Integrand size = 24, antiderivative size = 99 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^8} \, dx=-\frac {a^2 \left (c+d x^2\right )^{3/2}}{7 c x^7}-\frac {2 a (7 b c-2 a d) \left (c+d x^2\right )^{3/2}}{35 c^2 x^5}-\frac {\left (35 b^2 c^2-4 a d (7 b c-2 a d)\right ) \left (c+d x^2\right )^{3/2}}{105 c^3 x^3} \]
-1/7*a^2*(d*x^2+c)^(3/2)/c/x^7-2/35*a*(-2*a*d+7*b*c)*(d*x^2+c)^(3/2)/c^2/x ^5-1/105*(35*b^2*c^2-4*a*d*(-2*a*d+7*b*c))*(d*x^2+c)^(3/2)/c^3/x^3
Time = 0.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^8} \, dx=-\frac {\left (c+d x^2\right )^{3/2} \left (35 b^2 c^2 x^4+14 a b c x^2 \left (3 c-2 d x^2\right )+a^2 \left (15 c^2-12 c d x^2+8 d^2 x^4\right )\right )}{105 c^3 x^7} \]
-1/105*((c + d*x^2)^(3/2)*(35*b^2*c^2*x^4 + 14*a*b*c*x^2*(3*c - 2*d*x^2) + a^2*(15*c^2 - 12*c*d*x^2 + 8*d^2*x^4)))/(c^3*x^7)
Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {365, 359, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^8} \, dx\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\int \frac {\left (7 b^2 c x^2+2 a (7 b c-2 a d)\right ) \sqrt {d x^2+c}}{x^6}dx}{7 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{7 c x^7}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {\left (35 b^2 c^2-4 a d (7 b c-2 a d)\right ) \int \frac {\sqrt {d x^2+c}}{x^4}dx}{5 c}-\frac {2 a \left (c+d x^2\right )^{3/2} (7 b c-2 a d)}{5 c x^5}}{7 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{7 c x^7}\) |
| \(\Big \downarrow \) 242 |
| \(\displaystyle \frac {-\frac {\left (c+d x^2\right )^{3/2} \left (35 b^2 c^2-4 a d (7 b c-2 a d)\right )}{15 c^2 x^3}-\frac {2 a \left (c+d x^2\right )^{3/2} (7 b c-2 a d)}{5 c x^5}}{7 c}-\frac {a^2 \left (c+d x^2\right )^{3/2}}{7 c x^7}\) |
-1/7*(a^2*(c + d*x^2)^(3/2))/(c*x^7) + ((-2*a*(7*b*c - 2*a*d)*(c + d*x^2)^ (3/2))/(5*c*x^5) - ((35*b^2*c^2 - 4*a*d*(7*b*c - 2*a*d))*(c + d*x^2)^(3/2) )/(15*c^2*x^3))/(7*c)
3.7.10.3.1 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[((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]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Time = 2.88 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70
| method | result | size |
| pseudoelliptic | \(-\frac {\left (\left (\frac {7}{3} b^{2} x^{4}+\frac {14}{5} a b \,x^{2}+a^{2}\right ) c^{2}-\frac {4 x^{2} \left (\frac {7 b \,x^{2}}{3}+a \right ) d a c}{5}+\frac {8 a^{2} d^{2} x^{4}}{15}\right ) \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{7 x^{7} c^{3}}\) | \(69\) |
| gosper | \(-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} \left (8 a^{2} d^{2} x^{4}-28 x^{4} a b c d +35 b^{2} c^{2} x^{4}-12 a^{2} c d \,x^{2}+42 a b \,c^{2} x^{2}+15 a^{2} c^{2}\right )}{105 x^{7} c^{3}}\) | \(78\) |
