3.124 \(\int \frac {1}{(a+a x^2) (b-2 b \tan ^{-1}(x))} \, dx\)

Optimal. Leaf size=17 \[ -\frac {\log \left (1-2 \tan ^{-1}(x)\right )}{2 a b} \]

[Out]

-1/2*ln(1-2*arctan(x))/a/b

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {4882} \[ -\frac {\log \left (1-2 \tan ^{-1}(x)\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*x^2)*(b - 2*b*ArcTan[x])),x]

[Out]

-Log[1 - 2*ArcTan[x]]/(2*a*b)

Rule 4882

Int[1/(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[Log[RemoveContent[a + b*Ar
cTan[c*x], x]]/(b*c*d), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 17, normalized size = 1.00 \[ -\frac {\log \left (2 \tan ^{-1}(x)-1\right )}{2 a b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*x^2)*(b - 2*b*ArcTan[x])),x]

[Out]

-1/2*Log[-1 + 2*ArcTan[x]]/(a*b)

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fricas [A]  time = 0.58, size = 15, normalized size = 0.88 \[ -\frac {\log \left (2 \, \arctan \relax (x) - 1\right )}{2 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(2*arctan(x) - 1)/(a*b)

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giac [A]  time = 0.11, size = 16, normalized size = 0.94 \[ -\frac {\log \left ({\left | 2 \, \arctan \relax (x) - 1 \right |}\right )}{2 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(abs(2*arctan(x) - 1))/(a*b)

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maple [A]  time = 0.08, size = 19, normalized size = 1.12 \[ -\frac {\ln \left (2 b \arctan \relax (x )-b \right )}{2 a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a*ln(2*b*arctan(x)-b)/b

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maxima [A]  time = 0.33, size = 16, normalized size = 0.94 \[ -\frac {\log \left ({\left | 2 \, \arctan \relax (x) - 1 \right |}\right )}{2 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(abs(2*arctan(x) - 1))/(a*b)

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mupad [B]  time = 0.13, size = 15, normalized size = 0.88 \[ -\frac {\ln \left (2\,\mathrm {atan}\relax (x)-1\right )}{2\,a\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(2*atan(x) - 1)/(2*a*b)

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sympy [A]  time = 0.51, size = 14, normalized size = 0.82 \[ - \frac {\log {\left (\operatorname {atan}{\relax (x )} - \frac {1}{2} \right )}}{2 a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(atan(x) - 1/2)/(2*a*b)

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