Integrand size = 22, antiderivative size = 84 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{10}} \, dx=-\frac {A \left (a+b x^2\right )^{5/2}}{9 a x^9}+\frac {(4 A b-9 a B) \left (a+b x^2\right )^{5/2}}{63 a^2 x^7}-\frac {2 b (4 A b-9 a B) \left (a+b x^2\right )^{5/2}}{315 a^3 x^5} \]
-1/9*A*(b*x^2+a)^(5/2)/a/x^9+1/63*(4*A*b-9*B*a)*(b*x^2+a)^(5/2)/a^2/x^7-2/ 315*b*(4*A*b-9*B*a)*(b*x^2+a)^(5/2)/a^3/x^5
Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{10}} \, dx=\frac {\left (a+b x^2\right )^{5/2} \left (-35 a^2 A+20 a A b x^2-45 a^2 B x^2-8 A b^2 x^4+18 a b B x^4\right )}{315 a^3 x^9} \]
((a + b*x^2)^(5/2)*(-35*a^2*A + 20*a*A*b*x^2 - 45*a^2*B*x^2 - 8*A*b^2*x^4 + 18*a*b*B*x^4))/(315*a^3*x^9)
Time = 0.19 (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 = {359, 245, 242}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{10}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(4 A b-9 a B) \int \frac {\left (b x^2+a\right )^{3/2}}{x^8}dx}{9 a}-\frac {A \left (a+b x^2\right )^{5/2}}{9 a x^9}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {(4 A b-9 a B) \left (-\frac {2 b \int \frac {\left (b x^2+a\right )^{3/2}}{x^6}dx}{7 a}-\frac {\left (a+b x^2\right )^{5/2}}{7 a x^7}\right )}{9 a}-\frac {A \left (a+b x^2\right )^{5/2}}{9 a x^9}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle -\frac {\left (\frac {2 b \left (a+b x^2\right )^{5/2}}{35 a^2 x^5}-\frac {\left (a+b x^2\right )^{5/2}}{7 a x^7}\right ) (4 A b-9 a B)}{9 a}-\frac {A \left (a+b x^2\right )^{5/2}}{9 a x^9}\) |
-1/9*(A*(a + b*x^2)^(5/2))/(a*x^9) - ((4*A*b - 9*a*B)*(-1/7*(a + b*x^2)^(5 /2)/(a*x^7) + (2*b*(a + b*x^2)^(5/2))/(35*a^2*x^5)))/(9*a)
3.6.36.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Time = 2.85 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(-\frac {\left (\left (\frac {9 x^{2} B}{7}+A \right ) a^{2}-\frac {4 x^{2} b \left (\frac {9 x^{2} B}{10}+A \right ) a}{7}+\frac {8 A \,b^{2} x^{4}}{35}\right ) \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{9 x^{9} a^{3}}\) | \(55\) |
gosper | \(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (8 A \,b^{2} x^{4}-18 B a b \,x^{4}-20 a A b \,x^{2}+45 a^{2} B \,x^{2}+35 a^{2} A \right )}{315 x^{9} a^{3}}\) | \(59\) |
default | \(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 a \,x^{7}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 a^{2} x^{5}}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{9 a \,x^{9}}-\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 a \,x^{7}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 a^{2} x^{5}}\right )}{9 a}\right )\) | \(102\) |
trager | \(-\frac {\left (8 A \,b^{4} x^{8}-18 B a \,b^{3} x^{8}-4 A a \,b^{3} x^{6}+9 B \,a^{2} b^{2} x^{6}+3 A \,a^{2} b^{2} x^{4}+72 B \,a^{3} b \,x^{4}+50 A \,a^{3} b \,x^{2}+45 B \,a^{4} x^{2}+35 A \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{315 x^{9} a^{3}}\) | \(107\) |
risch | \(-\frac {\left (8 A \,b^{4} x^{8}-18 B a \,b^{3} x^{8}-4 A a \,b^{3} x^{6}+9 B \,a^{2} b^{2} x^{6}+3 A \,a^{2} b^{2} x^{4}+72 B \,a^{3} b \,x^{4}+50 A \,a^{3} b \,x^{2}+45 B \,a^{4} x^{2}+35 A \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{315 x^{9} a^{3}}\) | \(107\) |