| trager | \(-\frac {\left (8 a^{2} d^{3} x^{6}-28 x^{6} d^{2} a b c +35 b^{2} c^{2} d \,x^{6}-4 a^{2} c \,d^{2} x^{4}+14 a b \,c^{2} d \,x^{4}+35 b^{2} c^{3} x^{4}+3 a^{2} c^{2} d \,x^{2}+42 a b \,c^{3} x^{2}+15 a^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{105 x^{7} c^{3}}\) | \(117\) |
| risch | \(-\frac {\left (8 a^{2} d^{3} x^{6}-28 x^{6} d^{2} a b c +35 b^{2} c^{2} d \,x^{6}-4 a^{2} c \,d^{2} x^{4}+14 a b \,c^{2} d \,x^{4}+35 b^{2} c^{3} x^{4}+3 a^{2} c^{2} d \,x^{2}+42 a b \,c^{3} x^{2}+15 a^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{105 x^{7} c^{3}}\) | \(117\) |
| default | \(a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{7 c \,x^{7}}-\frac {4 d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 c \,x^{5}}+\frac {2 d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 c^{2} x^{3}}\right )}{7 c}\right )-\frac {b^{2} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 c \,x^{3}}+2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{5 c \,x^{5}}+\frac {2 d \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{15 c^{2} x^{3}}\right )\) | \(126\) |
-1/7*((7/3*b^2*x^4+14/5*a*b*x^2+a^2)*c^2-4/5*x^2*(7/3*b*x^2+a)*d*a*c+8/15* a^2*d^2*x^4)*(d*x^2+c)^(3/2)/x^7/c^3
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^8} \, dx=-\frac {{\left ({\left (35 \, b^{2} c^{2} d - 28 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{6} + 15 \, a^{2} c^{3} + {\left (35 \, b^{2} c^{3} + 14 \, a b c^{2} d - 4 \, a^{2} c d^{2}\right )} x^{4} + 3 \, {\left (14 \, a b c^{3} + a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{105 \, c^{3} x^{7}} \]
-1/105*((35*b^2*c^2*d - 28*a*b*c*d^2 + 8*a^2*d^3)*x^6 + 15*a^2*c^3 + (35*b ^2*c^3 + 14*a*b*c^2*d - 4*a^2*c*d^2)*x^4 + 3*(14*a*b*c^3 + a^2*c^2*d)*x^2) *sqrt(d*x^2 + c)/(c^3*x^7)
Leaf count of result is larger than twice the leaf count of optimal. 510 vs. \(2 (95) = 190\).
Time = 1.74 (sec) , antiderivative size = 510, normalized size of antiderivative = 5.15 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^8} \, dx=- \frac {15 a^{2} c^{5} d^{\frac {9}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{105 c^{5} d^{4} x^{6} + 210 c^{4} d^{5} x^{8} + 105 c^{3} d^{6} x^{10}} - \frac {33 a^{2} c^{4} d^{\frac {11}{2}} x^{2} \sqrt {\frac {c}{d x^{2}} + 1}}{105 c^{5} d^{4} x^{6} + 210 c^{4} d^{5} x^{8} + 105 c^{3} d^{6} x^{10}} - \frac {17 a^{2} c^{3} d^{\frac {13}{2}} x^{4} \sqrt {\frac {c}{d x^{2}} + 1}}{105 c^{5} d^{4} x^{6} + 210 c^{4} d^{5} x^{8} + 105 c^{3} d^{6} x^{10}} - \frac {3 a^{2} c^{2} d^{\frac {15}{2}} x^{6} \sqrt {\frac {c}{d x^{2}} + 1}}{105 c^{5} d^{4} x^{6} + 