Integrand size = 22, antiderivative size = 84 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^6 \, dx=-\frac {2 d (7 b c-4 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^3}{105 c^3}+\frac {(7 b c-4 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^5}{35 c^2}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^7}{7 c} \]
-2/105*d*(-4*a*d+7*b*c)*(c+d/x^2)^(3/2)*x^3/c^3+1/35*(-4*a*d+7*b*c)*(c+d/x ^2)^(3/2)*x^5/c^2+1/7*a*(c+d/x^2)^(3/2)*x^7/c
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.76 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^6 \, dx=\frac {\sqrt {c+\frac {d}{x^2}} x \left (d+c x^2\right ) \left (-14 b c d+8 a d^2+21 b c^2 x^2-12 a c d x^2+15 a c^2 x^4\right )}{105 c^3} \]
(Sqrt[c + d/x^2]*x*(d + c*x^2)*(-14*b*c*d + 8*a*d^2 + 21*b*c^2*x^2 - 12*a* c*d*x^2 + 15*a*c^2*x^4))/(105*c^3)
Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 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.
\(\displaystyle \int x^6 \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle \frac {(7 b c-4 a d) \int \sqrt {c+\frac {d}{x^2}} x^4dx}{7 c}+\frac {a x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {(7 b c-4 a d) \left (\frac {x^5 \left (c+\frac {d}{x^2}\right )^{3/2}}{5 c}-\frac {2 d \int \sqrt {c+\frac {d}{x^2}} x^2dx}{5 c}\right )}{7 c}+\frac {a x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {\left (\frac {x^5 \left (c+\frac {d}{x^2}\right )^{3/2}}{5 c}-\frac {2 d x^3 \left (c+\frac {d}{x^2}\right )^{3/2}}{15 c^2}\right ) (7 b c-4 a d)}{7 c}+\frac {a x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c}\) |
(a*(c + d/x^2)^(3/2)*x^7)/(7*c) + ((7*b*c - 4*a*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)
3.10.39.3.1 Defintions of rubi rules used
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 = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77
method | result | size |
gosper | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (15 a \,x^{4} c^{2}-12 a c d \,x^{2}+21 b \,c^{2} x^{2}+8 a \,d^{2}-14 b c d \right ) \left (c \,x^{2}+d \right )}{105 c^{3}}\) | \(65\) |
default | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (15 a \,x^{4} c^{2}-12 a c d \,x^{2}+21 b \,c^{2} x^{2}+8 a \,d^{2}-14 b c d \right ) \left (c \,x^{2}+d \right )}{105 c^{3}}\) | \(65\) |
risch | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (15 x^{6} a \,c^{3}+3 a \,c^{2} d \,x^{4}+21 b \,c^{3} x^{4}-4 a c \,d^{2} x^{2}+7 b \,c^{2} d \,x^{2}+8 a \,d^{3}-14 b c \,d^{2}\right )}{105 c^{3}}\) | \(82\) |
trager | \(\frac {\left (15 x^{6} a \,c^{3}+3 a \,c^{2} d \,x^{4}+21 b \,c^{3} x^{4}-4 a c \,d^{2} x^{2}+7 b \,c^{2} d \,x^{2}+8 a \,d^{3}-14 b c \,d^{2}\right ) x \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{105 c^{3}}\) | \(86\) |
1/105*((c*x^2+d)/x^2)^(1/2)*x*(15*a*c^2*x^4-12*a*c*d*x^2+21*b*c^2*x^2+8*a* d^2-14*b*c*d)*(c*x^2+d)/c^3
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^6 \, dx=\frac {{\left (15 \, a c^{3} x^{7} + 3 \, {\left (7 \, b c^{3} + a c^{2} d\right )} x^{5} + {\left (7 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{3} - 2 \, {\left (7 \, b c d^{2} - 4 \, a d^{3}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{105 \, c^{3}} \]
1/105*(15*a*c^3*x^7 + 3*(7*b*c^3 + a*c^2*d)*x^5 + (7*b*c^2*d - 4*a*c*d^2)* x^3 - 2*(7*b*c*d^2 - 4*a*d^3)*x)*sqrt((c*x^2 + d)/x^2)/c^3
Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (78) = 156\).
Time = 1.65 (sec) , antiderivative size = 422, normalized size of antiderivative = 5.02 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^6 \, dx=\frac {15 a c^{5} d^{\frac {9}{2}} x^{10} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {33 a c^{4} d^{\frac {11}{2}} x^{8} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {17 a c^{3} d^{\frac {13}{2}} x^{6} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {3 a c^{2} d^{\frac {15}{2}} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {12 a c d^{\frac {17}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {8 a d^{\frac {19}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {b \sqrt {d} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {b d^{\frac {3}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c} - \frac {2 b d^{\frac {5}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{2}} \]
15*a*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*a*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*a*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 + 1 05*c**3*d**6) + 3*a*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*a*c*d**(17/2)*x**2*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) + 8* a*d**(19/2)*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) + b*sqrt(d)*x**4*sqrt(c*x**2/d + 1)/5 + b*d**(3/2)*x**2*sqr t(c*x**2/d + 1)/(15*c) - 2*b*d**(5/2)*sqrt(c*x**2/d + 1)/(15*c**2)
Time = 0.20 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^6 \, dx=\frac {{\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5} - 5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d x^{3}\right )} b}{15 \, c^{2}} + \frac {{\left (15 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} x^{7} - 42 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d x^{5} + 35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} x^{3}\right )} a}{105 \, c^{3}} \]
1/15*(3*(c + d/x^2)^(5/2)*x^5 - 5*(c + d/x^2)^(3/2)*d*x^3)*b/c^2 + 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)*a/c^3
Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.25 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^6 \, dx=\frac {2 \, {\left (7 \, b c d^{\frac {5}{2}} - 4 \, a d^{\frac {7}{2}}\right )} \mathrm {sgn}\left (x\right )}{105 \, c^{3}} + \frac {15 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} a \mathrm {sgn}\left (x\right ) + 21 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c \mathrm {sgn}\left (x\right ) - 42 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a d \mathrm {sgn}\left (x\right ) - 35 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c d \mathrm {sgn}\left (x\right ) + 35 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a d^{2} \mathrm {sgn}\left (x\right )}{105 \, c^{3}} \]
2/105*(7*b*c*d^(5/2) - 4*a*d^(7/2))*sgn(x)/c^3 + 1/105*(15*(c*x^2 + d)^(7/ 2)*a*sgn(x) + 21*(c*x^2 + d)^(5/2)*b*c*sgn(x) - 42*(c*x^2 + d)^(5/2)*a*d*s gn(x) - 35*(c*x^2 + d)^(3/2)*b*c*d*sgn(x) + 35*(c*x^2 + d)^(3/2)*a*d^2*sgn (x))/c^3
Time = 9.05 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^6 \, dx=\sqrt {c+\frac {d}{x^2}}\,\left (\frac {a\,x^7}{7}+\frac {x\,\left (8\,a\,d^3-14\,b\,c\,d^2\right )}{105\,c^3}+\frac {x^5\,\left (21\,b\,c^3+3\,a\,d\,c^2\right )}{105\,c^3}-\frac {d\,x^3\,\left (4\,a\,d-7\,b\,c\right )}{105\,c^2}\right ) \]