x^2}{35 a^3}+\frac {x^4}{70 a}-\frac {a x^6}{42}+\frac {1}{5} x^5 \text {arctanh}(a x)-\frac {1}{7} a^2 x^7 \text {arctanh}(a x)+\frac {\log \left (1-a^2 x^2\right )}{35 a^5} \] Input:
Integrate[x^4*(1 - a^2*x^2)*ArcTanh[a*x],x]
Output:
x^2/(35*a^3) + x^4/(70*a) - (a*x^6)/42 + (x^5*ArcTanh[a*x])/5 - (a^2*x^7*A rcTanh[a*x])/7 + Log[1 - a^2*x^2]/(35*a^5)
Time = 0.41 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.65, 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.
\(\displaystyle \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx\) |
\(\Big \downarrow \) 6576 |
\(\displaystyle \int x^4 \text {arctanh}(a x)dx-a^2 \int x^6 \text {arctanh}(a x)dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle -a^2 \left (\frac {1}{7} x^7 \text {arctanh}(a x)-\frac {1}{7} a \int \frac {x^7}{1-a^2 x^2}dx\right )-\frac {1}{5} a \int \frac {x^5}{1-a^2 x^2}dx+\frac {1}{5} x^5 \text {arctanh}(a x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle -a^2 \left (\frac {1}{7} x^7 \text {arctanh}(a x)-\frac {1}{14} a \int \frac {x^6}{1-a^2 x^2}dx^2\right )-\frac {1}{10} a \int \frac {x^4}{1-a^2 x^2}dx^2+\frac {1}{5} x^5 \text {arctanh}(a x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {1}{10} a \int \left (-\frac {x^2}{a^2}-\frac {1}{a^4 \left (a^2 x^2-1\right )}-\frac {1}{a^4}\right )dx^2-a^2 \left (\frac {1}{7} x^7 \text {arctanh}(a x)-\frac {1}{14} a \int \left (-\frac {x^4}{a^2}-\frac {x^2}{a^4}-\frac {1}{a^6 \left (a^2 x^2-1\right )}-\frac {1}{a^6}\right )dx^2\right )+\frac {1}{5} x^5 \text {arctanh}(a x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{10} a \left (-\frac {x^2}{a^4}-\frac {x^4}{2 a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^6}\right )-a^2 \left (\frac {1}{7} x^7 \text {arctanh}(a x)-\frac {1}{14} a \left (-\frac {x^2}{a^6}-\frac {x^4}{2 a^4}-\frac {x^6}{3 a^2}-\frac {\log \left (1-a^2 x^2\right )}{a^8}\right )\right )+\frac {1}{5} x^5 \text {arctanh}(a x)\) |
Input:
Int[x^4*(1 - a^2*x^2)*ArcTanh[a*x],x]
Output:
(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 - a^2*((x^7*ArcTanh[a*x])/7 - (a*(-(x^2/a^6) - x^4/(2*a^4) - x^6/(3* a^2) - Log[1 - a^2*x^2]/a^8))/14)
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 = 0.42 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.96
method | result | size |
parts | \(-\frac {a^{2} x^{7} \operatorname {arctanh}\left (a x \right )}{7}+\frac {x^{5} \operatorname {arctanh}\left (a x \right )}{5}-\frac {a \left (\frac {\frac {5}{3} a^{4} x^{6}-a^{2} x^{4}-2 x^{2}}{2 a^{4}}-\frac {\ln \left (a^{2} x^{2}-1\right )}{a^{6}}\right )}{35}\) | \(69\) |
derivativedivides | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right ) a^{7} x^{7}}{7}+\frac {\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}}{5}-\frac {a^{6} x^{6}}{42}+\frac {a^{4} x^{4}}{70}+\frac {a^{2} x^{2}}{35}+\frac {\ln \left (a x -1\right )}{35}+\frac {\ln \left (a x +1\right )}{35}}{a^{5}}\) | \(70\) |
default | \(\frac {-\frac {\operatorname {arctanh}\left (a x \right ) a^{7} x^{7}}{7}+\frac {\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}}{5}-\frac {a^{6} x^{6}}{42}+\frac {a^{4} x^{4}}{70}+\frac {a^{2} x^{2}}{35}+\frac {\ln \left (a x -1\right )}{35}+\frac {\ln \left (a x +1\right )}{35}}{a^{5}}\) | \(70\) |
