\(\int \frac {(a+\frac {b}{x^2}) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx\) [143]

Optimal result

Integrand size = 22, antiderivative size = 104 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx=\frac {c^2 (b c-a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d^4}-\frac {c (3 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{5/2}}{5 d^4}+\frac {(3 b c-a d) \left (c+\frac {d}{x^2}\right )^{7/2}}{7 d^4}-\frac {b \left (c+\frac {d}{x^2}\right )^{9/2}}{9 d^4} \] Output:

1/3*c^2*(-a*d+b*c)*(c+d/x^2)^(3/2)/d^4-1/5*c*(-2*a*d+3*b*c)*(c+d/x^2)^(5/2 
)/d^4+1/7*(-a*d+3*b*c)*(c+d/x^2)^(7/2)/d^4-1/9*b*(c+d/x^2)^(9/2)/d^4
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx=\frac {\sqrt {c+\frac {d}{x^2}} \left (d+c x^2\right ) \left (-35 b d^3+30 b c d^2 x^2-45 a d^3 x^2-24 b c^2 d x^4+36 a c d^2 x^4+16 b c^3 x^6-24 a c^2 d x^6\right )}{315 d^4 x^8} \] Input:

Integrate[((a + b/x^2)*Sqrt[c + d/x^2])/x^7,x]
 

Output:

(Sqrt[c + d/x^2]*(d + c*x^2)*(-35*b*d^3 + 30*b*c*d^2*x^2 - 45*a*d^3*x^2 - 
24*b*c^2*d*x^4 + 36*a*c*d^2*x^4 + 16*b*c^3*x^6 - 24*a*c^2*d*x^6))/(315*d^4 
*x^8)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04, 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.

\(\displaystyle \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle -\frac {1}{2} \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^4}d\frac {1}{x^2}\)

\(\Big \downarrow \) 86

\(\displaystyle -\frac {1}{2} \int \left (\frac {b \left (c+\frac {d}{x^2}\right )^{7/2}}{d^3}+\frac {(a d-3 b c) \left (c+\frac {d}{x^2}\right )^{5/2}}{d^3}+\frac {c (3 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{d^3}-\frac {c^2 (b c-a d) \sqrt {c+\frac {d}{x^2}}}{d^3}\right )d\frac {1}{x^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {2 c^2 \left (c+\frac {d}{x^2}\right )^{3/2} (b c-a d)}{3 d^4}+\frac {2 \left (c+\frac {d}{x^2}\right )^{7/2} (3 b c-a d)}{7 d^4}-\frac {2 c \left (c+\frac {d}{x^2}\right )^{5/2} (3 b c-2 a d)}{5 d^4}-\frac {2 b \left (c+\frac {d}{x^2}\right )^{9/2}}{9 d^4}\right )\)

Input:

Int[((a + b/x^2)*Sqrt[c + d/x^2])/x^7,x]
 

Output:

((2*c^2*(b*c - a*d)*(c + d/x^2)^(3/2))/(3*d^4) - (2*c*(3*b*c - 2*a*d)*(c + 
 d/x^2)^(5/2))/(5*d^4) + (2*(3*b*c - a*d)*(c + d/x^2)^(7/2))/(7*d^4) - (2* 
b*(c + d/x^2)^(9/2))/(9*d^4))/2
 

Defintions of rubi rules used

rule 86

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]))))
 

rule 948

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]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90

