Integrand size = 26, antiderivative size = 96 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^9} \, dx=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}-\frac {(7 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{35 b^2 x^8}+\frac {2 c (7 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^6} \]
-1/7*A*(c*x^4+b*x^2)^(3/2)/b/x^10-1/35*(-4*A*c+7*B*b)*(c*x^4+b*x^2)^(3/2)/ b^2/x^8+2/105*c*(-4*A*c+7*B*b)*(c*x^4+b*x^2)^(3/2)/b^3/x^6
Time = 0.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^9} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (7 b B x^2 \left (-3 b+2 c x^2\right )+A \left (-15 b^2+12 b c x^2-8 c^2 x^4\right )\right )}{105 b^3 x^{10}} \]
((x^2*(b + c*x^2))^(3/2)*(7*b*B*x^2*(-3*b + 2*c*x^2) + A*(-15*b^2 + 12*b*c *x^2 - 8*c^2*x^4)))/(105*b^3*x^10)
Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1940, 1220, 1129, 1123}
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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^9} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \sqrt {c x^4+b x^2}}{x^{10}}dx^2\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {1}{2} \left (\frac {(7 b B-4 A c) \int \frac {\sqrt {c x^4+b x^2}}{x^8}dx^2}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )\) |
\(\Big \downarrow \) 1129 |
\(\displaystyle \frac {1}{2} \left (\frac {(7 b B-4 A c) \left (-\frac {2 c \int \frac {\sqrt {c x^4+b x^2}}{x^6}dx^2}{5 b}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}\right )}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (\frac {4 c \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6}-\frac {2 \left (b x^2+c x^4\right )^{3/2}}{5 b x^8}\right ) (7 b B-4 A c)}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )\) |
((-2*A*(b*x^2 + c*x^4)^(3/2))/(7*b*x^10) + ((7*b*B - 4*A*c)*((-2*(b*x^2 + c*x^4)^(3/2))/(5*b*x^8) + (4*c*(b*x^2 + c*x^4)^(3/2))/(15*b^2*x^6)))/(7*b) )/2
3.1.97.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b *e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d - b*e) )) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d , e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[Simplify[m + 2*p + 2], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) ^(n_))^(q_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1) *(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] && !IntegerQ[p] && NeQ[k, j] && I ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]
Time = 1.81 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(-\frac {\left (\left (\frac {7 x^{2} B}{5}+A \right ) b^{2}-\frac {4 \left (\frac {7 x^{2} B}{6}+A \right ) x^{2} c b}{5}+\frac {8 A \,c^{2} x^{4}}{15}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (c \,x^{2}+b \right )}{7 b^{3} x^{8}}\) | \(66\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 A \,c^{2} x^{4}-14 x^{4} B b c -12 A b c \,x^{2}+21 b^{2} B \,x^{2}+15 b^{2} A \right ) \sqrt {x^{4} c +b \,x^{2}}}{105 b^{3} x^{8}}\) | \(70\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 A \,c^{2} x^{4}-14 x^{4} B b c -12 A b c \,x^{2}+21 b^{2} B \,x^{2}+15 b^{2} A \right ) \sqrt {x^{4} c +b \,x^{2}}}{105 b^{3} x^{8}}\) | \(70\) |
trager | \(-\frac {\left (8 A \,c^{3} x^{6}-14 x^{6} B b \,c^{2}-4 A b \,c^{2} x^{4}+7 x^{4} B \,b^{2} c +3 A \,b^{2} c \,x^{2}+21 b^{3} B \,x^{2}+15 b^{3} A \right ) \sqrt {x^{4} c +b \,x^{2}}}{105 b^{3} x^{8}}\) | \(87\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (8 A \,c^{3} x^{6}-14 x^{6} B b \,c^{2}-4 A b \,c^{2} x^{4}+7 x^{4} B \,b^{2} c +3 A \,b^{2} c \,x^{2}+21 b^{3} B \,x^{2}+15 b^{3} A \right )}{105 x^{8} b^{3}}\) | \(87\) |
-1/7*((7/5*x^2*B+A)*b^2-4/5*(7/6*x^2*B+A)*x^2*c*b+8/15*A*c^2*x^4)*(x^2*(c* x^2+b))^(1/2)*(c*x^2+b)/b^3/x^8
Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^9} \, dx=\frac {{\left (2 \, {\left (7 \, B b c^{2} - 4 \, A c^{3}\right )} x^{6} - {\left (7 \, B b^{2} c - 4 \, A b c^{2}\right )} x^{4} - 15 \, A b^{3} - 3 \, {\left (7 \, B b^{3} + A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{105 \, b^{3} x^{8}} \]
