Integrand size = 22, antiderivative size = 117 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^8 \, dx=\frac {8 d^2 (3 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^3}{315 c^4}-\frac {4 d (3 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^5}{105 c^3}+\frac {(3 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^7}{21 c^2}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^9}{9 c} \] Output:
8/315*d^2*(-2*a*d+3*b*c)*(c+d/x^2)^(3/2)*x^3/c^4-4/105*d*(-2*a*d+3*b*c)*(c +d/x^2)^(3/2)*x^5/c^3+1/21*(-2*a*d+3*b*c)*(c+d/x^2)^(3/2)*x^7/c^2+1/9*a*(c +d/x^2)^(3/2)*x^9/c
Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^8 \, dx=\frac {\sqrt {c+\frac {d}{x^2}} x \left (d+c x^2\right ) \left (24 b c d^2-16 a d^3-36 b c^2 d x^2+24 a c d^2 x^2+45 b c^3 x^4-30 a c^2 d x^4+35 a c^3 x^6\right )}{315 c^4} \] Input:
Integrate[(a + b/x^2)*Sqrt[c + d/x^2]*x^8,x]
Output:
(Sqrt[c + d/x^2]*x*(d + c*x^2)*(24*b*c*d^2 - 16*a*d^3 - 36*b*c^2*d*x^2 + 2 4*a*c*d^2*x^2 + 45*b*c^3*x^4 - 30*a*c^2*d*x^4 + 35*a*c^3*x^6))/(315*c^4)
Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {955, 803, 803, 796}
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.
\(\displaystyle \int x^8 \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle \frac {(3 b c-2 a d) \int \sqrt {c+\frac {d}{x^2}} x^6dx}{3 c}+\frac {a x^9 \left (c+\frac {d}{x^2}\right )^{3/2}}{9 c}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}-\frac {4 d \int \sqrt {c+\frac {d}{x^2}} x^4dx}{7 c}\right )}{3 c}+\frac {a x^9 \left (c+\frac {d}{x^2}\right )^{3/2}}{9 c}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {(3 b c-2 a d) \left (\frac {x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}-\frac {4 d \left (\frac {x^5 \left (c+\frac {d}{x^2}\right )^{3/2}}{5 c}-\frac {2 d \int \sqrt {c+\frac {d}{x^2}} x^2dx}{5 c}\right )}{7 c}\right )}{3 c}+\frac {a x^9 \left (c+\frac {d}{x^2}\right )^{3/2}}{9 c}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {\left (\frac {x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}-\frac {4 d \left (\frac {x^5 \left (c+\frac {d}{x^2}\right )^{3/2}}{5 c}-\frac {2 d x^3 \left (c+\frac {d}{x^2}\right )^{3/2}}{15 c^2}\right )}{7 c}\right ) (3 b c-2 a d)}{3 c}+\frac {a x^9 \left (c+\frac {d}{x^2}\right )^{3/2}}{9 c}\) |
Input:
Int[(a + b/x^2)*Sqrt[c + d/x^2]*x^8,x]
Output:
(a*(c + d/x^2)^(3/2)*x^9)/(9*c) + ((3*b*c - 2*a*d)*(((c + d/x^2)^(3/2)*x^7 )/(7*c) - (4*d*((-2*d*(c + d/x^2)^(3/2)*x^3)/(15*c^2) + ((c + d/x^2)^(3/2) *x^5)/(5*c)))/(7*c)))/(3*c)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.76
method | result | size |
gosper | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (35 a \,x^{6} c^{3}-30 a \,c^{2} d \,x^{4}+45 b \,c^{3} x^{4}+24 a c \,d^{2} x^{2}-36 b \,c^{2} d \,x^{2}-16 a \,d^{3}+24 b c \,d^{2}\right ) \left (c \,x^{2}+d \right )}{315 c^{4}}\) | \(89\) |
default | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (35 a \,x^{6} c^{3}-30 a \,c^{2} d \,x^{4}+45 b \,c^{3} x^{4}+24 a c \,d^{2} x^{2}-36 b \,c^{2} d \,x^{2}-16 a \,d^{3}+24 b c \,d^{2}\right ) \left (c \,x^{2}+d \right )}{315 c^{4}}\) | \(89\) |
orering | \(\frac {\left (35 a \,x^{6} c^{3}-30 a \,c^{2} d \,x^{4}+45 b \,c^{3} x^{4}+24 a c \,d^{2} x^{2}-36 b \,c^{2} d \,x^{2}-16 a \,d^{3}+24 b c \,d^{2}\right ) \left (c \,x^{2}+d \right ) x^{3} \left (a +\frac {b}{x^{2}}\right ) \sqrt {c +\frac {d}{x^{2}}}}{315 c^{4} \left (a \,x^{2}+b \right )}\) | \(103\) |
