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

Optimal. Leaf size=77 \[ -\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac{b \sqrt{a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0230154, antiderivative size = 77, normalized size of antiderivative = 1., 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 = {1112, 14} \[ -\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac{b \sqrt{a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1112

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

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]]

Rubi steps

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

Mathematica [A]  time = 0.0069249, size = 37, normalized size = 0.48 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (a+3 b x^2\right )}{3 x^3 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 34, normalized size = 0.4 \begin{align*} -{\frac{3\,b{x}^{2}+a}{3\,{x}^{3} \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^4,x)

[Out]

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

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Maxima [A]  time = 1.01073, size = 18, normalized size = 0.23 \begin{align*} -\frac{3 \, b x^{2} + a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^4,x, algorithm="maxima")

[Out]

-1/3*(3*b*x^2 + a)/x^3

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Fricas [A]  time = 1.42792, size = 32, normalized size = 0.42 \begin{align*} -\frac{3 \, b x^{2} + a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

-1/3*(3*b*x^2 + a)/x^3

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Sympy [A]  time = 0.286232, size = 14, normalized size = 0.18 \begin{align*} - \frac{a + 3 b x^{2}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x**2+a)**2)**(1/2)/x**4,x)

[Out]

-(a + 3*b*x**2)/(3*x**3)

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Giac [A]  time = 1.13392, size = 41, normalized size = 0.53 \begin{align*} -\frac{3 \, b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + a \mathrm{sgn}\left (b x^{2} + a\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((b*x^2+a)^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

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