3.43 \(\int x^2 \tan ^{-1}(\cot (a+b x)) \, dx\)

Optimal. Leaf size=23 \[ \frac{1}{3} x^3 \tan ^{-1}(\cot (a+b x))+\frac{b x^4}{12} \]

[Out]

(b*x^4)/12 + (x^3*ArcTan[Cot[a + b*x]])/3

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Rubi [A]  time = 0.0083937, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2168, 30} \[ \frac{1}{3} x^3 \tan ^{-1}(\cot (a+b x))+\frac{b x^4}{12} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTan[Cot[a + b*x]],x]

[Out]

(b*x^4)/12 + (x^3*ArcTan[Cot[a + b*x]])/3

Rule 2168

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^
n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

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^2 \tan ^{-1}(\cot (a+b x)) \, dx &=\frac{1}{3} x^3 \tan ^{-1}(\cot (a+b x))+\frac{1}{3} b \int x^3 \, dx\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \tan ^{-1}(\cot (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.0151806, size = 20, normalized size = 0.87 \[ \frac{1}{12} x^3 \left (4 \tan ^{-1}(\cot (a+b x))+b x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTan[Cot[a + b*x]],x]

[Out]

(x^3*(b*x + 4*ArcTan[Cot[a + b*x]]))/12

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Maple [B]  time = 0.061, size = 65, normalized size = 2.8 \begin{align*}{\frac{\pi \,{x}^{3}}{6}}-{\frac{{x}^{3}{\rm arccot} \left (\cot \left ( bx+a \right ) \right )}{3}}-{\frac{1}{3\,{b}^{3}} \left ( -{\frac{ \left ( bx+a \right ) ^{4}}{4}}+a \left ( bx+a \right ) ^{3}-{\frac{3\,{a}^{2} \left ( bx+a \right ) ^{2}}{2}}+ \left ( bx+a \right ){a}^{3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*Pi*x^3-1/3*x^3*arccot(cot(b*x+a))-1/3/b^3*(-1/4*(b*x+a)^4+a*(b*x+a)^3-3/2*a^2*(b*x+a)^2+(b*x+a)*a^3)

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Maxima [A]  time = 0.964157, size = 23, normalized size = 1. \begin{align*} -\frac{1}{4} \, b x^{4} + \frac{1}{6} \,{\left (\pi - 2 \, a\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*x^4 + 1/6*(pi - 2*a)*x^3

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Fricas [A]  time = 2.07415, size = 45, normalized size = 1.96 \begin{align*} -\frac{1}{4} \, b x^{4} + \frac{1}{6} \,{\left (\pi - 2 \, a\right )} x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*x^4 + 1/6*(pi - 2*a)*x^3

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Sympy [A]  time = 0.782186, size = 68, normalized size = 2.96 \begin{align*} \begin{cases} \frac{\pi x^{3}}{6} - \frac{x^{2} \operatorname{acot}^{2}{\left (\cot{\left (a + b x \right )} \right )}}{2 b} + \frac{x \operatorname{acot}^{3}{\left (\cot{\left (a + b x \right )} \right )}}{3 b^{2}} - \frac{\operatorname{acot}^{4}{\left (\cot{\left (a + b x \right )} \right )}}{12 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \left (- \operatorname{acot}{\left (\cot{\left (a \right )} \right )} + \frac{\pi }{2}\right )}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((pi*x**3/6 - x**2*acot(cot(a + b*x))**2/(2*b) + x*acot(cot(a + b*x))**3/(3*b**2) - acot(cot(a + b*x)
)**4/(12*b**3), Ne(b, 0)), (x**3*(-acot(cot(a)) + pi/2)/3, True))

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Giac [A]  time = 1.15063, size = 26, normalized size = 1.13 \begin{align*} -\frac{1}{4} \, b x^{4} + \frac{1}{6} \, \pi x^{3} - \frac{1}{3} \, a x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*x^4 + 1/6*pi*x^3 - 1/3*a*x^3