Integrand size = 19, antiderivative size = 113 \[ \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {1}{7 b x^6 \sqrt {b x^2+c x^4}}+\frac {8 c}{35 b^2 x^4 \sqrt {b x^2+c x^4}}-\frac {16 c^2}{35 b^3 x^2 \sqrt {b x^2+c x^4}}+\frac {64 c^3 \left (b+2 c x^2\right )}{35 b^5 \sqrt {b x^2+c x^4}} \] Output:
-1/7/b/x^6/(c*x^4+b*x^2)^(1/2)+8/35*c/b^2/x^4/(c*x^4+b*x^2)^(1/2)-16/35*c^ 2/b^3/x^2/(c*x^4+b*x^2)^(1/2)+64/35*c^3*(2*c*x^2+b)/b^5/(c*x^4+b*x^2)^(1/2 )
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-5 b^4+8 b^3 c x^2-16 b^2 c^2 x^4+64 b c^3 x^6+128 c^4 x^8}{35 b^5 x^6 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[1/(x^5*(b*x^2 + c*x^4)^(3/2)),x]
Output:
(-5*b^4 + 8*b^3*c*x^2 - 16*b^2*c^2*x^4 + 64*b*c^3*x^6 + 128*c^4*x^8)/(35*b ^5*x^6*Sqrt[x^2*(b + c*x^2)])
Time = 0.47 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1423, 1423, 1423, 1424, 1088}
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.
\(\displaystyle \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {8 c \int \frac {1}{x^3 \left (c x^4+b x^2\right )^{3/2}}dx}{7 b}-\frac {1}{7 b x^6 \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {8 c \left (-\frac {6 c \int \frac {1}{x \left (c x^4+b x^2\right )^{3/2}}dx}{5 b}-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {1}{7 b x^6 \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {8 c \left (-\frac {6 c \left (-\frac {4 c \int \frac {x}{\left (c x^4+b x^2\right )^{3/2}}dx}{3 b}-\frac {1}{3 b x^2 \sqrt {b x^2+c x^4}}\right )}{5 b}-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {1}{7 b x^6 \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle -\frac {8 c \left (-\frac {6 c \left (-\frac {2 c \int \frac {1}{\left (c x^4+b x^2\right )^{3/2}}dx^2}{3 b}-\frac {1}{3 b x^2 \sqrt {b x^2+c x^4}}\right )}{5 b}-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {1}{7 b x^6 \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle -\frac {8 c \left (-\frac {6 c \left (\frac {4 c \left (b+2 c x^2\right )}{3 b^3 \sqrt {b x^2+c x^4}}-\frac {1}{3 b x^2 \sqrt {b x^2+c x^4}}\right )}{5 b}-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}\right )}{7 b}-\frac {1}{7 b x^6 \sqrt {b x^2+c x^4}}\) |
Input:
Int[1/(x^5*(b*x^2 + c*x^4)^(3/2)),x]
Output:
-1/7*1/(b*x^6*Sqrt[b*x^2 + c*x^4]) - (8*c*(-1/5*1/(b*x^4*Sqrt[b*x^2 + c*x^ 4]) - (6*c*(-1/3*1/(b*x^2*Sqrt[b*x^2 + c*x^4]) + (4*c*(b + 2*c*x^2))/(3*b^ 3*Sqrt[b*x^2 + c*x^4])))/(5*b)))/(7*b)
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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]
Int[(x_)^(m_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{b, c, m, p}, x] && !IntegerQ[p] && IntegerQ[(m - 1)/2]
Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.58
method | result | size |
pseudoelliptic | \(\frac {128 c^{4} x^{8}+64 b \,c^{3} x^{6}-16 b^{2} c^{2} x^{4}+8 x^{2} b^{3} c -5 b^{4}}{35 x^{6} b^{5} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(65\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-128 c^{4} x^{8}-64 b \,c^{3} x^{6}+16 b^{2} c^{2} x^{4}-8 x^{2} b^{3} c +5 b^{4}\right )}{35 x^{4} b^{5} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(72\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-128 c^{4} x^{8}-64 b \,c^{3} x^{6}+16 b^{2} c^{2} x^{4}-8 x^{2} b^{3} c +5 b^{4}\right )}{35 x^{4} b^{5} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(72\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (-128 c^{4} x^{8}-64 b \,c^{3} x^{6}+16 b^{2} c^{2} x^{4}-8 x^{2} b^{3} c +5 b^{4}\right )}{35 x^{4} b^{5} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(72\) |
trager | \(-\frac {\left (-128 c^{4} x^{8}-64 b \,c^{3} x^{6}+16 b^{2} c^{2} x^{4}-8 x^{2} b^{3} c +5 b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{35 \left (c \,x^{2}+b \right ) b^{5} x^{8}}\) | \(74\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (-93 c^{3} x^{6}+29 b \,c^{2} x^{4}-13 b^{2} c \,x^{2}+5 b^{3}\right )}{35 b^{5} x^{6} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {x^{2} c^{4}}{b^{5} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(85\) |
Input:
int(1/x^5/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/35*(128*c^4*x^8+64*b*c^3*x^6-16*b^2*c^2*x^4+8*b^3*c*x^2-5*b^4)/x^6/b^5/( x^2*(c*x^2+b))^(1/2)
