Integrand size = 27, antiderivative size = 140 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=\frac {e^3 \sqrt {d^2-e^2 x^2}}{4 d x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}-\frac {7 e^2 \left (d^2-e^2 x^2\right )^{3/2}}{15 d^2 x^3}-\frac {e^5 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{4 d^2} \]
-1/5*(-e^2*x^2+d^2)^(3/2)/x^5+1/2*e*(-e^2*x^2+d^2)^(3/2)/d/x^4-7/15*e^2*(- e^2*x^2+d^2)^(3/2)/d^2/x^3-1/4*e^5*arctanh((-e^2*x^2+d^2)^(1/2)/d)/d^2+1/4 *e^3*(-e^2*x^2+d^2)^(1/2)/d/x^2
Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.94 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-12 d^4+30 d^3 e x-16 d^2 e^2 x^2-15 d e^3 x^3+28 e^4 x^4\right )}{60 d^2 x^5}-\frac {\sqrt {d^2} e^5 \log (x)}{4 d^3}+\frac {\sqrt {d^2} e^5 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{4 d^3} \]
(Sqrt[d^2 - e^2*x^2]*(-12*d^4 + 30*d^3*e*x - 16*d^2*e^2*x^2 - 15*d*e^3*x^3 + 28*e^4*x^4))/(60*d^2*x^5) - (Sqrt[d^2]*e^5*Log[x])/(4*d^3) + (Sqrt[d^2] *e^5*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(4*d^3)
Time = 0.30 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {570, 540, 27, 539, 27, 534, 243, 51, 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.
| \(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx\) |
| \(\Big \downarrow \) 570 |
| \(\displaystyle \int \frac {(d-e x)^2 \sqrt {d^2-e^2 x^2}}{x^6}dx\) |
| \(\Big \downarrow \) 540 |
| \(\displaystyle -\frac {\int \frac {d^2 e (10 d-7 e x) \sqrt {d^2-e^2 x^2}}{x^5}dx}{5 d^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{5} e \int \frac {(10 d-7 e x) \sqrt {d^2-e^2 x^2}}{x^5}dx-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle -\frac {1}{5} e \left (-\frac {\int \frac {2 d e (14 d-5 e x) \sqrt {d^2-e^2 x^2}}{x^4}dx}{4 d^2}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{5} e \left (-\frac {e \int \frac {(14 d-5 e x) \sqrt {d^2-e^2 x^2}}{x^4}dx}{2 d}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 534 |
| \(\displaystyle -\frac {1}{5} e \left (-\frac {e \left (-5 e \int \frac {\sqrt {d^2-e^2 x^2}}{x^3}dx-\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )}{2 d}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle -\frac {1}{5} e \left (-\frac {e \left (-\frac {5}{2} e \int \frac {\sqrt {d^2-e^2 x^2}}{x^4}dx^2-\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )}{2 d}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle -\frac {1}{5} e \left (-\frac {e \left (-\frac {5}{2} e \left (-\frac {1}{2} e^2 \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )}{2 d}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {1}{5} e \left (-\frac {e \left (-\frac {5}{2} e \left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )}{2 d}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {1}{5} e \left (-\frac {e \left (-\frac {5}{2} e \left (\frac {e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {14 \left (d^2-e^2 x^2\right )^{3/2}}{3 d x^3}\right )}{2 d}-\frac {5 \left (d^2-e^2 x^2\right )^{3/2}}{2 d x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{5 x^5}\) |
-1/5*(d^2 - e^2*x^2)^(3/2)/x^5 - (e*((-5*(d^2 - e^2*x^2)^(3/2))/(2*d*x^4) - (e*((-14*(d^2 - e^2*x^2)^(3/2))/(3*d*x^3) - (5*e*(-(Sqrt[d^2 - e^2*x^2]/ x^2) + (e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d))/2))/(2*d)))/5
3.2.68.3.1 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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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_))*((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[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol ] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) , x] + Simp[1/(a*(m + 1)) Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
