Integrand size = 24, antiderivative size = 195 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=-\frac {(1-a x)^2}{a^2 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x}-\frac {2 (1-a x)^3}{5 a^3 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^2}+\frac {2 (1-a x)^3 (1+a x)}{15 a^4 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^3}-\frac {2 (1-a x)^3 (1+a x)^2 (28+13 a x)}{15 a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5}-\frac {2 (1-a x)^{5/2} (1+a x)^{5/2} \arcsin (a x)}{a^6 \left (c-\frac {c}{a^2 x^2}\right )^{5/2} x^5} \]
-(-a*x+1)^2/a^2/(c-c/a^2/x^2)^(5/2)/x-2/5*(-a*x+1)^3/a^3/(c-c/a^2/x^2)^(5/ 2)/x^2+2/15*(-a*x+1)^3*(a*x+1)/a^4/(c-c/a^2/x^2)^(5/2)/x^3-2/15*(-a*x+1)^3 *(a*x+1)^2*(13*a*x+28)/a^6/(c-c/a^2/x^2)^(5/2)/x^5-2*(-a*x+1)^(5/2)*(a*x+1 )^(5/2)*arcsin(a*x)/a^6/(c-c/a^2/x^2)^(5/2)/x^5
Time = 0.16 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.54 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\frac {-56-82 a x+32 a^2 x^2+76 a^3 x^3+15 a^4 x^4-30 (1+a x)^2 \sqrt {-1+a^2 x^2} \log \left (a x+\sqrt {-1+a^2 x^2}\right )}{15 a^2 c^2 \sqrt {c-\frac {c}{a^2 x^2}} x (1+a x)^2} \]
(-56 - 82*a*x + 32*a^2*x^2 + 76*a^3*x^3 + 15*a^4*x^4 - 30*(1 + a*x)^2*Sqrt [-1 + a^2*x^2]*Log[a*x + Sqrt[-1 + a^2*x^2]])/(15*a^2*c^2*Sqrt[c - c/(a^2* x^2)]*x*(1 + a*x)^2)
Time = 0.67 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.82, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6717, 6709, 570, 529, 25, 2166, 2345, 27, 455, 223}
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.
| \(\displaystyle \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6717 |
| \(\displaystyle -\int \frac {e^{-2 \text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 6709 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {x^5}{(a x+1)^2 \left (1-a^2 x^2\right )^{3/2}}dx}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
| \(\Big \downarrow \) 570 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \int \frac {x^5 (1-a x)^2}{\left (1-a^2 x^2\right )^{7/2}}dx}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
| \(\Big \downarrow \) 529 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {(1-a x)^2}{5 a^6 \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{5} \int -\frac {(1-a x) \left (\frac {5 x^4}{a}-\frac {5 x^3}{a^2}+\frac {5 x^2}{a^3}-\frac {5 x}{a^4}+\frac {2}{a^5}\right )}{\left (1-a^2 x^2\right )^{5/2}}dx\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {1}{5} \int \frac {(1-a x) \left (\frac {5 x^4}{a}-\frac {5 x^3}{a^2}+\frac {5 x^2}{a^3}-\frac {5 x}{a^4}+\frac {2}{a^5}\right )}{\left (1-a^2 x^2\right )^{5/2}}dx+\frac {(1-a x)^2}{5 a^6 \left (1-a^2 x^2\right )^{5/2}}\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
| \(\Big \downarrow \) 2166 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {1}{5} \left (-\frac {1}{3} \int \frac {-\frac {15 x^3}{a^2}+\frac {30 x^2}{a^3}-\frac {45 x}{a^4}+\frac {16}{a^5}}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {22 (1-a x)}{3 a^6 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {(1-a x)^2}{5 a^6 \left (1-a^2 x^2\right )^{5/2}}\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\int \frac {15 (2-a x)}{a^5 \sqrt {1-a^2 x^2}}dx+\frac {2 (30-23 a x)}{a^6 \sqrt {1-a^2 x^2}}\right )-\frac {22 (1-a x)}{3 a^6 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {(1-a x)^2}{5 a^6 \left (1-a^2 x^2\right )^{5/2}}\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {15 \int \frac {2-a x}{\sqrt {1-a^2 x^2}}dx}{a^5}+\frac {2 (30-23 a x)}{a^6 \sqrt {1-a^2 x^2}}\right )-\frac {22 (1-a x)}{3 a^6 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {(1-a