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

Optimal. Leaf size=23 \[ \frac{1}{2} x^2 \tan ^{-1}(\tan (a+b x))-\frac{b x^3}{6} \]

[Out]

-(b*x^3)/6 + (x^2*ArcTan[Tan[a + b*x]])/2

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Rubi [A]  time = 0.0070475, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {5171, 30} \[ \frac{1}{2} x^2 \tan ^{-1}(\tan (a+b x))-\frac{b x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTan[Tan[a + b*x]],x]

[Out]

-(b*x^3)/6 + (x^2*ArcTan[Tan[a + b*x]])/2

Rule 5171

Int[ArcTan[(c_.) + (d_.)*Tan[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcTan[c + d*Tan[a + b*x]])/(f*(m + 1)), x] - Dist[(I*b)/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c + I*d + c*
E^(2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c + I*d)^2, -1]

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 x \tan ^{-1}(\tan (a+b x)) \, dx &=\frac{1}{2} x^2 \tan ^{-1}(\tan (a+b x))-\frac{1}{2} b \int x^2 \, dx\\ &=-\frac{b x^3}{6}+\frac{1}{2} x^2 \tan ^{-1}(\tan (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.0148217, size = 20, normalized size = 0.87 \[ -\frac{1}{6} x^2 \left (b x-3 \tan ^{-1}(\tan (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTan[Tan[a + b*x]],x]

[Out]

-(x^2*(b*x - 3*ArcTan[Tan[a + b*x]]))/6

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Maple [A]  time = 0.048, size = 20, normalized size = 0.9 \begin{align*} -{\frac{b{x}^{3}}{6}}+{\frac{{x}^{2}\arctan \left ( \tan \left ( bx+a \right ) \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*x^3+1/2*x^2*arctan(tan(b*x+a))

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Maxima [B]  time = 1.00108, size = 77, normalized size = 3.35 \begin{align*} \frac{\frac{3 \,{\left ({\left (b x + a\right )}^{2} - 2 \,{\left (b x + a\right )} a\right )} \arctan \left (\tan \left (b x + a\right )\right )}{b} - \frac{{\left (b x + a\right )}^{3} - 3 \,{\left (b x + a\right )}^{2} a}{b}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*((b*x + a)^2 - 2*(b*x + a)*a)*arctan(tan(b*x + a))/b - ((b*x + a)^3 - 3*(b*x + a)^2*a)/b)/b

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Fricas [A]  time = 1.74936, size = 31, normalized size = 1.35 \begin{align*} \frac{1}{3} \, b x^{3} + \frac{1}{2} \, a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3 + 1/2*a*x^2

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Sympy [A]  time = 0.241458, size = 32, normalized size = 1.39 \begin{align*} - \frac{b x^{3}}{6} + \frac{x^{2} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*x**3/6 + x**2*(atan(tan(a + b*x)) + pi*floor((a + b*x - pi/2)/pi))/2

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Giac [A]  time = 1.11709, size = 36, normalized size = 1.57 \begin{align*} \frac{1}{3} \, b x^{3} - \frac{1}{2} \, \pi x^{2} \left \lfloor \frac{a}{\pi } + \frac{1}{2} \right \rfloor + \frac{1}{2} \, a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3 - 1/2*pi*x^2*floor(a/pi + 1/2) + 1/2*a*x^2