Integrand size = 24, antiderivative size = 169 \[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx=-\frac {a^2 \sqrt {1-a^2 x^2}}{3 x}-\frac {a \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{3 x^2}-\frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^2}{3 x^3}+\frac {2}{3} a^3 \text {arctanh}(a x) \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )-\frac {1}{3} a^3 \operatorname {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right )+\frac {1}{3} a^3 \operatorname {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \] Output:
-1/3*a^2*(-a^2*x^2+1)^(1/2)/x-1/3*a*(-a^2*x^2+1)^(1/2)*arctanh(a*x)/x^2-1/ 3*(-a^2*x^2+1)^(3/2)*arctanh(a*x)^2/x^3+2/3*a^3*arctanh(a*x)*arctanh((-a*x +1)^(1/2)/(a*x+1)^(1/2))-1/3*a^3*polylog(2,-(-a*x+1)^(1/2)/(a*x+1)^(1/2))+ 1/3*a^3*polylog(2,(-a*x+1)^(1/2)/(a*x+1)^(1/2))
Time = 1.38 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx=-\frac {1}{3} a^3 \operatorname {PolyLog}\left (2,-e^{-\text {arctanh}(a x)}\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \left (4 \text {arctanh}(a x)^2+2 (-1+\cosh (2 \text {arctanh}(a x)))-\frac {4 a^3 x^3 \operatorname {PolyLog}\left (2,e^{-\text {arctanh}(a x)}\right )}{\left (1-a^2 x^2\right )^{3/2}}+\text {arctanh}(a x) \left (2 \sinh (2 \text {arctanh}(a x))+\frac {\left (\log \left (1-e^{-\text {arctanh}(a x)}\right )-\log \left (1+e^{-\text {arctanh}(a x)}\right )\right ) \left (-3 a x+\sqrt {1-a^2 x^2} \sinh (3 \text {arctanh}(a x))\right )}{\sqrt {1-a^2 x^2}}\right )\right )}{12 x^3} \] Input:
Integrate[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/x^4,x]
Output:
-1/3*(a^3*PolyLog[2, -E^(-ArcTanh[a*x])]) - ((1 - a^2*x^2)^(3/2)*(4*ArcTan h[a*x]^2 + 2*(-1 + Cosh[2*ArcTanh[a*x]]) - (4*a^3*x^3*PolyLog[2, E^(-ArcTa nh[a*x])])/(1 - a^2*x^2)^(3/2) + ArcTanh[a*x]*(2*Sinh[2*ArcTanh[a*x]] + (( Log[1 - E^(-ArcTanh[a*x])] - Log[1 + E^(-ArcTanh[a*x])])*(-3*a*x + Sqrt[1 - a^2*x^2]*Sinh[3*ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2])))/(12*x^3)
Time = 0.73 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6570, 6572, 242, 6588, 242, 6580}
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.
\(\displaystyle \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx\) |
\(\Big \downarrow \) 6570 |
\(\displaystyle \frac {2}{3} a \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{x^3}dx-\frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^2}{3 x^3}\) |
\(\Big \downarrow \) 6572 |
\(\displaystyle \frac {2}{3} a \left (-\int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}}dx+a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{x^2}\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^2}{3 x^3}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {2}{3} a \left (-\int \frac {\text {arctanh}(a x)}{x^3 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{x^2}-\frac {a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^2}{3 x^3}\) |
\(\Big \downarrow \) 6588 |
\(\displaystyle \frac {2}{3} a \left (-\frac {1}{2} a^2 \int \frac {\text {arctanh}(a x)}{x \sqrt {1-a^2 x^2}}dx-\frac {1}{2} a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 x^2}-\frac {a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^2}{3 x^3}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {2}{3} a \left (-\frac {1}{2} a^2 \int \frac {\text {arctanh}(a x)}{x \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 x^2}-\frac {a \sqrt {1-a^2 x^2}}{2 x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^2}{3 x^3}\) |
\(\Big \downarrow \) 6580 |
\(\displaystyle \frac {2}{3} a \left (-\frac {1}{2} a^2 \left (-2 \text {arctanh}(a x) \text {arctanh}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )+\operatorname {PolyLog}\left (2,-\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )-\operatorname {PolyLog}\left (2,\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right )\right )-\frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 x^2}-\frac {a \sqrt {1-a^2 x^2}}{2 x}\right )-\frac {\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)^2}{3 x^3}\) |
Input:
Int[(Sqrt[1 - a^2*x^2]*ArcTanh[a*x]^2)/x^4,x]
