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

Optimal. Leaf size=65 \[ \frac{b \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}-\frac{a \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.018244, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {646, 43} \[ \frac{b \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}-\frac{a \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0096034, size = 31, normalized size = 0.48 \[ \frac{\sqrt{(a+b x)^2} (b x \log (x)-a)}{x (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.235, size = 22, normalized size = 0.3 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) \left ( \ln \left ( bx \right ) bx-a \right ) }{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

csgn(b*x+a)*(ln(b*x)*b*x-a)/x

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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+a)^2)^(1/2)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83424, size = 27, normalized size = 0.42 \begin{align*} \frac{b x \log \left (x\right ) - a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x*log(x) - a)/x

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Sympy [A]  time = 0.310047, size = 7, normalized size = 0.11 \begin{align*} - \frac{a}{x} + b \log{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/x + b*log(x)

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Giac [A]  time = 1.26161, size = 32, normalized size = 0.49 \begin{align*} b \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{a \mathrm{sgn}\left (b x + a\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*log(abs(x))*sgn(b*x + a) - a*sgn(b*x + a)/x