Integrand size = 22, antiderivative size = 72 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx=-\frac {c (b c-a d) \sqrt {c+\frac {d}{x^2}}}{d^3}+\frac {(2 b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^3} \] Output:
-c*(-a*d+b*c)*(c+d/x^2)^(1/2)/d^3+1/3*(-a*d+2*b*c)*(c+d/x^2)^(3/2)/d^3-1/5 *b*(c+d/x^2)^(5/2)/d^3
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (-5 a d x^2 \left (d-2 c x^2\right )+b \left (-3 d^2+4 c d x^2-8 c^2 x^4\right )\right )}{15 d^3 x^4} \] Input:
Integrate[(a + b/x^2)/(Sqrt[c + d/x^2]*x^5),x]
Output:
(Sqrt[c + d/x^2]*(-5*a*d*x^2*(d - 2*c*x^2) + b*(-3*d^2 + 4*c*d*x^2 - 8*c^2 *x^4)))/(15*d^3*x^4)
Time = 0.35 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int \frac {a+\frac {b}{x^2}}{x^5 \sqrt {c+\frac {d}{x^2}}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {1}{2} \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^2}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{d^2}+\frac {(a d-2 b c) \sqrt {c+\frac {d}{x^2}}}{d^2}+\frac {c (b c-a d)}{d^2 \sqrt {c+\frac {d}{x^2}}}\right )d\frac {1}{x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+\frac {d}{x^2}\right )^{3/2} (2 b c-a d)}{3 d^3}-\frac {2 c \sqrt {c+\frac {d}{x^2}} (b c-a d)}{d^3}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^3}\right )\) |
Input:
Int[(a + b/x^2)/(Sqrt[c + d/x^2]*x^5),x]
Output:
((-2*c*(b*c - a*d)*Sqrt[c + d/x^2])/d^3 + (2*(2*b*c - a*d)*(c + d/x^2)^(3/ 2))/(3*d^3) - (2*b*(c + d/x^2)^(5/2))/(5*d^3))/2
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 = 67, normalized size of antiderivative = 0.93
method | result | size |
trager | \(\frac {\left (10 a c d \,x^{4}-8 b \,c^{2} x^{4}-5 a \,d^{2} x^{2}+4 b c d \,x^{2}-3 b \,d^{2}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{15 x^{4} d^{3}}\) | \(67\) |
gosper | \(\frac {\left (10 a c d \,x^{4}-8 b \,c^{2} x^{4}-5 a \,d^{2} x^{2}+4 b c d \,x^{2}-3 b \,d^{2}\right ) \left (c \,x^{2}+d \right )}{15 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{3} x^{6}}\) | \(70\) |
default | \(\frac {\left (10 a c d \,x^{4}-8 b \,c^{2} x^{4}-5 a \,d^{2} x^{2}+4 b c d \,x^{2}-3 b \,d^{2}\right ) \left (c \,x^{2}+d \right )}{15 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{3} x^{6}}\) | \(70\) |
risch | \(\frac {\left (10 a c d \,x^{4}-8 b \,c^{2} x^{4}-5 a \,d^{2} x^{2}+4 b c d \,x^{2}-3 b \,d^{2}\right ) \left (c \,x^{2}+d \right )}{15 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{3} x^{6}}\) | \(70\) |
orering | \(\frac {\left (10 a c d \,x^{4}-8 b \,c^{2} x^{4}-5 a \,d^{2} x^{2}+4 b c d \,x^{2}-3 b \,d^{2}\right ) \left (c \,x^{2}+d \right ) \left (a +\frac {b}{x^{2}}\right )}{15 d^{3} \left (a \,x^{2}+b \right ) x^{4} \sqrt {c +\frac {d}{x^{2}}}}\) | \(82\) |
Input:
int((a+b/x^2)/(c+d/x^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)
Output:
1/15/x^4*(10*a*c*d*x^4-8*b*c^2*x^4-5*a*d^2*x^2+4*b*c*d*x^2-3*b*d^2)/d^3*(- (-c*x^2-d)/x^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx=-\frac {{\left (2 \, {\left (4 \, b c^{2} - 5 \, a c d\right )} x^{4} + 3 \, b d^{2} - {\left (4 \, b c d - 5 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, d^{3} x^{4}} \] Input:
integrate((a+b/x^2)/(c+d/x^2)^(1/2)/x^5,x, algorithm="fricas")
Output:
-1/15*(2*(4*b*c^2 - 5*a*c*d)*x^4 + 3*b*d^2 - (4*b*c*d - 5*a*d^2)*x^2)*sqrt ((c*x^2 + d)/x^2)/(d^3*x^4)
