Integrand size = 22, antiderivative size = 126 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^9} \, dx=\frac {c^3 (b c-a d)}{d^5 \sqrt {c+\frac {d}{x^2}}}+\frac {c^2 (4 b c-3 a d) \sqrt {c+\frac {d}{x^2}}}{d^5}-\frac {c (2 b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{d^5}+\frac {(4 b c-a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^5}-\frac {b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^5} \]
-c*(-a*d+2*b*c)*(c+d/x^2)^(3/2)/d^5+1/5*(-a*d+4*b*c)*(c+d/x^2)^(5/2)/d^5-1 /7*b*(c+d/x^2)^(7/2)/d^5+c^3*(-a*d+b*c)/d^5/(c+d/x^2)^(1/2)+c^2*(-3*a*d+4* b*c)*(c+d/x^2)^(1/2)/d^5
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.83 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^9} \, dx=\frac {-7 a d x^2 \left (d^3-2 c d^2 x^2+8 c^2 d x^4+16 c^3 x^6\right )+b \left (-5 d^4+8 c d^3 x^2-16 c^2 d^2 x^4+64 c^3 d x^6+128 c^4 x^8\right )}{35 d^5 \sqrt {c+\frac {d}{x^2}} x^8} \]
(-7*a*d*x^2*(d^3 - 2*c*d^2*x^2 + 8*c^2*d*x^4 + 16*c^3*x^6) + b*(-5*d^4 + 8 *c*d^3*x^2 - 16*c^2*d^2*x^4 + 64*c^3*d*x^6 + 128*c^4*x^8))/(35*d^5*Sqrt[c + d/x^2]*x^8)
Time = 0.26 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {948, 86, 2009}
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.
\(\displaystyle \int \frac {a+\frac {b}{x^2}}{x^9 \left (c+\frac {d}{x^2}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {1}{2} \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {(b c-a d) c^3}{d^4 \left (c+\frac {d}{x^2}\right )^{3/2}}-\frac {(4 b c-3 a d) c^2}{d^4 \sqrt {c+\frac {d}{x^2}}}+\frac {3 (2 b c-a d) \sqrt {c+\frac {d}{x^2}} c}{d^4}+\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{d^4}+\frac {(a d-4 b c) \left (c+\frac {d}{x^2}\right )^{3/2}}{d^4}\right )d\frac {1}{x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 c^3 (b c-a d)}{d^5 \sqrt {c+\frac {d}{x^2}}}+\frac {2 c^2 \sqrt {c+\frac {d}{x^2}} (4 b c-3 a d)}{d^5}+\frac {2 \left (c+\frac {d}{x^2}\right )^{5/2} (4 b c-a d)}{5 d^5}-\frac {2 c \left (c+\frac {d}{x^2}\right )^{3/2} (2 b c-a d)}{d^5}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^5}\right )\) |
((2*c^3*(b*c - a*d))/(d^5*Sqrt[c + d/x^2]) + (2*c^2*(4*b*c - 3*a*d)*Sqrt[c + d/x^2])/d^5 - (2*c*(2*b*c - a*d)*(c + d/x^2)^(3/2))/d^5 + (2*(4*b*c - a *d)*(c + d/x^2)^(5/2))/(5*d^5) - (2*b*(c + d/x^2)^(7/2))/(7*d^5))/2
3.10.79.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94
method | result | size |
gosper | \(-\frac {\left (112 a \,c^{3} d \,x^{8}-128 b \,c^{4} x^{8}+56 a \,c^{2} d^{2} x^{6}-64 b \,c^{3} d \,x^{6}-14 a c \,d^{3} x^{4}+16 b \,c^{2} d^{2} x^{4}+7 a \,d^{4} x^{2}-8 b c \,d^{3} x^{2}+5 b \,d^{4}\right ) \left (c \,x^{2}+d \right )}{35 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} d^{5} x^{10}}\) | \(118\) |
default | \(-\frac {\left (112 a \,c^{3} d \,x^{8}-128 b \,c^{4} x^{8}+56 a \,c^{2} d^{2} x^{6}-64 b \,c^{3} d \,x^{6}-14 a c \,d^{3} x^{4}+16 b \,c^{2} d^{2} x^{4}+7 a \,d^{4} x^{2}-8 b c \,d^{3} x^{2}+5 b \,d^{4}\right ) \left (c \,x^{2}+d \right )}{35 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} d^{5} x^{10}}\) | \(118\) |
trager | \(-\frac {\left (112 a \,c^{3} d \,x^{8}-128 b \,c^{4} x^{8}+56 a \,c^{2} d^{2} x^{6}-64 b \,c^{3} d \,x^{6}-14 a c \,d^{3} x^{4}+16 b \,c^{2} d^{2} x^{4}+7 a \,d^{4} x^{2}-8 b c \,d^{3} x^{2}+5 b \,d^{4}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{35 x^{6} d^{5} \left (c \,x^{2}+d \right )}\) | \(124\) |