parallelrisch | \(-\frac {30 \,\operatorname {arctanh}\left (a x \right ) a^{7} x^{7}+5 a^{6} x^{6}-42 \,\operatorname {arctanh}\left (a x \right ) a^{5} x^{5}-3 a^{4} x^{4}-6-6 a^{2} x^{2}-12 \ln \left (a x -1\right )-12 \,\operatorname {arctanh}\left (a x \right )}{210 a^{5}}\) | \(70\) |
risch | \(\left (-\frac {1}{14} a^{2} x^{7}+\frac {1}{10} x^{5}\right ) \ln \left (a x +1\right )+\frac {a^{2} x^{7} \ln \left (-a x +1\right )}{14}-\frac {a \,x^{6}}{42}-\frac {x^{5} \ln \left (-a x +1\right )}{10}+\frac {x^{4}}{70 a}+\frac {x^{2}}{35 a^{3}}+\frac {\ln \left (a^{2} x^{2}-1\right )}{35 a^{5}}\) | \(87\) |
meijerg | \(-\frac {\frac {x^{2} a^{2} \left (4 a^{4} x^{4}+6 a^{2} x^{2}+12\right )}{42}-\frac {2 x^{8} a^{8} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{7 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{7}}{4 a^{5}}-\frac {-\frac {a^{2} x^{2} \left (3 a^{2} x^{2}+6\right )}{15}+\frac {2 a^{6} x^{6} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{5}}{4 a^{5}}\) | \(176\) |
Input:
int(x^4*(-a^2*x^2+1)*arctanh(a*x),x,method=_RETURNVERBOSE)
Output:
-1/7*a^2*x^7*arctanh(a*x)+1/5*x^5*arctanh(a*x)-1/35*a*(1/2/a^4*(5/3*a^4*x^ 6-a^2*x^4-2*x^2)-1/a^6*ln(a^2*x^2-1))
Time = 0.07 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {5 \, a^{6} x^{6} - 3 \, a^{4} x^{4} - 6 \, a^{2} x^{2} + 3 \, {\left (5 \, a^{7} x^{7} - 7 \, a^{5} x^{5}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 6 \, \log \left (a^{2} x^{2} - 1\right )}{210 \, a^{5}} \] Input:
integrate(x^4*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="fricas")
Output:
-1/210*(5*a^6*x^6 - 3*a^4*x^4 - 6*a^2*x^2 + 3*(5*a^7*x^7 - 7*a^5*x^5)*log( -(a*x + 1)/(a*x - 1)) - 6*log(a^2*x^2 - 1))/a^5
Time = 0.53 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\begin {cases} - \frac {a^{2} x^{7} \operatorname {atanh}{\left (a x \right )}}{7} - \frac {a x^{6}}{42} + \frac {x^{5} \operatorname {atanh}{\left (a x \right )}}{5} + \frac {x^{4}}{70 a} + \frac {x^{2}}{35 a^{3}} + \frac {2 \log {\left (x - \frac {1}{a} \right )}}{35 a^{5}} + \frac {2 \operatorname {atanh}{\left (a x \right )}}{35 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate(x**4*(-a**2*x**2+1)*atanh(a*x),x)
Output:
Piecewise((-a**2*x**7*atanh(a*x)/7 - a*x**6/42 + x**5*atanh(a*x)/5 + x**4/ (70*a) + x**2/(35*a**3) + 2*log(x - 1/a)/(35*a**5) + 2*atanh(a*x)/(35*a**5 ), Ne(a, 0)), (0, True))
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {1}{210} \, a {\left (\frac {5 \, a^{4} x^{6} - 3 \, a^{2} x^{4} - 6 \, x^{2}}{a^{4}} - \frac {6 \, \log \left (a x + 1\right )}{a^{6}} - \frac {6 \, \log \left (a x - 1\right )}{a^{6}}\right )} - \frac {1}{35} \, {\left (5 \, a^{2} x^{7} - 7 \, x^{5}\right )} \operatorname {artanh}\left (a x\right ) \] Input:
integrate(x^4*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="maxima")
Output:
-1/210*a*((5*a^4*x^6 - 3*a^2*x^4 - 6*x^2)/a^4 - 6*log(a*x + 1)/a^6 - 6*log (a*x - 1)/a^6) - 1/35*(5*a^2*x^7 - 7*x^5)*arctanh(a*x)
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (60) = 120\).