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{10}} \, dx=\frac {{\left (2 \, {\left (9 \, B a b^{3} - 4 \, A b^{4}\right )} x^{8} - {\left (9 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{6} - 35 \, A a^{4} - 3 \, {\left (24 \, B a^{3} b + A a^{2} b^{2}\right )} x^{4} - 5 \, {\left (9 \, B a^{4} + 10 \, A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{315 \, a^{3} x^{9}} \]
1/315*(2*(9*B*a*b^3 - 4*A*b^4)*x^8 - (9*B*a^2*b^2 - 4*A*a*b^3)*x^6 - 35*A* a^4 - 3*(24*B*a^3*b + A*a^2*b^2)*x^4 - 5*(9*B*a^4 + 10*A*a^3*b)*x^2)*sqrt( b*x^2 + a)/(a^3*x^9)
Leaf count of result is larger than twice the leaf count of optimal. 1408 vs. \(2 (78) = 156\).
Time = 3.17 (sec) , antiderivative size = 1408, normalized size of antiderivative = 16.76 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{10}} \, dx=\text {Too large to display} \]
-35*A*a**8*b**(19/2)*sqrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a**6*b **10*x**10 + 945*a**5*b**11*x**12 + 315*a**4*b**12*x**14) - 110*A*a**7*b** (21/2)*x**2*sqrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a**6*b**10*x**1 0 + 945*a**5*b**11*x**12 + 315*a**4*b**12*x**14) - 114*A*a**6*b**(23/2)*x* *4*sqrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a**6*b**10*x**10 + 945*a **5*b**11*x**12 + 315*a**4*b**12*x**14) - 40*A*a**5*b**(25/2)*x**6*sqrt(a/ (b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a**6*b**10*x**10 + 945*a**5*b**11* x**12 + 315*a**4*b**12*x**14) - 15*A*a**5*b**(11/2)*sqrt(a/(b*x**2) + 1)/( 105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10) + 5*A*a**4* b**(27/2)*x**8*sqrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a**6*b**10*x **10 + 945*a**5*b**11*x**12 + 315*a**4*b**12*x**14) - 33*A*a**4*b**(13/2)* x**2*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a **3*b**6*x**10) + 30*A*a**3*b**(29/2)*x**10*sqrt(a/(b*x**2) + 1)/(315*a**7 *b**9*x**8 + 945*a**6*b**10*x**10 + 945*a**5*b**11*x**12 + 315*a**4*b**12* x**14) - 17*A*a**3*b**(15/2)*x**4*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10) + 40*A*a**2*b**(31/2)*x**12*s qrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a**6*b**10*x**10 + 945*a**5* b**11*x**12 + 315*a**4*b**12*x**14) - 3*A*a**2*b**(17/2)*x**6*sqrt(a/(b*x* *2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10) + 16*A*a*b**(33/2)*x**14*sqrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*...
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{10}} \, dx=\frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b}{35 \, a^{2} x^{5}} - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2}}{315 \, a^{3} x^{5}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{7 \, a x^{7}} + \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{63 \, a^{2} x^{7}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{9 \, a x^{9}} \]
2/35*(b*x^2 + a)^(5/2)*B*b/(a^2*x^5) - 8/315*(b*x^2 + a)^(5/2)*A*b^2/(a^3* x^5) - 1/7*(b*x^2 + a)^(5/2)*B/(a*x^7) + 4/63*(b*x^2 + a)^(5/2)*A*b/(a^2*x ^7) - 1/9*(b*x^2 + a)^(5/2)*A/(a*x^9)
Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (72) = 144\).
