Integrand size = 22, antiderivative size = 101 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^7} \, dx=\frac {c^2 (b c-a d) \sqrt {c+\frac {d}{x^2}}}{d^4}-\frac {c (3 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^4}+\frac {(3 b c-a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^4}-\frac {b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^4} \]
-1/3*c*(-2*a*d+3*b*c)*(c+d/x^2)^(3/2)/d^4+1/5*(-a*d+3*b*c)*(c+d/x^2)^(5/2) /d^4-1/7*b*(c+d/x^2)^(7/2)/d^4+c^2*(-a*d+b*c)*(c+d/x^2)^(1/2)/d^4
Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^7} \, dx=\frac {\left (d+c x^2\right ) \left (-15 b d^3+18 b c d^2 x^2-21 a d^3 x^2-24 b c^2 d x^4+28 a c d^2 x^4+48 b c^3 x^6-56 a c^2 d x^6\right )}{105 d^4 \sqrt {c+\frac {d}{x^2}} x^8} \]
((d + c*x^2)*(-15*b*d^3 + 18*b*c*d^2*x^2 - 21*a*d^3*x^2 - 24*b*c^2*d*x^4 + 28*a*c*d^2*x^4 + 48*b*c^3*x^6 - 56*a*c^2*d*x^6))/(105*d^4*Sqrt[c + d/x^2] *x^8)
Time = 0.24 (sec) , antiderivative size = 106, 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.
\(\displaystyle \int \frac {a+\frac {b}{x^2}}{x^7 \sqrt {c+\frac {d}{x^2}}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {1}{2} \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^4}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{d^3}+\frac {(a d-3 b c) \left (c+\frac {d}{x^2}\right )^{3/2}}{d^3}+\frac {c (3 b c-2 a d) \sqrt {c+\frac {d}{x^2}}}{d^3}-\frac {c^2 (b c-a d)}{d^3 \sqrt {c+\frac {d}{x^2}}}\right )d\frac {1}{x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 c^2 \sqrt {c+\frac {d}{x^2}} (b c-a d)}{d^4}+\frac {2 \left (c+\frac {d}{x^2}\right )^{5/2} (3 b c-a d)}{5 d^4}-\frac {2 c \left (c+\frac {d}{x^2}\right )^{3/2} (3 b c-2 a d)}{3 d^4}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^4}\right )\) |
((2*c^2*(b*c - a*d)*Sqrt[c + d/x^2])/d^4 - (2*c*(3*b*c - 2*a*d)*(c + d/x^2 )^(3/2))/(3*d^4) + (2*(3*b*c - a*d)*(c + d/x^2)^(5/2))/(5*d^4) - (2*b*(c + d/x^2)^(7/2))/(7*d^4))/2
3.10.67.3.1 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 = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90
method | result | size |
trager | \(-\frac {\left (56 a \,c^{2} d \,x^{6}-48 b \,c^{3} x^{6}-28 a c \,d^{2} x^{4}+24 b \,c^{2} d \,x^{4}+21 a \,d^{3} x^{2}-18 b c \,d^{2} x^{2}+15 b \,d^{3}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{105 x^{6} d^{4}}\) | \(91\) |
gosper | \(-\frac {\left (56 a \,c^{2} d \,x^{6}-48 b \,c^{3} x^{6}-28 a c \,d^{2} x^{4}+24 b \,c^{2} d \,x^{4}+21 a \,d^{3} x^{2}-18 b c \,d^{2} x^{2}+15 b \,d^{3}\right ) \left (c \,x^{2}+d \right )}{105 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{4} x^{8}}\) | \(94\) |
default | \(-\frac {\left (56 a \,c^{2} d \,x^{6}-48 b \,c^{3} x^{6}-28 a c \,d^{2} x^{4}+24 b \,c^{2} d \,x^{4}+21 a \,d^{3} x^{2}-18 b c \,d^{2} x^{2}+15 b \,d^{3}\right ) \left (c \,x^{2}+d \right )}{105 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{4} x^{8}}\) | \(94\) |
risch | \(-\frac {\left (56 a \,c^{2} d \,x^{6}-48 b \,c^{3} x^{6}-28 a c \,d^{2} x^{4}+24 b \,c^{2} d \,x^{4}+21 a \,d^{3} x^{2}-18 b c \,d^{2} x^{2}+15 b \,d^{3}\right ) \left (c \,x^{2}+d \right )}{105 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{4} x^{8}}\) | \(94\) |
-1/105/x^6*(56*a*c^2*d*x^6-48*b*c^3*x^6-28*a*c*d^2*x^4+24*b*c^2*d*x^4+21*a *d^3*x^2-18*b*c*d^2*x^2+15*b*d^3)/d^4*(-(-c*x^2-d)/x^2)^(1/2)
Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.85 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^7} \, dx=\frac {{\left (8 \, {\left (6 \, b c^{3} - 7 \, a c^{2} d\right )} x^{6} - 4 \, {\left (6 \, b c^{2} d - 7 \, a c d^{2}\right )} x^{4} - 15 \, b d^{3} + 3 \, {\left (6 \, b c d^{2} - 7 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{105 \, d^{4} x^{6}} \]
