Integrand size = 22, antiderivative size = 46 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx=\frac {(b c-a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^2}-\frac {b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^2} \] Output:
1/5*(-a*d+b*c)*(c+d/x^2)^(5/2)/d^2-1/7*b*(c+d/x^2)^(7/2)/d^2
Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx=-\frac {\sqrt {c+\frac {d}{x^2}} \left (d+c x^2\right )^2 \left (5 b d-2 b c x^2+7 a d x^2\right )}{35 d^2 x^6} \] Input:
Integrate[((a + b/x^2)*(c + d/x^2)^(3/2))/x^3,x]
Output:
-1/35*(Sqrt[c + d/x^2]*(d + c*x^2)^2*(5*b*d - 2*b*c*x^2 + 7*a*d*x^2))/(d^2 *x^6)
Time = 0.33 (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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx\) |
\(\Big \downarrow \) 946 |
\(\displaystyle -\frac {1}{2} \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{d}+\frac {(a d-b c) \left (c+\frac {d}{x^2}\right )^{3/2}}{d}\right )d\frac {1}{x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+\frac {d}{x^2}\right )^{5/2} (b c-a d)}{5 d^2}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^2}\right )\) |
Input:
Int[((a + b/x^2)*(c + d/x^2)^(3/2))/x^3,x]
Output:
((2*(b*c - a*d)*(c + d/x^2)^(5/2))/(5*d^2) - (2*b*(c + d/x^2)^(7/2))/(7*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 = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(-\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (7 a d \,x^{2}-2 b c \,x^{2}+5 b d \right ) \left (c \,x^{2}+d \right )}{35 d^{2} x^{4}}\) | \(48\) |
default | \(-\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (7 a d \,x^{2}-2 b c \,x^{2}+5 b d \right ) \left (c \,x^{2}+d \right )}{35 d^{2} x^{4}}\) | \(48\) |
orering | \(-\frac {\left (7 a d \,x^{2}-2 b c \,x^{2}+5 b d \right ) \left (c \,x^{2}+d \right ) \left (a +\frac {b}{x^{2}}\right ) \left (c +\frac {d}{x^{2}}\right )^{\frac {3}{2}}}{35 d^{2} \left (a \,x^{2}+b \right ) x^{2}}\) | \(60\) |
risch | \(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (7 a \,c^{2} d \,x^{6}-2 b \,c^{3} x^{6}+14 a c \,d^{2} x^{4}+b \,c^{2} d \,x^{4}+7 a \,d^{3} x^{2}+8 b c \,d^{2} x^{2}+5 b \,d^{3}\right )}{35 x^{6} d^{2}}\) | \(86\) |
trager | \(-\frac {\left (7 a \,c^{2} d \,x^{6}-2 b \,c^{3} x^{6}+14 a c \,d^{2} x^{4}+b \,c^{2} d \,x^{4}+7 a \,d^{3} x^{2}+8 b c \,d^{2} x^{2}+5 b \,d^{3}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{35 x^{6} d^{2}}\) | \(90\) |
Input:
int((a+b/x^2)*(c+d/x^2)^(3/2)/x^3,x,method=_RETURNVERBOSE)
Output:
-1/35*((c*x^2+d)/x^2)^(3/2)*(7*a*d*x^2-2*b*c*x^2+5*b*d)*(c*x^2+d)/d^2/x^4
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (38) = 76\).
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.83 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx=\frac {{\left ({\left (2 \, b c^{3} - 7 \, a c^{2} d\right )} x^{6} - {\left (b c^{2} d + 14 \, a c d^{2}\right )} x^{4} - 5 \, b d^{3} - {\left (8 \, b c d^{2} + 7 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{35 \, d^{2} x^{6}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^3,x, algorithm="fricas")
Output:
1/35*((2*b*c^3 - 7*a*c^2*d)*x^6 - (b*c^2*d + 14*a*c*d^2)*x^4 - 5*b*d^3 - ( 8*b*c*d^2 + 7*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2)/(d^2*x^6)
Time = 4.65 (sec) , antiderivative size = 189, normalized size of antiderivative = 4.11 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx=- \frac {a c \left (\begin {cases} \frac {2 \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3 d} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{x^{2}} & \text {otherwise} \end {cases}\right )}{2} - \frac {a d \left (\begin {cases} \frac {2 \left (- \frac {c \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{4}} & \text {otherwise} \end {cases}\right )}{2} - \frac {b c \left (\begin {cases} \frac {2 \left (- \frac {c \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{4}} & \text {otherwise} \end {cases}\right )}{2} - \frac {b d \left (\begin {cases} \frac {2 \left (\frac {c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{6}} & \text {otherwise} \end {cases}\right )}{2} \] Input:
integrate((a+b/x**2)*(c+d/x**2)**(3/2)/x**3,x)
Output:
-a*c*Piecewise((2*(c + d/x**2)**(3/2)/(3*d), Ne(d, 0)), (sqrt(c)/x**2, Tru e))/2 - a*d*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 - b*c*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 - b*d*Piecewise((2*(c**2*(c + d/x**2)**(3/2)/3 - 2*c*(c + d/ x**2)**(5/2)/5 + (c + d/x**2)**(7/2)/7)/d**3, Ne(d, 0)), (sqrt(c)/(3*x**6) , True))/2
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx=-\frac {a {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{5 \, d} - \frac {1}{35} \, {\left (\frac {5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}}}{d^{2}} - \frac {7 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c}{d^{2}}\right )} b \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^3,x, algorithm="maxima")
Output:
-1/5*a*(c + d/x^2)^(5/2)/d - 1/35*(5*(c + d/x^2)^(7/2)/d^2 - 7*(c + d/x^2) ^(5/2)*c/d^2)*b
Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (38) = 76\).
