3.3.28 \(\int \frac {\sqrt {b x^2+c x^4}}{x^9} \, dx\) [228]

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

[Out]

-1/7*(c*x^4+b*x^2)^(3/2)/b/x^10+4/35*c*(c*x^4+b*x^2)^(3/2)/b^2/x^8-8/105*c^2*(c*x^4+b*x^2)^(3/2)/b^3/x^6

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Rubi [A]
time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 = {2041, 2039} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^9,x]

[Out]

-1/7*(b*x^2 + c*x^4)^(3/2)/(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 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*} \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*}

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Mathematica [A]
time = 0.06, size = 57, normalized size = 0.71 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-15 b^3-3 b^2 c x^2+4 b c^2 x^4-8 c^3 x^6\right )}{105 b^3 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^9,x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(-15*b^3 - 3*b^2*c*x^2 + 4*b*c^2*x^4 - 8*c^3*x^6))/(105*b^3*x^8)

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Maple [A]
time = 0.08, size = 50, normalized size = 0.62

method result size
gosper \(-\frac {\left (c \,x^{2}+b \right ) \left (8 c^{2} x^{4}-12 b c \,x^{2}+15 b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 x^{8} b^{3}}\) \(50\)
default \(-\frac {\left (c \,x^{2}+b \right ) \left (8 c^{2} x^{4}-12 b c \,x^{2}+15 b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 x^{8} b^{3}}\) \(50\)
trager \(-\frac {\left (8 c^{3} x^{6}-4 b \,c^{2} x^{4}+3 b^{2} c \,x^{2}+15 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 x^{8} b^{3}}\) \(54\)
risch \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (8 c^{3} x^{6}-4 b \,c^{2} x^{4}+3 b^{2} c \,x^{2}+15 b^{3}\right )}{105 x^{8} b^{3}}\) \(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 [A]
time = 0.30, size = 89, normalized size = 1.11 \begin {gather*} -\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{105 \, b^{3} x^{2}} + \frac {4 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{105 \, b^{2} x^{4}} - \frac {\sqrt {c x^{4} + b x^{2}} c}{35 \, b x^{6}} - \frac {\sqrt {c x^{4} + b x^{2}}}{7 \, x^{8}} \end {gather*}

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]

-8/105*sqrt(c*x^4 + b*x^2)*c^3/(b^3*x^2) + 4/105*sqrt(c*x^4 + b*x^2)*c^2/(b^2*x^4) - 1/35*sqrt(c*x^4 + b*x^2)*
c/(b*x^6) - 1/7*sqrt(c*x^4 + b*x^2)/x^8

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Fricas [A]
time = 0.37, size = 53, normalized size = 0.66 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{9}}\, dx \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (68) = 136\).
time = 7.60, size = 148, normalized size = 1.85 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 4.34, size = 89, normalized size = 1.11 \begin {gather*} \frac {4\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b^2\,x^4}-\frac {c\,\sqrt {c\,x^4+b\,x^2}}{35\,b\,x^6}-\frac {\sqrt {c\,x^4+b\,x^2}}{7\,x^8}-\frac {8\,c^3\,\sqrt {c\,x^4+b\,x^2}}{105\,b^3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(1/2)/x^9,x)

[Out]

(4*c^2*(b*x^2 + c*x^4)^(1/2))/(105*b^2*x^4) - (c*(b*x^2 + c*x^4)^(1/2))/(35*b*x^6) - (b*x^2 + c*x^4)^(1/2)/(7*
x^8) - (8*c^3*(b*x^2 + c*x^4)^(1/2))/(105*b^3*x^2)

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