Integrand size = 21, antiderivative size = 212 \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=-\frac {a b d^2}{(b+a c)^3 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}-\frac {(3 b-4 a c) d \left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{8 (b+a c)^3 x^2}-\frac {\left (c+d x^2\right )^2 \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{4 (b+a c)^2 x^4}-\frac {3 b (b-4 a c) d^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {b+a c}}\right )}{8 \sqrt {c} (b+a c)^{7/2}} \]
-3/8*b*(-4*a*c+b)*d^2*arctanh(c^(1/2)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(a *c+b)^(1/2))/(a*c+b)^(7/2)/c^(1/2)-a*b*d^2/(a*c+b)^3/((a*d*x^2+a*c+b)/(d*x ^2+c))^(1/2)-1/8*(-4*a*c+3*b)*d*(d*x^2+c)*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2 )/(a*c+b)^3/x^2-1/4*(d*x^2+c)^2*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)/(a*c+b)^ 2/x^4
Time = 0.33 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=-\frac {\left (c+d x^2\right ) \sqrt {\frac {b+a c+a d x^2}{c+d x^2}} \left (b^2 \left (2 c+5 d x^2\right )+2 a^2 c \left (c^2-d^2 x^4\right )+a b \left (4 c^2+5 c d x^2+13 d^2 x^4\right )\right )}{8 (b+a c)^3 x^4 \left (b+a \left (c+d x^2\right )\right )}-\frac {3 b (b-4 a c) d^2 \arctan \left (\frac {\sqrt {c} \sqrt {\frac {b+a c+a d x^2}{c+d x^2}}}{\sqrt {-b-a c}}\right )}{8 \sqrt {c} (-b-a c)^{7/2}} \]
-1/8*((c + d*x^2)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)]*(b^2*(2*c + 5*d*x^ 2) + 2*a^2*c*(c^2 - d^2*x^4) + a*b*(4*c^2 + 5*c*d*x^2 + 13*d^2*x^4)))/((b + a*c)^3*x^4*(b + a*(c + d*x^2))) - (3*b*(b - 4*a*c)*d^2*ArcTan[(Sqrt[c]*S qrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[-b - a*c]])/(8*Sqrt[c]*(-b - a* c)^(7/2))
Time = 0.44 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.94, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {2057, 2053, 2052, 25, 27, 361, 25, 361, 25, 359, 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.
| \(\displaystyle \int \frac {1}{x^5 \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2057 |
| \(\displaystyle \int \frac {1}{x^5 \left (\frac {a c+a d x^2+b}{c+d x^2}\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 2053 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^6 \left (\frac {a d x^2+b+a c}{d x^2+c}\right )^{3/2}}dx^2\) |
| \(\Big \downarrow \) 2052 |
| \(\displaystyle -b d \int -\frac {d \left (a-x^4\right )}{x^4 \left (-c x^4+b+a c\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b d \int \frac {d \left (a-x^4\right )}{x^4 \left (-c x^4+b+a c\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle b d^2 \int \frac {a-x^4}{x^4 \left (-c x^4+b+a c\right )^3}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle b d^2 \left (-\frac {1}{4} \int -\frac {\frac {4 a}{b+a c}-\frac {3 b x^4}{(b+a c)^2}}{x^4 \left (-c x^4+b+a c\right )^2}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}-\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 (a c+b)^2 \left (a c+b-c x^4\right )^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b d^2 \left (\frac {1}{4} \int \frac {\frac {4 a}{b+a c}-\frac {3 b x^4}{(b+a c)^2}}{x^4 \left (-c x^4+b+a c\right )^2}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}-\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 (a c+b)^2 \left (a c+b-c x^4\right )^2}\right )\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle b d^2 \left (\frac {1}{4} \left (-\frac {1}{2} \int -\frac {\frac {8 a}{(b+a c)^2}-\frac {(3 b-4 a c) x^4}{(b+a c)^3}}{x^4 \left (-c x^4+b+a c\right )}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}-\frac {(3 b-4 a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 (a c+b)^3 \left (a c+b-c x^4\right )}\right )-\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 (a c+b)^2 \left (a c+b-c x^4\right )^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle b d^2 \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {\frac {8 a}{(b+a c)^2}-\frac {(3 b-4 a c) x^4}{(b+a c)^3}}{x^4 \left (-c x^4+b+a c\right )}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}-\frac {(3 b-4 a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 (a c+b)^3 \left (a c+b-c x^4\right )}\right )-\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 (a c+b)^2 \left (a c+b-c x^4\right )^2}\right )\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle b d^2 \left (\frac {1}{4} \left (\frac {1}{2} \left (-\frac {3 (b-4 a c) \int \frac {1}{-c x^4+b+a c}d\sqrt {\frac {a d x^2+b+a c}{d x^2+c}}}{(a c+b)^3}-\frac {8 a}{x^2 (a c+b)^3}\right )-\frac {(3 b-4 a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 (a c+b)^3 \left (a c+b-c x^4\right )}\right )-\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 (a c+b)^2 \left (a c+b-c x^4\right )^2}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle b d^2 \left (\frac {1}{4} \left (\frac {1}{2} \left (-\frac {3 (b-4 a c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{\sqrt {a c+b}}\right )}{\sqrt {c} (a c+b)^{7/2}}-\frac {8 a}{x^2 (a c+b)^3}\right )-\frac {(3 b-4 a c) \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{2 (a c+b)^3 \left (a c+b-c x^4\right )}\right )-\frac {b \sqrt {\frac {a c+a d x^2+b}{c+d x^2}}}{4 (a c+b)^2 \left (a c+b-c x^4\right )^2}\right )\) |
b*d^2*(-1/4*(b*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/((b + a*c)^2*(b + a* c - c*x^4)^2) + (-1/2*((3*b - 4*a*c)*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)] )/((b + a*c)^3*(b + a*c - c*x^4)) + ((-8*a)/((b + a*c)^3*x^2) - (3*(b - 4* a*c)*ArcTanh[(Sqrt[c]*Sqrt[(b + a*c + a*d*x^2)/(c + d*x^2)])/Sqrt[b + a*c] ])/(Sqrt[c]*(b + a*c)^(7/2)))/2)/4)
3.4.60.3.1 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_)^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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_S ymbol] :> With[{q = Denominator[p]}, Simp[q*e*(b*c - a*d) Subst[Int[x^(q* (p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a + b* x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]
Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.) ))^(p_), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(e*( (a + b*x)/(c + d*x)))^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[(u_.)*((a_) + (b_.)/((c_) + (d_.)*(x_)^(n_)))^(p_), x_Symbol] :> Int[u* ((b + a*c + a*d*x^n)/(c + d*x^n))^p, x] /; FreeQ[{a, b, c, d, n, p}, x]
Time = 0.25 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.32
| method | result | size |
| risch | \(-\frac {\left (a d \,x^{2}+a c +b \right ) \left (-2 a c d \,x^{2}+5 b d \,x^{2}+2 a \,c^{2}+2 b c \right )}{8 \left (a c +b \right )^{3} x^{4} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}}-\frac {d^{2} b \left (-\frac {\left (12 a c -3 b \right ) \ln \left (\frac {2 a \,c^{2}+2 b c +\left (2 a c d +b d \right ) x^{2}+2 \sqrt {a \,c^{2}+b c}\, \sqrt {a \,c^{2}+b c +\left (2 a c d +b d \right ) x^{2}+a \,d^{2} x^{4}}}{x^{2}}\right )}{2 \sqrt {a \,c^{2}+b c}}+\frac {8 a \left (d \,x^{2}+c \right )}{\sqrt {a \,d^{2} x^{4}+2 a c d \,x^{2}+b d \,x^{2}+a \,c^{2}+b c}}\right ) \sqrt {\left (a d \,x^{2}+a c +b \right ) \left (d \,x^{2}+c \right )}}{8 \left (a c +b \right )^{3} \sqrt {\frac {a d \,x^{2}+a c +b}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right )}\) | \(279\) |
| default | \(\text {Expression too large to display}\) | \(1947\) |
-1/8*(a*d*x^2+a*c+b)*(-2*a*c*d*x^2+5*b*d*x^2+2*a*c^2+2*b*c)/(a*c+b)^3/x^4/ ((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)-1/8*d^2*b/(a*c+b)^3*(-1/2*(12*a*c-3*b)/( a*c^2+b*c)^(1/2)*ln((2*a*c^2+2*b*c+(2*a*c*d+b*d)*x^2+2*(a*c^2+b*c)^(1/2)*( a*c^2+b*c+(2*a*c*d+b*d)*x^2+a*d^2*x^4)^(1/2))/x^2)+8*a*(d*x^2+c)/(a*d^2*x^ 4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2))/((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)* ((a*d*x^2+a*c+b)*(d*x^2+c))^(1/2)/(d*x^2+c)
Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (192) = 384\).