Time = 1.21 (sec) , antiderivative size = 370, normalized size of antiderivative = 8.04 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx=\frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{12} a c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} b c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} a c^{\frac {5}{2}} d \mathrm {sgn}\left (x\right ) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} b c^{\frac {7}{2}} d \mathrm {sgn}\left (x\right ) + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {5}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} b c^{\frac {7}{2}} d^{2} \mathrm {sgn}\left (x\right ) - 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {5}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 28 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {7}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 77 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {5}{2}} d^{4} \mathrm {sgn}\left (x\right ) + 14 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {7}{2}} d^{4} \mathrm {sgn}\left (x\right ) - 14 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {5}{2}} d^{5} \mathrm {sgn}\left (x\right ) - 2 \, b c^{\frac {7}{2}} d^{5} \mathrm {sgn}\left (x\right ) + 7 \, a c^{\frac {5}{2}} d^{6} \mathrm {sgn}\left (x\right )\right )}}{35 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{7}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^3,x, algorithm="giac")
Output:
2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + d))^12*a*c^(5/2)*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + d))^10*b*c^(7/2)*sgn(x) - 70*(sqrt(c)*x - sqrt(c*x^2 + d)) ^10*a*c^(5/2)*d*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + d))^8*b*c^(7/2)*d*sg n(x) + 105*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(5/2)*d^2*sgn(x) + 140*(sqr t(c)*x - sqrt(c*x^2 + d))^6*b*c^(7/2)*d^2*sgn(x) - 140*(sqrt(c)*x - sqrt(c *x^2 + d))^6*a*c^(5/2)*d^3*sgn(x) + 28*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c ^(7/2)*d^3*sgn(x) + 77*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(5/2)*d^4*sgn(x ) + 14*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(7/2)*d^4*sgn(x) - 14*(sqrt(c)* x - sqrt(c*x^2 + d))^2*a*c^(5/2)*d^5*sgn(x) - 2*b*c^(7/2)*d^5*sgn(x) + 7*a *c^(5/2)*d^6*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^7
Time = 4.50 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.65 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx=\frac {2\,b\,c^3\,\sqrt {c+\frac {d}{x^2}}}{35\,d^2}-\frac {a\,c^2\,\sqrt {c+\frac {d}{x^2}}}{5\,d}-\frac {2\,a\,c\,\sqrt {c+\frac {d}{x^2}}}{5\,x^2}-\frac {a\,d\,\sqrt {c+\frac {d}{x^2}}}{5\,x^4}-\frac {8\,b\,c\,\sqrt {c+\frac {d}{x^2}}}{35\,x^4}-\frac {b\,d\,\sqrt {c+\frac {d}{x^2}}}{7\,x^6}-\frac {b\,c^2\,\sqrt {c+\frac {d}{x^2}}}{35\,d\,x^2} \] Input:
int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^3,x)
Output:
(2*b*c^3*(c + d/x^2)^(1/2))/(35*d^2) - (a*c^2*(c + d/x^2)^(1/2))/(5*d) - ( 2*a*c*(c + d/x^2)^(1/2))/(5*x^2) - (a*d*(c + d/x^2)^(1/2))/(5*x^4) - (8*b* c*(c + d/x^2)^(1/2))/(35*x^4) - (b*d*(c + d/x^2)^(1/2))/(7*x^6) - (b*c^2*( c + d/x^2)^(1/2))/(35*d*x^2)
Time = 0.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 3.30 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^3} \, dx=\frac {-7 \sqrt {c \,x^{2}+d}\, a \,c^{2} d \,x^{6}-14 \sqrt {c \,x^{2}+d}\, a c \,d^{2} x^{4}-7 \sqrt {c \,x^{2}+d}\, a \,d^{3} x^{2}+2 \sqrt {c \,x^{2}+d}\, b \,c^{3} x^{6}-\sqrt {c \,x^{2}+d}\, b \,c^{2} d \,x^{4}-8 \sqrt {c \,x^{2}+d}\, b c \,d^{2} x^{2}-5 \sqrt {c \,x^{2}+d}\, b \,d^{3}-3 \sqrt {c}\, a \,c^{2} d \,x^{7}-2 \sqrt {c}\, b \,c^{3} x^{7}}{35 d^{2} x^{7}} \] Input:
int((a+b/x^2)*(c+d/x^2)^(3/2)/x^3,x)
Output:
( - 7*sqrt(c*x**2 + d)*a*c**2*d*x**6 - 14*sqrt(c*x**2 + d)*a*c*d**2*x**4 - 7*sqrt(c*x**2 + d)*a*d**3*x**2 + 2*sqrt(c*x**2 + d)*b*c**3*x**6 - sqrt(c* x**2 + d)*b*c**2*d*x**4 - 8*sqrt(c*x**2 + d)*b*c*d**2*x**2 - 5*sqrt(c*x**2 + d)*b*d**3 - 3*sqrt(c)*a*c**2*d*x**7 - 2*sqrt(c)*b*c**3*x**7)/(35*d**2*x **7)