Output:
-1/3*((1 - a^2*x^2)^(3/2)*ArcTanh[a*x]^2)/x^3 + (2*a*(-1/2*(a*Sqrt[1 - a^2 *x^2])/x - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(2*x^2) - (a^2*(-2*ArcTanh[a*x ]*ArcTanh[Sqrt[1 - a*x]/Sqrt[1 + a*x]] + PolyLog[2, -(Sqrt[1 - a*x]/Sqrt[1 + a*x])] - PolyLog[2, Sqrt[1 - a*x]/Sqrt[1 + a*x]]))/2))/3
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e _.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])^p/(d*(m + 1))), x] - Simp[b*c*(p/(m + 1)) Int[(f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[c^2*d + e, 0] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.) *(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcTanh[c *x])/(f*(m + 2))), x] + (Simp[d/(m + 2) Int[(f*x)^m*((a + b*ArcTanh[c*x]) /Sqrt[d + e*x^2]), x], x] - Simp[b*c*(d/(f*(m + 2))) Int[(f*x)^(m + 1)/Sq rt[d + e*x^2], x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && NeQ[m, -2]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x _Symbol] :> Simp[(-2/Sqrt[d])*(a + b*ArcTanh[c*x])*ArcTanh[Sqrt[1 - c*x]/Sq rt[1 + c*x]], x] + (Simp[(b/Sqrt[d])*PolyLog[2, -Sqrt[1 - c*x]/Sqrt[1 + c*x ]], x] - Simp[(b/Sqrt[d])*PolyLog[2, Sqrt[1 - c*x]/Sqrt[1 + c*x]], x]) /; F reeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*A rcTanh[c*x])^p/(d*f*(m + 1))), x] + (-Simp[b*c*(p/(f*(m + 1))) Int[(f*x)^ (m + 1)*((a + b*ArcTanh[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] + Simp[c^2*( (m + 2)/(f^2*(m + 1))) Int[(f*x)^(m + 2)*((a + b*ArcTanh[c*x])^p/Sqrt[d + e*x^2]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && G tQ[p, 0] && LtQ[m, -1] && NeQ[m, -2]
Time = 0.58 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (a^{2} x^{2} \operatorname {arctanh}\left (a x \right )^{2}-a^{2} x^{2}-a x \,\operatorname {arctanh}\left (a x \right )-\operatorname {arctanh}\left (a x \right )^{2}\right )}{3 x^{3}}-\frac {a^{3} \operatorname {arctanh}\left (a x \right ) \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}-\frac {a^{3} \operatorname {polylog}\left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}+\frac {a^{3} \operatorname {arctanh}\left (a x \right ) \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}+\frac {a^{3} \operatorname {polylog}\left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}\) | \(171\) |
Input:
int((-a^2*x^2+1)^(1/2)*arctanh(a*x)^2/x^4,x,method=_RETURNVERBOSE)
Output:
1/3*(-(a*x-1)*(a*x+1))^(1/2)*(a^2*x^2*arctanh(a*x)^2-a^2*x^2-a*x*arctanh(a *x)-arctanh(a*x)^2)/x^3-1/3*a^3*arctanh(a*x)*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/ 2))-1/3*a^3*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+1/3*a^3*arctanh(a*x)*ln( 1+(a*x+1)/(-a^2*x^2+1)^(1/2))+1/3*a^3*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2 ))
\[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}}{x^{4}} \,d x } \] Input:
integrate((-a^2*x^2+1)^(1/2)*arctanh(a*x)^2/x^4,x, algorithm="fricas")
Output:
integral(sqrt(-a^2*x^2 + 1)*arctanh(a*x)^2/x^4, x)
\[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \operatorname {atanh}^{2}{\left (a x \right )}}{x^{4}}\, dx \] Input:
integrate((-a**2*x**2+1)**(1/2)*atanh(a*x)**2/x**4,x)
Output:
Integral(sqrt(-(a*x - 1)*(a*x + 1))*atanh(a*x)**2/x**4, x)
\[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1} \operatorname {artanh}\left (a x\right )^{2}}{x^{4}} \,d x } \] Input:
integrate((-a^2*x^2+1)^(1/2)*arctanh(a*x)^2/x^4,x, algorithm="maxima")
Output:
integrate(sqrt(-a^2*x^2 + 1)*arctanh(a*x)^2/x^4, x)
Exception generated. \[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((-a^2*x^2+1)^(1/2)*arctanh(a*x)^2/x^4,x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx=\int \frac {{\mathrm {atanh}\left (a\,x\right )}^2\,\sqrt {1-a^2\,x^2}}{x^4} \,d x \] Input:
int((atanh(a*x)^2*(1 - a^2*x^2)^(1/2))/x^4,x)
Output:
int((atanh(a*x)^2*(1 - a^2*x^2)^(1/2))/x^4, x)
\[ \int \frac {\sqrt {1-a^2 x^2} \text {arctanh}(a x)^2}{x^4} \, dx=\int \frac {\sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right )^{2}}{x^{4}}d x \] Input:
int((-a^2*x^2+1)^(1/2)*atanh(a*x)^2/x^4,x)
Output:
int((sqrt( - a**2*x**2 + 1)*atanh(a*x)**2)/x**4,x)