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])
Time = 0.54 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.79
| method | result | size |
| risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (-28 e^{4} x^{4}+15 d \,e^{3} x^{3}+16 d^{2} e^{2} x^{2}-30 d^{3} e x +12 d^{4}\right )}{60 x^{5} d^{2}}-\frac {e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{4 d \sqrt {d^{2}}}\) | \(110\) |
| default | \(\text {Expression too large to display}\) | \(1349\) |
-1/60*(-e^2*x^2+d^2)^(1/2)*(-28*e^4*x^4+15*d*e^3*x^3+16*d^2*e^2*x^2-30*d^3 *e*x+12*d^4)/x^5/d^2-1/4/d*e^5/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x ^2+d^2)^(1/2))/x)
Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.69 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=\frac {15 \, e^{5} x^{5} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (28 \, e^{4} x^{4} - 15 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} + 30 \, d^{3} e x - 12 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{60 \, d^{2} x^{5}} \]
1/60*(15*e^5*x^5*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (28*e^4*x^4 - 15*d*e ^3*x^3 - 16*d^2*e^2*x^2 + 30*d^3*e*x - 12*d^4)*sqrt(-e^2*x^2 + d^2))/(d^2* x^5)
Result contains complex when optimal does not.
Time = 4.88 (sec) , antiderivative size = 660, normalized size of antiderivative = 4.71 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=d^{2} \left (\begin {cases} \frac {3 i d^{3} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} - \frac {4 i d e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} + \frac {2 i e^{6} x^{6} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{5} x^{5} + 15 d^{3} e^{2} x^{7}} - \frac {i e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{3} x^{5} + 15 d e^{2} x^{7}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {3 d^{3} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} - \frac {4 d e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{2} x^{5} + 15 e^{2} x^{7}} + \frac {2 e^{6} x^{6} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{5} x^{5} + 15 d^{3} e^{2} x^{7}} - \frac {e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{- 15 d^{3} x^{5} + 15 d e^{2} x^{7}} & \text {otherwise} \end {cases}\right ) - 2 d e \left (\begin {cases} - \frac {d^{2}}{4 e x^{5} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {3 e}{8 x^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - \frac {e^{3}}{8 d^{2} x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} + \frac {e^{4} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{4 e x^{5} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {3 i e}{8 x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + \frac {i e^{3}}{8 d^{2} x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{4} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{8 d^{3}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 x^{2}} + \frac {e^{3} \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{3 d^{2}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i e \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 x^{2}} + \frac {i e^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}}{3 d^{2}} & \text {otherwise} \end {cases}\right ) \]
