Integrand size = 27, antiderivative size = 108 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx=-\frac {3 e^3 \sqrt {d^2-e^2 x^2}}{8 x^2}+\frac {e \left (d^2-e^2 x^2\right )^{3/2}}{4 x^4}-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}+\frac {3 e^5 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{8 d} \]
1/4*e*(-e^2*x^2+d^2)^(3/2)/x^4-1/5*(-e^2*x^2+d^2)^(5/2)/d/x^5+3/8*e^5*arct anh((-e^2*x^2+d^2)^(1/2)/d)/d-3/8*e^3*(-e^2*x^2+d^2)^(1/2)/x^2
Time = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.13 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx=\frac {1}{40} \left (\frac {\sqrt {d^2-e^2 x^2} \left (-8 d^4+10 d^3 e x+16 d^2 e^2 x^2-25 d e^3 x^3-8 e^4 x^4\right )}{d x^5}+\frac {15 e^5 \log (x)}{\sqrt {d^2}}-\frac {15 e^5 \log \left (\sqrt {d^2}-\sqrt {d^2-e^2 x^2}\right )}{\sqrt {d^2}}\right ) \]
((Sqrt[d^2 - e^2*x^2]*(-8*d^4 + 10*d^3*e*x + 16*d^2*e^2*x^2 - 25*d*e^3*x^3 - 8*e^4*x^4))/(d*x^5) + (15*e^5*Log[x])/Sqrt[d^2] - (15*e^5*Log[Sqrt[d^2] - Sqrt[d^2 - e^2*x^2]])/Sqrt[d^2])/40
Time = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {566, 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.
| \(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx\) |
| \(\Big \downarrow \) 566 |
| \(\displaystyle \int \frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{x^6}dx\) |
| \(\Big \downarrow \) 534 |
| \(\displaystyle -e \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^5}dx-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle -\frac {1}{2} e \int \frac {\left (d^2-e^2 x^2\right )^{3/2}}{x^6}dx^2-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle -\frac {1}{2} e \left (-\frac {3}{4} e^2 \int \frac {\sqrt {d^2-e^2 x^2}}{x^4}dx^2-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle -\frac {1}{2} e \left (-\frac {3}{4} e^2 \left (-\frac {1}{2} e^2 \int \frac {1}{x^2 \sqrt {d^2-e^2 x^2}}dx^2-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {1}{2} e \left (-\frac {3}{4} e^2 \left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^4}{e^2}}d\sqrt {d^2-e^2 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {1}{2} e \left (-\frac {3}{4} e^2 \left (\frac {e^2 \text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d}-\frac {\sqrt {d^2-e^2 x^2}}{x^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{2 x^4}\right )-\frac {\left (d^2-e^2 x^2\right )^{5/2}}{5 d x^5}\) |
-1/5*(d^2 - e^2*x^2)^(5/2)/(d*x^5) - (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
3.2.13.3.1 Defintions of rubi rules used
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_)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] : > Int[x^m*(a/c + b*(x/d))*(a + b*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0]
Time = 0.46 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99
| method | result | size |
| risch | \(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (8 e^{4} x^{4}+25 d \,e^{3} x^{3}-16 d^{2} e^{2} x^{2}-10 d^{3} e x +8 d^{4}\right )}{40 x^{5} d}+\frac {3 e^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{8 \sqrt {d^{2}}}\) | \(107\) |
| default | \(\text {Expression too large to display}\) | \(1100\) |
-1/40*(-e^2*x^2+d^2)^(1/2)*(8*e^4*x^4+25*d*e^3*x^3-16*d^2*e^2*x^2-10*d^3*e *x+8*d^4)/x^5/d+3/8*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 = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.90 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx=-\frac {15 \, e^{5} x^{5} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (8 \, e^{4} x^{4} + 25 \, d e^{3} x^{3} - 16 \, d^{2} e^{2} x^{2} - 10 \, d^{3} e x + 8 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{40 \, d x^{5}} \]
-1/40*(15*e^5*x^5*log(-(d - sqrt(-e^2*x^2 + d^2))/x) + (8*e^4*x^4 + 25*d*e ^3*x^3 - 16*d^2*e^2*x^2 - 10*d^3*e*x + 8*d^4)*sqrt(-e^2*x^2 + d^2))/(d*x^5 )
Result contains complex when optimal does not.
