Integrand size = 19, antiderivative size = 108 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}+\frac {2 c \left (b x^2+c x^4\right )^{5/2}}{33 b^2 x^{14}}-\frac {8 c^2 \left (b x^2+c x^4\right )^{5/2}}{231 b^3 x^{12}}+\frac {16 c^3 \left (b x^2+c x^4\right )^{5/2}}{1155 b^4 x^{10}} \]
-1/11*(c*x^4+b*x^2)^(5/2)/b/x^16+2/33*c*(c*x^4+b*x^2)^(5/2)/b^2/x^14-8/231 *c^2*(c*x^4+b*x^2)^(5/2)/b^3/x^12+16/1155*c^3*(c*x^4+b*x^2)^(5/2)/b^4/x^10
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.53 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (-105 b^3+70 b^2 c x^2-40 b c^2 x^4+16 c^3 x^6\right )}{1155 b^4 x^{16}} \]
((x^2*(b + c*x^2))^(5/2)*(-105*b^3 + 70*b^2*c*x^2 - 40*b*c^2*x^4 + 16*c^3* x^6))/(1155*b^4*x^16)
Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {1423, 1423, 1423, 1422}
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.
\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {6 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{13}}dx}{11 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {6 c \left (-\frac {4 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{11}}dx}{9 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )}{11 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {6 c \left (-\frac {4 c \left (-\frac {2 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^9}dx}{7 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )}{9 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )}{11 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}\) |
\(\Big \downarrow \) 1422 |
\(\displaystyle -\frac {6 c \left (-\frac {4 c \left (\frac {2 c \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}}-\frac {\left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )}{9 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )}{11 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}\) |
-1/11*(b*x^2 + c*x^4)^(5/2)/(b*x^16) - (6*c*(-1/9*(b*x^2 + c*x^4)^(5/2)/(b *x^14) - (4*c*(-1/7*(b*x^2 + c*x^4)^(5/2)/(b*x^12) + (2*c*(b*x^2 + c*x^4)^ (5/2))/(35*b^2*x^10)))/(9*b)))/(11*b)
3.3.47.3.1 Defintions of rubi rules used
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(-d)*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(2*b*(p + 1))), x] /; FreeQ[{b , c, d, m, p}, x] && !IntegerQ[p] && EqQ[m + 4*p + 3, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( (m + 4*p + 3)/(b*d^2*(m + 2*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && ILtQ[Simplify[(m + 4*p + 3)/2], 0] && NeQ[m + 2*p + 1, 0]
Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+40 b \,c^{2} x^{4}-70 b^{2} c \,x^{2}+105 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{1155 x^{14} b^{4}}\) | \(61\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+40 b \,c^{2} x^{4}-70 b^{2} c \,x^{2}+105 b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{1155 x^{14} b^{4}}\) | \(61\) |
pseudoelliptic | \(-\frac {\left (c \,x^{2}+b \right )^{2} \left (-16 c^{3} x^{6}+40 b \,c^{2} x^{4}-70 b^{2} c \,x^{2}+105 b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1155 x^{12} b^{4}}\) | \(63\) |
trager | \(-\frac {\left (-16 c^{5} x^{10}+8 c^{4} x^{8} b -6 b^{2} c^{3} x^{6}+5 c^{2} x^{4} b^{3}+140 x^{2} c \,b^{4}+105 b^{5}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{1155 b^{4} x^{12}}\) | \(76\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-16 c^{5} x^{10}+8 c^{4} x^{8} b -6 b^{2} c^{3} x^{6}+5 c^{2} x^{4} b^{3}+140 x^{2} c \,b^{4}+105 b^{5}\right )}{1155 x^{12} b^{4}}\) | \(76\) |
-1/1155*(c*x^2+b)*(-16*c^3*x^6+40*b*c^2*x^4-70*b^2*c*x^2+105*b^3)*(c*x^4+b *x^2)^(3/2)/x^14/b^4
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.69 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\frac {{\left (16 \, c^{5} x^{10} - 8 \, b c^{4} x^{8} + 6 \, b^{2} c^{3} x^{6} - 5 \, b^{3} c^{2} x^{4} - 140 \, b^{4} c x^{2} - 105 \, b^{5}\right )} \sqrt {c x^{4} + b x^{2}}}{1155 \, b^{4} x^{12}} \]
1/1155*(16*c^5*x^10 - 8*b*c^4*x^8 + 6*b^2*c^3*x^6 - 5*b^3*c^2*x^4 - 140*b^ 4*c*x^2 - 105*b^5)*sqrt(c*x^4 + b*x^2)/(b^4*x^12)
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{15}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.42 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{1155 \, b^{4} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{1155 \, b^{3} x^{4}} + \frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{385 \, b^{2} x^{6}} - \frac {\sqrt {c x^{4} + b x^{2}} c^{2}}{231 \, b x^{8}} + \frac {\sqrt {c x^{4} + b x^{2}} c}{264 \, x^{10}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} b}{88 \, x^{12}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{8 \, x^{14}} \]
16/1155*sqrt(c*x^4 + b*x^2)*c^5/(b^4*x^2) - 8/1155*sqrt(c*x^4 + b*x^2)*c^4 /(b^3*x^4) + 2/385*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^6) - 1/231*sqrt(c*x^4 + b*x^2)*c^2/(b*x^8) + 1/264*sqrt(c*x^4 + b*x^2)*c/x^10 + 3/88*sqrt(c*x^4 + b*x^2)*b/x^12 - 1/8*(c*x^4 + b*x^2)^(3/2)/x^14
Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (92) = 184\).
Time = 0.33 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.19 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\frac {32 \, {\left (1155 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 2079 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} b c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 2541 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} b^{2} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 825 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b^{3} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 165 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{4} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) - 55 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{5} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 11 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{6} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) - b^{7} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right )\right )}}{1155 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{11}} \]
32/1155*(1155*(sqrt(c)*x - sqrt(c*x^2 + b))^14*c^(11/2)*sgn(x) + 2079*(sqr t(c)*x - sqrt(c*x^2 + b))^12*b*c^(11/2)*sgn(x) + 2541*(sqrt(c)*x - sqrt(c* x^2 + b))^10*b^2*c^(11/2)*sgn(x) + 825*(sqrt(c)*x - sqrt(c*x^2 + b))^8*b^3 *c^(11/2)*sgn(x) + 165*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b^4*c^(11/2)*sgn(x) - 55*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^5*c^(11/2)*sgn(x) + 11*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^6*c^(11/2)*sgn(x) - b^7*c^(11/2)*sgn(x))/((sqrt(c)* x - sqrt(c*x^2 + b))^2 - b)^11
Time = 13.96 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.25 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\frac {2\,c^3\,\sqrt {c\,x^4+b\,x^2}}{385\,b^2\,x^6}-\frac {4\,c\,\sqrt {c\,x^4+b\,x^2}}{33\,x^{10}}-\frac {c^2\,\sqrt {c\,x^4+b\,x^2}}{231\,b\,x^8}-\frac {b\,\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {8\,c^4\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^4}+\frac {16\,c^5\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^4\,x^2} \]