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]

1/12*b*x^4+1/3*x^3*(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 = 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 30

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

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]))

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.44, size = 17, normalized size = 0.74 \[ -\frac {1}{4} \, b x^{4} + \frac {1}{6} \, {\left (\pi - 2 \, a\right )} x^{3} \]

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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giac [A]  time = 0.13, size = 19, normalized size = 0.83 \[ -\frac {1}{4} \, b x^{4} + \frac {1}{6} \, \pi x^{3} - \frac {1}{3} \, a x^{3} \]

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

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maple [B]  time = 0.43, size = 65, normalized size = 2.83 \[ \frac {\pi \,x^{3}}{6}-\frac {x^{3} \mathrm {arccot}\left (\cot \left (b x +a \right )\right )}{3}-\frac {-\frac {\left (b x +a \right )^{4}}{4}+a \left (b x +a \right )^{3}-\frac {3 \left (b x +a \right )^{2} a^{2}}{2}+\left (b x +a \right ) a^{3}}{3 b^{3}} \]

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*(b*x+a)^2*a^2+(b*x+a)*a^3)

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maxima [A]  time = 0.32, size = 17, normalized size = 0.74 \[ -\frac {1}{4} \, b x^{4} + \frac {1}{6} \, {\left (\pi - 2 \, a\right )} x^{3} \]

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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mupad [B]  time = 0.54, size = 25, normalized size = 1.09 \[ \frac {\Pi \,x^3}{6}+\frac {b\,x^4}{12}-\frac {x^3\,\mathrm {acot}\left (\mathrm {cot}\left (a+b\,x\right )\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Pi*x^3)/6 + (b*x^4)/12 - (x^3*acot(cot(a + b*x)))/3

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sympy [A]  time = 0.39, size = 26, normalized size = 1.13 \[ \frac {b x^{4}}{12} - \frac {x^{3} \operatorname {acot}{\left (\cot {\left (a + b x \right )} \right )}}{3} + \frac {\pi x^{3}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x**4/12 - x**3*acot(cot(a + b*x))/3 + pi*x**3/6

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