Integrand size = 22, antiderivative size = 53 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^6 \, dx=\frac {(7 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{5/2} x^5}{35 c^2}+\frac {a \left (c+\frac {d}{x^2}\right )^{5/2} x^7}{7 c} \] Output:
1/35*(-2*a*d+7*b*c)*(c+d/x^2)^(5/2)*x^5/c^2+1/7*a*(c+d/x^2)^(5/2)*x^7/c
Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^6 \, dx=\frac {\sqrt {c+\frac {d}{x^2}} x \left (d+c x^2\right )^2 \left (7 b c-2 a d+5 a c x^2\right )}{35 c^2} \] Input:
Integrate[(a + b/x^2)*(c + d/x^2)^(3/2)*x^6,x]
Output:
(Sqrt[c + d/x^2]*x*(d + c*x^2)^2*(7*b*c - 2*a*d + 5*a*c*x^2))/(35*c^2)
Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {955, 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.
\(\displaystyle \int x^6 \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle \frac {(7 b c-2 a d) \int \left (c+\frac {d}{x^2}\right )^{3/2} x^4dx}{7 c}+\frac {a x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {x^5 \left (c+\frac {d}{x^2}\right )^{5/2} (7 b c-2 a d)}{35 c^2}+\frac {a x^7 \left (c+\frac {d}{x^2}\right )^{5/2}}{7 c}\) |
Input:
Int[(a + b/x^2)*(c + d/x^2)^(3/2)*x^6,x]
Output:
((7*b*c - 2*a*d)*(c + d/x^2)^(5/2)*x^5)/(35*c^2) + (a*(c + d/x^2)^(5/2)*x^ 7)/(7*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[((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 = 45, normalized size of antiderivative = 0.85
method | result | size |
gosper | \(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} x^{3} \left (5 a c \,x^{2}-2 a d +7 c b \right ) \left (c \,x^{2}+d \right )}{35 c^{2}}\) | \(45\) |
default | \(\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} x^{3} \left (5 a c \,x^{2}-2 a d +7 c b \right ) \left (c \,x^{2}+d \right )}{35 c^{2}}\) | \(45\) |
orering | \(\frac {\left (5 a c \,x^{2}-2 a d +7 c b \right ) \left (c \,x^{2}+d \right ) x^{5} \left (a +\frac {b}{x^{2}}\right ) \left (c +\frac {d}{x^{2}}\right )^{\frac {3}{2}}}{35 c^{2} \left (a \,x^{2}+b \right )}\) | \(57\) |
risch | \(\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, x \left (5 a \,x^{6} c^{3}+8 a \,c^{2} d \,x^{4}+7 b \,c^{3} x^{4}+a c \,d^{2} x^{2}+14 b \,c^{2} d \,x^{2}-2 a \,d^{3}+7 b c \,d^{2}\right )}{35 c^{2}}\) | \(81\) |
trager | \(\frac {\left (5 a \,x^{6} c^{3}+8 a \,c^{2} d \,x^{4}+7 b \,c^{3} x^{4}+a c \,d^{2} x^{2}+14 b \,c^{2} d \,x^{2}-2 a \,d^{3}+7 b c \,d^{2}\right ) x \sqrt {-\frac {-c \,x^{2}-d}{x^{2}}}}{35 c^{2}}\) | \(85\) |
Input:
int((a+b/x^2)*(c+d/x^2)^(3/2)*x^6,x,method=_RETURNVERBOSE)
Output:
1/35*((c*x^2+d)/x^2)^(3/2)*x^3*(5*a*c*x^2-2*a*d+7*b*c)*(c*x^2+d)/c^2
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^6 \, dx=\frac {{\left (5 \, a c^{3} x^{7} + {\left (7 \, b c^{3} + 8 \, a c^{2} d\right )} x^{5} + {\left (14 \, b c^{2} d + a c d^{2}\right )} x^{3} + {\left (7 \, b c d^{2} - 2 \, a d^{3}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{35 \, c^{2}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)*x^6,x, algorithm="fricas")
Output:
1/35*(5*a*c^3*x^7 + (7*b*c^3 + 8*a*c^2*d)*x^5 + (14*b*c^2*d + a*c*d^2)*x^3 + (7*b*c*d^2 - 2*a*d^3)*x)*sqrt((c*x^2 + d)/x^2)/c^2
Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (46) = 92\).