Time = 0.73 (sec) , antiderivative size = 961, normalized size of antiderivative = 4.53 \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (4 \, a^{2} b c - a b^{2}\right )} d^{3} x^{6} + {\left (4 \, a^{2} b c^{2} + 3 \, a b^{2} c - b^{3}\right )} d^{2} x^{4}\right )} \sqrt {a c^{2} + b c} \log \left (\frac {{\left (8 \, a^{2} c^{2} + 8 \, a b c + b^{2}\right )} d^{2} x^{4} + 8 \, a^{2} c^{4} + 16 \, a b c^{3} + 8 \, b^{2} c^{2} + 8 \, {\left (2 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c\right )} d x^{2} + 4 \, {\left ({\left (2 \, a c + b\right )} d^{2} x^{4} + 2 \, a c^{3} + {\left (4 \, a c^{2} + 3 \, b c\right )} d x^{2} + 2 \, b c^{2}\right )} \sqrt {a c^{2} + b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{x^{4}}\right ) + 4 \, {\left ({\left (2 \, a^{3} c^{3} - 11 \, a^{2} b c^{2} - 13 \, a b^{2} c\right )} d^{3} x^{6} - 2 \, a^{3} c^{6} - 6 \, a^{2} b c^{5} - 6 \, a b^{2} c^{4} + {\left (2 \, a^{3} c^{4} - 16 \, a^{2} b c^{3} - 23 \, a b^{2} c^{2} - 5 \, b^{3} c\right )} d^{2} x^{4} - 2 \, b^{3} c^{3} - {\left (2 \, a^{3} c^{5} + 11 \, a^{2} b c^{4} + 16 \, a b^{2} c^{3} + 7 \, b^{3} c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{32 \, {\left ({\left (a^{5} c^{5} + 4 \, a^{4} b c^{4} + 6 \, a^{3} b^{2} c^{3} + 4 \, a^{2} b^{3} c^{2} + a b^{4} c\right )} d x^{6} + {\left (a^{5} c^{6} + 5 \, a^{4} b c^{5} + 10 \, a^{3} b^{2} c^{4} + 10 \, a^{2} b^{3} c^{3} + 5 \, a b^{4} c^{2} + b^{5} c\right )} x^{4}\right )}}, -\frac {3 \, {\left ({\left (4 \, a^{2} b c - a b^{2}\right )} d^{3} x^{6} + {\left (4 \, a^{2} b c^{2} + 3 \, a b^{2} c - b^{3}\right )} d^{2} x^{4}\right )} \sqrt {-a c^{2} - b c} \arctan \left (\frac {{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt {-a c^{2} - b c} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{2 \, {\left (a^{2} c^{3} + 2 \, a b c^{2} + {\left (a^{2} c^{2} + a b c\right )} d x^{2} + b^{2} c\right )}}\right ) - 2 \, {\left ({\left (2 \, a^{3} c^{3} - 11 \, a^{2} b c^{2} - 13 \, a b^{2} c\right )} d^{3} x^{6} - 2 \, a^{3} c^{6} - 6 \, a^{2} b c^{5} - 6 \, a b^{2} c^{4} + {\left (2 \, a^{3} c^{4} - 16 \, a^{2} b c^{3} - 23 \, a b^{2} c^{2} - 5 \, b^{3} c\right )} d^{2} x^{4} - 2 \, b^{3} c^{3} - {\left (2 \, a^{3} c^{5} + 11 \, a^{2} b c^{4} + 16 \, a b^{2} c^{3} + 7 \, b^{3} c^{2}\right )} d x^{2}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}}{16 \, {\left ({\left (a^{5} c^{5} + 4 \, a^{4} b c^{4} + 6 \, a^{3} b^{2} c^{3} + 4 \, a^{2} b^{3} c^{2} + a b^{4} c\right )} d x^{6} + {\left (a^{5} c^{6} + 5 \, a^{4} b c^{5} + 10 \, a^{3} b^{2} c^{4} + 10 \, a^{2} b^{3} c^{3} + 5 \, a b^{4} c^{2} + b^{5} c\right )} x^{4}\right )}}\right ] \]
[1/32*(3*((4*a^2*b*c - a*b^2)*d^3*x^6 + (4*a^2*b*c^2 + 3*a*b^2*c - b^3)*d^ 2*x^4)*sqrt(a*c^2 + b*c)*log(((8*a^2*c^2 + 8*a*b*c + b^2)*d^2*x^4 + 8*a^2* c^4 + 16*a*b*c^3 + 8*b^2*c^2 + 8*(2*a^2*c^3 + 3*a*b*c^2 + b^2*c)*d*x^2 + 4 *((2*a*c + b)*d^2*x^4 + 2*a*c^3 + (4*a*c^2 + 3*b*c)*d*x^2 + 2*b*c^2)*sqrt( a*c^2 + b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/x^4) + 4*((2*a^3*c^3 - 11*a^2*b*c^2 - 13*a*b^2*c)*d^3*x^6 - 2*a^3*c^6 - 6*a^2*b*c^5 - 6*a*b^2*c^ 4 + (2*a^3*c^4 - 16*a^2*b*c^3 - 23*a*b^2*c^2 - 5*b^3*c)*d^2*x^4 - 2*b^3*c^ 3 - (2*a^3*c^5 + 11*a^2*b*c^4 + 16*a*b^2*c^3 + 7*b^3*c^2)*d*x^2)*sqrt((a*d *x^2 + a*c + b)/(d*x^2 + c)))/((a^5*c^5 + 4*a^4*b*c^4 + 6*a^3*b^2*c^3 + 4* a^2*b^3*c^2 + a*b^4*c)*d*x^6 + (a^5*c^6 + 5*a^4*b*c^5 + 10*a^3*b^2*c^4 + 1 0*a^2*b^3*c^3 + 5*a*b^4*c^2 + b^5*c)*x^4), -1/16*(3*((4*a^2*b*c - a*b^2)*d ^3*x^6 + (4*a^2*b*c^2 + 3*a*b^2*c - b^3)*d^2*x^4)*sqrt(-a*c^2 - b*c)*arcta n(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 + 2*b*c)*sqrt(-a*c^2 - b*c)*sqrt((a*d*x ^2 + a*c + b)/(d*x^2 + c))/(a^2*c^3 + 2*a*b*c^2 + (a^2*c^2 + a*b*c)*d*x^2 + b^2*c)) - 2*((2*a^3*c^3 - 11*a^2*b*c^2 - 13*a*b^2*c)*d^3*x^6 - 2*a^3*c^6 - 6*a^2*b*c^5 - 6*a*b^2*c^4 + (2*a^3*c^4 - 16*a^2*b*c^3 - 23*a*b^2*c^2 - 5*b^3*c)*d^2*x^4 - 2*b^3*c^3 - (2*a^3*c^5 + 11*a^2*b*c^4 + 16*a*b^2*c^3 + 7*b^3*c^2)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a^5*c^5 + 4*a^4 *b*c^4 + 6*a^3*b^2*c^3 + 4*a^2*b^3*c^2 + a*b^4*c)*d*x^6 + (a^5*c^6 + 5*a^4 *b*c^5 + 10*a^3*b^2*c^4 + 10*a^2*b^3*c^3 + 5*a*b^4*c^2 + b^5*c)*x^4)]
\[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {1}{x^{5} \left (\frac {a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (192) = 384\).
