3.228 \(\int \frac{\sqrt{b x^2+c x^4}}{x^9} \, dx\)

Optimal. Leaf size=80 \[ -\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^6}+\frac{4 c \left (b x^2+c x^4\right )^{3/2}}{35 b^2 x^8}-\frac{\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}} \]

[Out]

-(b*x^2 + c*x^4)^(3/2)/(7*b*x^10) + (4*c*(b*x^2 + c*x^4)^(3/2))/(35*b^2*x^8) - (8*c^2*(b*x^2 + c*x^4)^(3/2))/(
105*b^3*x^6)

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Rubi [A]  time = 0.119807, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2014} \[ -\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^6}+\frac{4 c \left (b x^2+c x^4\right )^{3/2}}{35 b^2 x^8}-\frac{\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^9,x]

[Out]

-(b*x^2 + c*x^4)^(3/2)/(7*b*x^10) + (4*c*(b*x^2 + c*x^4)^(3/2))/(35*b^2*x^8) - (8*c^2*(b*x^2 + c*x^4)^(3/2))/(
105*b^3*x^6)

Rule 2016

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])

Rule 2014

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] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{b x^2+c x^4}}{x^9} \, dx &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}-\frac{(4 c) \int \frac{\sqrt{b x^2+c x^4}}{x^7} \, dx}{7 b}\\ &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}+\frac{4 c \left (b x^2+c x^4\right )^{3/2}}{35 b^2 x^8}+\frac{\left (8 c^2\right ) \int \frac{\sqrt{b x^2+c x^4}}{x^5} \, dx}{35 b^2}\\ &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{7 b x^{10}}+\frac{4 c \left (b x^2+c x^4\right )^{3/2}}{35 b^2 x^8}-\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0119416, size = 46, normalized size = 0.57 \[ -\frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (15 b^2-12 b c x^2+8 c^2 x^4\right )}{105 b^3 x^{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^9,x]

[Out]

-((x^2*(b + c*x^2))^(3/2)*(15*b^2 - 12*b*c*x^2 + 8*c^2*x^4))/(105*b^3*x^10)

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Maple [A]  time = 0.046, size = 50, normalized size = 0.6 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( 8\,{c}^{2}{x}^{4}-12\,bc{x}^{2}+15\,{b}^{2} \right ) }{105\,{x}^{8}{b}^{3}}\sqrt{c{x}^{4}+b{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(1/2)/x^9,x)

[Out]

-1/105*(c*x^2+b)*(8*c^2*x^4-12*b*c*x^2+15*b^2)*(c*x^4+b*x^2)^(1/2)/x^8/b^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54625, size = 117, normalized size = 1.46 \begin{align*} -\frac{{\left (8 \, c^{3} x^{6} - 4 \, b c^{2} x^{4} + 3 \, b^{2} c x^{2} + 15 \, b^{3}\right )} \sqrt{c x^{4} + b x^{2}}}{105 \, b^{3} x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^9,x, algorithm="fricas")

[Out]

-1/105*(8*c^3*x^6 - 4*b*c^2*x^4 + 3*b^2*c*x^2 + 15*b^3)*sqrt(c*x^4 + b*x^2)/(b^3*x^8)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} \left (b + c x^{2}\right )}}{x^{9}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**9,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**9, x)

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Giac [B]  time = 1.28066, size = 200, normalized size = 2.5 \begin{align*} \frac{16 \,{\left (70 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 35 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} b c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + 21 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} b^{2} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) - 7 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} b^{3} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right ) + b^{4} c^{\frac{7}{2}} \mathrm{sgn}\left (x\right )\right )}}{105 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^9,x, algorithm="giac")

[Out]

16/105*(70*(sqrt(c)*x - sqrt(c*x^2 + b))^8*c^(7/2)*sgn(x) + 35*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b*c^(7/2)*sgn(x
) + 21*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^2*c^(7/2)*sgn(x) - 7*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^3*c^(7/2)*sgn(
x) + b^4*c^(7/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^7