Integrand size = 22, antiderivative size = 68 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^5} \, dx=\frac {c (b c-a d)}{d^3 \sqrt {c+\frac {d}{x^2}}}+\frac {(2 b c-a d) \sqrt {c+\frac {d}{x^2}}}{d^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^3} \] Output:
c*(-a*d+b*c)/d^3/(c+d/x^2)^(1/2)+(-a*d+2*b*c)*(c+d/x^2)^(1/2)/d^3-1/3*b*(c +d/x^2)^(3/2)/d^3
Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.88 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^5} \, dx=\frac {-3 a d x^2 \left (d+2 c x^2\right )+b \left (-d^2+4 c d x^2+8 c^2 x^4\right )}{3 d^3 \sqrt {c+\frac {d}{x^2}} x^4} \] Input:
Integrate[(a + b/x^2)/((c + d/x^2)^(3/2)*x^5),x]
Output:
(-3*a*d*x^2*(d + 2*c*x^2) + b*(-d^2 + 4*c*d*x^2 + 8*c^2*x^4))/(3*d^3*Sqrt[ c + d/x^2]*x^4)
Time = 0.36 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, 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.
\(\displaystyle \int \frac {a+\frac {b}{x^2}}{x^5 \left (c+\frac {d}{x^2}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {1}{2} \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^2}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {\sqrt {c+\frac {d}{x^2}} b}{d^2}+\frac {a d-2 b c}{d^2 \sqrt {c+\frac {d}{x^2}}}+\frac {c (b c-a d)}{d^2 \left (c+\frac {d}{x^2}\right )^{3/2}}\right )d\frac {1}{x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \sqrt {c+\frac {d}{x^2}} (2 b c-a d)}{d^3}+\frac {2 c (b c-a d)}{d^3 \sqrt {c+\frac {d}{x^2}}}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^3}\right )\) |
Input:
Int[(a + b/x^2)/((c + d/x^2)^(3/2)*x^5),x]
Output:
((2*c*(b*c - a*d))/(d^3*Sqrt[c + d/x^2]) + (2*(2*b*c - a*d)*Sqrt[c + d/x^2 ])/d^3 - (2*b*(c + d/x^2)^(3/2))/(3*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 = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01
method | result | size |
gosper | \(-\frac {\left (6 a c d \,x^{4}-8 b \,c^{2} x^{4}+3 a \,d^{2} x^{2}-4 b c d \,x^{2}+b \,d^{2}\right ) \left (c \,x^{2}+d \right )}{3 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} d^{3} x^{6}}\) | \(69\) |
default | \(-\frac {\left (6 a c d \,x^{4}-8 b \,c^{2} x^{4}+3 a \,d^{2} x^{2}-4 b c d \,x^{2}+b \,d^{2}\right ) \left (c \,x^{2}+d \right )}{3 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} d^{3} x^{6}}\) | \(69\) |
trager | \(-\frac {\left (6 a c d \,x^{4}-8 b \,c^{2} x^{4}+3 a \,d^{2} x^{2}-4 b c d \,x^{2}+b \,d^{2}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{3 x^{2} d^{3} \left (c \,x^{2}+d \right )}\) | \(75\) |
risch | \(-\frac {\left (c \,x^{2}+d \right ) \left (3 a d \,x^{2}-5 b c \,x^{2}+b d \right )}{3 d^{3} x^{4} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}-\frac {\left (a d -c b \right ) c}{d^{3} \sqrt {\frac {c \,x^{2}+d}{x^{2}}}}\) | \(75\) |
orering | \(-\frac {\left (6 a c d \,x^{4}-8 b \,c^{2} x^{4}+3 a \,d^{2} x^{2}-4 b c d \,x^{2}+b \,d^{2}\right ) \left (c \,x^{2}+d \right ) \left (a +\frac {b}{x^{2}}\right )}{3 d^{3} \left (a \,x^{2}+b \right ) x^{4} \left (c +\frac {d}{x^{2}}\right )^{\frac {3}{2}}}\) | \(81\) |
Input:
int((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/3*(6*a*c*d*x^4-8*b*c^2*x^4+3*a*d^2*x^2-4*b*c*d*x^2+b*d^2)*(c*x^2+d)/((c *x^2+d)/x^2)^(3/2)/d^3/x^6
Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^5} \, dx=\frac {{\left (2 \, {\left (4 \, b c^{2} - 3 \, a c d\right )} x^{4} - b d^{2} + {\left (4 \, b c d - 3 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{3 \, {\left (c d^{3} x^{4} + d^{4} x^{2}\right )}} \] Input:
integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x, algorithm="fricas")
Output:
1/3*(2*(4*b*c^2 - 3*a*c*d)*x^4 - b*d^2 + (4*b*c*d - 3*a*d^2)*x^2)*sqrt((c* x^2 + d)/x^2)/(c*d^3*x^4 + d^4*x^2)
