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]

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

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, 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 30

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

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]

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*}

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Mathematica [A]  time = 0.01, 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]

-1/2*(b*x^2) + x*ArcTan[Tan[a + b*x]]

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fricas [A]  time = 0.76, size = 10, normalized size = 0.62 \[ \frac {1}{2} \, b x^{2} + a x \]

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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giac [A]  time = 0.13, size = 26, normalized size = 1.62 \[ \frac {1}{2} \, b x^{2} - \pi x \left \lfloor \frac {b x + a}{\pi } + \frac {1}{2} \right \rfloor + a x \]

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

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maple [A]  time = 0.04, size = 15, normalized size = 0.94 \[ \frac {\arctan \left (\tan \left (b x +a \right )\right )^{2}}{2 b} \]

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 = 0.32, size = 12, normalized size = 0.75 \[ \frac {{\left (b x + a\right )}^{2}}{2 \, b} \]

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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mupad [B]  time = 0.04, size = 16, normalized size = 1.00 \[ x\,\mathrm {atan}\left (\mathrm {tan}\left (a+b\,x\right )\right )-\frac {b\,x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(tan(a + b*x)),x)

[Out]

x*atan(tan(a + b*x)) - (b*x^2)/2

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sympy [A]  time = 0.14, size = 42, normalized size = 2.62 \[ \begin {cases} \frac {\left (\operatorname {atan}{\left (\tan {\left (a + b x \right )} \right )} + \pi \left \lfloor {\frac {a + b x - \frac {\pi }{2}}{\pi }}\right \rfloor \right )^{2}}{2 b} & \text {for}\: b \neq 0 \\x \left (\operatorname {atan}{\left (\tan {\relax (a )} \right )} + \pi \left \lfloor {\frac {a - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) & \text {otherwise} \end {cases} \]

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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