Time = 0.32 (sec) , antiderivative size = 400, normalized size of antiderivative = 4.76 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{10}} \, dx=\frac {4 \, {\left (315 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{14} B b^{\frac {7}{2}} - 315 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} B a b^{\frac {7}{2}} + 840 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} A b^{\frac {9}{2}} + 315 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} B a^{2} b^{\frac {7}{2}} + 1260 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} A a b^{\frac {9}{2}} - 819 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B a^{3} b^{\frac {7}{2}} + 1764 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} A a^{2} b^{\frac {9}{2}} + 441 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a^{4} b^{\frac {7}{2}} + 504 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} A a^{3} b^{\frac {9}{2}} - 9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{5} b^{\frac {7}{2}} + 144 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{4} b^{\frac {9}{2}} + 81 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{6} b^{\frac {7}{2}} - 36 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{5} b^{\frac {9}{2}} - 9 \, B a^{7} b^{\frac {7}{2}} + 4 \, A a^{6} b^{\frac {9}{2}}\right )}}{315 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{9}} \]
4/315*(315*(sqrt(b)*x - sqrt(b*x^2 + a))^14*B*b^(7/2) - 315*(sqrt(b)*x - s qrt(b*x^2 + a))^12*B*a*b^(7/2) + 840*(sqrt(b)*x - sqrt(b*x^2 + a))^12*A*b^ (9/2) + 315*(sqrt(b)*x - sqrt(b*x^2 + a))^10*B*a^2*b^(7/2) + 1260*(sqrt(b) *x - sqrt(b*x^2 + a))^10*A*a*b^(9/2) - 819*(sqrt(b)*x - sqrt(b*x^2 + a))^8 *B*a^3*b^(7/2) + 1764*(sqrt(b)*x - sqrt(b*x^2 + a))^8*A*a^2*b^(9/2) + 441* (sqrt(b)*x - sqrt(b*x^2 + a))^6*B*a^4*b^(7/2) + 504*(sqrt(b)*x - sqrt(b*x^ 2 + a))^6*A*a^3*b^(9/2) - 9*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^5*b^(7/2) + 144*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^4*b^(9/2) + 81*(sqrt(b)*x - sqrt (b*x^2 + a))^2*B*a^6*b^(7/2) - 36*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^5*b^ (9/2) - 9*B*a^7*b^(7/2) + 4*A*a^6*b^(9/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^ 2 - a)^9
Time = 6.47 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.02 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^{10}} \, dx=\frac {4\,A\,b^3\,\sqrt {b\,x^2+a}}{315\,a^2\,x^3}-\frac {10\,A\,b\,\sqrt {b\,x^2+a}}{63\,x^7}-\frac {B\,a\,\sqrt {b\,x^2+a}}{7\,x^7}-\frac {8\,B\,b\,\sqrt {b\,x^2+a}}{35\,x^5}-\frac {A\,b^2\,\sqrt {b\,x^2+a}}{105\,a\,x^5}-\frac {A\,a\,\sqrt {b\,x^2+a}}{9\,x^9}-\frac {8\,A\,b^4\,\sqrt {b\,x^2+a}}{315\,a^3\,x}-\frac {B\,b^2\,\sqrt {b\,x^2+a}}{35\,a\,x^3}+\frac {2\,B\,b^3\,\sqrt {b\,x^2+a}}{35\,a^2\,x} \]
(4*A*b^3*(a + b*x^2)^(1/2))/(315*a^2*x^3) - (10*A*b*(a + b*x^2)^(1/2))/(63 *x^7) - (B*a*(a + b*x^2)^(1/2))/(7*x^7) - (8*B*b*(a + b*x^2)^(1/2))/(35*x^ 5) - (A*b^2*(a + b*x^2)^(1/2))/(105*a*x^5) - (A*a*(a + b*x^2)^(1/2))/(9*x^ 9) - (8*A*b^4*(a + b*x^2)^(1/2))/(315*a^3*x) - (B*b^2*(a + b*x^2)^(1/2))/( 35*a*x^3) + (2*B*b^3*(a + b*x^2)^(1/2))/(35*a^2*x)