Integrand size = 19, antiderivative size = 108 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx=-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}+\frac {2 c \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}-\frac {8 c^2 \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}+\frac {16 c^3 \left (b x^2+c x^4\right )^{3/2}}{315 b^4 x^6} \] Output:
-1/9*(c*x^4+b*x^2)^(3/2)/b/x^12+2/21*c*(c*x^4+b*x^2)^(3/2)/b^2/x^10-8/105* c^2*(c*x^4+b*x^2)^(3/2)/b^3/x^8+16/315*c^3*(c*x^4+b*x^2)^(3/2)/b^4/x^6
Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-35 b^4-5 b^3 c x^2+6 b^2 c^2 x^4-8 b c^3 x^6+16 c^4 x^8\right )}{315 b^4 x^{10}} \] Input:
Integrate[Sqrt[b*x^2 + c*x^4]/x^11,x]
Output:
(Sqrt[x^2*(b + c*x^2)]*(-35*b^4 - 5*b^3*c*x^2 + 6*b^2*c^2*x^4 - 8*b*c^3*x^ 6 + 16*c^4*x^8))/(315*b^4*x^10)
Time = 0.43 (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.
\(\displaystyle \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {2 c \int \frac {\sqrt {c x^4+b x^2}}{x^9}dx}{3 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {2 c \left (-\frac {4 c \int \frac {\sqrt {c x^4+b x^2}}{x^7}dx}{7 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )}{3 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {2 c \left (-\frac {4 c \left (-\frac {2 c \int \frac {\sqrt {c x^4+b x^2}}{x^5}dx}{5 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{5 b x^8}\right )}{7 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )}{3 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}\) |
\(\Big \downarrow \) 1422 |
\(\displaystyle -\frac {2 c \left (-\frac {4 c \left (\frac {2 c \left (b x^2+c x^4\right )^{3/2}}{15 b^2 x^6}-\frac {\left (b x^2+c x^4\right )^{3/2}}{5 b x^8}\right )}{7 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}\right )}{3 b}-\frac {\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}\) |
Input:
Int[Sqrt[b*x^2 + c*x^4]/x^11,x]
Output:
-1/9*(b*x^2 + c*x^4)^(3/2)/(b*x^12) - (2*c*(-1/7*(b*x^2 + c*x^4)^(3/2)/(b* x^10) - (4*c*(-1/5*(b*x^2 + c*x^4)^(3/2)/(b*x^8) + (2*c*(b*x^2 + c*x^4)^(3 /2))/(15*b^2*x^6)))/(7*b)))/(3*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 = 61, normalized size of antiderivative = 0.56
method | result | size |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+24 b \,c^{2} x^{4}-30 b^{2} c \,x^{2}+35 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 x^{10} b^{4}}\) | \(61\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+24 b \,c^{2} x^{4}-30 b^{2} c \,x^{2}+35 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 x^{10} b^{4}}\) | \(61\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-16 c^{3} x^{6}+24 b \,c^{2} x^{4}-30 b^{2} c \,x^{2}+35 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 x^{10} b^{4}}\) | \(61\) |
trager | \(-\frac {\left (-16 c^{4} x^{8}+8 b \,c^{3} x^{6}-6 b^{2} c^{2} x^{4}+5 x^{2} b^{3} c +35 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 x^{10} b^{4}}\) | \(65\) |
risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-16 c^{4} x^{8}+8 b \,c^{3} x^{6}-6 b^{2} c^{2} x^{4}+5 x^{2} b^{3} c +35 b^{4}\right )}{315 x^{10} b^{4}}\) | \(65\) |
pseudoelliptic | \(\frac {\left (16 c^{4} x^{8}-8 b \,c^{3} x^{6}+6 b^{2} c^{2} x^{4}-5 x^{2} b^{3} c -35 b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{315 x^{10} b^{4}}\) | \(65\) |
Input:
int((c*x^4+b*x^2)^(1/2)/x^11,x,method=_RETURNVERBOSE)
Output:
-1/315*(c*x^2+b)*(-16*c^3*x^6+24*b*c^2*x^4-30*b^2*c*x^2+35*b^3)*(c*x^4+b*x ^2)^(1/2)/x^10/b^4
Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.59 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx=\frac {{\left (16 \, c^{4} x^{8} - 8 \, b c^{3} x^{6} + 6 \, b^{2} c^{2} x^{4} - 5 \, b^{3} c x^{2} - 35 \, b^{4}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, b^{4} x^{10}} \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^11,x, algorithm="fricas")
Output:
1/315*(16*c^4*x^8 - 8*b*c^3*x^6 + 6*b^2*c^2*x^4 - 5*b^3*c*x^2 - 35*b^4)*sq rt(c*x^4 + b*x^2)/(b^4*x^10)
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{11}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(1/2)/x**11,x)
Output:
Integral(sqrt(x**2*(b + c*x**2))/x**11, x)
Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx=\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{315 \, b^{4} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{315 \, b^{3} x^{4}} + \frac {2 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{105 \, b^{2} x^{6}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{63 \, b x^{8}} - \frac {\sqrt {c x^{4} + b x^{2}}}{9 \, x^{10}} \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^11,x, algorithm="maxima")
Output:
16/315*sqrt(c*x^4 + b*x^2)*c^4/(b^4*x^2) - 8/315*sqrt(c*x^4 + b*x^2)*c^3/( b^3*x^4) + 2/105*sqrt(c*x^4 + b*x^2)*c^2/(b^2*x^6) - 1/63*sqrt(c*x^4 + b*x ^2)*c/(b*x^8) - 1/9*sqrt(c*x^4 + b*x^2)/x^10
Time = 0.19 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.65 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx=\frac {32 \, {\left (315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 189 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 84 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{2} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 36 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{3} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{4} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - b^{5} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right )\right )}}{315 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{9}} \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^11,x, algorithm="giac")
Output:
32/315*(315*(sqrt(c)*x - sqrt(c*x^2 + b))^10*c^(9/2)*sgn(x) + 189*(sqrt(c) *x - sqrt(c*x^2 + b))^8*b*c^(9/2)*sgn(x) + 84*(sqrt(c)*x - sqrt(c*x^2 + b) )^6*b^2*c^(9/2)*sgn(x) - 36*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^3*c^(9/2)*sg n(x) + 9*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^4*c^(9/2)*sgn(x) - b^5*c^(9/2)* sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^9
Time = 18.82 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx=\frac {2\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b^2\,x^6}-\frac {c\,\sqrt {c\,x^4+b\,x^2}}{63\,b\,x^8}-\frac {\sqrt {c\,x^4+b\,x^2}}{9\,x^{10}}-\frac {8\,c^3\,\sqrt {c\,x^4+b\,x^2}}{315\,b^3\,x^4}+\frac {16\,c^4\,\sqrt {c\,x^4+b\,x^2}}{315\,b^4\,x^2} \] Input:
int((b*x^2 + c*x^4)^(1/2)/x^11,x)
Output:
(2*c^2*(b*x^2 + c*x^4)^(1/2))/(105*b^2*x^6) - (c*(b*x^2 + c*x^4)^(1/2))/(6 3*b*x^8) - (b*x^2 + c*x^4)^(1/2)/(9*x^10) - (8*c^3*(b*x^2 + c*x^4)^(1/2))/ (315*b^3*x^4) + (16*c^4*(b*x^2 + c*x^4)^(1/2))/(315*b^4*x^2)
Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11}} \, dx=\frac {-35 \sqrt {c \,x^{2}+b}\, b^{4}-5 \sqrt {c \,x^{2}+b}\, b^{3} c \,x^{2}+6 \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{4}-8 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{6}+16 \sqrt {c \,x^{2}+b}\, c^{4} x^{8}-16 \sqrt {c}\, c^{4} x^{9}}{315 b^{4} x^{9}} \] Input:
int((c*x^4+b*x^2)^(1/2)/x^11,x)
Output:
( - 35*sqrt(b + c*x**2)*b**4 - 5*sqrt(b + c*x**2)*b**3*c*x**2 + 6*sqrt(b + c*x**2)*b**2*c**2*x**4 - 8*sqrt(b + c*x**2)*b*c**3*x**6 + 16*sqrt(b + c*x **2)*c**4*x**8 - 16*sqrt(c)*c**4*x**9)/(315*b**4*x**9)