3.1.20 \(\int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^9} \, dx\)

Optimal. Leaf size=79 \[ -\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 x^5 \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1355, 14} \begin {gather*} -\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 x^5 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^9,x]

[Out]

-(a*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(8*x^8*(a + b*x^3)) - (b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(5*x^5*(a + b*x
^3))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{x^9} \, dx &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {a b+b^2 x^3}{x^9} \, dx}{a b+b^2 x^3}\\ &=\frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a b}{x^9}+\frac {b^2}{x^6}\right ) \, dx}{a b+b^2 x^3}\\ &=-\frac {a \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 x^8 \left (a+b x^3\right )}-\frac {b \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 x^5 \left (a+b x^3\right )}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.49 \begin {gather*} -\frac {\sqrt {\left (a+b x^3\right )^2} \left (5 a+8 b x^3\right )}{40 x^8 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^9,x]

[Out]

-1/40*(Sqrt[(a + b*x^3)^2]*(5*a + 8*b*x^3))/(x^8*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 19.24, size = 39, normalized size = 0.49 \begin {gather*} \frac {\left (-5 a-8 b x^3\right ) \sqrt {\left (a+b x^3\right )^2}}{40 x^8 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]/x^9,x]

[Out]

((-5*a - 8*b*x^3)*Sqrt[(a + b*x^3)^2])/(40*x^8*(a + b*x^3))

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fricas [A]  time = 1.07, size = 15, normalized size = 0.19 \begin {gather*} -\frac {8 \, b x^{3} + 5 \, a}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^9,x, algorithm="fricas")

[Out]

-1/40*(8*b*x^3 + 5*a)/x^8

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giac [A]  time = 0.38, size = 31, normalized size = 0.39 \begin {gather*} -\frac {8 \, b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 5 \, a \mathrm {sgn}\left (b x^{3} + a\right )}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^9,x, algorithm="giac")

[Out]

-1/40*(8*b*x^3*sgn(b*x^3 + a) + 5*a*sgn(b*x^3 + a))/x^8

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maple [A]  time = 0.00, size = 36, normalized size = 0.46 \begin {gather*} -\frac {\left (8 b \,x^{3}+5 a \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{40 \left (b \,x^{3}+a \right ) x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^3+a)^2)^(1/2)/x^9,x)

[Out]

-1/40*(8*b*x^3+5*a)*((b*x^3+a)^2)^(1/2)/x^8/(b*x^3+a)

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maxima [A]  time = 0.45, size = 15, normalized size = 0.19 \begin {gather*} -\frac {8 \, b x^{3} + 5 \, a}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^3+a)^2)^(1/2)/x^9,x, algorithm="maxima")

[Out]

-1/40*(8*b*x^3 + 5*a)/x^8

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mupad [B]  time = 1.16, size = 35, normalized size = 0.44 \begin {gather*} -\frac {\left (8\,b\,x^3+5\,a\right )\,\sqrt {{\left (b\,x^3+a\right )}^2}}{40\,x^8\,\left (b\,x^3+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^3)^2)^(1/2)/x^9,x)

[Out]

-((5*a + 8*b*x^3)*((a + b*x^3)^2)^(1/2))/(40*x^8*(a + b*x^3))

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sympy [A]  time = 0.20, size = 15, normalized size = 0.19 \begin {gather*} \frac {- 5 a - 8 b x^{3}}{40 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x**3+a)**2)**(1/2)/x**9,x)

[Out]

(-5*a - 8*b*x**3)/(40*x**8)

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