3.229 \(\int \frac{\sqrt{b x^2+c x^4}}{x^{11}} \, dx\)

Optimal. Leaf size=108 \[ \frac{16 c^3 \left (b x^2+c x^4\right )^{3/2}}{315 b^4 x^6}-\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}+\frac{2 c \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}-\frac{\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}} \]

[Out]

-(b*x^2 + c*x^4)^(3/2)/(9*b*x^12) + (2*c*(b*x^2 + c*x^4)^(3/2))/(21*b^2*x^10) - (8*c^2*(b*x^2 + c*x^4)^(3/2))/
(105*b^3*x^8) + (16*c^3*(b*x^2 + c*x^4)^(3/2))/(315*b^4*x^6)

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Rubi [A]  time = 0.163684, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2014} \[ \frac{16 c^3 \left (b x^2+c x^4\right )^{3/2}}{315 b^4 x^6}-\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}+\frac{2 c \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}-\frac{\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^11,x]

[Out]

-(b*x^2 + c*x^4)^(3/2)/(9*b*x^12) + (2*c*(b*x^2 + c*x^4)^(3/2))/(21*b^2*x^10) - (8*c^2*(b*x^2 + c*x^4)^(3/2))/
(105*b^3*x^8) + (16*c^3*(b*x^2 + c*x^4)^(3/2))/(315*b^4*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^{11}} \, dx &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}-\frac{(2 c) \int \frac{\sqrt{b x^2+c x^4}}{x^9} \, dx}{3 b}\\ &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}+\frac{2 c \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}+\frac{\left (8 c^2\right ) \int \frac{\sqrt{b x^2+c x^4}}{x^7} \, dx}{21 b^2}\\ &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}+\frac{2 c \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}-\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}-\frac{\left (16 c^3\right ) \int \frac{\sqrt{b x^2+c x^4}}{x^5} \, dx}{105 b^3}\\ &=-\frac{\left (b x^2+c x^4\right )^{3/2}}{9 b x^{12}}+\frac{2 c \left (b x^2+c x^4\right )^{3/2}}{21 b^2 x^{10}}-\frac{8 c^2 \left (b x^2+c x^4\right )^{3/2}}{105 b^3 x^8}+\frac{16 c^3 \left (b x^2+c x^4\right )^{3/2}}{315 b^4 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0143128, size = 57, normalized size = 0.53 \[ \frac{\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (30 b^2 c x^2-35 b^3-24 b c^2 x^4+16 c^3 x^6\right )}{315 b^4 x^{12}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^11,x]

[Out]

((x^2*(b + c*x^2))^(3/2)*(-35*b^3 + 30*b^2*c*x^2 - 24*b*c^2*x^4 + 16*c^3*x^6))/(315*b^4*x^12)

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Maple [A]  time = 0.046, size = 61, normalized size = 0.6 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ) \left ( -16\,{c}^{3}{x}^{6}+24\,b{c}^{2}{x}^{4}-30\,{b}^{2}c{x}^{2}+35\,{b}^{3} \right ) }{315\,{x}^{10}{b}^{4}}\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^11,x)

[Out]

-1/315*(c*x^2+b)*(-16*c^3*x^6+24*b*c^2*x^4-30*b^2*c*x^2+35*b^3)*(c*x^4+b*x^2)^(1/2)/x^10/b^4

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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^11,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58623, size = 140, normalized size = 1.3 \begin{align*} \frac{{\left (16 \, c^{4} x^{8} - 8 \, b c^{3} x^{6} + 6 \, b^{2} c^{2} x^{4} - 5 \, b^{3} c x^{2} - 35 \, b^{4}\right )} \sqrt{c x^{4} + b x^{2}}}{315 \, b^{4} x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^11,x, algorithm="fricas")

[Out]

1/315*(16*c^4*x^8 - 8*b*c^3*x^6 + 6*b^2*c^2*x^4 - 5*b^3*c*x^2 - 35*b^4)*sqrt(c*x^4 + b*x^2)/(b^4*x^10)

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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^{11}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**11,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**11, x)

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Giac [A]  time = 1.3714, size = 240, normalized size = 2.22 \begin{align*} \frac{32 \,{\left (315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{10} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 189 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{8} b c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 84 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{6} b^{2} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - 36 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{4} b^{3} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) + 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} b^{4} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right ) - b^{5} c^{\frac{9}{2}} \mathrm{sgn}\left (x\right )\right )}}{315 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b}\right )}^{2} - b\right )}^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^11,x, algorithm="giac")

[Out]

32/315*(315*(sqrt(c)*x - sqrt(c*x^2 + b))^10*c^(9/2)*sgn(x) + 189*(sqrt(c)*x - sqrt(c*x^2 + b))^8*b*c^(9/2)*sg
n(x) + 84*(sqrt(c)*x - sqrt(c*x^2 + b))^6*b^2*c^(9/2)*sgn(x) - 36*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b^3*c^(9/2)*
sgn(x) + 9*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^4*c^(9/2)*sgn(x) - b^5*c^(9/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 +
b))^2 - b)^9