Integrand size = 19, antiderivative size = 80 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}+\frac {4 c \left (b x^2+c x^4\right )^{5/2}}{63 b^2 x^{12}}-\frac {8 c^2 \left (b x^2+c x^4\right )^{5/2}}{315 b^3 x^{10}} \] Output:
-1/9*(c*x^4+b*x^2)^(5/2)/b/x^14+4/63*c*(c*x^4+b*x^2)^(5/2)/b^2/x^12-8/315* c^2*(c*x^4+b*x^2)^(5/2)/b^3/x^10
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (-35 b^2+20 b c x^2-8 c^2 x^4\right )}{315 b^3 x^{14}} \] Input:
Integrate[(b*x^2 + c*x^4)^(3/2)/x^13,x]
Output:
((x^2*(b + c*x^2))^(5/2)*(-35*b^2 + 20*b*c*x^2 - 8*c^2*x^4))/(315*b^3*x^14 )
Time = 0.37 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {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.
\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {4 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{11}}dx}{9 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {4 c \left (-\frac {2 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^9}dx}{7 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )}{9 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\) |
\(\Big \downarrow \) 1422 |
\(\displaystyle -\frac {4 c \left (\frac {2 c \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}}-\frac {\left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )}{9 b}-\frac {\left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\) |
Input:
Int[(b*x^2 + c*x^4)^(3/2)/x^13,x]
Output:
-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)
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 = 0.20 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 c^{2} x^{4}-20 b c \,x^{2}+35 b^{2}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 x^{12} b^{3}}\) | \(50\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 c^{2} x^{4}-20 b c \,x^{2}+35 b^{2}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 x^{12} b^{3}}\) | \(50\) |
pseudoelliptic | \(-\frac {\left (\frac {8}{35} c^{2} x^{4}-\frac {4}{7} b c \,x^{2}+b^{2}\right ) \left (c \,x^{2}+b \right )^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{9 x^{10} b^{3}}\) | \(50\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (8 c^{2} x^{4}-20 b c \,x^{2}+35 b^{2}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 x^{12} b^{3}}\) | \(50\) |
trager | \(-\frac {\left (8 c^{4} x^{8}-4 b \,c^{3} x^{6}+3 b^{2} c^{2} x^{4}+50 x^{2} b^{3} c +35 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 b^{3} x^{10}}\) | \(65\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (8 c^{4} x^{8}-4 b \,c^{3} x^{6}+3 b^{2} c^{2} x^{4}+50 x^{2} b^{3} c +35 b^{4}\right )}{315 x^{10} b^{3}}\) | \(65\) |
Input:
int((c*x^4+b*x^2)^(3/2)/x^13,x,method=_RETURNVERBOSE)
Output:
-1/315*(c*x^2+b)*(8*c^2*x^4-20*b*c*x^2+35*b^2)*(c*x^4+b*x^2)^(3/2)/x^12/b^ 3
Time = 0.17 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=-\frac {{\left (8 \, c^{4} x^{8} - 4 \, b c^{3} x^{6} + 3 \, b^{2} c^{2} x^{4} + 50 \, b^{3} c x^{2} + 35 \, b^{4}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, b^{3} x^{10}} \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="fricas")
Output:
-1/315*(8*c^4*x^8 - 4*b*c^3*x^6 + 3*b^2*c^2*x^4 + 50*b^3*c*x^2 + 35*b^4)*s qrt(c*x^4 + b*x^2)/(b^3*x^10)
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{13}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(3/2)/x**13,x)
Output:
Integral((x**2*(b + c*x**2))**(3/2)/x**13, x)
Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.61 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=-\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{315 \, b^{3} x^{2}} + \frac {4 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{315 \, b^{2} x^{4}} - \frac {\sqrt {c x^{4} + b x^{2}} c^{2}}{105 \, b x^{6}} + \frac {\sqrt {c x^{4} + b x^{2}} c}{126 \, x^{8}} + \frac {\sqrt {c x^{4} + b x^{2}} b}{18 \, x^{10}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{6 \, x^{12}} \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="maxima")
Output:
-8/315*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^2) + 4/315*sqrt(c*x^4 + b*x^2)*c^3/( b^2*x^4) - 1/105*sqrt(c*x^4 + b*x^2)*c^2/(b*x^6) + 1/126*sqrt(c*x^4 + b*x^ 2)*c/x^8 + 1/18*sqrt(c*x^4 + b*x^2)*b/x^10 - 1/6*(c*x^4 + b*x^2)^(3/2)/x^1 2
Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (68) = 136\).
Time = 0.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.58 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {16 \, {\left (210 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} b c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 441 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b^{2} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 126 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{3} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 36 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{4} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{5} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + b^{6} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right )\right )}}{315 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{9}} \] Input:
integrate((c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="giac")
Output:
16/315*(210*(sqrt(c)*x - sqrt(c*x^2 + b))^12*c^(9/2)*sgn(x) + 315*(sqrt(c) *x - sqrt(c*x^2 + b))^10*b*c^(9/2)*sgn(x) + 441*(sqrt(c)*x - sqrt(c*x^2 + b))^8*b^2*c^(9/2)*sgn(x) + 126*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b^3*c^(9/2) *sgn(x) + 36*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^4*c^(9/2)*sgn(x) - 9*(sqrt( c)*x - sqrt(c*x^2 + b))^2*b^5*c^(9/2)*sgn(x) + b^6*c^(9/2)*sgn(x))/((sqrt( c)*x - sqrt(c*x^2 + b))^2 - b)^9
Time = 19.31 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.39 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {4\,c^3\,\sqrt {c\,x^4+b\,x^2}}{315\,b^2\,x^4}-\frac {10\,c\,\sqrt {c\,x^4+b\,x^2}}{63\,x^8}-\frac {c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b\,x^6}-\frac {b\,\sqrt {c\,x^4+b\,x^2}}{9\,x^{10}}-\frac {8\,c^4\,\sqrt {c\,x^4+b\,x^2}}{315\,b^3\,x^2} \] Input:
int((b*x^2 + c*x^4)^(3/2)/x^13,x)
Output:
(4*c^3*(b*x^2 + c*x^4)^(1/2))/(315*b^2*x^4) - (10*c*(b*x^2 + c*x^4)^(1/2)) /(63*x^8) - (c^2*(b*x^2 + c*x^4)^(1/2))/(105*b*x^6) - (b*(b*x^2 + c*x^4)^( 1/2))/(9*x^10) - (8*c^4*(b*x^2 + c*x^4)^(1/2))/(315*b^3*x^2)
Time = 0.17 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.26 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {-35 \sqrt {c \,x^{2}+b}\, b^{4}-50 \sqrt {c \,x^{2}+b}\, b^{3} c \,x^{2}-3 \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{4}+4 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{6}-8 \sqrt {c \,x^{2}+b}\, c^{4} x^{8}+8 \sqrt {c}\, c^{4} x^{9}}{315 b^{3} x^{9}} \] Input:
int((c*x^4+b*x^2)^(3/2)/x^13,x)
Output:
( - 35*sqrt(b + c*x**2)*b**4 - 50*sqrt(b + c*x**2)*b**3*c*x**2 - 3*sqrt(b + c*x**2)*b**2*c**2*x**4 + 4*sqrt(b + c*x**2)*b*c**3*x**6 - 8*sqrt(b + c*x **2)*c**4*x**8 + 8*sqrt(c)*c**4*x**9)/(315*b**3*x**9)