\(\int \frac {1}{x^3 (b x^2+c x^4)^{3/2}} \, dx\) [279]

Optimal result

Integrand size = 19, antiderivative size = 102 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{b x^4 \sqrt {b x^2+c x^4}}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}+\frac {8 c \sqrt {b x^2+c x^4}}{5 b^3 x^4}-\frac {16 c^2 \sqrt {b x^2+c x^4}}{5 b^4 x^2} \]

[Out]

1/b/x^4/(c*x^4+b*x^2)^(1/2)-6/5*(c*x^4+b*x^2)^(1/2)/b^2/x^6+8/5*c*(c*x^4+b*x^2)^(1/2)/b^3/x^4-16/5*c^2*(c*x^4+
b*x^2)^(1/2)/b^4/x^2

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2040, 2041, 2039} \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {16 c^2 \sqrt {b x^2+c x^4}}{5 b^4 x^2}+\frac {8 c \sqrt {b x^2+c x^4}}{5 b^3 x^4}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}+\frac {1}{b x^4 \sqrt {b x^2+c x^4}} \]

[In]

Int[1/(x^3*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

1/(b*x^4*Sqrt[b*x^2 + c*x^4]) - (6*Sqrt[b*x^2 + c*x^4])/(5*b^2*x^6) + (8*c*Sqrt[b*x^2 + c*x^4])/(5*b^3*x^4) -
(16*c^2*Sqrt[b*x^2 + c*x^4])/(5*b^4*x^2)

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2040

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] + Dist[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))
, Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n,
j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{b x^4 \sqrt {b x^2+c x^4}}+\frac {6 \int \frac {1}{x^5 \sqrt {b x^2+c x^4}} \, dx}{b} \\ & = \frac {1}{b x^4 \sqrt {b x^2+c x^4}}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}-\frac {(24 c) \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx}{5 b^2} \\ & = \frac {1}{b x^4 \sqrt {b x^2+c x^4}}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}+\frac {8 c \sqrt {b x^2+c x^4}}{5 b^3 x^4}+\frac {\left (16 c^2\right ) \int \frac {1}{x \sqrt {b x^2+c x^4}} \, dx}{5 b^3} \\ & = \frac {1}{b x^4 \sqrt {b x^2+c x^4}}-\frac {6 \sqrt {b x^2+c x^4}}{5 b^2 x^6}+\frac {8 c \sqrt {b x^2+c x^4}}{5 b^3 x^4}-\frac {16 c^2 \sqrt {b x^2+c x^4}}{5 b^4 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-b^3+2 b^2 c x^2-8 b c^2 x^4-16 c^3 x^6}{5 b^4 x^4 \sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

Integrate[1/(x^3*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(-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)])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.51

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	\(-\frac {\left (c \,x^{2}+b \right ) \left (16 c^{3} x^{6}+8 b \,c^{2} x^{4}-2 b^{2} c \,x^{2}+b^{3}\right )}{5 x^{2} b^{4} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\)	\(59\)
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	\(-\frac {\left (c \,x^{2}+b \right ) \left (11 c^{2} x^{4}-3 b c \,x^{2}+b^{2}\right )}{5 b^{4} x^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {x^{2} c^{3}}{b^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(73\)

[In]

int(1/x^3/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {{\left (16 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} - 2 \, b^{2} c x^{2} + b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{5 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \]

[In]

integrate(1/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

-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)

Sympy [F]

\[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/x**3/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(1/(x**3*(x**2*(b + c*x**2))**(3/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {16 \, c^{3} x^{2}}{5 \, \sqrt {c x^{4} + b x^{2}} b^{4}} - \frac {8 \, c^{2}}{5 \, \sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {2 \, c}{5 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}} - \frac {1}{5 \, \sqrt {c x^{4} + b x^{2}} b x^{4}} \]

[In]

integrate(1/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.66 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {c^{3} x}{\sqrt {c x^{2} + b} b^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} c^{\frac {5}{2}} - 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} b c^{\frac {5}{2}} + 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b^{2} c^{\frac {5}{2}} - 50 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{3} c^{\frac {5}{2}} + 11 \, b^{4} c^{\frac {5}{2}}\right )}}{5 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{5} b^{3} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(1/x^3/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

-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))

Mupad [B] (verification not implemented)

Time = 13.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^3 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\sqrt {c\,x^4+b\,x^2}\,\left (b^3-2\,b^2\,c\,x^2+8\,b\,c^2\,x^4+16\,c^3\,x^6\right )}{5\,b^4\,x^6\,\left (c\,x^2+b\right )} \]

[In]

int(1/(x^3*(b*x^2 + c*x^4)^(3/2)),x)

[Out]

-((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))