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

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

[Out]

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

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Rubi [A]  time = 0.0382253, antiderivative size = 39, normalized size of antiderivative = 1., 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} \[ -\frac{\left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}{4 a x^4} \]

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 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])

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 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]

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*}

Mathematica [A]  time = 0.008054, size = 37, normalized size = 0.95 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (a+2 b x^2\right )}{4 x^4 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 34, normalized size = 0.9 \begin{align*} -{\frac{2\,b{x}^{2}+a}{4\,{x}^{4} \left ( b{x}^{2}+a \right ) }\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}} \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.48193, size = 32, normalized size = 0.82 \begin{align*} -\frac{2 \, b x^{2} + a}{4 \, x^{4}} \end{align*}

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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Sympy [A]  time = 0.291191, size = 14, normalized size = 0.36 \begin{align*} - \frac{a + 2 b x^{2}}{4 x^{4}} \end{align*}

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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Giac [A]  time = 1.15952, size = 41, normalized size = 1.05 \begin{align*} -\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{align*}

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