\(\int \frac {1}{x (b x^2+c x^4)^{3/2}} \, dx\) [278]

Optimal result

Integrand size = 19, antiderivative size = 74 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{b x^2 \sqrt {b x^2+c x^4}}-\frac {4 \sqrt {b x^2+c x^4}}{3 b^2 x^4}+\frac {8 c \sqrt {b x^2+c x^4}}{3 b^3 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2040, 2041, 2039} \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {8 c \sqrt {b x^2+c x^4}}{3 b^3 x^2}-\frac {4 \sqrt {b x^2+c x^4}}{3 b^2 x^4}+\frac {1}{b x^2 \sqrt {b x^2+c x^4}} \]

[In]

Int[1/(x*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

1/(b*x^2*Sqrt[b*x^2 + c*x^4]) - (4*Sqrt[b*x^2 + c*x^4])/(3*b^2*x^4) + (8*c*Sqrt[b*x^2 + c*x^4])/(3*b^3*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^2 \sqrt {b x^2+c x^4}}+\frac {4 \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx}{b} \\ & = \frac {1}{b x^2 \sqrt {b x^2+c x^4}}-\frac {4 \sqrt {b x^2+c x^4}}{3 b^2 x^4}-\frac {(8 c) \int \frac {1}{x \sqrt {b x^2+c x^4}} \, dx}{3 b^2} \\ & = \frac {1}{b x^2 \sqrt {b x^2+c x^4}}-\frac {4 \sqrt {b x^2+c x^4}}{3 b^2 x^4}+\frac {8 c \sqrt {b x^2+c x^4}}{3 b^3 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-b^2+4 b c x^2+8 c^2 x^4}{3 b^3 x^2 \sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55

method	result	size
pseudoelliptic	\(-\frac {-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}}{3 b^{3} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(41\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}\right )}{3 b^{3} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\)	\(45\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}\right )}{3 b^{3} \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\)	\(45\)
trager	\(-\frac {\left (-8 c^{2} x^{4}-4 b c \,x^{2}+b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3 \left (c \,x^{2}+b \right ) b^{3} x^{4}}\)	\(50\)
risch	\(-\frac {\left (c \,x^{2}+b \right ) \left (-5 c \,x^{2}+b \right )}{3 b^{3} x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {x^{2} c^{2}}{b^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (8 \, c^{2} x^{4} + 4 \, b c x^{2} - b^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {8 \, c^{2} x^{2}}{3 \, \sqrt {c x^{4} + b x^{2}} b^{3}} + \frac {4 \, c}{3 \, \sqrt {c x^{4} + b x^{2}} b^{2}} - \frac {1}{3 \, \sqrt {c x^{4} + b x^{2}} b x^{2}} \]

[In]

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

[Out]

8/3*c^2*x^2/(sqrt(c*x^4 + b*x^2)*b^3) + 4/3*c/(sqrt(c*x^4 + b*x^2)*b^2) - 1/3/(sqrt(c*x^4 + b*x^2)*b*x^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.54 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {c^{2} x}{\sqrt {c x^{2} + b} b^{3} \mathrm {sgn}\left (x\right )} - \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} c^{\frac {3}{2}} - 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b c^{\frac {3}{2}} + 5 \, b^{2} c^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{3} b^{2} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

c^2*x/(sqrt(c*x^2 + b)*b^3*sgn(x)) - 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b))^4*c^(3/2) - 12*(sqrt(c)*x - sqrt(c*x
^2 + b))^2*b*c^(3/2) + 5*b^2*c^(3/2))/(((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^3*b^2*sgn(x))

Mupad [B] (verification not implemented)

Time = 12.98 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.69 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-b^2+4\,b\,c\,x^2+8\,c^2\,x^4\right )}{3\,b^3\,x^4\,\left (c\,x^2+b\right )} \]

[In]

int(1/(x*(b*x^2 + c*x^4)^(3/2)),x)

[Out]

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