Time = 3.70 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.25 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^5} \, dx=\begin {cases} \frac {2 \left (- \frac {b \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{6 d^{2}} - \frac {c \left (a d - b c\right )}{2 d^{2} \sqrt {c + \frac {d}{x^{2}}}} - \frac {\sqrt {c + \frac {d}{x^{2}}} \left (a d - 2 b c\right )}{2 d^{2}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {- \frac {a}{2 x^{4}} - \frac {b}{3 x^{6}}}{2 c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b/x**2)/(c+d/x**2)**(3/2)/x**5,x)
Output:
Piecewise((2*(-b*(c + d/x**2)**(3/2)/(6*d**2) - c*(a*d - b*c)/(2*d**2*sqrt (c + d/x**2)) - sqrt(c + d/x**2)*(a*d - 2*b*c)/(2*d**2))/d, Ne(d, 0)), ((- a/(2*x**4) - b/(3*x**6))/(2*c**(3/2)), True))
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.19 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^5} \, dx=-\frac {1}{3} \, b {\left (\frac {{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}}}{d^{3}} - \frac {6 \, \sqrt {c + \frac {d}{x^{2}}} c}{d^{3}} - \frac {3 \, c^{2}}{\sqrt {c + \frac {d}{x^{2}}} d^{3}}\right )} - a {\left (\frac {\sqrt {c + \frac {d}{x^{2}}}}{d^{2}} + \frac {c}{\sqrt {c + \frac {d}{x^{2}}} d^{2}}\right )} \] Input:
integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x, algorithm="maxima")
Output:
-1/3*b*((c + d/x^2)^(3/2)/d^3 - 6*sqrt(c + d/x^2)*c/d^3 - 3*c^2/(sqrt(c + d/x^2)*d^3)) - a*(sqrt(c + d/x^2)/d^2 + c/(sqrt(c + d/x^2)*d^2))
Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (60) = 120\).
Time = 0.40 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.76 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^5} \, dx=\frac {{\left (b c^{2} - a c d\right )} x}{\sqrt {c x^{2} + d} d^{3} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a \sqrt {c} d - 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {3}{2}} d + 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a \sqrt {c} d^{2} + 5 \, b c^{\frac {3}{2}} d^{2} - 3 \, a \sqrt {c} d^{3}\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{3} d^{2} \mathrm {sgn}\left (x\right )} \] Input:
integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x, algorithm="giac")
Output:
(b*c^2 - a*c*d)*x/(sqrt(c*x^2 + d)*d^3*sgn(x)) - 2/3*(3*(sqrt(c)*x - sqrt( c*x^2 + d))^4*b*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*sqrt(c)*d - 12*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(3/2)*d + 6*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*sqrt(c)*d^2 + 5*b*c^(3/2)*d^2 - 3*a*sqrt(c)*d^3)/(((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^3*d^2*sgn(x))
Time = 3.73 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^5} \, dx=-\frac {\sqrt {c+\frac {d}{x^2}}\,\left (-8\,b\,c^2\,x^4+6\,a\,c\,d\,x^4-4\,b\,c\,d\,x^2+3\,a\,d^2\,x^2+b\,d^2\right )}{3\,d^3\,x^2\,\left (c\,x^2+d\right )} \] Input:
int((a + b/x^2)/(x^5*(c + d/x^2)^(3/2)),x)
Output:
-((c + d/x^2)^(1/2)*(b*d^2 + 3*a*d^2*x^2 - 8*b*c^2*x^4 + 6*a*c*d*x^4 - 4*b *c*d*x^2))/(3*d^3*x^2*(d + c*x^2))
Time = 0.21 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.06 \[ \int \frac {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^5} \, dx=\frac {-6 \sqrt {c \,x^{2}+d}\, a c d \,x^{4}-3 \sqrt {c \,x^{2}+d}\, a \,d^{2} x^{2}+8 \sqrt {c \,x^{2}+d}\, b \,c^{2} x^{4}+4 \sqrt {c \,x^{2}+d}\, b c d \,x^{2}-\sqrt {c \,x^{2}+d}\, b \,d^{2}+6 \sqrt {c}\, a c d \,x^{5}+6 \sqrt {c}\, a \,d^{2} x^{3}-8 \sqrt {c}\, b \,c^{2} x^{5}-8 \sqrt {c}\, b c d \,x^{3}}{3 d^{3} x^{3} \left (c \,x^{2}+d \right )} \] Input:
int((a+b/x^2)/(c+d/x^2)^(3/2)/x^5,x)
Output:
( - 6*sqrt(c*x**2 + d)*a*c*d*x**4 - 3*sqrt(c*x**2 + d)*a*d**2*x**2 + 8*sqr t(c*x**2 + d)*b*c**2*x**4 + 4*sqrt(c*x**2 + d)*b*c*d*x**2 - sqrt(c*x**2 + d)*b*d**2 + 6*sqrt(c)*a*c*d*x**5 + 6*sqrt(c)*a*d**2*x**3 - 8*sqrt(c)*b*c** 2*x**5 - 8*sqrt(c)*b*c*d*x**3)/(3*d**3*x**3*(c*x**2 + d))