method	result	size
gosper	\(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (24 a \,c^{2} d \,x^{6}-16 b \,c^{3} x^{6}-36 a c \,d^{2} x^{4}+24 b \,c^{2} d \,x^{4}+45 a \,d^{3} x^{2}-30 b c \,d^{2} x^{2}+35 b \,d^{3}\right ) \left (c \,x^{2}+d \right )}{315 d^{4} x^{8}}\)	\(94\)
default	\(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (24 a \,c^{2} d \,x^{6}-16 b \,c^{3} x^{6}-36 a c \,d^{2} x^{4}+24 b \,c^{2} d \,x^{4}+45 a \,d^{3} x^{2}-30 b c \,d^{2} x^{2}+35 b \,d^{3}\right ) \left (c \,x^{2}+d \right )}{315 d^{4} x^{8}}\)	\(94\)
orering	\(-\frac {\left (24 a \,c^{2} d \,x^{6}-16 b \,c^{3} x^{6}-36 a c \,d^{2} x^{4}+24 b \,c^{2} d \,x^{4}+45 a \,d^{3} x^{2}-30 b c \,d^{2} x^{2}+35 b \,d^{3}\right ) \left (c \,x^{2}+d \right ) \left (a +\frac {b}{x^{2}}\right ) \sqrt {c +\frac {d}{x^{2}}}}{315 d^{4} \left (a \,x^{2}+b \right ) x^{6}}\)	\(106\)
risch	\(-\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (24 a \,c^{3} d \,x^{8}-16 b \,c^{4} x^{8}-12 a \,c^{2} d^{2} x^{6}+8 b \,c^{3} d \,x^{6}+9 a c \,d^{3} x^{4}-6 b \,c^{2} d^{2} x^{4}+45 a \,d^{4} x^{2}+5 b c \,d^{3} x^{2}+35 b \,d^{4}\right )}{315 x^{8} d^{4}}\)	\(111\)
trager	\(-\frac {\left (24 a \,c^{3} d \,x^{8}-16 b \,c^{4} x^{8}-12 a \,c^{2} d^{2} x^{6}+8 b \,c^{3} d \,x^{6}+9 a c \,d^{3} x^{4}-6 b \,c^{2} d^{2} x^{4}+45 a \,d^{4} x^{2}+5 b c \,d^{3} x^{2}+35 b \,d^{4}\right ) \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{315 x^{8} d^{4}}\)	\(115\)

Input:

int((a+b/x^2)*(c+d/x^2)^(1/2)/x^7,x,method=_RETURNVERBOSE)
 

Output:

-1/315*((c*x^2+d)/x^2)^(1/2)*(24*a*c^2*d*x^6-16*b*c^3*x^6-36*a*c*d^2*x^4+2 
4*b*c^2*d*x^4+45*a*d^3*x^2-30*b*c*d^2*x^2+35*b*d^3)*(c*x^2+d)/d^4/x^8
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx=\frac {{\left (8 \, {\left (2 \, b c^{4} - 3 \, a c^{3} d\right )} x^{8} - 4 \, {\left (2 \, b c^{3} d - 3 \, a c^{2} d^{2}\right )} x^{6} - 35 \, b d^{4} + 3 \, {\left (2 \, b c^{2} d^{2} - 3 \, a c d^{3}\right )} x^{4} - 5 \, {\left (b c d^{3} + 9 \, a d^{4}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{315 \, d^{4} x^{8}} \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^7,x, algorithm="fricas")
 

Output:

1/315*(8*(2*b*c^4 - 3*a*c^3*d)*x^8 - 4*(2*b*c^3*d - 3*a*c^2*d^2)*x^6 - 35* 
b*d^4 + 3*(2*b*c^2*d^2 - 3*a*c*d^3)*x^4 - 5*(b*c*d^3 + 9*a*d^4)*x^2)*sqrt( 
(c*x^2 + d)/x^2)/(d^4*x^8)
 

Sympy [A] (verification not implemented)

Time = 2.23 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.38 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx=- \frac {a \left (\begin {cases} \frac {2 \left (\frac {c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} - \frac {2 c \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7}\right )}{d^{3}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{3 x^{6}} & \text {otherwise} \end {cases}\right )}{2} - \frac {b \left (\begin {cases} \frac {2 \left (- \frac {c^{3} \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3} + \frac {3 c^{2} \left (c + \frac {d}{x^{2}}\right )^{\frac {5}{2}}}{5} - \frac {3 c \left (c + \frac {d}{x^{2}}\right )^{\frac {7}{2}}}{7} + \frac {\left (c + \frac {d}{x^{2}}\right )^{\frac {9}{2}}}{9}\right )}{d^{4}} & \text {for}\: d \neq 0 \\\frac {\sqrt {c}}{4 x^{8}} & \text {otherwise} \end {cases}\right )}{2} \] Input:

integrate((a+b/x**2)*(c+d/x**2)**(1/2)/x**7,x)
 

