Integrand size = 25, antiderivative size = 201 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx=\frac {3 e^6 \sqrt {d^2-e^2 x^2}}{128 d^3 x^2}-\frac {e^4 \left (d^2-e^2 x^2\right )^{3/2}}{64 d^3 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}-\frac {e \left (d^2-e^2 x^2\right )^{5/2}}{7 d^2 x^7}-\frac {e^2 \left (d^2-e^2 x^2\right )^{5/2}}{16 d^3 x^6}-\frac {2 e^3 \left (d^2-e^2 x^2\right )^{5/2}}{35 d^4 x^5}-\frac {3 e^8 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{128 d^4} \]
-1/64*e^4*(-e^2*x^2+d^2)^(3/2)/d^3/x^4-1/8*(-e^2*x^2+d^2)^(5/2)/d/x^8-1/7* e*(-e^2*x^2+d^2)^(5/2)/d^2/x^7-1/16*e^2*(-e^2*x^2+d^2)^(5/2)/d^3/x^6-2/35* e^3*(-e^2*x^2+d^2)^(5/2)/d^4/x^5-3/128*e^8*arctanh((-e^2*x^2+d^2)^(1/2)/d) /d^4+3/128*e^6*(-e^2*x^2+d^2)^(1/2)/d^3/x^2
Time = 0.33 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.82 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-560 d^7-640 d^6 e x+840 d^5 e^2 x^2+1024 d^4 e^3 x^3-70 d^3 e^4 x^4-128 d^2 e^5 x^5-105 d e^6 x^6-256 e^7 x^7\right )}{4480 d^4 x^8}-\frac {3 \sqrt {d^2} e^8 \log (x)}{128 d^5}+\frac {3 \sqrt {d^2} e^8 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{128 d^5} \]
(Sqrt[d^2 - e^2*x^2]*(-560*d^7 - 640*d^6*e*x + 840*d^5*e^2*x^2 + 1024*d^4* e^3*x^3 - 70*d^3*e^4*x^4 - 128*d^2*e^5*x^5 - 105*d*e^6*x^6 - 256*e^7*x^7)) /(4480*d^4*x^8) - (3*Sqrt[d^2]*e^8*Log[x])/(128*d^5) + (3*Sqrt[d^2]*e^8*Lo g[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/(128*d^5)
Time = 0.33 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {539, 25, 27, 539, 25, 27, 539, 27, 534, 243, 51, 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.
| \(\displaystyle \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle -\frac {\int -\frac {d e (8 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8}dx}{8 d^2}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {d e (8 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8}dx}{8 d^2}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e \int \frac {(8 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^8}dx}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle \frac {e \left (-\frac {\int -\frac {d e (21 d+16 e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^7}dx}{7 d^2}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e \left (\frac {\int \frac {d e (21 d+16 e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^7}dx}{7 d^2}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e \left (\frac {e \int \frac {(21 d+16 e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^7}dx}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 539 |
| \(\displaystyle \frac {e \left (\frac {e \left (-\frac {\int -\frac {3 d e (32 d+7 e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^6}dx}{6 d^2}-\frac {7 \left (d^2-e^2 x^2\right )^{5/2}}{2 d x^6}\right )}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e \left (\frac {e \left (\frac {e \int \frac {(32 d+7 e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^6}dx}{2 d}-\frac {7 \left (d^2-e^2 x^2\right )^{5/2}}{2 d x^6}\right )}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 534 |
| \(\displaystyle \frac {e \left (\frac {e \left (\frac {e \left (7 e \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^5}dx-\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\right )}{2 d}-\frac {7 \left (d^2-e^2 x^2\right )^{5/2}}{2 d x^6}\right )}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {e \left (\frac {e \left (\frac {e \left (\frac {7}{2} e \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^6}dx^2-\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\right )}{2 d}-\frac {7 \left (d^2-e^2 x^2\right )^{5/2}}{2 d x^6}\right )}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {e \left (\frac {e \left (\frac {e \left (\frac {7}{2} e \left (-\frac {3}{4} e^2 \int \frac {\sqrt {d^2-e^2 x^2}}{x^4}dx^2-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\right )}{2 d}-\frac {7 \left (d^2-e^2 x^2\right )^{5/2}}{2 d x^6}\right )}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {e \left (\frac {e \left (\frac {e \left (\frac {7}{2} e \left (-\frac {3}{4} e^2 \left (-\frac {1}{2} e^2 \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\right )}{2 d}-\frac {7 \left (d^2-e^2 x^2\right )^{5/2}}{2 d x^6}\right )}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {e \left (\frac {e \left (\frac {e \left (\frac {7}{2} e \left (-\frac {3}{4} e^2 \left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\right )}{2 d}-\frac {7 \left (d^2-e^2 x^2\right )^{5/2}}{2 d x^6}\right )}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {e \left (\frac {e \left (\frac {e \left (\frac {7}{2} e \left (-\frac {3}{4} e^2 \left (\frac {e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\right )}{2 d}-\frac {7 \left (d^2-e^2 x^2\right )^{5/2}}{2 d x^6}\right )}{7 d}-\frac {8 \left (d^2-e^2 x^2\right )^{5/2}}{7 d x^7}\right )}{8 d}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{8 d x^8}\) |
-1/8*(d^2 - e^2*x^2)^(5/2)/(d*x^8) + (e*((-8*(d^2 - e^2*x^2)^(5/2))/(7*d*x ^7) + (e*((-7*(d^2 - e^2*x^2)^(5/2))/(2*d*x^6) + (e*((-32*(d^2 - e^2*x^2)^ (5/2))/(5*d*x^5) + (7*e*(-1/2*(d^2 - e^2*x^2)^(3/2)/x^4 - (3*e^2*(-(Sqrt[d ^2 - e^2*x^2]/x^2) + (e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d))/4))/2))/(2*d )))/(7*d)))/(8*d)
3.1.15.3.1 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]
Time = 0.40 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.71
| method | result | size |
| risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (256 e^{7} x^{7}+105 d \,e^{6} x^{6}+128 d^{2} e^{5} x^{5}+70 d^{3} e^{4} x^{4}-1024 d^{4} e^{3} x^{3}-840 d^{5} e^{2} x^{2}+640 d^{6} e x +560 d^{7}\right )}{4480 x^{8} d^{4}}-\frac {3 e^{8} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{128 d^{3} \sqrt {d^{2}}}\) | \(143\) |
| default | \(e \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{7 d^{2} x^{7}}-\frac {2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{35 d^{4} x^{5}}\right )+d \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{8 d^{2} x^{8}}+\frac {3 e^{2} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6 d^{2} x^{6}}+\frac {e^{2} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{4 d^{2} x^{4}}-\frac {e^{2} \left (-\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{2 d^{2} x^{2}}-\frac {3 e^{2} \left (\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{3}+d^{2} \left (\sqrt {-e^{2} x^{2}+d^{2}}-\frac {d^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}\right )\right )}{2 d^{2}}\right )}{4 d^{2}}\right )}{6 d^{2}}\right )}{8 d^{2}}\right )\) | \(256\) |
-1/4480*(-e^2*x^2+d^2)^(1/2)*(256*e^7*x^7+105*d*e^6*x^6+128*d^2*e^5*x^5+70 *d^3*e^4*x^4-1024*d^4*e^3*x^3-840*d^5*e^2*x^2+640*d^6*e*x+560*d^7)/x^8/d^4 -3/128/d^3*e^8/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x )
Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.65 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx=\frac {105 \, e^{8} x^{8} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (256 \, e^{7} x^{7} + 105 \, d e^{6} x^{6} + 128 \, d^{2} e^{5} x^{5} + 70 \, d^{3} e^{4} x^{4} - 1024 \, d^{4} e^{3} x^{3} - 840 \, d^{5} e^{2} x^{2} + 640 \, d^{6} e x + 560 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{4480 \, d^{4} x^{8}} \]
1/4480*(105*e^8*x^8*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (256*e^7*x^7 + 10 5*d*e^6*x^6 + 128*d^2*e^5*x^5 + 70*d^3*e^4*x^4 - 1024*d^4*e^3*x^3 - 840*d^ 5*e^2*x^2 + 640*d^6*e*x + 560*d^7)*sqrt(-e^2*x^2 + d^2))/(d^4*x^8)
Result contains complex when optimal does not.