210 c^{4} d^{5} x^{8} + 105 c^{3} d^{6} x^{10}} - \frac {12 a^{2} c d^{\frac {17}{2}} x^{8} \sqrt {\frac {c}{d x^{2}} + 1}}{105 c^{5} d^{4} x^{6} + 210 c^{4} d^{5} x^{8} + 105 c^{3} d^{6} x^{10}} - \frac {8 a^{2} d^{\frac {19}{2}} x^{10} \sqrt {\frac {c}{d x^{2}} + 1}}{105 c^{5} d^{4} x^{6} + 210 c^{4} d^{5} x^{8} + 105 c^{3} d^{6} x^{10}} - \frac {2 a b \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{5 x^{4}} - \frac {2 a b d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{15 c x^{2}} + \frac {4 a b d^{\frac {5}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{15 c^{2}} - \frac {b^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{3 x^{2}} - \frac {b^{2} d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{3 c} \]
-15*a**2*c**5*d**(9/2)*sqrt(c/(d*x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4 *d**5*x**8 + 105*c**3*d**6*x**10) - 33*a**2*c**4*d**(11/2)*x**2*sqrt(c/(d* x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4*d**5*x**8 + 105*c**3*d**6*x**10) - 17*a**2*c**3*d**(13/2)*x**4*sqrt(c/(d*x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4*d**5*x**8 + 105*c**3*d**6*x**10) - 3*a**2*c**2*d**(15/2)*x**6*sqr t(c/(d*x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4*d**5*x**8 + 105*c**3*d**6 *x**10) - 12*a**2*c*d**(17/2)*x**8*sqrt(c/(d*x**2) + 1)/(105*c**5*d**4*x** 6 + 210*c**4*d**5*x**8 + 105*c**3*d**6*x**10) - 8*a**2*d**(19/2)*x**10*sqr t(c/(d*x**2) + 1)/(105*c**5*d**4*x**6 + 210*c**4*d**5*x**8 + 105*c**3*d**6 *x**10) - 2*a*b*sqrt(d)*sqrt(c/(d*x**2) + 1)/(5*x**4) - 2*a*b*d**(3/2)*sqr t(c/(d*x**2) + 1)/(15*c*x**2) + 4*a*b*d**(5/2)*sqrt(c/(d*x**2) + 1)/(15*c* *2) - b**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/(3*x**2) - b**2*d**(3/2)*sqrt(c/(d *x**2) + 1)/(3*c)
Time = 0.21 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.25 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^8} \, dx=-\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2}}{3 \, c x^{3}} + \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d}{15 \, c^{2} x^{3}} - \frac {8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2}}{105 \, c^{3} x^{3}} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b}{5 \, c x^{5}} + \frac {4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d}{35 \, c^{2} x^{5}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2}}{7 \, c x^{7}} \]
-1/3*(d*x^2 + c)^(3/2)*b^2/(c*x^3) + 4/15*(d*x^2 + c)^(3/2)*a*b*d/(c^2*x^3 ) - 8/105*(d*x^2 + c)^(3/2)*a^2*d^2/(c^3*x^3) - 2/5*(d*x^2 + c)^(3/2)*a*b/ (c*x^5) + 4/35*(d*x^2 + c)^(3/2)*a^2*d/(c^2*x^5) - 1/7*(d*x^2 + c)^(3/2)*a ^2/(c*x^7)
Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (87) = 174\).