Output:

-a*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 - b*Pi 
ecewise((2*(-c**3*(c + d/x**2)**(3/2)/3 + 3*c**2*(c + d/x**2)**(5/2)/5 - 3 
*c*(c + d/x**2)**(7/2)/7 + (c + d/x**2)**(9/2)/9)/d**4, Ne(d, 0)), (sqrt(c 
)/(4*x**8), True))/2
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.13 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx=-\frac {1}{315} \, b {\left (\frac {35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}}}{d^{4}} - \frac {135 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} c}{d^{4}} + \frac {189 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{2}}{d^{4}} - \frac {105 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{3}}{d^{4}}\right )} - \frac {1}{105} \, a {\left (\frac {15 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}}}{d^{3}} - \frac {42 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c}{d^{3}} + \frac {35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2}}{d^{3}}\right )} \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^7,x, algorithm="maxima")
 

Output:

-1/315*b*(35*(c + d/x^2)^(9/2)/d^4 - 135*(c + d/x^2)^(7/2)*c/d^4 + 189*(c 
+ d/x^2)^(5/2)*c^2/d^4 - 105*(c + d/x^2)^(3/2)*c^3/d^4) - 1/105*a*(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)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (88) = 176\).

Time = 0.95 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.56 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx=\frac {16 \, {\left (210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{12} a c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 630 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} b c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{10} a c^{\frac {7}{2}} d \mathrm {sgn}\left (x\right ) + 378 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} b c^{\frac {9}{2}} d \mathrm {sgn}\left (x\right ) + 63 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{8} a c^{\frac {7}{2}} d^{2} \mathrm {sgn}\left (x\right ) + 168 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} b c^{\frac {9}{2}} d^{2} \mathrm {sgn}\left (x\right ) - 42 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{6} a c^{\frac {7}{2}} d^{3} \mathrm {sgn}\left (x\right ) - 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {9}{2}} d^{3} \mathrm {sgn}\left (x\right ) + 108 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a c^{\frac {7}{2}} d^{4} \mathrm {sgn}\left (x\right ) + 18 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} b c^{\frac {9}{2}} d^{4} \mathrm {sgn}\left (x\right ) - 27 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a c^{\frac {7}{2}} d^{5} \mathrm {sgn}\left (x\right ) - 2 \, b c^{\frac {9}{2}} d^{5} \mathrm {sgn}\left (x\right ) + 3 \, a c^{\frac {7}{2}} d^{6} \mathrm {sgn}\left (x\right )\right )}}{315 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{9}} \] Input:

integrate((a+b/x^2)*(c+d/x^2)^(1/2)/x^7,x, algorithm="giac")
 

Output:

16/315*(210*(sqrt(c)*x - sqrt(c*x^2 + d))^12*a*c^(7/2)*sgn(x) + 630*(sqrt( 
c)*x - sqrt(c*x^2 + d))^10*b*c^(9/2)*sgn(x) - 315*(sqrt(c)*x - sqrt(c*x^2 
+ d))^10*a*c^(7/2)*d*sgn(x) + 378*(sqrt(c)*x - sqrt(c*x^2 + d))^8*b*c^(9/2 
)*d*sgn(x) + 63*(sqrt(c)*x - sqrt(c*x^2 + d))^8*a*c^(7/2)*d^2*sgn(x) + 168 
*(sqrt(c)*x - sqrt(c*x^2 + d))^6*b*c^(9/2)*d^2*sgn(x) - 42*(sqrt(c)*x - sq 
rt(c*x^2 + d))^6*a*c^(7/2)*d^3*sgn(x) - 72*(sqrt(c)*x - sqrt(c*x^2 + d))^4 
*b*c^(9/2)*d^3*sgn(x) + 108*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*c^(7/2)*d^4* 
sgn(x) + 18*(sqrt(c)*x - sqrt(c*x^2 + d))^2*b*c^(9/2)*d^4*sgn(x) - 27*(sqr 
t(c)*x - sqrt(c*x^2 + d))^2*a*c^(7/2)*d^5*sgn(x) - 2*b*c^(9/2)*d^5*sgn(x) 
+ 3*a*c^(7/2)*d^6*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^9
 

