3.545 \(\int \frac{\sqrt{a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx\)

Optimal. Leaf size=75 \[ \frac{b \log (x) \sqrt{a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0218993, antiderivative size = 75, 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} \[ \frac{b \log (x) \sqrt{a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^3,x]

[Out]

-(a*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^2*(a + b*x^2)) + (b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*Log[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^3} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \frac{a b+b^2 x^2}{x^3} \, 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^3}+\frac{b^2}{x}\right ) \, dx}{a b+b^2 x^2}\\ &=-\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \log (x)}{a+b x^2}\\ \end{align*}

Mathematica [A]  time = 0.0109884, size = 39, normalized size = 0.52 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (a-2 b x^2 \log (x)\right )}{2 x^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]/x^3,x]

[Out]

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

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Maple [A]  time = 0.173, size = 38, normalized size = 0.5 \begin{align*}{\frac{2\,b\ln \left ( x \right ){x}^{2}-a}{2\,{x}^{2} \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^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50795, size = 41, normalized size = 0.55 \begin{align*} \frac{2 \, b x^{2} \log \left (x\right ) - a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*x^2*log(x) - a)/x^2

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Sympy [A]  time = 0.274169, size = 10, normalized size = 0.13 \begin{align*} - \frac{a}{2 x^{2}} + b \log{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(2*x**2) + b*log(x)

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Giac [A]  time = 1.15268, size = 61, normalized size = 0.81 \begin{align*} \frac{1}{2} \, b \log \left (x^{2}\right ) \mathrm{sgn}\left (b x^{2} + a\right ) - \frac{b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + a \mathrm{sgn}\left (b x^{2} + a\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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