Integrand size = 22, antiderivative size = 46 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx=\frac {(b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^2}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^2} \] Output:
1/3*(-a*d+b*c)*(c+d/x^2)^(3/2)/d^2-1/5*b*(c+d/x^2)^(5/2)/d^2
Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (d+c x^2\right ) \left (-3 b d+2 b c x^2-5 a d x^2\right )}{15 d^2 x^4} \] Input:
Integrate[((a + b/x^2)*Sqrt[c + d/x^2])/x^3,x]
Output:
(Sqrt[c + d/x^2]*(d + c*x^2)*(-3*b*d + 2*b*c*x^2 - 5*a*d*x^2))/(15*d^2*x^4 )
Time = 0.32 (sec) , antiderivative size = 50, 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 = {946, 53, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle -\frac {1}{2} \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{d}+\frac {(a d-b c) \sqrt {c+\frac {d}{x^2}}}{d}\right )d\frac {1}{x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+\frac {d}{x^2}\right )^{3/2} (b c-a d)}{3 d^2}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^2}\right )\) |
Input:
Int[((a + b/x^2)*Sqrt[c + d/x^2])/x^3,x]
Output:
((2*(b*c - a*d)*(c + d/x^2)^(3/2))/(3*d^2) - (2*b*(c + d/x^2)^(5/2))/(5*d^ 2))/2
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[(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] && EqQ[m - n + 1, 0]
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (5 a d \,x^{2}-2 b c \,x^{2}+3 b d \right ) \left (c \,x^{2}+d \right )}{15 d^{2} x^{4}}\) | \(48\) |
default | \(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (5 a d \,x^{2}-2 b c \,x^{2}+3 b d \right ) \left (c \,x^{2}+d \right )}{15 d^{2} x^{4}}\) | \(48\) |
orering | \(-\frac {\left (5 a d \,x^{2}-2 b c \,x^{2}+3 b d \right ) \left (c \,x^{2}+d \right ) \left (a +\frac {b}{x^{2}}\right ) \sqrt {c +\frac {d}{x^{2}}}}{15 d^{2} \left (a \,x^{2}+b \right ) x^{2}}\) | \(60\) |
risch | \(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (5 a c d \,x^{4}-2 b \,c^{2} x^{4}+5 a \,d^{2} x^{2}+b c d \,x^{2}+3 b \,d^{2}\right )}{15 x^{4} d^{2}}\) | \(62\) |
trager | \(-\frac {\left (5 a c d \,x^{4}-2 b \,c^{2} x^{4}+5 a \,d^{2} x^{2}+b c d \,x^{2}+3 b \,d^{2}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{15 x^{4} d^{2}}\) | \(66\) |
Input:
int((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/15*((c*x^2+d)/x^2)^(1/2)*(5*a*d*x^2-2*b*c*x^2+3*b*d)*(c*x^2+d)/d^2/x^4
Time = 0.08 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx=\frac {{\left ({\left (2 \, b c^{2} - 5 \, a c d\right )} x^{4} - 3 \, b d^{2} - {\left (b c d + 5 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, d^{2} x^{4}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x, algorithm="fricas")
Output:
1/15*((2*b*c^2 - 5*a*c*d)*x^4 - 3*b*d^2 - (b*c*d + 5*a*d^2)*x^2)*sqrt((c*x ^2 + d)/x^2)/(d^2*x^4)
Time = 1.99 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx=- \frac {a \left (\begin {cases} \frac {2 \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{x^{2}} & \text {otherwise} \end {cases}\right )}{2} - \frac {b \left (\begin {cases} \frac {2 \left (- \frac {c \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{4}} & \text {otherwise} \end {cases}\right )}{2} \] Input:
integrate((a+b/x**2)*(c+d/x**2)**(1/2)/x**3,x)
Output:
-a*Piecewise((2*(c + d/x**2)**(3/2)/(3*d), Ne(d, 0)), (sqrt(c)/x**2, True) )/2 - b*Piecewise((2*(-c*(c + d/x**2)**(3/2)/3 + (c + d/x**2)**(5/2)/5)/d* *2, Ne(d, 0)), (sqrt(c)/(2*x**4), True))/2
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx=-\frac {1}{15} \, b {\left (\frac {3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{d^{2}} - \frac {5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c}{d^{2}}\right )} - \frac {a {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}}}{3 \, d} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x, algorithm="maxima")
Output:
-1/15*b*(3*(c + d/x^2)^(5/2)/d^2 - 5*(c + d/x^2)^(3/2)*c/d^2) - 1/3*a*(c + d/x^2)^(3/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (38) = 76\).
