\(\int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx\) [265]

Optimal result

Integrand size = 19, antiderivative size = 52 \[ \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx=-\frac {\sqrt {b x^2+c x^4}}{3 b x^4}+\frac {2 c \sqrt {b x^2+c x^4}}{3 b^2 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2041, 2039} \[ \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx=\frac {2 c \sqrt {b x^2+c x^4}}{3 b^2 x^2}-\frac {\sqrt {b x^2+c x^4}}{3 b x^4} \]

[In]

Int[1/(x^3*Sqrt[b*x^2 + c*x^4]),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-b+2 c x^2\right )}{3 b^2 x^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.58

method	result	size
trager	\(-\frac {\left (-2 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3 b^{2} x^{4}}\)	\(30\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (-2 c \,x^{2}+b \right )}{3 x^{2} b^{2} \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(37\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (-2 c \,x^{2}+b \right )}{3 x^{2} b^{2} \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(37\)
risch	\(-\frac {\left (c \,x^{2}+b \right ) \left (-2 c \,x^{2}+b \right )}{3 x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b^{2}}\)	\(37\)
pseudoelliptic	\(-\frac {\left (c \,x^{2}+b \right ) \left (-2 c \,x^{2}+b \right )}{3 x^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b^{2}}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.60 \[ \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c x^{4} + b x^{2}} {\left (2 \, c x^{2} - b\right )}}{3 \, b^{2} x^{4}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx=\frac {2 \, \sqrt {c x^{4} + b x^{2}} c}{3 \, b^{2} x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}}}{3 \, b x^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx=\frac {4 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{3} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.79 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^3 \sqrt {b x^2+c x^4}} \, dx=-\frac {\left (b-2\,c\,x^2\right )\,\sqrt {c\,x^4+b\,x^2}}{3\,b^2\,x^4} \]

[In]

int(1/(x^3*(b*x^2 + c*x^4)^(1/2)),x)

[Out]

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