3.4.74 \(\int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx\)

Optimal. Leaf size=39 \[ -\frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 a x^4} \]

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Rubi [A]  time = 0.04, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1111, 646, 37} \begin {gather*} -\frac {\left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^5,x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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

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Mathematica [A]  time = 0.01, size = 37, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (a+2 b x^2\right )}{4 x^4 \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^5,x]

[Out]

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

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IntegrateAlgebraic [B]  time = 0.38, size = 118, normalized size = 3.03 \begin {gather*} \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-a b-2 b^2 x^2\right )+\sqrt {b^2} \left (a^2+3 a b x^2+2 b^2 x^4\right )}{4 x^4 \left (a b+b^2 x^2\right )-4 \sqrt {b^2} x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^5,x]

[Out]

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

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fricas [A]  time = 0.91, size = 13, normalized size = 0.33 \begin {gather*} -\frac {2 \, b x^{2} + a}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

-1/4*(2*b*x^2 + a)/x^4

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giac [A]  time = 0.16, size = 30, normalized size = 0.77 \begin {gather*} -\frac {2 \, b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a \mathrm {sgn}\left (b x^{2} + a\right )}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 34, normalized size = 0.87 \begin {gather*} -\frac {\left (2 b \,x^{2}+a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 \left (b \,x^{2}+a \right ) x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^2+a)^2)^(1/2)/x^5,x)

[Out]

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

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maxima [A]  time = 1.33, size = 13, normalized size = 0.33 \begin {gather*} -\frac {2 \, b x^{2} + a}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^5,x, algorithm="maxima")

[Out]

-1/4*(2*b*x^2 + a)/x^4

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mupad [B]  time = 4.21, size = 33, normalized size = 0.85 \begin {gather*} -\frac {\left (2\,b\,x^2+a\right )\,\sqrt {{\left (b\,x^2+a\right )}^2}}{4\,x^4\,\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^5,x)

[Out]

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

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sympy [A]  time = 0.17, size = 14, normalized size = 0.36 \begin {gather*} \frac {- a - 2 b x^{2}}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x**2+a)**2)**(1/2)/x**5,x)

[Out]

(-a - 2*b*x**2)/(4*x**4)

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