Integrand size = 25, antiderivative size = 153 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d+e x}{5 d^2 x \left (d^2-e^2 x^2\right )^{5/2}}+\frac {6 d+5 e x}{15 d^4 x \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 d+5 e x}{5 d^6 x \sqrt {d^2-e^2 x^2}}-\frac {16 \sqrt {d^2-e^2 x^2}}{5 d^7 x}-\frac {e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^7} \]
1/5*(e*x+d)/d^2/x/(-e^2*x^2+d^2)^(5/2)+1/15*(5*e*x+6*d)/d^4/x/(-e^2*x^2+d^ 2)^(3/2)-e*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^7+1/5*(5*e*x+8*d)/d^6/x/(-e^2 *x^2+d^2)^(1/2)-16/5*(-e^2*x^2+d^2)^(1/2)/d^7/x
Time = 0.41 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.88 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (15 d^5-38 d^4 e x-52 d^3 e^2 x^2+87 d^2 e^3 x^3+33 d e^4 x^4-48 e^5 x^5\right )}{x (-d+e x)^3 (d+e x)^2}+30 e \text {arctanh}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^7} \]
((Sqrt[d^2 - e^2*x^2]*(15*d^5 - 38*d^4*e*x - 52*d^3*e^2*x^2 + 87*d^2*e^3*x ^3 + 33*d*e^4*x^4 - 48*e^5*x^5))/(x*(-d + e*x)^3*(d + e*x)^2) + 30*e*ArcTa nh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(15*d^7)
Time = 0.50 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {532, 25, 2336, 27, 2336, 27, 534, 243, 73, 221}
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.
| \(\displaystyle \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int -\frac {\frac {4 e^2 x^2}{d}+5 e x+5 d}{x^2 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\frac {4 e^2 x^2}{d}+5 e x+5 d}{x^2 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {e (5 d+9 e x)}{3 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int -\frac {3 \left (\frac {6 e^2 x^2}{d}+5 e x+5 d\right )}{x^2 \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\frac {6 e^2 x^2}{d}+5 e x+5 d}{x^2 \left (d^2-e^2 x^2\right )^{3/2}}dx}{d^2}+\frac {e (5 d+9 e x)}{3 d^3 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {\frac {e (5 d+11 e x)}{d^3 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {5 (d+e x)}{x^2 \sqrt {d^2-e^2 x^2}}dx}{d^2}}{d^2}+\frac {e (5 d+9 e x)}{3 d^3 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {5 \int \frac {d+e x}{x^2 \sqrt {d^2-e^2 x^2}}dx}{d^2}+\frac {e (5 d+11 e x)}{d^3 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+9 e x)}{3 d^3 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 534 |
| \(\displaystyle \frac {\frac {\frac {5 \left (e \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )}{d^2}+\frac {e (5 d+11 e x)}{d^3 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+9 e x)}{3 d^3 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {\frac {5 \left (\frac {1}{2} e \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )}{d^2}+\frac {e (5 d+11 e x)}{d^3 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+9 e x)}{3 d^3 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {\frac {5 \left (-\frac {\int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{e}-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )}{d^2}+\frac {e (5 d+11 e x)}{d^3 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+9 e x)}{3 d^3 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {5 \left (-\frac {e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d x}\right )}{d^2}+\frac {e (5 d+11 e x)}{d^3 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e (5 d+9 e x)}{3 d^3 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e (d+e x)}{5 d^3 \left (d^2-e^2 x^2\right )^{5/2}}\) |
(e*(d + e*x))/(5*d^3*(d^2 - e^2*x^2)^(5/2)) + ((e*(5*d + 9*e*x))/(3*d^3*(d ^2 - e^2*x^2)^(3/2)) + ((e*(5*d + 11*e*x))/(d^3*Sqrt[d^2 - e^2*x^2]) + (5* (-(Sqrt[d^2 - e^2*x^2]/(d*x)) - (e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d))/d^2 )/d^2)/(5*d^2)
3.1.28.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[x^m *(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*p]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d Int[ x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*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)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
Time = 0.41 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.39
| method | result | size |
| default | \(e \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )+d \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )\) | \(213\) |
| risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d^{7} x}-\frac {e \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{6} \sqrt {d^{2}}}-\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{24 d^{6} e \left (x +\frac {d}{e}\right )^{2}}-\frac {23 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{48 d^{7} \left (x +\frac {d}{e}\right )}+\frac {17 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{60 d^{6} e \left (x -\frac {d}{e}\right )^{2}}-\frac {413 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{240 d^{7} \left (x -\frac {d}{e}\right )}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{20 d^{5} e^{2} \left (x -\frac {d}{e}\right )^{3}}\) | \(295\) |