x)^2}{5 a^6 \left (1-a^2 x^2\right )^{5/2}}\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {15 \left (2 \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^5}+\frac {2 (30-23 a x)}{a^6 \sqrt {1-a^2 x^2}}\right )-\frac {22 (1-a x)}{3 a^6 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {(1-a x)^2}{5 a^6 \left (1-a^2 x^2\right )^{5/2}}\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle -\frac {\left (1-a^2 x^2\right )^{5/2} \left (\frac {(1-a x)^2}{5 a^6 \left (1-a^2 x^2\right )^{5/2}}+\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 (30-23 a x)}{a^6 \sqrt {1-a^2 x^2}}+\frac {15 \left (\frac {\sqrt {1-a^2 x^2}}{a}+\frac {2 \arcsin (a x)}{a}\right )}{a^5}\right )-\frac {22 (1-a x)}{3 a^6 \left (1-a^2 x^2\right )^{3/2}}\right )\right )}{x^5 \left (c-\frac {c}{a^2 x^2}\right )^{5/2}}\) |
-(((1 - a^2*x^2)^(5/2)*((1 - a*x)^2/(5*a^6*(1 - a^2*x^2)^(5/2)) + ((-22*(1 - a*x))/(3*a^6*(1 - a^2*x^2)^(3/2)) + ((2*(30 - 23*a*x))/(a^6*Sqrt[1 - a^ 2*x^2]) + (15*(Sqrt[1 - a^2*x^2]/a + (2*ArcSin[a*x])/a))/a^5)/3)/5))/((c - c/(a^2*x^2))^(5/2)*x^5))
3.9.69.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ (2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b* x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* c^2 + a*d^2, 0]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c^(2*n)/a^n Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I LtQ[n, -1] && !(IGtQ[m, 0] && ILtQ[m + n, 0] && !GtQ[p, 1])
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e *(p + 1))), x] + Simp[d/(2*a*(p + 1)) Int[(d + e*x)^(m - 1)*(a + b*x^2)^( p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ [{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 0] && GtQ[m, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbo l] :> Simp[x^(2*p)*((c + d/x^2)^p/(1 - a^2*x^2)^p) Int[u*((1 + a*x)^n/(x^ (2*p)*(1 - a^2*x^2)^(n/2 - p))), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a^2*d, 0] && !IntegerQ[p] && IntegerQ[n/2] && !GtQ[c, 0]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2) Int[ u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
Time = 0.55 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.47
| method | result | size |
| risch | \(\frac {a^{2} x^{2}-1}{a^{2} c^{2} x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}+\frac {\left (-\frac {2 \ln \left (\frac {a^{2} c x}{\sqrt {a^{2} c}}+\sqrt {a^{2} c \,x^{2}-c}\right )}{a^{5} \sqrt {a^{2} c}}-\frac {41 \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-2 \left (x +\frac {1}{a}\right ) a c}}{60 a^{8} c \left (x +\frac {1}{a}\right )^{2}}+\frac {383 \sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-2 \left (x +\frac {1}{a}\right ) a c}}{120 a^{7} c \left (x +\frac {1}{a}\right )}-\frac {\sqrt {a^{2} c \left (x -\frac {1}{a}\right )^{2}+2 \left (x -\frac {1}{a}\right ) a c}}{8 a^{7} c \left (x -\frac {1}{a}\right )}+\frac {\sqrt {a^{2} c \left (x +\frac {1}{a}\right )^{2}-2 \left (x +\frac {1}{a}\right ) a c}}{10 a^{9} c \left (x +\frac {1}{a}\right )^{3}}\right ) a^{4} \sqrt {c \left (a^{2} x^{2}-1\right )}}{c^{2} x \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}\) | \(286\) |
