Integrand size = 25, antiderivative size = 184 \[ \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}-\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^8} \]
1/5*(e*x+d)/d^2/x^2/(-e^2*x^2+d^2)^(5/2)+1/15*(6*e*x+7*d)/d^4/x^2/(-e^2*x^ 2+d^2)^(3/2)-7/2*e^2*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^8+1/15*(24*e*x+35*d )/d^6/x^2/(-e^2*x^2+d^2)^(1/2)-7/2*(-e^2*x^2+d^2)^(1/2)/d^7/x^2-16/5*e*(-e ^2*x^2+d^2)^(1/2)/d^8/x
Time = 0.47 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.80 \[ \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (15 d^6+15 d^5 e x-176 d^4 e^2 x^2-4 d^3 e^3 x^3+249 d^2 e^4 x^4-9 d e^5 x^5-96 e^6 x^6\right )}{x^2 (-d+e x)^3 (d+e x)^2}+210 e^2 \text {arctanh}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{30 d^8} \]
((Sqrt[d^2 - e^2*x^2]*(15*d^6 + 15*d^5*e*x - 176*d^4*e^2*x^2 - 4*d^3*e^3*x ^3 + 249*d^2*e^4*x^4 - 9*d*e^5*x^5 - 96*e^6*x^6))/(x^2*(-d + e*x)^3*(d + e *x)^2) + 210*e^2*ArcTanh[(Sqrt[-e^2]*x - Sqrt[d^2 - e^2*x^2])/d])/(30*d^8)
Time = 0.67 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {532, 25, 2336, 27, 2336, 27, 2338, 25, 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.
| \(\displaystyle \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 532 |
| \(\displaystyle \frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int -\frac {\frac {4 e^3 x^3}{d^2}+\frac {5 e^2 x^2}{d}+5 e x+5 d}{x^3 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\frac {4 e^3 x^3}{d^2}+\frac {5 e^2 x^2}{d}+5 e x+5 d}{x^3 \left (d^2-e^2 x^2\right )^{5/2}}dx}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int -\frac {3 \left (\frac {6 e^3 x^3}{d^2}+\frac {10 e^2 x^2}{d}+5 e x+5 d\right )}{x^3 \left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d^2}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\frac {6 e^3 x^3}{d^2}+\frac {10 e^2 x^2}{d}+5 e x+5 d}{x^3 \left (d^2-e^2 x^2\right )^{3/2}}dx}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {5 \left (\frac {3 e^2 x^2}{d}+e x+d\right )}{x^3 \sqrt {d^2-e^2 x^2}}dx}{d^2}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {5 \int \frac {\frac {3 e^2 x^2}{d}+e x+d}{x^3 \sqrt {d^2-e^2 x^2}}dx}{d^2}+\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 2338 |
| \(\displaystyle \frac {\frac {\frac {5 \left (-\frac {\int -\frac {d e (2 d+7 e x)}{x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d^2}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\right )}{d^2}+\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {5 \left (\frac {\int \frac {d e (2 d+7 e x)}{x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d^2}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\right )}{d^2}+\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {5 \left (\frac {e \int \frac {2 d+7 e x}{x^2 \sqrt {d^2-e^2 x^2}}dx}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\right )}{d^2}+\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 534 |
| \(\displaystyle \frac {\frac {\frac {5 \left (\frac {e \left (7 e \int \frac {1}{x \sqrt {d^2-e^2 x^2}}dx-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\right )}{d^2}+\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\frac {\frac {5 \left (\frac {e \left (\frac {7}{2} e \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\right )}{d^2}+\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {\frac {5 \left (\frac {e \left (-\frac {7 \int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}}{e}-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\right )}{d^2}+\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {5 \left (\frac {e \left (-\frac {7 e \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {2 \sqrt {d^2-e^2 x^2}}{d x}\right )}{2 d}-\frac {\sqrt {d^2-e^2 x^2}}{2 d x^2}\right )}{d^2}+\frac {e^2 (15 d+11 e x)}{d^4 \sqrt {d^2-e^2 x^2}}}{d^2}+\frac {e^2 (10 d+9 e x)}{3 d^4 \left (d^2-e^2 x^2\right )^{3/2}}}{5 d^2}+\frac {e^2 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}\) |
