Integrand size = 22, antiderivative size = 74 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx=-\frac {c (b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^3}+\frac {(2 b c-a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^3} \] Output:
-1/3*c*(-a*d+b*c)*(c+d/x^2)^(3/2)/d^3+1/5*(-a*d+2*b*c)*(c+d/x^2)^(5/2)/d^3 -1/7*b*(c+d/x^2)^(7/2)/d^3
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (d+c x^2\right ) \left (-15 b d^2+12 b c d x^2-21 a d^2 x^2-8 b c^2 x^4+14 a c d x^4\right )}{105 d^3 x^6} \] Input:
Integrate[((a + b/x^2)*Sqrt[c + d/x^2])/x^5,x]
Output:
(Sqrt[c + d/x^2]*(d + c*x^2)*(-15*b*d^2 + 12*b*c*d*x^2 - 21*a*d^2*x^2 - 8* b*c^2*x^4 + 14*a*c*d*x^4))/(105*d^3*x^6)
Time = 0.35 (sec) , antiderivative size = 78, 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.
\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle -\frac {1}{2} \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^2}d\frac {1}{x^2}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{d^2}+\frac {(a d-2 b c) \left (c+\frac {d}{x^2}\right )^{3/2}}{d^2}+\frac {c (b c-a d) \sqrt {c+\frac {d}{x^2}}}{d^2}\right )d\frac {1}{x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {2 \left (c+\frac {d}{x^2}\right )^{5/2} (2 b c-a d)}{5 d^3}-\frac {2 c \left (c+\frac {d}{x^2}\right )^{3/2} (b c-a d)}{3 d^3}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^3}\right )\) |
Input:
Int[((a + b/x^2)*Sqrt[c + d/x^2])/x^5,x]
Output:
((-2*c*(b*c - a*d)*(c + d/x^2)^(3/2))/(3*d^3) + (2*(2*b*c - a*d)*(c + d/x^ 2)^(5/2))/(5*d^3) - (2*b*(c + d/x^2)^(7/2))/(7*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 = 70, normalized size of antiderivative = 0.95
method | result | size |
gosper | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (14 a c d \,x^{4}-8 b \,c^{2} x^{4}-21 a \,d^{2} x^{2}+12 b c d \,x^{2}-15 b \,d^{2}\right ) \left (c \,x^{2}+d \right )}{105 d^{3} x^{6}}\) | \(70\) |
default | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (14 a c d \,x^{4}-8 b \,c^{2} x^{4}-21 a \,d^{2} x^{2}+12 b c d \,x^{2}-15 b \,d^{2}\right ) \left (c \,x^{2}+d \right )}{105 d^{3} x^{6}}\) | \(70\) |
orering | \(\frac {\left (14 a c d \,x^{4}-8 b \,c^{2} x^{4}-21 a \,d^{2} x^{2}+12 b c d \,x^{2}-15 b \,d^{2}\right ) \left (c \,x^{2}+d \right ) \left (a +\frac {b}{x^{2}}\right ) \sqrt {c +\frac {d}{x^{2}}}}{105 d^{3} \left (a \,x^{2}+b \right ) x^{4}}\) | \(82\) |
risch | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (14 a \,c^{2} d \,x^{6}-8 b \,c^{3} x^{6}-7 a c \,d^{2} x^{4}+4 b \,c^{2} d \,x^{4}-21 a \,d^{3} x^{2}-3 b c \,d^{2} x^{2}-15 b \,d^{3}\right )}{105 x^{6} d^{3}}\) | \(87\) |
trager | \(\frac {\left (14 a \,c^{2} d \,x^{6}-8 b \,c^{3} x^{6}-7 a c \,d^{2} x^{4}+4 b \,c^{2} d \,x^{4}-21 a \,d^{3} x^{2}-3 b c \,d^{2} x^{2}-15 b \,d^{3}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{105 x^{6} d^{3}}\) | \(91\) |
Input:
int((a+b/x^2)*(c+d/x^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)
Output:
1/105*((c*x^2+d)/x^2)^(1/2)*(14*a*c*d*x^4-8*b*c^2*x^4-21*a*d^2*x^2+12*b*c* d*x^2-15*b*d^2)*(c*x^2+d)/d^3/x^6
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx=-\frac {{\left (2 \, {\left (4 \, b c^{3} - 7 \, a c^{2} d\right )} x^{6} - {\left (4 \, b c^{2} d - 7 \, a c d^{2}\right )} x^{4} + 15 \, b d^{3} + 3 \, {\left (b c d^{2} + 7 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{105 \, d^{3} x^{6}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^5,x, algorithm="fricas")
Output:
-1/105*(2*(4*b*c^3 - 7*a*c^2*d)*x^6 - (4*b*c^2*d - 7*a*c*d^2)*x^4 + 15*b*d ^3 + 3*(b*c*d^2 + 7*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2)/(d^3*x^6)
Time = 2.72 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx=- \frac {a \left (\begin {cases} \frac {2 \left (- \frac {c \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5}\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{2 x^{4}} & \text {otherwise} \end {cases}\right )}{2} - \frac {b \left (\begin {cases} \frac {2 \left (\frac {c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{6}} & \text {otherwise} \end {cases}\right )}{2} \] Input:
integrate((a+b/x**2)*(c+d/x**2)**(1/2)/x**5,x)
Output:
-a*Piecewise((2*(-c*(c + d/x**2)**(3/2)/3 + (c + d/x**2)**(5/2)/5)/d**2, N e(d, 0)), (sqrt(c)/(2*x**4), True))/2 - b*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 = 84, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx=-\frac {1}{105} \, b {\left (\frac {15 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}}}{d^{3}} - \frac {42 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c}{d^{3}} + \frac {35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2}}{d^{3}}\right )} - \frac {1}{15} \, a {\left (\frac {3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}}}{d^{2}} - \frac {5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c}{d^{2}}\right )} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^5,x, algorithm="maxima")
Output:
-1/105*b*(15*(c + d/x^2)^(7/2)/d^3 - 42*(c + d/x^2)^(5/2)*c/d^3 + 35*(c + d/x^2)^(3/2)*c^2/d^3) - 1/15*a*(3*(c + d/x^2)^(5/2)/d^2 - 5*(c + d/x^2)^(3 /2)*c/d^2)
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (62) = 124\).