1/105*(2*(7*B*b*c^2 - 4*A*c^3)*x^6 - (7*B*b^2*c - 4*A*b*c^2)*x^4 - 15*A*b^ 3 - 3*(7*B*b^3 + A*b^2*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^3*x^8)
\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^9} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{9}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.68 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^9} \, dx=\frac {1}{15} \, B {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{b x^{4}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}}}{x^{6}}\right )} - \frac {1}{105} \, A {\left (\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{3} x^{2}} - \frac {4 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{2} x^{4}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} c}{b x^{6}} + \frac {15 \, \sqrt {c x^{4} + b x^{2}}}{x^{8}}\right )} \]
1/15*B*(2*sqrt(c*x^4 + b*x^2)*c^2/(b^2*x^2) - sqrt(c*x^4 + b*x^2)*c/(b*x^4 ) - 3*sqrt(c*x^4 + b*x^2)/x^6) - 1/105*A*(8*sqrt(c*x^4 + b*x^2)*c^3/(b^3*x ^2) - 4*sqrt(c*x^4 + b*x^2)*c^2/(b^2*x^4) + 3*sqrt(c*x^4 + b*x^2)*c/(b*x^6 ) + 15*sqrt(c*x^4 + b*x^2)/x^8)
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (84) = 168\).
Time = 0.89 (sec) , antiderivative size = 310, normalized size of antiderivative = 3.23 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^9} \, dx=\frac {4 \, {\left (105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 175 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 280 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{2} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 42 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{3} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 84 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{2} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 49 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{4} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 28 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{3} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 7 \, B b^{5} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 4 \, A b^{4} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )\right )}}{105 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{7}} \]
4/105*(105*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*c^(5/2)*sgn(x) - 175*(sqrt(c )*x - sqrt(c*x^2 + b))^8*B*b*c^(5/2)*sgn(x) + 280*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*c^(7/2)*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^2*c^(5/2 )*sgn(x) + 140*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b*c^(7/2)*sgn(x) - 42*(sq rt(c)*x - sqrt(c*x^2 + b))^4*B*b^3*c^(5/2)*sgn(x) + 84*(sqrt(c)*x - sqrt(c *x^2 + b))^4*A*b^2*c^(7/2)*sgn(x) + 49*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b ^4*c^(5/2)*sgn(x) - 28*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^3*c^(7/2)*sgn(x ) - 7*B*b^5*c^(5/2)*sgn(x) + 4*A*b^4*c^(7/2)*sgn(x))/((sqrt(c)*x - sqrt(c* x^2 + b))^2 - b)^7
Time = 9.43 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.67 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^9} \, dx=\frac {4\,A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b^2\,x^4}-\frac {B\,\sqrt {c\,x^4+b\,x^2}}{5\,x^6}-\frac {A\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,b\,x^6}-\frac {B\,c\,\sqrt {c\,x^4+b\,x^2}}{15\,b\,x^4}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{7\,x^8}-\frac {8\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{105\,b^3\,x^2}+\frac {2\,B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{15\,b^2\,x^2} \]
(4*A*c^2*(b*x^2 + c*x^4)^(1/2))/(105*b^2*x^4) - (B*(b*x^2 + c*x^4)^(1/2))/ (5*x^6) - (A*c*(b*x^2 + c*x^4)^(1/2))/(35*b*x^6) - (B*c*(b*x^2 + c*x^4)^(1 /2))/(15*b*x^4) - (A*(b*x^2 + c*x^4)^(1/2))/(7*x^8) - (8*A*c^3*(b*x^2 + c* x^4)^(1/2))/(105*b^3*x^2) + (2*B*c^2*(b*x^2 + c*x^4)^(1/2))/(15*b^2*x^2)