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

Optimal. Leaf size=79 \[ -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 646, 43} \begin {gather*} -\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^9,x]

[Out]

-(a*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*x^8*(a + b*x^2)) - (b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*x^6*(a + b*x
^2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps

\begin {align*} \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^9} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx,x,x^2\right )\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \frac {a b+b^2 x}{x^5} \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \operatorname {Subst}\left (\int \left (\frac {a b}{x^5}+\frac {b^2}{x^4}\right ) \, dx,x,x^2\right )}{2 \left (a b+b^2 x^2\right )}\\ &=-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\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^2\right )^2} \left (3 a+4 b x^2\right )}{24 x^8 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^9,x]

[Out]

-1/24*(Sqrt[(a + b*x^2)^2]*(3*a + 4*b*x^2))/(x^8*(a + b*x^2))

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IntegrateAlgebraic [B]  time = 0.58, size = 266, normalized size = 3.37 \begin {gather*} \frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-3 a^4 b-13 a^3 b^2 x^2-21 a^2 b^3 x^4-15 a b^4 x^6-4 b^5 x^8\right )+\sqrt {b^2} b^3 \left (3 a^5+16 a^4 b x^2+34 a^3 b^2 x^4+36 a^2 b^3 x^6+19 a b^4 x^8+4 b^5 x^{10}\right )}{3 \sqrt {b^2} x^8 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (-8 a^3 b^3-24 a^2 b^4 x^2-24 a b^5 x^4-8 b^6 x^6\right )+3 x^8 \left (8 a^4 b^4+32 a^3 b^5 x^2+48 a^2 b^6 x^4+32 a b^7 x^6+8 b^8 x^8\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^9,x]

[Out]

(b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-3*a^4*b - 13*a^3*b^2*x^2 - 21*a^2*b^3*x^4 - 15*a*b^4*x^6 - 4*b^5*x^8) +
b^3*Sqrt[b^2]*(3*a^5 + 16*a^4*b*x^2 + 34*a^3*b^2*x^4 + 36*a^2*b^3*x^6 + 19*a*b^4*x^8 + 4*b^5*x^10))/(3*Sqrt[b
^2]*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-8*a^3*b^3 - 24*a^2*b^4*x^2 - 24*a*b^5*x^4 - 8*b^6*x^6) + 3*x^8*(8*a^
4*b^4 + 32*a^3*b^5*x^2 + 48*a^2*b^6*x^4 + 32*a*b^7*x^6 + 8*b^8*x^8))

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fricas [A]  time = 0.82, size = 15, normalized size = 0.19 \begin {gather*} -\frac {4 \, b x^{2} + 3 \, a}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(4*b*x^2 + 3*a)/x^8

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giac [A]  time = 0.16, size = 31, normalized size = 0.39 \begin {gather*} -\frac {4 \, b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a \mathrm {sgn}\left (b x^{2} + a\right )}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(4*b*x^2*sgn(b*x^2 + a) + 3*a*sgn(b*x^2 + a))/x^8

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maple [A]  time = 0.00, size = 36, normalized size = 0.46 \begin {gather*} -\frac {\left (4 b \,x^{2}+3 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{24 \left (b \,x^{2}+a \right ) x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(4*b*x^2+3*a)*((b*x^2+a)^2)^(1/2)/x^8/(b*x^2+a)

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maxima [A]  time = 1.35, size = 15, normalized size = 0.19 \begin {gather*} -\frac {4 \, b x^{2} + 3 \, a}{24 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(4*b*x^2 + 3*a)/x^8

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mupad [B]  time = 4.24, size = 35, normalized size = 0.44 \begin {gather*} -\frac {\left (4\,b\,x^2+3\,a\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{24\,x^8\,\left (b\,x^2+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)^2)^(1/2)/x^9,x)

[Out]

-((3*a + 4*b*x^2)*((a + b*x^2)^2)^(1/2))/(24*x^8*(a + b*x^2))

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sympy [A]  time = 0.20, size = 15, normalized size = 0.19 \begin {gather*} \frac {- 3 a - 4 b x^{2}}{24 x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*a - 4*b*x**2)/(24*x**8)

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