risch | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (35 a \,x^{8} c^{4}+5 a \,c^{3} d \,x^{6}+45 b \,c^{4} x^{6}-6 a \,c^{2} d^{2} x^{4}+9 b \,c^{3} d \,x^{4}+8 a c \,d^{3} x^{2}-12 b \,c^{2} d^{2} x^{2}-16 a \,d^{4}+24 b c \,d^{3}\right )}{315 c^{4}}\) | \(106\) |
trager | \(\frac {\left (35 a \,x^{8} c^{4}+5 a \,c^{3} d \,x^{6}+45 b \,c^{4} x^{6}-6 a \,c^{2} d^{2} x^{4}+9 b \,c^{3} d \,x^{4}+8 a c \,d^{3} x^{2}-12 b \,c^{2} d^{2} x^{2}-16 a \,d^{4}+24 b c \,d^{3}\right ) x \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{315 c^{4}}\) | \(110\) |
Input:
int((a+b/x^2)*(c+d/x^2)^(1/2)*x^8,x,method=_RETURNVERBOSE)
Output:
1/315*((c*x^2+d)/x^2)^(1/2)*x*(35*a*c^3*x^6-30*a*c^2*d*x^4+45*b*c^3*x^4+24 *a*c*d^2*x^2-36*b*c^2*d*x^2-16*a*d^3+24*b*c*d^2)*(c*x^2+d)/c^4
Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^8 \, dx=\frac {{\left (35 \, a c^{4} x^{9} + 5 \, {\left (9 \, b c^{4} + a c^{3} d\right )} x^{7} + 3 \, {\left (3 \, b c^{3} d - 2 \, a c^{2} d^{2}\right )} x^{5} - 4 \, {\left (3 \, b c^{2} d^{2} - 2 \, a c d^{3}\right )} x^{3} + 8 \, {\left (3 \, b c d^{3} - 2 \, a d^{4}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{315 \, c^{4}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)*x^8,x, algorithm="fricas")
Output:
1/315*(35*a*c^4*x^9 + 5*(9*b*c^4 + a*c^3*d)*x^7 + 3*(3*b*c^3*d - 2*a*c^2*d ^2)*x^5 - 4*(3*b*c^2*d^2 - 2*a*c*d^3)*x^3 + 8*(3*b*c*d^3 - 2*a*d^4)*x)*sqr t((c*x^2 + d)/x^2)/c^4
Leaf count of result is larger than twice the leaf count of optimal. 910 vs. \(2 (112) = 224\).
Time = 3.71 (sec) , antiderivative size = 910, normalized size of antiderivative = 7.78 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^8 \, dx =\text {Too large to display} \] Input:
integrate((a+b/x**2)*(c+d/x**2)**(1/2)*x**8,x)
Output:
35*a*c**7*d**(19/2)*x**14*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c** 6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 110*a*c**6*d**(21/2 )*x**12*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945 *c**5*d**11*x**2 + 315*c**4*d**12) + 114*a*c**5*d**(23/2)*x**10*sqrt(c*x** 2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 40*a*c**4*d**(25/2)*x**8*sqrt(c*x**2/d + 1)/(315*c**7*d **9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) - 5 *a*c**3*d**(27/2)*x**6*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d **10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) - 30*a*c**2*d**(29/2)*x* *4*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5 *d**11*x**2 + 315*c**4*d**12) - 40*a*c*d**(31/2)*x**2*sqrt(c*x**2/d + 1)/( 315*c**7*d**9*x**6 + 945*c**6*d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4* d**12) - 16*a*d**(33/2)*sqrt(c*x**2/d + 1)/(315*c**7*d**9*x**6 + 945*c**6* d**10*x**4 + 945*c**5*d**11*x**2 + 315*c**4*d**12) + 15*b*c**5*d**(9/2)*x* *10*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3 *d**6) + 33*b*c**4*d**(11/2)*x**8*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 17*b*c**3*d**(13/2)*x**6*sqrt(c*x** 2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 3*b*c **2*d**(15/2)*x**4*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5* x**2 + 105*c**3*d**6) + 12*b*c*d**(17/2)*x**2*sqrt(c*x**2/d + 1)/(105*c...
Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^8 \, dx=\frac {{\left (15 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} x^{7} - 42 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d x^{5} + 35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} x^{3}\right )} b}{105 \, c^{3}} + \frac {{\left (35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {9}{2}} x^{9} - 135 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} d x^{7} + 189 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d^{2} x^{5} - 105 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{3} x^{3}\right )} a}{315 \, c^{4}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)*x^8,x, algorithm="maxima")
Output:
1/105*(15*(c + d/x^2)^(7/2)*x^7 - 42*(c + d/x^2)^(5/2)*d*x^5 + 35*(c + d/x ^2)^(3/2)*d^2*x^3)*b/c^3 + 1/315*(35*(c + d/x^2)^(9/2)*x^9 - 135*(c + d/x^ 2)^(7/2)*d*x^7 + 189*(c + d/x^2)^(5/2)*d^2*x^5 - 105*(c + d/x^2)^(3/2)*d^3 *x^3)*a/c^4
Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.20 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^8 \, dx=-\frac {8 \, {\left (3 \, b c d^{\frac {7}{2}} - 2 \, a d^{\frac {9}{2}}\right )} \mathrm {sgn}\left (x\right )}{315 \, c^{4}} + \frac {35 \, {\left (c x^{2} + d\right )}^{\frac {9}{2}} a \mathrm {sgn}\left (x\right ) + 45 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} b c \mathrm {sgn}\left (x\right ) - 135 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} a d \mathrm {sgn}\left (x\right ) - 126 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c d \mathrm {sgn}\left (x\right ) + 189 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a d^{2} \mathrm {sgn}\left (x\right ) + 105 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c d^{2} \mathrm {sgn}\left (x\right ) - 105 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a d^{3} \mathrm {sgn}\left (x\right )}{315 \, c^{4}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(1/2)*x^8,x, algorithm="giac")
Output:
-8/315*(3*b*c*d^(7/2) - 2*a*d^(9/2))*sgn(x)/c^4 + 1/315*(35*(c*x^2 + d)^(9 /2)*a*sgn(x) + 45*(c*x^2 + d)^(7/2)*b*c*sgn(x) - 135*(c*x^2 + d)^(7/2)*a*d *sgn(x) - 126*(c*x^2 + d)^(5/2)*b*c*d*sgn(x) + 189*(c*x^2 + d)^(5/2)*a*d^2 *sgn(x) + 105*(c*x^2 + d)^(3/2)*b*c*d^2*sgn(x) - 105*(c*x^2 + d)^(3/2)*a*d ^3*sgn(x))/c^4
Time = 4.54 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.83 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^8 \, dx=\sqrt {c+\frac {d}{x^2}}\,\left (\frac {a\,x^9}{9}-\frac {x\,\left (16\,a\,d^4-24\,b\,c\,d^3\right )}{315\,c^4}+\frac {x^7\,\left (45\,b\,c^4+5\,a\,d\,c^3\right )}{315\,c^4}-\frac {d\,x^5\,\left (2\,a\,d-3\,b\,c\right )}{105\,c^2}+\frac {4\,d^2\,x^3\,\left (2\,a\,d-3\,b\,c\right )}{315\,c^3}\right ) \] Input:
int(x^8*(a + b/x^2)*(c + d/x^2)^(1/2),x)
Output:
(c + d/x^2)^(1/2)*((a*x^9)/9 - (x*(16*a*d^4 - 24*b*c*d^3))/(315*c^4) + (x^ 7*(45*b*c^4 + 5*a*c^3*d))/(315*c^4) - (d*x^5*(2*a*d - 3*b*c))/(105*c^2) + (4*d^2*x^3*(2*a*d - 3*b*c))/(315*c^3))
Time = 0.18 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^8 \, dx=\frac {\sqrt {c \,x^{2}+d}\, \left (35 a \,c^{4} x^{8}+5 a \,c^{3} d \,x^{6}+45 b \,c^{4} x^{6}-6 a \,c^{2} d^{2} x^{4}+9 b \,c^{3} d \,x^{4}+8 a c \,d^{3} x^{2}-12 b \,c^{2} d^{2} x^{2}-16 a \,d^{4}+24 b c \,d^{3}\right )}{315 c^{4}} \] Input:
int((a+b/x^2)*(c+d/x^2)^(1/2)*x^8,x)
Output:
(sqrt(c*x**2 + d)*(35*a*c**4*x**8 + 5*a*c**3*d*x**6 - 6*a*c**2*d**2*x**4 + 8*a*c*d**3*x**2 - 16*a*d**4 + 45*b*c**4*x**6 + 9*b*c**3*d*x**4 - 12*b*c** 2*d**2*x**2 + 24*b*c*d**3))/(315*c**4)