Time = 0.31 (sec) , antiderivative size = 490, normalized size of antiderivative = 4.95 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^8} \, dx=\frac {2 \, {\left (105 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{12} b^{2} d^{\frac {3}{2}} - 420 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{10} b^{2} c d^{\frac {3}{2}} + 420 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{10} a b d^{\frac {5}{2}} + 665 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} b^{2} c^{2} d^{\frac {3}{2}} - 700 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a b c d^{\frac {5}{2}} + 560 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{8} a^{2} d^{\frac {7}{2}} - 560 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} b^{2} c^{3} d^{\frac {3}{2}} + 280 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a b c^{2} d^{\frac {5}{2}} + 280 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{6} a^{2} c d^{\frac {7}{2}} + 315 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b^{2} c^{4} d^{\frac {3}{2}} - 168 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a b c^{3} d^{\frac {5}{2}} + 168 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} a^{2} c^{2} d^{\frac {7}{2}} - 140 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{5} d^{\frac {3}{2}} + 196 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c^{4} d^{\frac {5}{2}} - 56 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} c^{3} d^{\frac {7}{2}} + 35 \, b^{2} c^{6} d^{\frac {3}{2}} - 28 \, a b c^{5} d^{\frac {5}{2}} + 8 \, a^{2} c^{4} d^{\frac {7}{2}}\right )}}{105 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c\right )}^{7}} \]
2/105*(105*(sqrt(d)*x - sqrt(d*x^2 + c))^12*b^2*d^(3/2) - 420*(sqrt(d)*x - sqrt(d*x^2 + c))^10*b^2*c*d^(3/2) + 420*(sqrt(d)*x - sqrt(d*x^2 + c))^10* a*b*d^(5/2) + 665*(sqrt(d)*x - sqrt(d*x^2 + c))^8*b^2*c^2*d^(3/2) - 700*(s qrt(d)*x - sqrt(d*x^2 + c))^8*a*b*c*d^(5/2) + 560*(sqrt(d)*x - sqrt(d*x^2 + c))^8*a^2*d^(7/2) - 560*(sqrt(d)*x - sqrt(d*x^2 + c))^6*b^2*c^3*d^(3/2) + 280*(sqrt(d)*x - sqrt(d*x^2 + c))^6*a*b*c^2*d^(5/2) + 280*(sqrt(d)*x - s qrt(d*x^2 + c))^6*a^2*c*d^(7/2) + 315*(sqrt(d)*x - sqrt(d*x^2 + c))^4*b^2* c^4*d^(3/2) - 168*(sqrt(d)*x - sqrt(d*x^2 + c))^4*a*b*c^3*d^(5/2) + 168*(s qrt(d)*x - sqrt(d*x^2 + c))^4*a^2*c^2*d^(7/2) - 140*(sqrt(d)*x - sqrt(d*x^ 2 + c))^2*b^2*c^5*d^(3/2) + 196*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c^4*d^ (5/2) - 56*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*c^3*d^(7/2) + 35*b^2*c^6*d^ (3/2) - 28*a*b*c^5*d^(5/2) + 8*a^2*c^4*d^(7/2))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^7
Time = 6.22 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.83 \[ \int \frac {\left (a+b x^2\right )^2 \sqrt {c+d x^2}}{x^8} \, dx=\frac {4\,a^2\,d^2\,\sqrt {d\,x^2+c}}{105\,c^2\,x^3}-\frac {b^2\,\sqrt {d\,x^2+c}}{3\,x^3}-\frac {2\,a\,b\,\sqrt {d\,x^2+c}}{5\,x^5}-\frac {a^2\,\sqrt {d\,x^2+c}}{7\,x^7}-\frac {8\,a^2\,d^3\,\sqrt {d\,x^2+c}}{105\,c^3\,x}-\frac {a^2\,d\,\sqrt {d\,x^2+c}}{35\,c\,x^5}-\frac {b^2\,d\,\sqrt {d\,x^2+c}}{3\,c\,x}+\frac {4\,a\,b\,d^2\,\sqrt {d\,x^2+c}}{15\,c^2\,x}-\frac {2\,a\,b\,d\,\sqrt {d\,x^2+c}}{15\,c\,x^3} \]
(4*a^2*d^2*(c + d*x^2)^(1/2))/(105*c^2*x^3) - (b^2*(c + d*x^2)^(1/2))/(3*x ^3) - (2*a*b*(c + d*x^2)^(1/2))/(5*x^5) - (a^2*(c + d*x^2)^(1/2))/(7*x^7) - (8*a^2*d^3*(c + d*x^2)^(1/2))/(105*c^3*x) - (a^2*d*(c + d*x^2)^(1/2))/(3 5*c*x^5) - (b^2*d*(c + d*x^2)^(1/2))/(3*c*x) + (4*a*b*d^2*(c + d*x^2)^(1/2 ))/(15*c^2*x) - (2*a*b*d*(c + d*x^2)^(1/2))/(15*c*x^3)