risch | \(-\frac {\left (c \,x^{2}+d \right ) \left (77 a \,c^{2} d \,x^{6}-93 b \,c^{3} x^{6}-21 a c \,d^{2} x^{4}+29 b \,c^{2} d \,x^{4}+7 a \,d^{3} x^{2}-13 b c \,d^{2} x^{2}+5 b \,d^{3}\right )}{35 d^{5} x^{8} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}-\frac {c^{3} \left (a d -b c \right )}{d^{5} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}\) | \(124\) |
-1/35*(112*a*c^3*d*x^8-128*b*c^4*x^8+56*a*c^2*d^2*x^6-64*b*c^3*d*x^6-14*a* c*d^3*x^4+16*b*c^2*d^2*x^4+7*a*d^4*x^2-8*b*c*d^3*x^2+5*b*d^4)*(c*x^2+d)/(( c*x^2+d)/x^2)^(3/2)/d^5/x^10
Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.96 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^9} \, dx=\frac {{\left (16 \, {\left (8 \, b c^{4} - 7 \, a c^{3} d\right )} x^{8} + 8 \, {\left (8 \, b c^{3} d - 7 \, a c^{2} d^{2}\right )} x^{6} - 5 \, b d^{4} - 2 \, {\left (8 \, b c^{2} d^{2} - 7 \, a c d^{3}\right )} x^{4} + {\left (8 \, b c d^{3} - 7 \, a d^{4}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{35 \, {\left (c d^{5} x^{8} + d^{6} x^{6}\right )}} \]
1/35*(16*(8*b*c^4 - 7*a*c^3*d)*x^8 + 8*(8*b*c^3*d - 7*a*c^2*d^2)*x^6 - 5*b *d^4 - 2*(8*b*c^2*d^2 - 7*a*c*d^3)*x^4 + (8*b*c*d^3 - 7*a*d^4)*x^2)*sqrt(( c*x^2 + d)/x^2)/(c*d^5*x^8 + d^6*x^6)
Time = 3.06 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.16 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^9} \, dx=\begin {cases} \frac {2 \left (- \frac {b \left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{14 d^{4}} - \frac {c^{3} \left (a d - b c\right )}{2 d^{4} \sqrt {c + \frac {d}{x^{2}}}} - \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}} \left (a d - 4 b c\right )}{10 d^{4}} - \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}} \left (- 3 a c d + 6 b c^{2}\right )}{6 d^{4}} - \frac {\sqrt {c + \frac {d}{x^{2}}} \cdot \left (3 a c^{2} d - 4 b c^{3}\right )}{2 d^{4}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {- \frac {a}{4 x^{8}} - \frac {b}{5 x^{10}}}{2 c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-b*(c + d/x**2)**(7/2)/(14*d**4) - c**3*(a*d - b*c)/(2*d**4* sqrt(c + d/x**2)) - (c + d/x**2)**(5/2)*(a*d - 4*b*c)/(10*d**4) - (c + d/x **2)**(3/2)*(-3*a*c*d + 6*b*c**2)/(6*d**4) - sqrt(c + d/x**2)*(3*a*c**2*d - 4*b*c**3)/(2*d**4))/d, Ne(d, 0)), ((-a/(4*x**8) - b/(5*x**10))/(2*c**(3/ 2)), True))
Time = 0.19 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.20 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^9} \, dx=-\frac {1}{35} \, b {\left (\frac {5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}}}{d^{5}} - \frac {28 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c}{d^{5}} + \frac {70 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2}}{d^{5}} - \frac {140 \, \sqrt {c + \frac {d}{x^{2}}} c^{3}}{d^{5}} - \frac {35 \, c^{4}}{\sqrt {c + \frac {d}{x^{2}}} d^{5}}\right )} - \frac {1}{5} \, a {\left (\frac {{\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{d^{4}} - \frac {5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c}{d^{4}} + \frac {15 \, \sqrt {c + \frac {d}{x^{2}}} c^{2}}{d^{4}} + \frac {5 \, c^{3}}{\sqrt {c + \frac {d}{x^{2}}} d^{4}}\right )} \]
-1/35*b*(5*(c + d/x^2)^(7/2)/d^5 - 28*(c + d/x^2)^(5/2)*c/d^5 + 70*(c + d/ x^2)^(3/2)*c^2/d^5 - 140*sqrt(c + d/x^2)*c^3/d^5 - 35*c^4/(sqrt(c + d/x^2) *d^5)) - 1/5*a*((c + d/x^2)^(5/2)/d^4 - 5*(c + d/x^2)^(3/2)*c/d^4 + 15*sqr t(c + d/x^2)*c^2/d^4 + 5*c^3/(sqrt(c + d/x^2)*d^4))
Leaf count of result is larger than twice the leaf count of optimal. 414 vs. \(2 (112) = 224\).