Time = 5.04 (sec) , antiderivative size = 774, normalized size of antiderivative = 7.17 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx=d^{3} \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 ) - d^{2} 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 ) - d 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 ) + e^{3} \left (\begin {cases} - \frac {e \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}}{2 x} + \frac {e^{2} \operatorname {acosh}{\left (\frac {d}{e x} \right )}}{2 d} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\\frac {i d^{2}}{2 e x^{3} \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e}{2 x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} - \frac {i e^{2} \operatorname {asin}{\left (\frac {d}{e x} \right )}}{2 d} & \text {otherwise} \end {cases}\right ) \]
d**3*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)) - d**2*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)) - d*e**2*Piecewise((- e*sqrt(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)) + e**3*Piecewis e((-e*sqrt(d**2/(e**2*x**2) - 1)/(2*x) + e**2*acosh(d/(e*x))/(2*d), Abs(d* *2/(e**2*x**2)) > 1), (I*d**2/(2*e*x**3*sqrt(-d**2/(e**2*x**2) + 1)) - I*e /(2*x*sqrt(-d**2/(e**2*x**2) + 1)) - I*e**2*asin(d/(e*x))/(2*d), True))
Time = 0.28 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.42 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx=\frac {3 \, e^{5} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{8 \, d} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} e^{5}}{8 \, d^{2}} - \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}}{8 \, d^{2} x^{2}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}}{5 \, d x^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e}{4 \, x^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d}{5 \, x^{5}} \]
3/8*e^5*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d - 3/8*sqrt(- e^2*x^2 + d^2)*e^5/d^2 - 3/8*(-e^2*x^2 + d^2)^(3/2)*e^3/(d^2*x^2) + 1/5*(- e^2*x^2 + d^2)^(3/2)*e^2/(d*x^3) + 1/4*(-e^2*x^2 + d^2)^(3/2)*e/x^4 - 1/5* (-e^2*x^2 + d^2)^(3/2)*d/x^5
Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (92) = 184\).
Time = 0.30 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.59 \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx=\frac {{\left (2 \, e^{6} - \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} e^{4}}{x} - \frac {10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} e^{2}}{x^{2}} + \frac {40 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{x^{3}} + \frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{2} x^{4}}\right )} e^{10} x^{5}}{320 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5} d {\left | e \right |}} + \frac {3 \, e^{6} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |} \right |}}{2 \, e^{2} {\left | x \right |}}\right )}{8 \, d {\left | e \right |}} - \frac {\frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{4} e^{8}}{x} + \frac {40 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{4} e^{6}}{x^{2}} - \frac {10 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d^{4} e^{4}}{x^{3}} - \frac {5 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} d^{4} e^{2}}{x^{4}} + \frac {2 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5} d^{4}}{x^{5}}}{320 \, d^{5} e^{4} {\left | e \right |}} \]
1/320*(2*e^6 - 5*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*e^4/x - 10*(d*e + sqr t(-e^2*x^2 + d^2)*abs(e))^2*e^2/x^2 + 40*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e ))^3/x^3 + 20*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/(e^2*x^4))*e^10*x^5/(( d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^5*d*abs(e)) + 3/8*e^6*log(1/2*abs(-2*d* e - 2*sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*abs(x)))/(d*abs(e)) - 1/320*(20*(d *e + sqrt(-e^2*x^2 + d^2)*abs(e))*d^4*e^8/x + 40*(d*e + sqrt(-e^2*x^2 + d^ 2)*abs(e))^2*d^4*e^6/x^2 - 10*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3*d^4*e^ 4/x^3 - 5*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4*d^4*e^2/x^4 + 2*(d*e + sqr t(-e^2*x^2 + d^2)*abs(e))^5*d^4/x^5)/(d^5*e^4*abs(e))
Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^6 (d+e x)} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{5/2}}{x^6\,\left (d+e\,x\right )} \,d x \]