d**2*Piecewise((3*I*d**3*sqrt(-1 + e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e** 2*x**7) - 4*I*d*e**2*x**2*sqrt(-1 + e**2*x**2/d**2)/(-15*d**2*x**5 + 15*e* *2*x**7) + 2*I*e**6*x**6*sqrt(-1 + e**2*x**2/d**2)/(-15*d**5*x**5 + 15*d** 3*e**2*x**7) - I*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(-15*d**3*x**5 + 15*d *e**2*x**7), Abs(e**2*x**2/d**2) > 1), (3*d**3*sqrt(1 - e**2*x**2/d**2)/(- 15*d**2*x**5 + 15*e**2*x**7) - 4*d*e**2*x**2*sqrt(1 - e**2*x**2/d**2)/(-15 *d**2*x**5 + 15*e**2*x**7) + 2*e**6*x**6*sqrt(1 - e**2*x**2/d**2)/(-15*d** 5*x**5 + 15*d**3*e**2*x**7) - e**4*x**4*sqrt(1 - e**2*x**2/d**2)/(-15*d**3 *x**5 + 15*d*e**2*x**7), True)) - 2*d*e*Piecewise((-d**2/(4*e*x**5*sqrt(d* *2/(e**2*x**2) - 1)) + 3*e/(8*x**3*sqrt(d**2/(e**2*x**2) - 1)) - e**3/(8*d **2*x*sqrt(d**2/(e**2*x**2) - 1)) + e**4*acosh(d/(e*x))/(8*d**3), Abs(d**2 /(e**2*x**2)) > 1), (I*d**2/(4*e*x**5*sqrt(-d**2/(e**2*x**2) + 1)) - 3*I*e /(8*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**3/(8*d**2*x*sqrt(-d**2/(e**2* x**2) + 1)) - I*e**4*asin(d/(e*x))/(8*d**3), True)) + e**2*Piecewise((-e*s qrt(d**2/(e**2*x**2) - 1)/(3*x**2) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(3*d* *2), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(3*x**2 ) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(3*d**2), True))
Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=-\frac {e^{5} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{4 \, d^{2}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} e^{5}}{4 \, d^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}}{4 \, d^{3} x^{2}} - \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{15 \, d^{2} x^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{2 \, d x^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{5 \, x^{5}} \]
-1/4*e^5*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^2 + 1/4*sqr t(-e^2*x^2 + d^2)*e^5/d^3 + 1/4*(-e^2*x^2 + d^2)^(3/2)*e^3/(d^3*x^2) - 7/1 5*(-e^2*x^2 + d^2)^(3/2)*e^2/(d^2*x^3) + 1/2*(-e^2*x^2 + d^2)^(3/2)*e/(d*x ^4) - 1/5*(-e^2*x^2 + d^2)^(3/2)/x^5
Result contains complex when optimal does not.
Time = 0.33 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.96 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=-\frac {1}{960} \, {\left (\frac {240 \, e^{4} \log \left (\sqrt {\frac {2 \, d}{e x + d} - 1} + 1\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2}} - \frac {240 \, e^{4} \log \left ({\left | \sqrt {\frac {2 \, d}{e x + d} - 1} - 1 \right |}\right ) \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2}} + \frac {8 \, {\left (15 \, e^{4} \log \left (2\right ) - 30 \, e^{4} \log \left (i + 1\right ) + 56 i \, e^{4}\right )} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2}} - \frac {15 \, e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {9}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 250 \, e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 128 \, e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) + 70 \, e^{4} {\left (\frac {2 \, d}{e x + d} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right ) - 15 \, e^{4} \sqrt {\frac {2 \, d}{e x + d} - 1} \mathrm {sgn}\left (\frac {1}{e x + d}\right ) \mathrm {sgn}\left (e\right )}{d^{2} {\left (\frac {d}{e x + d} - 1\right )}^{5}}\right )} {\left | e \right |} \]
-1/960*(240*e^4*log(sqrt(2*d/(e*x + d) - 1) + 1)*sgn(1/(e*x + d))*sgn(e)/d ^2 - 240*e^4*log(abs(sqrt(2*d/(e*x + d) - 1) - 1))*sgn(1/(e*x + d))*sgn(e) /d^2 + 8*(15*e^4*log(2) - 30*e^4*log(I + 1) + 56*I*e^4)*sgn(1/(e*x + d))*s gn(e)/d^2 - (15*e^4*(2*d/(e*x + d) - 1)^(9/2)*sgn(1/(e*x + d))*sgn(e) + 25 0*e^4*(2*d/(e*x + d) - 1)^(7/2)*sgn(1/(e*x + d))*sgn(e) - 128*e^4*(2*d/(e* x + d) - 1)^(5/2)*sgn(1/(e*x + d))*sgn(e) + 70*e^4*(2*d/(e*x + d) - 1)^(3/ 2)*sgn(1/(e*x + d))*sgn(e) - 15*e^4*sqrt(2*d/(e*x + d) - 1)*sgn(1/(e*x + d ))*sgn(e))/(d^2*(d/(e*x + d) - 1)^5))*abs(e)
Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)^2} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^6\,{\left (d+e\,x\right )}^2} \,d x \]