1/105*(8*(6*b*c^3 - 7*a*c^2*d)*x^6 - 4*(6*b*c^2*d - 7*a*c*d^2)*x^4 - 15*b* d^3 + 3*(6*b*c*d^2 - 7*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2)/(d^4*x^6)
Time = 1.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^7} \, dx=\frac {\begin {cases} \frac {- \frac {2 a \left (c^{2} \sqrt {c + \frac {d}{x^{2}}} - \frac {2 c \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} - \frac {2 b \left (- c^{3} \sqrt {c + \frac {d}{x^{2}}} + c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}} - \frac {3 c \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7}\right )}{d^{3}}}{d} & \text {for}\: d \neq 0 \\\frac {- \frac {a}{3 x^{6}} - \frac {b}{4 x^{8}}}{\sqrt {c}} & \text {otherwise} \end {cases}}{2} \]
Piecewise(((-2*a*(c**2*sqrt(c + d/x**2) - 2*c*(c + d/x**2)**(3/2)/3 + (c + d/x**2)**(5/2)/5)/d**2 - 2*b*(-c**3*sqrt(c + d/x**2) + c**2*(c + d/x**2)* *(3/2) - 3*c*(c + d/x**2)**(5/2)/5 + (c + d/x**2)**(7/2)/7)/d**3)/d, Ne(d, 0)), ((-a/(3*x**6) - b/(4*x**8))/sqrt(c), True))/2
Time = 0.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.17 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^7} \, dx=-\frac {1}{35} \, b {\left (\frac {5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}}}{d^{4}} - \frac {21 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c}{d^{4}} + \frac {35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2}}{d^{4}} - \frac {35 \, \sqrt {c + \frac {d}{x^{2}}} c^{3}}{d^{4}}\right )} - \frac {1}{15} \, a {\left (\frac {3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{d^{3}} - \frac {10 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c}{d^{3}} + \frac {15 \, \sqrt {c + \frac {d}{x^{2}}} c^{2}}{d^{3}}\right )} \]
-1/35*b*(5*(c + d/x^2)^(7/2)/d^4 - 21*(c + d/x^2)^(5/2)*c/d^4 + 35*(c + d/ x^2)^(3/2)*c^2/d^4 - 35*sqrt(c + d/x^2)*c^3/d^4) - 1/15*a*(3*(c + d/x^2)^( 5/2)/d^3 - 10*(c + d/x^2)^(3/2)*c/d^3 + 15*sqrt(c + d/x^2)*c^2/d^3)
Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (87) = 174\).
Time = 0.87 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.34 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^7} \, dx=\frac {16 \, {\left (70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {5}{2}} + 210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} b c^{\frac {7}{2}} - 175 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {5}{2}} d - 126 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {7}{2}} d + 147 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {5}{2}} d^{2} + 42 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {7}{2}} d^{2} - 49 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {5}{2}} d^{3} - 6 \, b c^{\frac {7}{2}} d^{3} + 7 \, a c^{\frac {5}{2}} d^{4}\right )}}{105 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{7} \mathrm {sgn}\left (x\right )} \]
16/105*(70*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(5/2) + 210*(sqrt(c)*x - sq rt(c*x^2 + d))^6*b*c^(7/2) - 175*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(5/2) *d - 126*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(7/2)*d + 147*(sqrt(c)*x - sq rt(c*x^2 + d))^4*a*c^(5/2)*d^2 + 42*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(7 /2)*d^2 - 49*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(5/2)*d^3 - 6*b*c^(7/2)*d ^3 + 7*a*c^(5/2)*d^4)/(((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^7*sgn(x))
Time = 9.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.01 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^7} \, dx=\frac {\sqrt {c+\frac {d}{x^2}}\,\left (48\,b\,c^3-56\,a\,c^2\,d\right )}{105\,d^4}-\frac {b\,\sqrt {c+\frac {d}{x^2}}}{7\,d\,x^6}-\frac {\sqrt {c+\frac {d}{x^2}}\,\left (24\,b\,c^2-28\,a\,c\,d\right )}{105\,d^3\,x^2}-\frac {\sqrt {c+\frac {d}{x^2}}\,\left (7\,a\,d-6\,b\,c\right )}{35\,d^2\,x^4} \]