Mupad [B] (verification not implemented)

Time = 4.18 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx=\frac {16\,b\,c^4\,\sqrt {c+\frac {d}{x^2}}}{315\,d^4}-\frac {b\,\sqrt {c+\frac {d}{x^2}}}{9\,x^8}-\frac {8\,a\,c^3\,\sqrt {c+\frac {d}{x^2}}}{105\,d^3}-\frac {a\,\sqrt {c+\frac {d}{x^2}}}{7\,x^6}-\frac {a\,c\,\sqrt {c+\frac {d}{x^2}}}{35\,d\,x^4}-\frac {b\,c\,\sqrt {c+\frac {d}{x^2}}}{63\,d\,x^6}+\frac {4\,a\,c^2\,\sqrt {c+\frac {d}{x^2}}}{105\,d^2\,x^2}+\frac {2\,b\,c^2\,\sqrt {c+\frac {d}{x^2}}}{105\,d^2\,x^4}-\frac {8\,b\,c^3\,\sqrt {c+\frac {d}{x^2}}}{315\,d^3\,x^2} \] Input:

int(((a + b/x^2)*(c + d/x^2)^(1/2))/x^7,x)
 

Output:

(16*b*c^4*(c + d/x^2)^(1/2))/(315*d^4) - (b*(c + d/x^2)^(1/2))/(9*x^8) - ( 
8*a*c^3*(c + d/x^2)^(1/2))/(105*d^3) - (a*(c + d/x^2)^(1/2))/(7*x^6) - (a* 
c*(c + d/x^2)^(1/2))/(35*d*x^4) - (b*c*(c + d/x^2)^(1/2))/(63*d*x^6) + (4* 
a*c^2*(c + d/x^2)^(1/2))/(105*d^2*x^2) + (2*b*c^2*(c + d/x^2)^(1/2))/(105* 
d^2*x^4) - (8*b*c^3*(c + d/x^2)^(1/2))/(315*d^3*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}}}{x^7} \, dx=\frac {-24 \sqrt {c \,x^{2}+d}\, a \,c^{3} d \,x^{8}+12 \sqrt {c \,x^{2}+d}\, a \,c^{2} d^{2} x^{6}-9 \sqrt {c \,x^{2}+d}\, a c \,d^{3} x^{4}-45 \sqrt {c \,x^{2}+d}\, a \,d^{4} x^{2}+16 \sqrt {c \,x^{2}+d}\, b \,c^{4} x^{8}-8 \sqrt {c \,x^{2}+d}\, b \,c^{3} d \,x^{6}+6 \sqrt {c \,x^{2}+d}\, b \,c^{2} d^{2} x^{4}-5 \sqrt {c \,x^{2}+d}\, b c \,d^{3} x^{2}-35 \sqrt {c \,x^{2}+d}\, b \,d^{4}+24 \sqrt {c}\, a \,c^{3} d \,x^{9}-16 \sqrt {c}\, b \,c^{4} x^{9}}{315 d^{4} x^{9}} \] Input:

int((a+b/x^2)*(c+d/x^2)^(1/2)/x^7,x)
 

Output:

( - 24*sqrt(c*x**2 + d)*a*c**3*d*x**8 + 12*sqrt(c*x**2 + d)*a*c**2*d**2*x* 
*6 - 9*sqrt(c*x**2 + d)*a*c*d**3*x**4 - 45*sqrt(c*x**2 + d)*a*d**4*x**2 + 
16*sqrt(c*x**2 + d)*b*c**4*x**8 - 8*sqrt(c*x**2 + d)*b*c**3*d*x**6 + 6*sqr 
t(c*x**2 + d)*b*c**2*d**2*x**4 - 5*sqrt(c*x**2 + d)*b*c*d**3*x**2 - 35*sqr 
t(c*x**2 + d)*b*d**4 + 24*sqrt(c)*a*c**3*d*x**9 - 16*sqrt(c)*b*c**4*x**9)/ 
(315*d**4*x**9)