Integrand size = 26, antiderivative size = 61 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}-\frac {(7 b B-2 A c) \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}} \]
Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (-5 A b-7 b B x^2+2 A c x^2\right )}{35 b^2 x^{12}} \]
Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1940, 1220, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx\) |
\(\Big \downarrow \) 1940 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^4+b x^2\right )^{3/2}}{x^{12}}dx^2\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {1}{2} \left (\frac {(7 b B-2 A c) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{10}}dx^2}{7 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )\) |
\(\Big \downarrow \) 1123 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 \left (b x^2+c x^4\right )^{5/2} (7 b B-2 A c)}{35 b^2 x^{10}}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )\) |
((-2*A*(b*x^2 + c*x^4)^(5/2))/(7*b*x^12) - (2*(7*b*B - 2*A*c)*(b*x^2 + c*x ^4)^(5/2))/(35*b^2*x^10))/2
3.2.14.3.1 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_)*((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 = 1.83 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-2 A c \,x^{2}+7 b B \,x^{2}+5 A b \right ) \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{10}}\) | \(48\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-2 A c \,x^{2}+7 b B \,x^{2}+5 A b \right ) \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{10}}\) | \(48\) |
pseudoelliptic | \(-\frac {\left (\left (\frac {7 x^{2} B}{5}+A \right ) b -\frac {2 A c \,x^{2}}{5}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (c \,x^{2}+b \right )^{2}}{7 x^{8} b^{2}}\) | \(49\) |
trager | \(-\frac {\left (-2 A \,c^{3} x^{6}+7 x^{6} B b \,c^{2}+A b \,c^{2} x^{4}+14 x^{4} B \,b^{2} c +8 A \,b^{2} c \,x^{2}+7 b^{3} B \,x^{2}+5 b^{3} A \right ) \sqrt {x^{4} c +b \,x^{2}}}{35 b^{2} x^{8}}\) | \(86\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-2 A \,c^{3} x^{6}+7 x^{6} B b \,c^{2}+A b \,c^{2} x^{4}+14 x^{4} B \,b^{2} c +8 A \,b^{2} c \,x^{2}+7 b^{3} B \,x^{2}+5 b^{3} A \right )}{35 x^{8} b^{2}}\) | \(86\) |
Time = 0.76 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.34 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {{\left ({\left (7 \, B b c^{2} - 2 \, A c^{3}\right )} x^{6} + {\left (14 \, B b^{2} c + A b c^{2}\right )} x^{4} + 5 \, A b^{3} + {\left (7 \, B b^{3} + 8 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, b^{2} x^{8}} \]
-1/35*((7*B*b*c^2 - 2*A*c^3)*x^6 + (14*B*b^2*c + A*b*c^2)*x^4 + 5*A*b^3 + (7*B*b^3 + 8*A*b^2*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^2*x^8)
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{11}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (53) = 106\).
Time = 0.23 (sec) , antiderivative size = 193, normalized size of antiderivative = 3.16 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=-\frac {1}{10} \, B {\left (\frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{x^{4}} - \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{x^{6}} + \frac {5 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{8}}\right )} + \frac {1}{140} \, A {\left (\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{2}} - \frac {4 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{4}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} c}{x^{6}} + \frac {15 \, \sqrt {c x^{4} + b x^{2}} b}{x^{8}} - \frac {35 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{10}}\right )} \]
-1/10*B*(2*sqrt(c*x^4 + b*x^2)*c^2/(b*x^2) - sqrt(c*x^4 + b*x^2)*c/x^4 - 3 *sqrt(c*x^4 + b*x^2)*b/x^6 + 5*(c*x^4 + b*x^2)^(3/2)/x^8) + 1/140*A*(8*sqr t(c*x^4 + b*x^2)*c^3/(b^2*x^2) - 4*sqrt(c*x^4 + b*x^2)*c^2/(b*x^4) + 3*sqr t(c*x^4 + b*x^2)*c/x^6 + 15*sqrt(c*x^4 + b*x^2)*b/x^8 - 35*(c*x^4 + b*x^2) ^(3/2)/x^10)
Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (53) = 106\).
Time = 1.19 (sec) , antiderivative size = 370, normalized size of antiderivative = 6.07 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 105 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{2} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 70 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{3} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 140 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{2} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 77 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{4} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 28 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{3} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 14 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{5} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) + 14 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{4} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 7 \, B b^{6} c^{\frac {5}{2}} \mathrm {sgn}\left (x\right ) - 2 \, A b^{5} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )\right )}}{35 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{7}} \]
2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*c^(5/2)*sgn(x) - 70*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b*c^(5/2)*sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + b ))^10*A*c^(7/2)*sgn(x) + 105*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b^2*c^(5/2) *sgn(x) + 70*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b*c^(7/2)*sgn(x) - 140*(sqr t(c)*x - sqrt(c*x^2 + b))^6*B*b^3*c^(5/2)*sgn(x) + 140*(sqrt(c)*x - sqrt(c *x^2 + b))^6*A*b^2*c^(7/2)*sgn(x) + 77*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b ^4*c^(5/2)*sgn(x) + 28*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^3*c^(7/2)*sgn(x ) - 14*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^5*c^(5/2)*sgn(x) + 14*(sqrt(c)* x - sqrt(c*x^2 + b))^2*A*b^4*c^(7/2)*sgn(x) + 7*B*b^6*c^(5/2)*sgn(x) - 2*A *b^5*c^(7/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^7
Time = 9.82 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.56 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{11}} \, dx=\frac {2\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{35\,b^2\,x^2}-\frac {8\,A\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,x^6}-\frac {B\,b\,\sqrt {c\,x^4+b\,x^2}}{5\,x^6}-\frac {2\,B\,c\,\sqrt {c\,x^4+b\,x^2}}{5\,x^4}-\frac {A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{35\,b\,x^4}-\frac {A\,b\,\sqrt {c\,x^4+b\,x^2}}{7\,x^8}-\frac {B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{5\,b\,x^2} \]