| default | \(\frac {\left (15 c^{\frac {5}{2}} \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}} a^{5} x^{5}+45 x^{4} c^{\frac {5}{2}} a^{4} \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}}+16 c^{\frac {5}{2}} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a^{4} x^{4}-60 c^{\frac {5}{2}} \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}} a^{3} x^{3}+16 c^{\frac {5}{2}} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a^{3} x^{3}-30 \ln \left (\sqrt {c}\, x +\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\right ) \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a^{4} c x -90 c^{\frac {5}{2}} \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}} a^{2} x^{2}-24 c^{\frac {5}{2}} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a^{2} x^{2}-30 \ln \left (\sqrt {c}\, x +\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}}\right ) \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a^{3} c +50 c^{\frac {5}{2}} \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}} a x -24 c^{\frac {5}{2}} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}} a x +50 c^{\frac {5}{2}} \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}}+6 c^{\frac {5}{2}} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2}}\right )}^{\frac {3}{2}}\right ) \left (a x -1\right )}{15 \left (\frac {c \left (a x -1\right ) \left (a x +1\right )}{a^{2}}\right )^{\frac {3}{2}} x^{5} {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {5}{2}} a^{6} c^{\frac {5}{2}}}\) | \(462\) |
1/a^2*(a^2*x^2-1)/c^2/x/(c*(a^2*x^2-1)/a^2/x^2)^(1/2)+(-2/a^5*ln(a^2*c*x/( a^2*c)^(1/2)+(a^2*c*x^2-c)^(1/2))/(a^2*c)^(1/2)-41/60/a^8/c/(x+1/a)^2*(a^2 *c*(x+1/a)^2-2*(x+1/a)*a*c)^(1/2)+383/120/a^7/c/(x+1/a)*(a^2*c*(x+1/a)^2-2 *(x+1/a)*a*c)^(1/2)-1/8/a^7/c/(x-1/a)*(a^2*c*(x-1/a)^2+2*(x-1/a)*a*c)^(1/2 )+1/10/a^9/c/(x+1/a)^3*(a^2*c*(x+1/a)^2-2*(x+1/a)*a*c)^(1/2))*a^4/c^2/x/(c *(a^2*x^2-1)/a^2/x^2)^(1/2)*(c*(a^2*x^2-1))^(1/2)
Time = 0.28 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.80 \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \sqrt {c} \log \left (2 \, a^{2} c x^{2} - 2 \, a^{2} \sqrt {c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right ) + {\left (15 \, a^{5} x^{5} + 76 \, a^{4} x^{4} + 32 \, a^{3} x^{3} - 82 \, a^{2} x^{2} - 56 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{15 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}}, \frac {30 \, {\left (a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1\right )} \sqrt {-c} \arctan \left (\frac {a^{2} \sqrt {-c} x^{2} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right ) + {\left (15 \, a^{5} x^{5} + 76 \, a^{4} x^{4} + 32 \, a^{3} x^{3} - 82 \, a^{2} x^{2} - 56 \, a x\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{15 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} - 2 \, a^{2} c^{3} x - a c^{3}\right )}}\right ] \]
[1/15*(15*(a^4*x^4 + 2*a^3*x^3 - 2*a*x - 1)*sqrt(c)*log(2*a^2*c*x^2 - 2*a^ 2*sqrt(c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - c) + (15*a^5*x^5 + 76*a^4* x^4 + 32*a^3*x^3 - 82*a^2*x^2 - 56*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/( a^5*c^3*x^4 + 2*a^4*c^3*x^3 - 2*a^2*c^3*x - a*c^3), 1/15*(30*(a^4*x^4 + 2* a^3*x^3 - 2*a*x - 1)*sqrt(-c)*arctan(a^2*sqrt(-c)*x^2*sqrt((a^2*c*x^2 - c) /(a^2*x^2))/(a^2*c*x^2 - c)) + (15*a^5*x^5 + 76*a^4*x^4 + 32*a^3*x^3 - 82* a^2*x^2 - 56*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)))/(a^5*c^3*x^4 + 2*a^4*c^ 3*x^3 - 2*a^2*c^3*x - a*c^3)]
\[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {a x - 1}{\left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}\, dx \]
\[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int { \frac {a x - 1}{{\left (a x + 1\right )} {\left (c - \frac {c}{a^{2} x^{2}}\right )}^{\frac {5}{2}}} \,d x } \]
Exception generated. \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {e^{-2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{5/2}} \, dx=\int \frac {a\,x-1}{{\left (c-\frac {c}{a^2\,x^2}\right )}^{5/2}\,\left (a\,x+1\right )} \,d x \]