Time = 1.19 (sec) , antiderivative size = 414, normalized size of antiderivative = 3.29 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^9} \, dx=\frac {{\left (b c^{4} - a c^{3} d\right )} x}{\sqrt {c x^{2} + d} d^{5} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{12} b c^{\frac {7}{2}} - 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{12} a c^{\frac {5}{2}} d - 280 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} b c^{\frac {7}{2}} d + 280 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} a c^{\frac {5}{2}} d^{2} + 1015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} b c^{\frac {7}{2}} d^{2} - 1015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {5}{2}} d^{3} - 2240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} b c^{\frac {7}{2}} d^{3} + 1680 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {5}{2}} d^{4} + 1673 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {7}{2}} d^{4} - 1337 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {5}{2}} d^{5} - 616 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {7}{2}} d^{5} + 504 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {5}{2}} d^{6} + 93 \, b c^{\frac {7}{2}} d^{6} - 77 \, a c^{\frac {5}{2}} d^{7}\right )}}{35 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{7} d^{4} \mathrm {sgn}\left (x\right )} \]
(b*c^4 - a*c^3*d)*x/(sqrt(c*x^2 + d)*d^5*sgn(x)) - 2/35*(35*(sqrt(c)*x - s qrt(c*x^2 + d))^12*b*c^(7/2) - 35*(sqrt(c)*x - sqrt(c*x^2 + d))^12*a*c^(5/ 2)*d - 280*(sqrt(c)*x - sqrt(c*x^2 + d))^10*b*c^(7/2)*d + 280*(sqrt(c)*x - sqrt(c*x^2 + d))^10*a*c^(5/2)*d^2 + 1015*(sqrt(c)*x - sqrt(c*x^2 + d))^8* b*c^(7/2)*d^2 - 1015*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(5/2)*d^3 - 2240* (sqrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(7/2)*d^3 + 1680*(sqrt(c)*x - sqrt(c*x ^2 + d))^6*a*c^(5/2)*d^4 + 1673*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(7/2)* d^4 - 1337*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(5/2)*d^5 - 616*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(7/2)*d^5 + 504*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a *c^(5/2)*d^6 + 93*b*c^(7/2)*d^6 - 77*a*c^(5/2)*d^7)/(((sqrt(c)*x - sqrt(c* x^2 + d))^2 - d)^7*d^4*sgn(x))
Time = 9.35 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.22 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^9} \, dx=\frac {c\,\sqrt {c+\frac {d}{x^2}}\,\left (21\,a\,d-29\,b\,c\right )}{35\,d^4\,x^2}-\frac {b\,\sqrt {c+\frac {d}{x^2}}}{7\,d^2\,x^6}-\frac {\sqrt {c+\frac {d}{x^2}}\,\left (7\,a\,d^2-13\,b\,c\,d\right )}{35\,d^4\,x^4}-\frac {\sqrt {c+\frac {d}{x^2}}\,\left (x^2\,\left (\frac {58\,b\,c^4-42\,a\,c^3\,d}{35\,d^5}+\frac {2\,c^3\,\left (77\,a\,d-93\,b\,c\right )}{35\,d^5}\right )+\frac {c^2\,\left (77\,a\,d-93\,b\,c\right )}{35\,d^4}\right )}{c\,x^2+d} \]
(c*(c + d/x^2)^(1/2)*(21*a*d - 29*b*c))/(35*d^4*x^2) - (b*(c + d/x^2)^(1/2 ))/(7*d^2*x^6) - ((c + d/x^2)^(1/2)*(7*a*d^2 - 13*b*c*d))/(35*d^4*x^4) - ( (c + d/x^2)^(1/2)*(x^2*((58*b*c^4 - 42*a*c^3*d)/(35*d^5) + (2*c^3*(77*a*d - 93*b*c))/(35*d^5)) + (c^2*(77*a*d - 93*b*c))/(35*d^4)))/(d + c*x^2)