Integrand size = 19, antiderivative size = 85 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}+\frac {2 c}{5 b^2 x^2 \sqrt {b x^2+c x^4}}-\frac {8 c^2 \left (b+2 c x^2\right )}{5 b^4 \sqrt {b x^2+c x^4}} \] Output:
-1/5/b/x^4/(c*x^4+b*x^2)^(1/2)+2/5*c/b^2/x^2/(c*x^4+b*x^2)^(1/2)-8/5*c^2*( 2*c*x^2+b)/b^4/(c*x^4+b*x^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-b^3+2 b^2 c x^2-8 b c^2 x^4-16 c^3 x^6}{5 b^4 x^4 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[1/(x^3*(b*x^2 + c*x^4)^(3/2)),x]
Output:
(-b^3 + 2*b^2*c*x^2 - 8*b*c^2*x^4 - 16*c^3*x^6)/(5*b^4*x^4*Sqrt[x^2*(b + c *x^2)])
Time = 0.40 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {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.
\(\displaystyle \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {6 c \int \frac {1}{x \left (c x^4+b x^2\right )^{3/2}}dx}{5 b}-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1423 |
\(\displaystyle -\frac {6 c \left (-\frac {4 c \int \frac {x}{\left (c x^4+b x^2\right )^{3/2}}dx}{3 b}-\frac {1}{3 b x^2 \sqrt {b x^2+c x^4}}\right )}{5 b}-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle -\frac {6 c \left (-\frac {2 c \int \frac {1}{\left (c x^4+b x^2\right )^{3/2}}dx^2}{3 b}-\frac {1}{3 b x^2 \sqrt {b x^2+c x^4}}\right )}{5 b}-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 1088 |
\(\displaystyle -\frac {6 c \left (\frac {4 c \left (b+2 c x^2\right )}{3 b^3 \sqrt {b x^2+c x^4}}-\frac {1}{3 b x^2 \sqrt {b x^2+c x^4}}\right )}{5 b}-\frac {1}{5 b x^4 \sqrt {b x^2+c x^4}}\) |
Input:
Int[1/(x^3*(b*x^2 + c*x^4)^(3/2)),x]
Output:
-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)
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 = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.61
method | result | size |
pseudoelliptic | \(-\frac {16 c^{3} x^{6}+8 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+b^{3}}{5 x^{4} b^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(52\) |
gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (16 c^{3} x^{6}+8 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+b^{3}\right )}{5 x^{2} b^{4} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(59\) |
default | \(-\frac {\left (c \,x^{2}+b \right ) \left (16 c^{3} x^{6}+8 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+b^{3}\right )}{5 x^{2} b^{4} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(59\) |
orering | \(-\frac {\left (c \,x^{2}+b \right ) \left (16 c^{3} x^{6}+8 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+b^{3}\right )}{5 x^{2} b^{4} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\) | \(59\) |
trager | \(-\frac {\left (16 c^{3} x^{6}+8 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{5 \left (c \,x^{2}+b \right ) b^{4} x^{6}}\) | \(61\) |
risch | \(-\frac {\left (c \,x^{2}+b \right ) \left (11 c^{2} x^{4}-3 b c \,x^{2}+b^{2}\right )}{5 b^{4} x^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {x^{2} c^{3}}{b^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(73\) |
Input:
int(1/x^3/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/5*(16*c^3*x^6+8*b*c^2*x^4-2*b^2*c*x^2+b^3)/x^4/b^4/(x^2*(c*x^2+b))^(1/2 )
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {{\left (16 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} - 2 \, b^{2} c x^{2} + b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{5 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \] Input:
integrate(1/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
Output:
-1/5*(16*c^3*x^6 + 8*b*c^2*x^4 - 2*b^2*c*x^2 + b^3)*sqrt(c*x^4 + b*x^2)/(b ^4*c*x^8 + b^5*x^6)
\[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x**3/(c*x**4+b*x**2)**(3/2),x)
Output:
Integral(1/(x**3*(x**2*(b + c*x**2))**(3/2)), x)
Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {16 \, c^{3} x^{2}}{5 \, \sqrt {c x^{4} + b x^{2}} b^{4}} - \frac {8 \, c^{2}}{5 \, \sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {2 \, c}{5 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}} - \frac {1}{5 \, \sqrt {c x^{4} + b x^{2}} b x^{4}} \] Input:
integrate(1/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
Output:
-16/5*c^3*x^2/(sqrt(c*x^4 + b*x^2)*b^4) - 8/5*c^2/(sqrt(c*x^4 + b*x^2)*b^3 ) + 2/5*c/(sqrt(c*x^4 + b*x^2)*b^2*x^2) - 1/5/(sqrt(c*x^4 + b*x^2)*b*x^4)
Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (73) = 146\).
Time = 0.13 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.99 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {c^{3} x}{\sqrt {c x^{2} + b} b^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} c^{\frac {5}{2}} - 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b c^{\frac {5}{2}} + 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{2} c^{\frac {5}{2}} - 50 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{3} c^{\frac {5}{2}} + 11 \, b^{4} c^{\frac {5}{2}}\right )}}{5 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5} b^{3} \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
Output:
-c^3*x/(sqrt(c*x^2 + b)*b^4*sgn(x)) + 2/5*(5*(sqrt(c)*x - sqrt(c*x^2 + b)) ^8*c^(5/2) - 30*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b*c^(5/2) + 80*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^2*c^(5/2) - 50*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^3* c^(5/2) + 11*b^4*c^(5/2))/(((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^5*b^3*sgn (x))
Time = 17.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\sqrt {c\,x^4+b\,x^2}\,\left (b^3-2\,b^2\,c\,x^2+8\,b\,c^2\,x^4+16\,c^3\,x^6\right )}{5\,b^4\,x^6\,\left (c\,x^2+b\right )} \] Input:
int(1/(x^3*(b*x^2 + c*x^4)^(3/2)),x)
Output:
-((b*x^2 + c*x^4)^(1/2)*(b^3 + 16*c^3*x^6 - 2*b^2*c*x^2 + 8*b*c^2*x^4))/(5 *b^4*x^6*(b + c*x^2))
Time = 0.18 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-\sqrt {c \,x^{2}+b}\, b^{3}+2 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{2}-8 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}-16 \sqrt {c \,x^{2}+b}\, c^{3} x^{6}+16 \sqrt {c}\, b \,c^{2} x^{5}+16 \sqrt {c}\, c^{3} x^{7}}{5 b^{4} x^{5} \left (c \,x^{2}+b \right )} \] Input:
int(1/x^3/(c*x^4+b*x^2)^(3/2),x)
Output:
( - sqrt(b + c*x**2)*b**3 + 2*sqrt(b + c*x**2)*b**2*c*x**2 - 8*sqrt(b + c* x**2)*b*c**2*x**4 - 16*sqrt(b + c*x**2)*c**3*x**6 + 16*sqrt(c)*b*c**2*x**5 + 16*sqrt(c)*c**3*x**7)/(5*b**4*x**5*(b + c*x**2))