Time = 3.18 (sec) , antiderivative size = 498, normalized size of antiderivative = 9.40 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^6 \, dx=\frac {15 a c^{6} d^{\frac {9}{2}} x^{10} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {33 a c^{5} d^{\frac {11}{2}} x^{8} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {17 a c^{4} d^{\frac {13}{2}} x^{6} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {3 a c^{3} d^{\frac {15}{2}} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {12 a c^{2} d^{\frac {17}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {8 a c d^{\frac {19}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {a d^{\frac {3}{2}} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {a d^{\frac {5}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c} - \frac {2 a d^{\frac {7}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{2}} + \frac {b c \sqrt {d} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {2 b d^{\frac {3}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {b d^{\frac {5}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{5 c} \] Input:
integrate((a+b/x**2)*(c+d/x**2)**(3/2)*x**6,x)
Output:
15*a*c**6*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**5*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**4*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**3*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**2*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*c*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) + a*d**(3/2)*x**4*sqrt(c*x**2/d + 1)/5 + a*d**(5/2)*x* *2*sqrt(c*x**2/d + 1)/(15*c) - 2*a*d**(7/2)*sqrt(c*x**2/d + 1)/(15*c**2) + b*c*sqrt(d)*x**4*sqrt(c*x**2/d + 1)/5 + 2*b*d**(3/2)*x**2*sqrt(c*x**2/d + 1)/5 + b*d**(5/2)*sqrt(c*x**2/d + 1)/(5*c)
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^6 \, dx=\frac {b {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5}}{5 \, c} + \frac {{\left (5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} x^{7} - 7 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d x^{5}\right )} a}{35 \, c^{2}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)*x^6,x, algorithm="maxima")
Output:
1/5*b*(c + d/x^2)^(5/2)*x^5/c + 1/35*(5*(c + d/x^2)^(7/2)*x^7 - 7*(c + d/x ^2)^(5/2)*d*x^5)*a/c^2
Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.36 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^6 \, dx=-\frac {{\left (7 \, b c d^{\frac {5}{2}} - 2 \, a d^{\frac {7}{2}}\right )} \mathrm {sgn}\left (x\right )}{35 \, c^{2}} + \frac {5 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} a \mathrm {sgn}\left (x\right ) + 7 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c \mathrm {sgn}\left (x\right ) - 7 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a d \mathrm {sgn}\left (x\right )}{35 \, c^{2}} \] Input:
integrate((a+b/x^2)*(c+d/x^2)^(3/2)*x^6,x, algorithm="giac")
Output:
-1/35*(7*b*c*d^(5/2) - 2*a*d^(7/2))*sgn(x)/c^2 + 1/35*(5*(c*x^2 + d)^(7/2) *a*sgn(x) + 7*(c*x^2 + d)^(5/2)*b*c*sgn(x) - 7*(c*x^2 + d)^(5/2)*a*d*sgn(x ))/c^2
Time = 3.91 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.45 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^6 \, dx=\sqrt {c+\frac {d}{x^2}}\,\left (\frac {x^5\,\left (7\,b\,c^3+8\,a\,d\,c^2\right )}{35\,c^2}-\frac {x\,\left (2\,a\,d^3-7\,b\,c\,d^2\right )}{35\,c^2}+\frac {a\,c\,x^7}{7}+\frac {d\,x^3\,\left (a\,d+14\,b\,c\right )}{35\,c}\right ) \] Input:
int(x^6*(a + b/x^2)*(c + d/x^2)^(3/2),x)
Output:
(c + d/x^2)^(1/2)*((x^5*(7*b*c^3 + 8*a*c^2*d))/(35*c^2) - (x*(2*a*d^3 - 7* b*c*d^2))/(35*c^2) + (a*c*x^7)/7 + (d*x^3*(a*d + 14*b*c))/(35*c))
Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40 \[ \int \left (a+\frac {b}{x^2}\right ) \left (c+\frac {d}{x^2}\right )^{3/2} x^6 \, dx=\frac {\sqrt {c \,x^{2}+d}\, \left (5 a \,c^{3} x^{6}+8 a \,c^{2} d \,x^{4}+7 b \,c^{3} x^{4}+a c \,d^{2} x^{2}+14 b \,c^{2} d \,x^{2}-2 a \,d^{3}+7 b c \,d^{2}\right )}{35 c^{2}} \] Input:
int((a+b/x^2)*(c+d/x^2)^(3/2)*x^6,x)
Output:
(sqrt(c*x**2 + d)*(5*a*c**3*x**6 + 8*a*c**2*d*x**4 + a*c*d**2*x**2 - 2*a*d **3 + 7*b*c**3*x**4 + 14*b*c**2*d*x**2 + 7*b*c*d**2))/(35*c**2)