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (128 \, c^{4} x^{8} + 64 \, b c^{3} x^{6} - 16 \, b^{2} c^{2} x^{4} + 8 \, b^{3} c x^{2} - 5 \, b^{4}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, {\left (b^{5} c x^{10} + b^{6} x^{8}\right )}} \] Input:
integrate(1/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
1/35*(128*c^4*x^8 + 64*b*c^3*x^6 - 16*b^2*c^2*x^4 + 8*b^3*c*x^2 - 5*b^4)*s qrt(c*x^4 + b*x^2)/(b^5*c*x^10 + b^6*x^8)
\[ \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{5} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x**5/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(1/(x**5*(x**2*(b + c*x**2))**(3/2)), x)
Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {128 \, c^{4} x^{2}}{35 \, \sqrt {c x^{4} + b x^{2}} b^{5}} + \frac {64 \, c^{3}}{35 \, \sqrt {c x^{4} + b x^{2}} b^{4}} - \frac {16 \, c^{2}}{35 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}} + \frac {8 \, c}{35 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{4}} - \frac {1}{7 \, \sqrt {c x^{4} + b x^{2}} b x^{6}} \] Input:
integrate(1/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
128/35*c^4*x^2/(sqrt(c*x^4 + b*x^2)*b^5) + 64/35*c^3/(sqrt(c*x^4 + b*x^2)* b^4) - 16/35*c^2/(sqrt(c*x^4 + b*x^2)*b^3*x^2) + 8/35*c/(sqrt(c*x^4 + b*x^ 2)*b^2*x^4) - 1/7/(sqrt(c*x^4 + b*x^2)*b*x^6)
Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (97) = 194\).
Time = 0.14 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.96 \[ \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {c^{4} x}{\sqrt {c x^{2} + b} b^{5} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (35 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} c^{\frac {7}{2}} - 280 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} b c^{\frac {7}{2}} + 1015 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} b^{2} c^{\frac {7}{2}} - 2240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b^{3} c^{\frac {7}{2}} + 1673 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{4} c^{\frac {7}{2}} - 616 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{5} c^{\frac {7}{2}} + 93 \, b^{6} c^{\frac {7}{2}}\right )}}{35 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{7} b^{4} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^5/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
c^4*x/(sqrt(c*x^2 + b)*b^5*sgn(x)) - 2/35*(35*(sqrt(c)*x - sqrt(c*x^2 + b) )^12*c^(7/2) - 280*(sqrt(c)*x - sqrt(c*x^2 + b))^10*b*c^(7/2) + 1015*(sqrt (c)*x - sqrt(c*x^2 + b))^8*b^2*c^(7/2) - 2240*(sqrt(c)*x - sqrt(c*x^2 + b) )^6*b^3*c^(7/2) + 1673*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^4*c^(7/2) - 616*( sqrt(c)*x - sqrt(c*x^2 + b))^2*b^5*c^(7/2) + 93*b^6*c^(7/2))/(((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^7*b^4*sgn(x))
Time = 18.55 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {13\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,b^3\,x^6}-\frac {\sqrt {c\,x^4+b\,x^2}}{7\,b^2\,x^8}-\frac {29\,c^2\,\sqrt {c\,x^4+b\,x^2}}{35\,b^4\,x^4}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {93\,c^3}{35\,b^4}+\frac {128\,c^4\,x^2}{35\,b^5}\right )}{x^2\,\left (c\,x^2+b\right )} \] Input:
int(1/(x^5*(b*x^2 + c*x^4)^(3/2)),x)
Output:
(13*c*(b*x^2 + c*x^4)^(1/2))/(35*b^3*x^6) - (b*x^2 + c*x^4)^(1/2)/(7*b^2*x ^8) - (29*c^2*(b*x^2 + c*x^4)^(1/2))/(35*b^4*x^4) + ((b*x^2 + c*x^4)^(1/2) *((93*c^3)/(35*b^4) + (128*c^4*x^2)/(35*b^5)))/(x^2*(b + c*x^2))
Time = 0.18 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^5 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-5 \sqrt {c \,x^{2}+b}\, b^{4}+8 \sqrt {c \,x^{2}+b}\, b^{3} c \,x^{2}-16 \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{4}+64 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{6}+128 \sqrt {c \,x^{2}+b}\, c^{4} x^{8}-128 \sqrt {c}\, b \,c^{3} x^{7}-128 \sqrt {c}\, c^{4} x^{9}}{35 b^{5} x^{7} \left (c \,x^{2}+b \right )} \] Input:
int(1/x^5/(c*x^4+b*x^2)^(3/2),x)
Output:
( - 5*sqrt(b + c*x**2)*b**4 + 8*sqrt(b + c*x**2)*b**3*c*x**2 - 16*sqrt(b + c*x**2)*b**2*c**2*x**4 + 64*sqrt(b + c*x**2)*b*c**3*x**6 + 128*sqrt(b + c *x**2)*c**4*x**8 - 128*sqrt(c)*b*c**3*x**7 - 128*sqrt(c)*c**4*x**9)/(35*b* *5*x**7*(b + c*x**2))