Time = 0.66 (sec) , antiderivative size = 250, normalized size of antiderivative = 5.43 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx=\frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {3}{2}} \mathrm {sgn}\left (x\right ) + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} b c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {3}{2}} d \mathrm {sgn}\left (x\right ) + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {5}{2}} d \mathrm {sgn}\left (x\right ) + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {3}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {5}{2}} d^{2} \mathrm {sgn}\left (x\right ) - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {3}{2}} d^{3} \mathrm {sgn}\left (x\right ) - 2 \, b c^{\frac {5}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 5 \, a c^{\frac {3}{2}} d^{4} \mathrm {sgn}\left (x\right )\right )}}{15 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{5}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x, algorithm="giac")
Output:
2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(3/2)*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(5/2)*sgn(x) - 30*(sqrt(c)*x - sqrt(c*x^2 + d))^6 *a*c^(3/2)*d*sgn(x) + 10*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(5/2)*d*sgn(x ) + 20*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(3/2)*d^2*sgn(x) + 10*(sqrt(c)* x - sqrt(c*x^2 + d))^2*b*c^(5/2)*d^2*sgn(x) - 10*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(3/2)*d^3*sgn(x) - 2*b*c^(5/2)*d^3*sgn(x) + 5*a*c^(3/2)*d^4*sgn (x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^5
Time = 3.66 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.98 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx=\frac {\sqrt {c+\frac {d}{x^2}}\,\left (b\,c^2+a\,d\,c\right )}{5\,d^2}-\frac {b\,\sqrt {c+\frac {d}{x^2}}}{5\,x^4}-\frac {\sqrt {c+\frac {d}{x^2}}\,\left (5\,a\,d^2+b\,c\,d\right )}{15\,d^2\,x^2}-\frac {c\,\sqrt {c+\frac {d}{x^2}}\,\left (8\,a\,d+b\,c\right )}{15\,d^2} \] Input:
int(((a + b/x^2)*(c + d/x^2)^(1/2))/x^3,x)
Output:
((c + d/x^2)^(1/2)*(b*c^2 + a*c*d))/(5*d^2) - (b*(c + d/x^2)^(1/2))/(5*x^4 ) - ((c + d/x^2)^(1/2)*(5*a*d^2 + b*c*d))/(15*d^2*x^2) - (c*(c + d/x^2)^(1 /2)*(8*a*d + b*c))/(15*d^2)
Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.39 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^3} \, dx=\frac {-5 \sqrt {c \,x^{2}+d}\, a c d \,x^{4}-5 \sqrt {c \,x^{2}+d}\, a \,d^{2} x^{2}+2 \sqrt {c \,x^{2}+d}\, b \,c^{2} x^{4}-\sqrt {c \,x^{2}+d}\, b c d \,x^{2}-3 \sqrt {c \,x^{2}+d}\, b \,d^{2}-\sqrt {c}\, a c d \,x^{5}-2 \sqrt {c}\, b \,c^{2} x^{5}}{15 d^{2} x^{5}} \] Input:
int((a+b/x^2)*(c+d/x^2)^(1/2)/x^3,x)
Output:
( - 5*sqrt(c*x**2 + d)*a*c*d*x**4 - 5*sqrt(c*x**2 + d)*a*d**2*x**2 + 2*sqr t(c*x**2 + d)*b*c**2*x**4 - sqrt(c*x**2 + d)*b*c*d*x**2 - 3*sqrt(c*x**2 + d)*b*d**2 - sqrt(c)*a*c*d*x**5 - 2*sqrt(c)*b*c**2*x**5)/(15*d**2*x**5)