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]

1/6*b*x^3+1/2*x^2*(1/2*Pi-arccot(cot(b*x+a)))

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, 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 30

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

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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.39, size = 17, normalized size = 0.74 \[ -\frac {1}{3} \, b x^{3} + \frac {1}{4} \, {\left (\pi - 2 \, a\right )} x^{2} \]

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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giac [A]  time = 0.14, size = 19, normalized size = 0.83 \[ -\frac {1}{3} \, b x^{3} + \frac {1}{4} \, \pi x^{2} - \frac {1}{2} \, a x^{2} \]

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

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maple [B]  time = 0.44, size = 54, normalized size = 2.35 \[ \frac {\pi \,x^{2}}{4}-\frac {x^{2} \mathrm {arccot}\left (\cot \left (b x +a \right )\right )}{2}-\frac {-\frac {\left (b x +a \right )^{3}}{3}+\left (b x +a \right )^{2} a -a^{2} \left (b x +a \right )}{2 b^{2}} \]

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 = 0.32, size = 17, normalized size = 0.74 \[ -\frac {1}{3} \, b x^{3} + \frac {1}{4} \, {\left (\pi - 2 \, a\right )} x^{2} \]

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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mupad [B]  time = 0.07, size = 25, normalized size = 1.09 \[ \frac {\Pi \,x^2}{4}+\frac {b\,x^3}{6}-\frac {x^2\,\mathrm {acot}\left (\mathrm {cot}\left (a+b\,x\right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(Pi/2 - acot(cot(a + b*x))),x)

[Out]

(Pi*x^2)/4 + (b*x^3)/6 - (x^2*acot(cot(a + b*x)))/2

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sympy [A]  time = 0.30, size = 49, normalized size = 2.13 \[ \begin {cases} \frac {\pi x^{2}}{4} - \frac {x \operatorname {acot}^{2}{\left (\cot {\left (a + b x \right )} \right )}}{2 b} + \frac {\operatorname {acot}^{3}{\left (\cot {\left (a + b x \right )} \right )}}{6 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \left (- \operatorname {acot}{\left (\cot {\relax (a )} \right )} + \frac {\pi }{2}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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