Time = 25.09 (sec) , antiderivative size = 1159, normalized size of antiderivative = 5.77 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx=\text {Too large to display} \]
d**3*Piecewise((-d**2/(8*e*x**9*sqrt(d**2/(e**2*x**2) - 1)) + 7*e/(48*x**7 *sqrt(d**2/(e**2*x**2) - 1)) + e**3/(192*d**2*x**5*sqrt(d**2/(e**2*x**2) - 1)) + 5*e**5/(384*d**4*x**3*sqrt(d**2/(e**2*x**2) - 1)) - 5*e**7/(128*d** 6*x*sqrt(d**2/(e**2*x**2) - 1)) + 5*e**8*acosh(d/(e*x))/(128*d**7), Abs(d* *2/(e**2*x**2)) > 1), (I*d**2/(8*e*x**9*sqrt(-d**2/(e**2*x**2) + 1)) - 7*I *e/(48*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**3/(192*d**2*x**5*sqrt(-d** 2/(e**2*x**2) + 1)) - 5*I*e**5/(384*d**4*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + 5*I*e**7/(128*d**6*x*sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e**8*asin(d/(e* x))/(128*d**7), True)) + d**2*e*Piecewise((-e*sqrt(d**2/(e**2*x**2) - 1)/( 7*x**6) + e**3*sqrt(d**2/(e**2*x**2) - 1)/(35*d**2*x**4) + 4*e**5*sqrt(d** 2/(e**2*x**2) - 1)/(105*d**4*x**2) + 8*e**7*sqrt(d**2/(e**2*x**2) - 1)/(10 5*d**6), Abs(d**2/(e**2*x**2)) > 1), (-I*e*sqrt(-d**2/(e**2*x**2) + 1)/(7* x**6) + I*e**3*sqrt(-d**2/(e**2*x**2) + 1)/(35*d**2*x**4) + 4*I*e**5*sqrt( -d**2/(e**2*x**2) + 1)/(105*d**4*x**2) + 8*I*e**7*sqrt(-d**2/(e**2*x**2) + 1)/(105*d**6), True)) - d*e**2*Piecewise((-d**2/(6*e*x**7*sqrt(d**2/(e**2 *x**2) - 1)) + 5*e/(24*x**5*sqrt(d**2/(e**2*x**2) - 1)) + e**3/(48*d**2*x* *3*sqrt(d**2/(e**2*x**2) - 1)) - e**5/(16*d**4*x*sqrt(d**2/(e**2*x**2) - 1 )) + e**6*acosh(d/(e*x))/(16*d**5), Abs(d**2/(e**2*x**2)) > 1), (I*d**2/(6 *e*x**7*sqrt(-d**2/(e**2*x**2) + 1)) - 5*I*e/(24*x**5*sqrt(-d**2/(e**2*x** 2) + 1)) - I*e**3/(48*d**2*x**3*sqrt(-d**2/(e**2*x**2) + 1)) + I*e**5/(...
Time = 0.28 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.14 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx=-\frac {3 \, e^{8} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{128 \, d^{4}} + \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{8}}{128 \, d^{5}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{8}}{128 \, d^{7}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}{128 \, d^{7} x^{2}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}}{64 \, d^{5} x^{4}} - \frac {2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}}{35 \, d^{4} x^{5}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}}{16 \, d^{3} x^{6}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e}{7 \, d^{2} x^{7}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}}{8 \, d x^{8}} \]
-3/128*e^8*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^4 + 3/128 *sqrt(-e^2*x^2 + d^2)*e^8/d^5 + 1/128*(-e^2*x^2 + d^2)^(3/2)*e^8/d^7 + 1/1 28*(-e^2*x^2 + d^2)^(5/2)*e^6/(d^7*x^2) - 1/64*(-e^2*x^2 + d^2)^(5/2)*e^4/ (d^5*x^4) - 2/35*(-e^2*x^2 + d^2)^(5/2)*e^3/(d^4*x^5) - 1/16*(-e^2*x^2 + d ^2)^(5/2)*e^2/(d^3*x^6) - 1/7*(-e^2*x^2 + d^2)^(5/2)*e/(d^2*x^7) - 1/8*(-e ^2*x^2 + d^2)^(5/2)/(d*x^8)
Leaf count of result is larger than twice the leaf count of optimal. 463 vs. \(2 (173) = 346\).