e*(1/5/d^2/(-e^2*x^2+d^2)^(5/2)+1/d^2*(1/3/d^2/(-e^2*x^2+d^2)^(3/2)+1/d^2* (1/d^2/(-e^2*x^2+d^2)^(1/2)-1/d^2/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^ 2*x^2+d^2)^(1/2))/x))))+d*(-1/d^2/x/(-e^2*x^2+d^2)^(5/2)+6*e^2/d^2*(1/5*x/ d^2/(-e^2*x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4 /(-e^2*x^2+d^2)^(1/2))))
Time = 0.29 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.76 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {23 \, e^{6} x^{6} - 23 \, d e^{5} x^{5} - 46 \, d^{2} e^{4} x^{4} + 46 \, d^{3} e^{3} x^{3} + 23 \, d^{4} e^{2} x^{2} - 23 \, d^{5} e x + 15 \, {\left (e^{6} x^{6} - d e^{5} x^{5} - 2 \, d^{2} e^{4} x^{4} + 2 \, d^{3} e^{3} x^{3} + d^{4} e^{2} x^{2} - d^{5} e x\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (48 \, e^{5} x^{5} - 33 \, d e^{4} x^{4} - 87 \, d^{2} e^{3} x^{3} + 52 \, d^{3} e^{2} x^{2} + 38 \, d^{4} e x - 15 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{7} e^{5} x^{6} - d^{8} e^{4} x^{5} - 2 \, d^{9} e^{3} x^{4} + 2 \, d^{10} e^{2} x^{3} + d^{11} e x^{2} - d^{12} x\right )}} \]
1/15*(23*e^6*x^6 - 23*d*e^5*x^5 - 46*d^2*e^4*x^4 + 46*d^3*e^3*x^3 + 23*d^4 *e^2*x^2 - 23*d^5*e*x + 15*(e^6*x^6 - d*e^5*x^5 - 2*d^2*e^4*x^4 + 2*d^3*e^ 3*x^3 + d^4*e^2*x^2 - d^5*e*x)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (48*e^ 5*x^5 - 33*d*e^4*x^4 - 87*d^2*e^3*x^3 + 52*d^3*e^2*x^2 + 38*d^4*e*x - 15*d ^5)*sqrt(-e^2*x^2 + d^2))/(d^7*e^5*x^6 - d^8*e^4*x^5 - 2*d^9*e^3*x^4 + 2*d ^10*e^2*x^3 + d^11*e*x^2 - d^12*x)
Result contains complex when optimal does not.
Time = 12.16 (sec) , antiderivative size = 2404, normalized size of antiderivative = 15.71 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\text {Too large to display} \]
d*Piecewise((5*d**6*e*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2 *x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 30*d**4*e**3*x**2*sqrt(d* *2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) + 40*d**2*e**5*x**4*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 16*e**7*x **6*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e **4*x**4 + 5*d**8*e**6*x**6), Abs(d**2/(e**2*x**2)) > 1), (5*I*d**6*e*sqrt (-d**2/(e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x** 4 + 5*d**8*e**6*x**6) - 30*I*d**4*e**3*x**2*sqrt(-d**2/(e**2*x**2) + 1)/(- 5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) + 40 *I*d**2*e**5*x**4*sqrt(-d**2/(e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x* *2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 16*I*e**7*x**6*sqrt(-d**2/(e **2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d** 8*e**6*x**6), True)) + e*Piecewise((-46*I*d**6*sqrt(-1 + e**2*x**2/d**2)/( -30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 15*d**6*log(e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4 *x**4 + 30*d**7*e**6*x**6) + 30*d**6*log(e*x/d)/(-30*d**13 + 90*d**11*e**2 *x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) - 30*I*d**6*asin(d/(e*x))/( -30*d**13 + 90*d**11*e**2*x**2 - 90*d**9*e**4*x**4 + 30*d**7*e**6*x**6) + 70*I*d**4*e**2*x**2*sqrt(-1 + e**2*x**2/d**2)/(-30*d**13 + 90*d**11*e**...
Time = 0.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.24 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {6 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {e}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {8 \, e^{2} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {e}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} - \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x} + \frac {16 \, e^{2} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} - \frac {e \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{7}} + \frac {e}{\sqrt {-e^{2} x^{2} + d^{2}} d^{6}} \]
6/5*e^2*x/((-e^2*x^2 + d^2)^(5/2)*d^3) + 1/5*e/((-e^2*x^2 + d^2)^(5/2)*d^2 ) + 8/5*e^2*x/((-e^2*x^2 + d^2)^(3/2)*d^5) + 1/3*e/((-e^2*x^2 + d^2)^(3/2) *d^4) - 1/((-e^2*x^2 + d^2)^(5/2)*d*x) + 16/5*e^2*x/(sqrt(-e^2*x^2 + d^2)* d^7) - e*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^7 + e/(sqrt (-e^2*x^2 + d^2)*d^6)
\[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {e x + d}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} x^{2}} \,d x } \]
Time = 12.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.92 \[ \int \frac {d+e x}{x^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {e}{5\,d^2}+\frac {e\,{\left (d^2-e^2\,x^2\right )}^2}{d^6}+\frac {e\,\left (d^2-e^2\,x^2\right )}{3\,d^4}}{{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {e\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{d^7}-\frac {d^6-6\,d^4\,e^2\,x^2+8\,d^2\,e^4\,x^4-\frac {16\,e^6\,x^6}{5}}{d^7\,x\,{\left (d^2-e^2\,x^2\right )}^{5/2}} \]