Time = 1.84 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx=\frac {\begin {cases} \frac {- \frac {2 a \left (- c \sqrt {c + \frac {d}{x^{2}}} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3}\right )}{d} - \frac {2 b \left (c^{2} \sqrt {c + \frac {d}{x^{2}}} - \frac {2 c \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}}}{d} & \text {for}\: d \neq 0 \\\frac {- \frac {a}{2 x^{4}} - \frac {b}{3 x^{6}}}{\sqrt {c}} & \text {otherwise} \end {cases}}{2} \] Input:
integrate((a+b/x**2)/(c+d/x**2)**(1/2)/x**5,x)
Output:
Piecewise(((-2*a*(-c*sqrt(c + d/x**2) + (c + d/x**2)**(3/2)/3)/d - 2*b*(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)/d, Ne(d, 0)), ((-a/(2*x**4) - b/(3*x**6))/sqrt(c), True))/2
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.15 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx=-\frac {1}{15} \, b {\left (\frac {3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{d^{3}} - \frac {10 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c}{d^{3}} + \frac {15 \, \sqrt {c + \frac {d}{x^{2}}} c^{2}}{d^{3}}\right )} - \frac {1}{3} \, a {\left (\frac {{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}}}{d^{2}} - \frac {3 \, \sqrt {c + \frac {d}{x^{2}}} c}{d^{2}}\right )} \] Input:
integrate((a+b/x^2)/(c+d/x^2)^(1/2)/x^5,x, algorithm="maxima")
Output:
-1/15*b*(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) - 1/3*a*((c + d/x^2)^(3/2)/d^2 - 3*sqrt(c + d/x^2)*c/d^2 )
Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (62) = 124\).
Time = 0.38 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.50 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx=\frac {4 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {3}{2}} + 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {5}{2}} - 35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {3}{2}} d - 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {5}{2}} d + 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {3}{2}} d^{2} + 4 \, b c^{\frac {5}{2}} d^{2} - 5 \, a c^{\frac {3}{2}} d^{3}\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{5} \mathrm {sgn}\left (x\right )} \] Input:
integrate((a+b/x^2)/(c+d/x^2)^(1/2)/x^5,x, algorithm="giac")
Output:
4/15*(15*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(3/2) + 40*(sqrt(c)*x - sqrt( c*x^2 + d))^4*b*c^(5/2) - 35*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(3/2)*d - 20*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(5/2)*d + 25*(sqrt(c)*x - sqrt(c*x ^2 + d))^2*a*c^(3/2)*d^2 + 4*b*c^(5/2)*d^2 - 5*a*c^(3/2)*d^3)/(((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^5*sgn(x))
Time = 3.88 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.81 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx=-\frac {\sqrt {c+\frac {d}{x^2}}\,\left (8\,b\,c^2\,x^4-10\,a\,c\,d\,x^4-4\,b\,c\,d\,x^2+5\,a\,d^2\,x^2+3\,b\,d^2\right )}{15\,d^3\,x^4} \] Input:
int((a + b/x^2)/(x^5*(c + d/x^2)^(1/2)),x)
Output:
-((c + d/x^2)^(1/2)*(3*b*d^2 + 5*a*d^2*x^2 + 8*b*c^2*x^4 - 10*a*c*d*x^4 - 4*b*c*d*x^2))/(15*d^3*x^4)
Time = 0.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.53 \[ \int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^5} \, dx=\frac {10 \sqrt {c \,x^{2}+d}\, a c d \,x^{4}-5 \sqrt {c \,x^{2}+d}\, a \,d^{2} x^{2}-8 \sqrt {c \,x^{2}+d}\, b \,c^{2} x^{4}+4 \sqrt {c \,x^{2}+d}\, b c d \,x^{2}-3 \sqrt {c \,x^{2}+d}\, b \,d^{2}-10 \sqrt {c}\, a c d \,x^{5}+8 \sqrt {c}\, b \,c^{2} x^{5}}{15 d^{3} x^{5}} \] Input:
int((a+b/x^2)/(c+d/x^2)^(1/2)/x^5,x)
Output:
(10*sqrt(c*x**2 + d)*a*c*d*x**4 - 5*sqrt(c*x**2 + d)*a*d**2*x**2 - 8*sqrt( c*x**2 + d)*b*c**2*x**4 + 4*sqrt(c*x**2 + d)*b*c*d*x**2 - 3*sqrt(c*x**2 + d)*b*d**2 - 10*sqrt(c)*a*c*d*x**5 + 8*sqrt(c)*b*c**2*x**5)/(15*d**3*x**5)