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

Optimal. Leaf size=16 \[ \frac{\tan ^{-1}(\tan (a+b x))^2}{2 b} \]

[Out]

ArcTan[Tan[a + b*x]]^2/(2*b)

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Rubi [A]  time = 0.002841, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2157, 30} \[ \frac{\tan ^{-1}(\tan (a+b x))^2}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[Tan[a + b*x]],x]

[Out]

ArcTan[Tan[a + b*x]]^2/(2*b)

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0074467, size = 18, normalized size = 1.12 \[ x \tan ^{-1}(\tan (a+b x))-\frac{b x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[Tan[a + b*x]],x]

[Out]

-(b*x^2)/2 + x*ArcTan[Tan[a + b*x]]

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Maple [A]  time = 0.034, size = 15, normalized size = 0.9 \begin{align*}{\frac{ \left ( \arctan \left ( \tan \left ( bx+a \right ) \right ) \right ) ^{2}}{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(tan(b*x+a)),x)

[Out]

1/2*arctan(tan(b*x+a))^2/b

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Maxima [A]  time = 0.972396, size = 16, normalized size = 1. \begin{align*} \frac{{\left (b x + a\right )}^{2}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(tan(b*x+a)),x, algorithm="maxima")

[Out]

1/2*(b*x + a)^2/b

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Fricas [A]  time = 1.75121, size = 23, normalized size = 1.44 \begin{align*} \frac{1}{2} \, b x^{2} + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(tan(b*x+a)),x, algorithm="fricas")

[Out]

1/2*b*x^2 + a*x

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Sympy [A]  time = 0.172969, size = 42, normalized size = 2.62 \begin{align*} \begin{cases} \frac{\left (\operatorname{atan}{\left (\tan{\left (a + b x \right )} \right )} + \pi \left \lfloor{\frac{a + b x - \frac{\pi }{2}}{\pi }}\right \rfloor \right )^{2}}{2 b} & \text{for}\: b \neq 0 \\x \left (\operatorname{atan}{\left (\tan{\left (a \right )} \right )} + \pi \left \lfloor{\frac{a - \frac{\pi }{2}}{\pi }}\right \rfloor \right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(tan(b*x+a)),x)

[Out]

Piecewise(((atan(tan(a + b*x)) + pi*floor((a + b*x - pi/2)/pi))**2/(2*b), Ne(b, 0)), (x*(atan(tan(a)) + pi*flo
or((a - pi/2)/pi)), True))

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Giac [A]  time = 1.07878, size = 35, normalized size = 2.19 \begin{align*} \frac{1}{2} \, b x^{2} - \pi x \left \lfloor \frac{b x + a}{\pi } + \frac{1}{2} \right \rfloor + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(tan(b*x+a)),x, algorithm="giac")

[Out]

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