(e^2*(d + e*x))/(5*d^4*(d^2 - e^2*x^2)^(5/2)) + ((e^2*(10*d + 9*e*x))/(3*d ^4*(d^2 - e^2*x^2)^(3/2)) + ((e^2*(15*d + 11*e*x))/(d^4*Sqrt[d^2 - e^2*x^2 ]) + (5*(-1/2*Sqrt[d^2 - e^2*x^2]/(d*x^2) + (e*((-2*Sqrt[d^2 - e^2*x^2])/( d*x) - (7*e*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d))/(2*d)))/d^2)/d^2)/(5*d^2)
3.1.29.3.1 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]
Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( m + 1)) Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
Time = 0.39 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.33
| method | result | size |
| default | \(d \left (-\frac {1}{2 d^{2} x^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {7 e^{2} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )}{2 d^{2}}\right )+e \left (-\frac {1}{d^{2} x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 e^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{d^{2}}\right )\) | \(244\) |
| risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (2 e x +d \right )}{2 d^{8} x^{2}}-\frac {7 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 d^{7} \sqrt {d^{2}}}+\frac {29 e \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{48 d^{8} \left (x +\frac {d}{e}\right )}+\frac {11 \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{30 d^{7} \left (x -\frac {d}{e}\right )^{2}}-\frac {673 e \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{240 d^{8} \left (x -\frac {d}{e}\right )}+\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{24 d^{7} \left (x +\frac {d}{e}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{20 d^{6} e \left (x -\frac {d}{e}\right )^{3}}\) | \(299\) |
d*(-1/2/d^2/x^2/(-e^2*x^2+d^2)^(5/2)+7/2*e^2/d^2*(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)))))+e* (-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 = 0.33 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.58 \[ \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {116 \, e^{7} x^{7} - 116 \, d e^{6} x^{6} - 232 \, d^{2} e^{5} x^{5} + 232 \, d^{3} e^{4} x^{4} + 116 \, d^{4} e^{3} x^{3} - 116 \, d^{5} e^{2} x^{2} + 105 \, {\left (e^{7} x^{7} - d e^{6} x^{6} - 2 \, d^{2} e^{5} x^{5} + 2 \, d^{3} e^{4} x^{4} + d^{4} e^{3} x^{3} - d^{5} e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (96 \, e^{6} x^{6} + 9 \, d e^{5} x^{5} - 249 \, d^{2} e^{4} x^{4} + 4 \, d^{3} e^{3} x^{3} + 176 \, d^{4} e^{2} x^{2} - 15 \, d^{5} e x - 15 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (d^{8} e^{5} x^{7} - d^{9} e^{4} x^{6} - 2 \, d^{10} e^{3} x^{5} + 2 \, d^{11} e^{2} x^{4} + d^{12} e x^{3} - d^{13} x^{2}\right )}} \]
1/30*(116*e^7*x^7 - 116*d*e^6*x^6 - 232*d^2*e^5*x^5 + 232*d^3*e^4*x^4 + 11 6*d^4*e^3*x^3 - 116*d^5*e^2*x^2 + 105*(e^7*x^7 - d*e^6*x^6 - 2*d^2*e^5*x^5 + 2*d^3*e^4*x^4 + d^4*e^3*x^3 - d^5*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 + d^ 2))/x) - (96*e^6*x^6 + 9*d*e^5*x^5 - 249*d^2*e^4*x^4 + 4*d^3*e^3*x^3 + 176 *d^4*e^2*x^2 - 15*d^5*e*x - 15*d^6)*sqrt(-e^2*x^2 + d^2))/(d^8*e^5*x^7 - d ^9*e^4*x^6 - 2*d^10*e^3*x^5 + 2*d^11*e^2*x^4 + d^12*e*x^3 - d^13*x^2)
Result contains complex when optimal does not.
