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

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

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Rubi [A]  time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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)} \end {gather*}

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 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])

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]

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*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.48 \begin {gather*} \frac {\sqrt {(a+b x)^2} (b x \log (x)-a)}{x (a+b x)} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.53, size = 649, normalized size = 9.98 \begin {gather*} \frac {-a b \sqrt {b^2} x \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-a b \sqrt {b^2} x \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )-\left (b^2\right )^{3/2} x^2 \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )+b^2 x \sqrt {a^2+2 a b x+b^2 x^2} \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-\left (b^2\right )^{3/2} x^2 \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )+b^2 x \sqrt {a^2+2 a b x+b^2 x^2} \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )+2 a^2 \sqrt {b^2}}{\left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right ) \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}+\frac {-2 a b \sqrt {a^2+2 a b x+b^2 x^2}-2 a b^2 x \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}{a}\right )+2 \sqrt {b^2} b x \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}{a}\right )-2 b^3 x^2 \tanh ^{-1}\left (\frac {\sqrt {a^2+2 a b x+b^2 x^2}-\sqrt {b^2} x}{a}\right )+2 a \sqrt {b^2} b x}{\left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right ) \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*Sqrt[b^2]*x - 2*a*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 2*a*b^2*x*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*
b*x + b^2*x^2])/a] - 2*b^3*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 2*b*Sqrt[b^2]*x*S
qrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/((-a - Sqrt[b^2]*x +
Sqrt[a^2 + 2*a*b*x + b^2*x^2])*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (2*a^2*Sqrt[b^2] - a*b*Sq
rt[b^2]*x*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (b^2)^(3/2)*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[
a^2 + 2*a*b*x + b^2*x^2]] + b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^
2*x^2]] - a*b*Sqrt[b^2]*x*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (b^2)^(3/2)*x^2*Log[a - Sqrt[
b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 +
2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*
x + b^2*x^2]))

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fricas [A]  time = 0.40, size = 13, normalized size = 0.20 \begin {gather*} \frac {b x \log \relax (x) - a}{x} \end {gather*}

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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giac [A]  time = 0.20, size = 24, normalized size = 0.37 \begin {gather*} 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 {gather*}

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

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maple [C]  time = 0.05, size = 22, normalized size = 0.34 \begin {gather*} \frac {\left (b x \ln \left (b x \right )-a \right ) \mathrm {csgn}\left (b x +a \right )}{x} \end {gather*}

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)*x*b-a)/x

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maxima [B]  time = 1.36, size = 87, normalized size = 1.34 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} b \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 103, normalized size = 1.58 \begin {gather*} \ln \left (a\,b+\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {b^2}+b^2\,x\right )\,\sqrt {b^2}-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}-\frac {a\,b\,\ln \left (a\,b+\frac {a^2}{x}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}\right )}{\sqrt {a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)/x^2,x)

[Out]

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

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sympy [A]  time = 0.12, size = 7, normalized size = 0.11 \begin {gather*} - \frac {a}{x} + b \log {\relax (x )} \end {gather*}

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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