3.1564 \(\int (a+\frac {b}{x})^2 \, dx\)

Optimal. Leaf size=20 \[ a^2 x+2 a b \log (x)-\frac {b^2}{x} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {193, 43} \[ a^2 x+2 a b \log (x)-\frac {b^2}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^2,x]

[Out]

-(b^2/x) + a^2*x + 2*a*b*Log[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 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.00 \[ a^2 x+2 a b \log (x)-\frac {b^2}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^2,x]

[Out]

-(b^2/x) + a^2*x + 2*a*b*Log[x]

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fricas [A]  time = 0.75, size = 24, normalized size = 1.20 \[ \frac {a^{2} x^{2} + 2 \, a b x \log \relax (x) - b^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*x^2 + 2*a*b*x*log(x) - b^2)/x

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giac [A]  time = 0.17, size = 21, normalized size = 1.05 \[ a^{2} x + 2 \, a b \log \left ({\left | x \right |}\right ) - \frac {b^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + 2*a*b*log(abs(x)) - b^2/x

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maple [A]  time = 0.00, size = 21, normalized size = 1.05 \[ a^{2} x +2 a b \ln \relax (x )-\frac {b^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 20, normalized size = 1.00 \[ a^{2} x + 2 \, a b \log \relax (x) - \frac {b^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + 2*a*b*log(x) - b^2/x

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mupad [B]  time = 0.03, size = 20, normalized size = 1.00 \[ a^2\,x-\frac {b^2}{x}+2\,a\,b\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^2,x)

[Out]

a^2*x - b^2/x + 2*a*b*log(x)

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sympy [A]  time = 0.12, size = 17, normalized size = 0.85 \[ a^{2} x + 2 a b \log {\relax (x )} - \frac {b^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x + 2*a*b*log(x) - b**2/x

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