Time = 0.12 (sec) , antiderivative size = 335, normalized size of antiderivative = 4.65 \[ \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {2}{105} \, a {\left (\frac {3 \, \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{6}} - \frac {3 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{6}} - \frac {\frac {3 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} + \frac {36 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {2 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {36 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {3 \, {\left (a x + 1\right )}}{a x - 1}}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{6}} - \frac {3 \, {\left (\frac {35 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} + \frac {35 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {70 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {14 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {7 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{7}}\right )} \] Input:
integrate(x^4*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="giac")
Output:
2/105*a*(3*log(abs(-a*x - 1)/abs(a*x - 1))/a^6 - 3*log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^6 - (3*(a*x + 1)^5/(a*x - 1)^5 + 36*(a*x + 1)^4/(a*x - 1)^4 + 2*(a*x + 1)^3/(a*x - 1)^3 + 36*(a*x + 1)^2/(a*x - 1)^2 + 3*(a*x + 1)/(a* x - 1))/(a^6*((a*x + 1)/(a*x - 1) - 1)^6) - 3*(35*(a*x + 1)^5/(a*x - 1)^5 + 35*(a*x + 1)^4/(a*x - 1)^4 + 70*(a*x + 1)^3/(a*x - 1)^3 + 14*(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^6*((a*x + 1)/(a*x - 1) - 1)^7))
Time = 3.72 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.85 \[ \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {\frac {\ln \left (a^2\,x^2-1\right )}{35}+\frac {a^2\,x^2}{35}+\frac {a^4\,x^4}{70}}{a^5}-\frac {a\,x^6}{42}+\frac {x^5\,\mathrm {atanh}\left (a\,x\right )}{5}-\frac {a^2\,x^7\,\mathrm {atanh}\left (a\,x\right )}{7} \] Input:
int(-x^4*atanh(a*x)*(a^2*x^2 - 1),x)
Output:
(log(a^2*x^2 - 1)/35 + (a^2*x^2)/35 + (a^4*x^4)/70)/a^5 - (a*x^6)/42 + (x^ 5*atanh(a*x))/5 - (a^2*x^7*atanh(a*x))/7
Time = 0.17 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int x^4 \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {-30 \mathit {atanh} \left (a x \right ) a^{7} x^{7}+42 \mathit {atanh} \left (a x \right ) a^{5} x^{5}+12 \mathit {atanh} \left (a x \right )+12 \,\mathrm {log}\left (a^{2} x -a \right )-5 a^{6} x^{6}+3 a^{4} x^{4}+6 a^{2} x^{2}}{210 a^{5}} \] Input:
int(x^4*(-a^2*x^2+1)*atanh(a*x),x)
Output:
( - 30*atanh(a*x)*a**7*x**7 + 42*atanh(a*x)*a**5*x**5 + 12*atanh(a*x) + 12 *log(a**2*x - a) - 5*a**6*x**6 + 3*a**4*x**4 + 6*a**2*x**2)/(210*a**5)