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

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

[Out]

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

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Rubi [A]  time = 0.0071134, 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 = {5173, 30} \[ \frac{1}{2} x^2 \tan ^{-1}(\cot (a+b x))+\frac{b x^3}{6} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTan[Cot[a + b*x]],x]

[Out]

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

Rule 5173

Int[ArcTan[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcTan[c + d*Cot[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}(\cot (a+b x)) \, dx &=\frac{1}{2} x^2 \tan ^{-1}(\cot (a+b x))+\frac{1}{2} b \int x^2 \, dx\\ &=\frac{b x^3}{6}+\frac{1}{2} x^2 \tan ^{-1}(\cot (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.0142949, size = 20, normalized size = 0.87 \[ \frac{1}{6} x^2 \left (3 \tan ^{-1}(\cot (a+b x))+b x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcTan[Cot[a + b*x]],x]

[Out]

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

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Maple [B]  time = 0.057, size = 54, normalized size = 2.4 \begin{align*}{\frac{\pi \,{x}^{2}}{4}}-{\frac{{x}^{2}{\rm arccot} \left (\cot \left ( bx+a \right ) \right )}{2}}-{\frac{1}{2\,{b}^{2}} \left ( -{\frac{ \left ( bx+a \right ) ^{3}}{3}}+ \left ( bx+a \right ) ^{2}a-{a}^{2} \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(1/2*Pi-arccot(cot(b*x+a))),x)

[Out]

1/4*Pi*x^2-1/2*x^2*arccot(cot(b*x+a))-1/2/b^2*(-1/3*(b*x+a)^3+(b*x+a)^2*a-a^2*(b*x+a))

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Maxima [A]  time = 0.991631, size = 23, normalized size = 1. \begin{align*} -\frac{1}{3} \, b x^{3} + \frac{1}{4} \,{\left (\pi - 2 \, a\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1/2*pi-arccot(cot(b*x+a))),x, algorithm="maxima")

[Out]

-1/3*b*x^3 + 1/4*(pi - 2*a)*x^2

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Fricas [A]  time = 2.04841, size = 45, normalized size = 1.96 \begin{align*} -\frac{1}{3} \, b x^{3} + \frac{1}{4} \,{\left (\pi - 2 \, a\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1/2*pi-arccot(cot(b*x+a))),x, algorithm="fricas")

[Out]

-1/3*b*x^3 + 1/4*(pi - 2*a)*x^2

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Sympy [A]  time = 0.381749, size = 65, normalized size = 2.83 \begin{align*} \begin{cases} \frac{b x^{3}}{6} - \frac{x^{2} \operatorname{acot}{\left (\cot{\left (a + b x \right )} \right )}}{2} + \frac{\pi x \operatorname{acot}{\left (\cot{\left (a + b x \right )} \right )}}{2 b} - \frac{\pi \operatorname{acot}^{2}{\left (\cot{\left (a + b x \right )} \right )}}{4 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \left (- \operatorname{acot}{\left (\cot{\left (a \right )} \right )} + \frac{\pi }{2}\right )}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1/2*pi-acot(cot(b*x+a))),x)

[Out]

Piecewise((b*x**3/6 - x**2*acot(cot(a + b*x))/2 + pi*x*acot(cot(a + b*x))/(2*b) - pi*acot(cot(a + b*x))**2/(4*
b**2), Ne(b, 0)), (x**2*(-acot(cot(a)) + pi/2)/2, True))

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Giac [A]  time = 1.11006, size = 26, normalized size = 1.13 \begin{align*} -\frac{1}{3} \, b x^{3} + \frac{1}{4} \, \pi x^{2} - \frac{1}{2} \, a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1/2*pi-arccot(cot(b*x+a))),x, algorithm="giac")

[Out]

-1/3*b*x^3 + 1/4*pi*x^2 - 1/2*a*x^2