Time = 0.29 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.30 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx=\frac {{\left (35 \, e^{9} + \frac {80 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e^{7}}{x} - \frac {112 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} e^{3}}{x^{3}} - \frac {280 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} e}{x^{4}} - \frac {560 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5}}{e x^{5}} + \frac {1680 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{7}}{e^{5} x^{7}}\right )} e^{16} x^{8}}{71680 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{8} d^{4} {\left | e \right |}} - \frac {3 \, e^{9} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{128 \, d^{4} {\left | e \right |}} - \frac {\frac {1680 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{28} e^{13} {\left | e \right |}}{x} - \frac {560 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d^{28} e^{9} {\left | e \right |}}{x^{3}} - \frac {280 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} d^{28} e^{7} {\left | e \right |}}{x^{4}} - \frac {112 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5} d^{28} e^{5} {\left | e \right |}}{x^{5}} + \frac {80 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{7} d^{28} e {\left | e \right |}}{x^{7}} + \frac {35 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{8} d^{28} {\left | e \right |}}{e x^{8}}}{71680 \, d^{32} e^{8}} \]
1/71680*(35*e^9 + 80*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*e^7/x - 112*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*e^3/x^3 - 280*(d*e + sqrt(-e^2*x^2 + d^2) *abs(e))^4*e/x^4 - 560*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5/(e*x^5) + 168 0*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^7/(e^5*x^7))*e^16*x^8/((d*e + sqrt(- e^2*x^2 + d^2)*abs(e))^8*d^4*abs(e)) - 3/128*e^9*log(1/2*abs(-2*d*e - 2*sq rt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d^4*abs(e)) - 1/71680*(1680*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d^28*e^13*abs(e)/x - 560*(d*e + sqrt(-e^2* x^2 + d^2)*abs(e))^3*d^28*e^9*abs(e)/x^3 - 280*(d*e + sqrt(-e^2*x^2 + d^2) *abs(e))^4*d^28*e^7*abs(e)/x^4 - 112*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5 *d^28*e^5*abs(e)/x^5 + 80*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^7*d^28*e*abs (e)/x^7 + 35*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^8*d^28*abs(e)/(e*x^8))/(d ^32*e^8)
Time = 14.95 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^9} \, dx=\frac {3\,d^3\,\sqrt {d^2-e^2\,x^2}}{128\,x^8}-\frac {11\,d\,{\left (d^2-e^2\,x^2\right )}^{3/2}}{128\,x^8}-\frac {11\,{\left (d^2-e^2\,x^2\right )}^{5/2}}{128\,d\,x^8}+\frac {3\,{\left (d^2-e^2\,x^2\right )}^{7/2}}{128\,d^3\,x^8}+\frac {8\,e^3\,\sqrt {d^2-e^2\,x^2}}{35\,x^5}-\frac {e^5\,\sqrt {d^2-e^2\,x^2}}{35\,d^2\,x^3}-\frac {2\,e^7\,\sqrt {d^2-e^2\,x^2}}{35\,d^4\,x}-\frac {d^2\,e\,\sqrt {d^2-e^2\,x^2}}{7\,x^7}+\frac {e^8\,\mathrm {atan}\left (\frac {\sqrt {d^2-e^2\,x^2}\,1{}\mathrm {i}}{d}\right )\,3{}\mathrm {i}}{128\,d^4} \]
(3*d^3*(d^2 - e^2*x^2)^(1/2))/(128*x^8) - (11*d*(d^2 - e^2*x^2)^(3/2))/(12 8*x^8) - (11*(d^2 - e^2*x^2)^(5/2))/(128*d*x^8) + (3*(d^2 - e^2*x^2)^(7/2) )/(128*d^3*x^8) + (8*e^3*(d^2 - e^2*x^2)^(1/2))/(35*x^5) + (e^8*atan(((d^2 - e^2*x^2)^(1/2)*1i)/d)*3i)/(128*d^4) - (e^5*(d^2 - e^2*x^2)^(1/2))/(35*d ^2*x^3) - (2*e^7*(d^2 - e^2*x^2)^(1/2))/(35*d^4*x) - (d^2*e*(d^2 - e^2*x^2 )^(1/2))/(7*x^7)