Time = 0.96 (sec) , antiderivative size = 310, normalized size of antiderivative = 4.19 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx=\frac {4 \, {\left (105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} a c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 280 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} b c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 175 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {5}{2}} d \mathrm {sgn}\left (x\right ) + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} b c^{\frac {7}{2}} d \mathrm {sgn}\left (x\right ) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {5}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 84 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {7}{2}} d^{2} \mathrm {sgn}\left (x\right ) - 42 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {5}{2}} d^{3} \mathrm {sgn}\left (x\right ) - 28 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {7}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 49 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {5}{2}} d^{4} \mathrm {sgn}\left (x\right ) + 4 \, b c^{\frac {7}{2}} d^{4} \mathrm {sgn}\left (x\right ) - 7 \, a c^{\frac {5}{2}} d^{5} \mathrm {sgn}\left (x\right )\right )}}{105 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{7}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^5,x, algorithm="giac")
Output:
4/105*(105*(sqrt(c)*x - sqrt(c*x^2 + d))^10*a*c^(5/2)*sgn(x) + 280*(sqrt(c )*x - sqrt(c*x^2 + d))^8*b*c^(7/2)*sgn(x) - 175*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(5/2)*d*sgn(x) + 140*(sqrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(7/2)*d *sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + d))^6*a*c^(5/2)*d^2*sgn(x) + 84*(sq rt(c)*x - sqrt(c*x^2 + d))^4*b*c^(7/2)*d^2*sgn(x) - 42*(sqrt(c)*x - sqrt(c *x^2 + d))^4*a*c^(5/2)*d^3*sgn(x) - 28*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c ^(7/2)*d^3*sgn(x) + 49*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*c^(5/2)*d^4*sgn(x ) + 4*b*c^(7/2)*d^4*sgn(x) - 7*a*c^(5/2)*d^5*sgn(x))/((sqrt(c)*x - sqrt(c* x^2 + d))^2 - d)^7
Time = 3.82 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.70 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx=\frac {2\,a\,c^2\,\sqrt {c+\frac {d}{x^2}}}{15\,d^2}-\frac {b\,\sqrt {c+\frac {d}{x^2}}}{7\,x^6}-\frac {a\,\sqrt {c+\frac {d}{x^2}}}{5\,x^4}-\frac {8\,b\,c^3\,\sqrt {c+\frac {d}{x^2}}}{105\,d^3}-\frac {a\,c\,\sqrt {c+\frac {d}{x^2}}}{15\,d\,x^2}-\frac {b\,c\,\sqrt {c+\frac {d}{x^2}}}{35\,d\,x^4}+\frac {4\,b\,c^2\,\sqrt {c+\frac {d}{x^2}}}{105\,d^2\,x^2} \] Input:
int(((a + b/x^2)*(c + d/x^2)^(1/2))/x^5,x)
Output:
(2*a*c^2*(c + d/x^2)^(1/2))/(15*d^2) - (b*(c + d/x^2)^(1/2))/(7*x^6) - (a* (c + d/x^2)^(1/2))/(5*x^4) - (8*b*c^3*(c + d/x^2)^(1/2))/(105*d^3) - (a*c* (c + d/x^2)^(1/2))/(15*d*x^2) - (b*c*(c + d/x^2)^(1/2))/(35*d*x^4) + (4*b* c^2*(c + d/x^2)^(1/2))/(105*d^2*x^2)
Time = 0.20 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.05 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^5} \, dx=\frac {14 \sqrt {c \,x^{2}+d}\, a \,c^{2} d \,x^{6}-7 \sqrt {c \,x^{2}+d}\, a c \,d^{2} x^{4}-21 \sqrt {c \,x^{2}+d}\, a \,d^{3} x^{2}-8 \sqrt {c \,x^{2}+d}\, b \,c^{3} x^{6}+4 \sqrt {c \,x^{2}+d}\, b \,c^{2} d \,x^{4}-3 \sqrt {c \,x^{2}+d}\, b c \,d^{2} x^{2}-15 \sqrt {c \,x^{2}+d}\, b \,d^{3}-14 \sqrt {c}\, a \,c^{2} d \,x^{7}+8 \sqrt {c}\, b \,c^{3} x^{7}}{105 d^{3} x^{7}} \] Input:
int((a+b/x^2)*(c+d/x^2)^(1/2)/x^5,x)
Output:
(14*sqrt(c*x**2 + d)*a*c**2*d*x**6 - 7*sqrt(c*x**2 + d)*a*c*d**2*x**4 - 21 *sqrt(c*x**2 + d)*a*d**3*x**2 - 8*sqrt(c*x**2 + d)*b*c**3*x**6 + 4*sqrt(c* x**2 + d)*b*c**2*d*x**4 - 3*sqrt(c*x**2 + d)*b*c*d**2*x**2 - 15*sqrt(c*x** 2 + d)*b*d**3 - 14*sqrt(c)*a*c**2*d*x**7 + 8*sqrt(c)*b*c**3*x**7)/(105*d** 3*x**7)