Time = 14.95 (sec) , antiderivative size = 2691, normalized size of antiderivative = 14.62 \[ \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\text {Too large to display} \]
d*Piecewise((30*I*d**8*sqrt(-1 + e**2*x**2/d**2)/(-60*d**15*x**2 + 180*d** 13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 322*I*d**6*e**2* x**2*sqrt(-1 + e**2*x**2/d**2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180 *d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 105*d**6*e**2*x**2*log(e**2*x**2/d **2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9 *e**6*x**8) + 210*d**6*e**2*x**2*log(e*x/d)/(-60*d**15*x**2 + 180*d**13*e* *2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 210*I*d**6*e**2*x**2* asin(d/(e*x))/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) + 490*I*d**4*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-60 *d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x** 8) + 315*d**4*e**4*x**4*log(e**2*x**2/d**2)/(-60*d**15*x**2 + 180*d**13*e* *2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 630*d**4*e**4*x**4*lo g(e*x/d)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60* d**9*e**6*x**8) + 630*I*d**4*e**4*x**4*asin(d/(e*x))/(-60*d**15*x**2 + 180 *d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 210*I*d**2*e **6*x**6*sqrt(-1 + e**2*x**2/d**2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 315*d**2*e**6*x**6*log(e**2*x* *2/d**2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60* d**9*e**6*x**8) + 630*d**2*e**6*x**6*log(e*x/d)/(-60*d**15*x**2 + 180*d**1 3*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 630*I*d**2*e**...
Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.20 \[ \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {6 \, e^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{4}} + \frac {7 \, e^{2}}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {8 \, e^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{6}} + \frac {7 \, e^{2}}{6 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} - \frac {e}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} x} + \frac {16 \, e^{3} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{8}} - \frac {7 \, e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d^{8}} + \frac {7 \, e^{2}}{2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} - \frac {1}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x^{2}} \]
6/5*e^3*x/((-e^2*x^2 + d^2)^(5/2)*d^4) + 7/10*e^2/((-e^2*x^2 + d^2)^(5/2)* d^3) + 8/5*e^3*x/((-e^2*x^2 + d^2)^(3/2)*d^6) + 7/6*e^2/((-e^2*x^2 + d^2)^ (3/2)*d^5) - e/((-e^2*x^2 + d^2)^(5/2)*d^2*x) + 16/5*e^3*x/(sqrt(-e^2*x^2 + d^2)*d^8) - 7/2*e^2*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/ d^8 + 7/2*e^2/(sqrt(-e^2*x^2 + d^2)*d^7) - 1/2/((-e^2*x^2 + d^2)^(5/2)*d*x ^2)
\[ \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {e x + d}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} x^{3}} \,d x } \]
Time = 12.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.98 \[ \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {161\,e^2}{30\,d^3\,{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {1}{2\,d\,x^2\,{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {7\,e^2\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{2\,d^8}-\frac {49\,e^4\,x^2}{6\,d^5\,{\left (d^2-e^2\,x^2\right )}^{5/2}}+\frac {7\,e^6\,x^4}{2\,d^7\,{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {e\,\left (5\,d^6-30\,d^4\,e^2\,x^2+40\,d^2\,e^4\,x^4-16\,e^6\,x^6\right )}{5\,d^8\,x\,{\left (d^2-e^2\,x^2\right )}^{5/2}} \]
(161*e^2)/(30*d^3*(d^2 - e^2*x^2)^(5/2)) - 1/(2*d*x^2*(d^2 - e^2*x^2)^(5/2 )) - (7*e^2*atanh((d^2 - e^2*x^2)^(1/2)/d))/(2*d^8) - (49*e^4*x^2)/(6*d^5* (d^2 - e^2*x^2)^(5/2)) + (7*e^6*x^4)/(2*d^7*(d^2 - e^2*x^2)^(5/2)) - (e*(5 *d^6 - 16*e^6*x^6 - 30*d^4*e^2*x^2 + 40*d^2*e^4*x^4))/(5*d^8*x*(d^2 - e^2* x^2)^(5/2))