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

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

[Out]

-1/12*b*x^4+1/3*x^3*arctan(tan(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} \[ \frac {1}{3} x^3 \tan ^{-1}(\tan (a+b x))-\frac {b x^4}{12} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTan[Tan[a + b*x]],x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.87 \[ -\frac {1}{12} x^3 \left (b x-4 \tan ^{-1}(\tan (a+b x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTan[Tan[a + b*x]],x]

[Out]

-1/12*(x^3*(b*x - 4*ArcTan[Tan[a + b*x]]))

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fricas [A]  time = 0.82, size = 13, normalized size = 0.57 \[ \frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(tan(b*x+a)),x, algorithm="fricas")

[Out]

1/4*b*x^4 + 1/3*a*x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(tan(b*x+a)),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.18, size = 20, normalized size = 0.87 \[ -\frac {b \,x^{4}}{12}+\frac {x^{3} \arctan \left (\tan \left (b x +a \right )\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctan(tan(b*x+a)),x)

[Out]

-1/12*b*x^4+1/3*x^3*arctan(tan(b*x+a))

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maxima [B]  time = 0.32, size = 81, normalized size = 3.52 \[ \frac {\frac {4 \, {\left ({\left (b x + a\right )}^{3} - 3 \, {\left (b x + a\right )}^{2} a + 3 \, {\left (b x + a\right )} a^{2}\right )} \arctan \left (\tan \left (b x + a\right )\right )}{b^{2}} - \frac {{\left (b x + a\right )}^{4} - 4 \, {\left (b x + a\right )}^{3} a + 6 \, {\left (b x + a\right )}^{2} a^{2}}{b^{2}}}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(tan(b*x+a)),x, algorithm="maxima")

[Out]

1/12*(4*((b*x + a)^3 - 3*(b*x + a)^2*a + 3*(b*x + a)*a^2)*arctan(tan(b*x + a))/b^2 - ((b*x + a)^4 - 4*(b*x + a
)^3*a + 6*(b*x + a)^2*a^2)/b^2)/b

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mupad [B]  time = 0.12, size = 19, normalized size = 0.83 \[ \frac {x^3\,\mathrm {atan}\left (\mathrm {tan}\left (a+b\,x\right )\right )}{3}-\frac {b\,x^4}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atan(tan(a + b*x)),x)

[Out]

(x^3*atan(tan(a + b*x)))/3 - (b*x^4)/12

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sympy [A]  time = 0.36, size = 32, normalized size = 1.39 \[ - \frac {b x^{4}}{12} + \frac {x^{3} \left (\operatorname {atan}{\left (\tan {\left (a + b x \right )} \right )} + \pi \left \lfloor {\frac {a + b x - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atan(tan(b*x+a)),x)

[Out]

-b*x**4/12 + x**3*(atan(tan(a + b*x)) + pi*floor((a + b*x - pi/2)/pi))/3

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