3.5.55 \(\int \sqrt {a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}} \, dx\) [455]

Optimal. Leaf size=73 \[ \frac {a \sqrt {a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}} x}{a+\frac {b}{x}}-\frac {b \sqrt {a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}} \log \left (\frac {1}{x}\right )}{a+\frac {b}{x}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1356, 660, 45} \begin {gather*} \frac {a x \sqrt {a^2+\frac {2 a b}{x}+\frac {b^2}{x^2}}}{a+\frac {b}{x}}-\frac {b \log \left (\frac {1}{x}\right ) \sqrt {a^2+\frac {2 a b}{x}+\frac {b^2}{x^2}}}{a+\frac {b}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a^2 + b^2/x^2 + (2*a*b)/x]*x)/(a + b/x) - (b*Sqrt[a^2 + b^2/x^2 + (2*a*b)/x]*Log[x^(-1)])/(a + b/x)

Rule 45

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 660

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 1356

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n + c/x^(2*n))^p/x^2,
x], x, 1/x] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \sqrt {a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}} \, dx &=-\text {Subst}\left (\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\sqrt {a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}} \text {Subst}\left (\int \frac {a b+b^2 x}{x^2} \, dx,x,\frac {1}{x}\right )}{a b+\frac {b^2}{x}}\\ &=-\frac {\sqrt {a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}} \text {Subst}\left (\int \left (\frac {a b}{x^2}+\frac {b^2}{x}\right ) \, dx,x,\frac {1}{x}\right )}{a b+\frac {b^2}{x}}\\ &=\frac {a \sqrt {a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}} x}{a+\frac {b}{x}}+\frac {b \sqrt {a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}} \log (x)}{a+\frac {b}{x}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 32, normalized size = 0.44 \begin {gather*} \frac {x \sqrt {\frac {(b+a x)^2}{x^2}} (a x+b \log (x))}{b+a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 40, normalized size = 0.55

method	result	size
default	\(\frac {\sqrt {\frac {a^{2} x^{2}+2 a b x +b^{2}}{x^{2}}}\, x \left (a x +b \ln \left (x \right )\right )}{a x +b}\)	\(40\)
risch	\(\frac {\sqrt {\frac {\left (a x +b \right )^{2}}{x^{2}}}\, x^{2} a}{a x +b}+\frac {\sqrt {\frac {\left (a x +b \right )^{2}}{x^{2}}}\, x b \ln \left (x \right )}{a x +b}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2*x^2+2*a*b*x+b^2)/x^2)^(1/2)/(a*x+b)*x*(a*x+b*ln(x))

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Maxima [A]
time = 0.30, size = 8, normalized size = 0.11 \begin {gather*} a x + b \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*log(x)

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Fricas [A]
time = 0.42, size = 8, normalized size = 0.11 \begin {gather*} a x + b \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*log(x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a^{2} + \frac {2 a b}{x} + \frac {b^{2}}{x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a**2 + 2*a*b/x + b**2/x**2), x)

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Giac [A]
time = 4.39, size = 29, normalized size = 0.40 \begin {gather*} a x \mathrm {sgn}\left (a x^{2} + b x\right ) + b \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x^{2} + b x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x*sgn(a*x^2 + b*x) + b*log(abs(x))*sgn(a*x^2 + b*x)

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Mupad [B]
time = 0.11, size = 134, normalized size = 1.84 \begin {gather*} x\,\sqrt {\frac {1}{x^2}}\,\sqrt {a^2\,x^2+2\,a\,b\,x+b^2}-x\,\ln \left (\frac {2\,\sqrt {b^2}\,\sqrt {a^2\,x^2+2\,a\,b\,x+b^2}+2\,b^2+2\,a\,b\,x}{x}\right )\,\sqrt {b^2}\,\sqrt {\frac {1}{x^2}}+\frac {a\,b\,x\,\ln \left (\frac {a\,b+\sqrt {a^2}\,\sqrt {a^2\,x^2+2\,a\,b\,x+b^2}+a^2\,x}{\sqrt {a^2}}\right )\,\sqrt {\frac {1}{x^2}}}{\sqrt {a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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