Time = 0.32 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.12 \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=-\frac {3 \, {\left (4 \, a b c - b^{2}\right )} d^{2} \log \left (\frac {c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} - \sqrt {{\left (a c + b\right )} c}}{c \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}} + \sqrt {{\left (a c + b\right )} c}}\right )}{16 \, {\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} + 3 \, a b^{2} c + b^{3}\right )} \sqrt {{\left (a c + b\right )} c}} - \frac {8 \, {\left (a^{3} b c^{2} + 2 \, a^{2} b^{2} c + a b^{3}\right )} d^{2} + \frac {3 \, {\left (4 \, a b c^{2} - b^{2} c\right )} {\left (a d x^{2} + a c + b\right )}^{2} d^{2}}{{\left (d x^{2} + c\right )}^{2}} - \frac {5 \, {\left (4 \, a^{2} b c^{2} + 3 \, a b^{2} c - b^{3}\right )} {\left (a d x^{2} + a c + b\right )} d^{2}}{d x^{2} + c}}{8 \, {\left ({\left (a^{3} c^{5} + 3 \, a^{2} b c^{4} + 3 \, a b^{2} c^{3} + b^{3} c^{2}\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {5}{2}} - 2 \, {\left (a^{4} c^{5} + 4 \, a^{3} b c^{4} + 6 \, a^{2} b^{2} c^{3} + 4 \, a b^{3} c^{2} + b^{4} c\right )} \left (\frac {a d x^{2} + a c + b}{d x^{2} + c}\right )^{\frac {3}{2}} + {\left (a^{5} c^{5} + 5 \, a^{4} b c^{4} + 10 \, a^{3} b^{2} c^{3} + 10 \, a^{2} b^{3} c^{2} + 5 \, a b^{4} c + b^{5}\right )} \sqrt {\frac {a d x^{2} + a c + b}{d x^{2} + c}}\right )}} \]
-3/16*(4*a*b*c - b^2)*d^2*log((c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)) - s qrt((a*c + b)*c))/(c*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)) + sqrt((a*c + b )*c)))/((a^3*c^3 + 3*a^2*b*c^2 + 3*a*b^2*c + b^3)*sqrt((a*c + b)*c)) - 1/8 *(8*(a^3*b*c^2 + 2*a^2*b^2*c + a*b^3)*d^2 + 3*(4*a*b*c^2 - b^2*c)*(a*d*x^2 + a*c + b)^2*d^2/(d*x^2 + c)^2 - 5*(4*a^2*b*c^2 + 3*a*b^2*c - b^3)*(a*d*x ^2 + a*c + b)*d^2/(d*x^2 + c))/((a^3*c^5 + 3*a^2*b*c^4 + 3*a*b^2*c^3 + b^3 *c^2)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(5/2) - 2*(a^4*c^5 + 4*a^3*b*c^4 + 6*a^2*b^2*c^3 + 4*a*b^3*c^2 + b^4*c)*((a*d*x^2 + a*c + b)/(d*x^2 + c))^(3 /2) + (a^5*c^5 + 5*a^4*b*c^4 + 10*a^3*b^2*c^3 + 10*a^2*b^3*c^2 + 5*a*b^4*c + b^5)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))
\[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (a + \frac {b}{d x^{2} + c}\right )}^{\frac {3}{2}} x^{5}} \,d x } \]
Timed out. \[ \int \frac {1}{x^5 \left (a+\frac {b}{c+d x^2}\right )^{3/2}} \, dx=\int \frac {1}{x^5\,{\left (a+\frac {b}{d\,x^2+c}\right )}^{3/2}} \,d x \]