Integrand size = 15, antiderivative size = 64 \[ \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {1-a^2 x^2}{6 a}+\frac {2}{3} x \text {arctanh}(a x)+\frac {1}{3} x \left (1-a^2 x^2\right ) \text {arctanh}(a x)+\frac {\log \left (1-a^2 x^2\right )}{3 a} \] Output:
1/6*(-a^2*x^2+1)/a+2/3*x*arctanh(a*x)+1/3*x*(-a^2*x^2+1)*arctanh(a*x)+1/3* ln(-a^2*x^2+1)/a
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.73 \[ \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {a x^2}{6}+x \text {arctanh}(a x)-\frac {1}{3} a^2 x^3 \text {arctanh}(a x)+\frac {\log \left (1-a^2 x^2\right )}{3 a} \] Input:
Integrate[(1 - a^2*x^2)*ArcTanh[a*x],x]
Output:
-1/6*(a*x^2) + x*ArcTanh[a*x] - (a^2*x^3*ArcTanh[a*x])/3 + Log[1 - a^2*x^2 ]/(3*a)
Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6504, 6436, 240}
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.
\(\displaystyle \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx\) |
\(\Big \downarrow \) 6504 |
\(\displaystyle \frac {2}{3} \int \text {arctanh}(a x)dx+\frac {1}{3} x \left (1-a^2 x^2\right ) \text {arctanh}(a x)+\frac {1-a^2 x^2}{6 a}\) |
\(\Big \downarrow \) 6436 |
\(\displaystyle \frac {2}{3} \left (x \text {arctanh}(a x)-a \int \frac {x}{1-a^2 x^2}dx\right )+\frac {1}{3} x \left (1-a^2 x^2\right ) \text {arctanh}(a x)+\frac {1-a^2 x^2}{6 a}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {1}{3} x \left (1-a^2 x^2\right ) \text {arctanh}(a x)+\frac {2}{3} \left (\frac {\log \left (1-a^2 x^2\right )}{2 a}+x \text {arctanh}(a x)\right )+\frac {1-a^2 x^2}{6 a}\) |
Input:
Int[(1 - a^2*x^2)*ArcTanh[a*x],x]
Output:
(1 - a^2*x^2)/(6*a) + (x*(1 - a^2*x^2)*ArcTanh[a*x])/3 + (2*(x*ArcTanh[a*x ] + Log[1 - a^2*x^2]/(2*a)))/3
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_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symb ol] :> Simp[b*((d + e*x^2)^q/(2*c*q*(2*q + 1))), x] + (Simp[x*(d + e*x^2)^q *((a + b*ArcTanh[c*x])/(2*q + 1)), x] + Simp[2*d*(q/(2*q + 1)) Int[(d + e *x^2)^(q - 1)*(a + b*ArcTanh[c*x]), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[q, 0]
Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.69
method | result | size |
parts | \(-\frac {x^{3} a^{2} \operatorname {arctanh}\left (a x \right )}{3}+x \,\operatorname {arctanh}\left (a x \right )-\frac {a \left (\frac {x^{2}}{2}-\frac {\ln \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}{3}\) | \(44\) |
parallelrisch | \(-\frac {2 a^{3} x^{3} \operatorname {arctanh}\left (a x \right )+a^{2} x^{2}-6 a x \,\operatorname {arctanh}\left (a x \right )-4 \ln \left (a x -1\right )-4 \,\operatorname {arctanh}\left (a x \right )}{6 a}\) | \(48\) |
derivativedivides | \(\frac {-\frac {a^{3} x^{3} \operatorname {arctanh}\left (a x \right )}{3}+a x \,\operatorname {arctanh}\left (a x \right )-\frac {a^{2} x^{2}}{6}+\frac {\ln \left (a x -1\right )}{3}+\frac {\ln \left (a x +1\right )}{3}}{a}\) | \(49\) |
default | \(\frac {-\frac {a^{3} x^{3} \operatorname {arctanh}\left (a x \right )}{3}+a x \,\operatorname {arctanh}\left (a x \right )-\frac {a^{2} x^{2}}{6}+\frac {\ln \left (a x -1\right )}{3}+\frac {\ln \left (a x +1\right )}{3}}{a}\) | \(49\) |
risch | \(\left (-\frac {1}{6} a^{2} x^{3}+\frac {1}{2} x \right ) \ln \left (a x +1\right )+\frac {a^{2} x^{3} \ln \left (-a x +1\right )}{6}-\frac {a \,x^{2}}{6}-\frac {x \ln \left (-a x +1\right )}{2}+\frac {\ln \left (a^{2} x^{2}-1\right )}{3 a}\) | \(67\) |
meijerg | \(-\frac {\frac {2 a^{2} x^{2} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (-a^{2} x^{2}+1\right )}{4 a}-\frac {\frac {2 a^{2} x^{2}}{3}-\frac {2 a^{4} x^{4} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{3 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{3}}{4 a}\) | \(140\) |
Input:
int((-a^2*x^2+1)*arctanh(a*x),x,method=_RETURNVERBOSE)
Output:
-1/3*x^3*a^2*arctanh(a*x)+x*arctanh(a*x)-1/3*a*(1/2*x^2-1/a^2*ln(a^2*x^2-1 ))
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {a^{2} x^{2} + {\left (a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 2 \, \log \left (a^{2} x^{2} - 1\right )}{6 \, a} \] Input:
integrate((-a^2*x^2+1)*arctanh(a*x),x, algorithm="fricas")
Output:
-1/6*(a^2*x^2 + (a^3*x^3 - 3*a*x)*log(-(a*x + 1)/(a*x - 1)) - 2*log(a^2*x^ 2 - 1))/a
Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.77 \[ \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\begin {cases} - \frac {a^{2} x^{3} \operatorname {atanh}{\left (a x \right )}}{3} - \frac {a x^{2}}{6} + x \operatorname {atanh}{\left (a x \right )} + \frac {2 \log {\left (x - \frac {1}{a} \right )}}{3 a} + \frac {2 \operatorname {atanh}{\left (a x \right )}}{3 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:
integrate((-a**2*x**2+1)*atanh(a*x),x)
Output:
Piecewise((-a**2*x**3*atanh(a*x)/3 - a*x**2/6 + x*atanh(a*x) + 2*log(x - 1 /a)/(3*a) + 2*atanh(a*x)/(3*a), Ne(a, 0)), (0, True))
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.73 \[ \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=-\frac {1}{6} \, {\left (x^{2} - \frac {2 \, \log \left (a x + 1\right )}{a^{2}} - \frac {2 \, \log \left (a x - 1\right )}{a^{2}}\right )} a - \frac {1}{3} \, {\left (a^{2} x^{3} - 3 \, x\right )} \operatorname {artanh}\left (a x\right ) \] Input:
integrate((-a^2*x^2+1)*arctanh(a*x),x, algorithm="maxima")
Output:
-1/6*(x^2 - 2*log(a*x + 1)/a^2 - 2*log(a*x - 1)/a^2)*a - 1/3*(a^2*x^3 - 3* x)*arctanh(a*x)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (54) = 108\).
Time = 0.12 (sec) , antiderivative size = 203, normalized size of antiderivative = 3.17 \[ \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {2}{3} \, a {\left (\frac {\log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{2}} - \frac {\log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{2}} - \frac {{\left (\frac {3 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{3}} - \frac {a x + 1}{{\left (a x - 1\right )} a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{2}}\right )} \] Input:
integrate((-a^2*x^2+1)*arctanh(a*x),x, algorithm="giac")
Output:
2/3*a*(log(abs(-a*x - 1)/abs(a*x - 1))/a^2 - log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^2 - (3*(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^2*((a*x + 1)/(a*x - 1) - 1)^3) - (a*x + 1)/((a*x - 1)*a^2*((a*x + 1)/(a*x - 1) - 1)^2))
Time = 3.59 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.62 \[ \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=x\,\mathrm {atanh}\left (a\,x\right )-\frac {a\,x^2}{6}+\frac {\ln \left (a^2\,x^2-1\right )}{3\,a}-\frac {a^2\,x^3\,\mathrm {atanh}\left (a\,x\right )}{3} \] Input:
int(-atanh(a*x)*(a^2*x^2 - 1),x)
Output:
x*atanh(a*x) - (a*x^2)/6 + log(a^2*x^2 - 1)/(3*a) - (a^2*x^3*atanh(a*x))/3
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.81 \[ \int \left (1-a^2 x^2\right ) \text {arctanh}(a x) \, dx=\frac {-2 \mathit {atanh} \left (a x \right ) a^{3} x^{3}+6 \mathit {atanh} \left (a x \right ) a x +4 \mathit {atanh} \left (a x \right )+4 \,\mathrm {log}\left (a^{2} x -a \right )-a^{2} x^{2}}{6 a} \] Input:
int((-a^2*x^2+1)*atanh(a*x),x)
Output:
( - 2*atanh(a*x)*a**3*x**3 + 6*atanh(a*x)*a*x + 4*atanh(a*x) + 4*log(a**2* x - a) - a**2*x**2)/(6*a)