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

Optimal. Leaf size=36 \[ \frac{x^{m+1} \tan ^{-1}(\cot (a+b x))}{m+1}+\frac{b x^{m+2}}{m^2+3 m+2} \]

[Out]

(b*x^(2 + m))/(2 + 3*m + m^2) + (x^(1 + m)*ArcTan[Cot[a + b*x]])/(1 + m)

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Rubi [A]  time = 0.0210559, antiderivative size = 36, 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} \[ \frac{x^{m+1} \tan ^{-1}(\cot (a+b x))}{m+1}+\frac{b x^{m+2}}{m^2+3 m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*ArcTan[Cot[a + b*x]],x]

[Out]

(b*x^(2 + m))/(2 + 3*m + m^2) + (x^(1 + m)*ArcTan[Cot[a + b*x]])/(1 + m)

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

Mathematica [A]  time = 0.0479835, size = 31, normalized size = 0.86 \[ \frac{x^{m+1} \left ((m+2) \tan ^{-1}(\cot (a+b x))+b x\right )}{(m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*ArcTan[Cot[a + b*x]],x]

[Out]

(x^(1 + m)*(b*x + (2 + m)*ArcTan[Cot[a + b*x]]))/((1 + m)*(2 + m))

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Maple [A]  time = 0.052, size = 56, normalized size = 1.6 \begin{align*}{\frac{\pi \,{x}^{1+m}}{2+2\,m}}-{\frac{b{x}^{2}{{\rm e}^{m\ln \left ( x \right ) }}}{2+m}}-{\frac{ \left ({\rm arccot} \left (\cot \left ( bx+a \right ) \right )-bx \right ) x{{\rm e}^{m\ln \left ( x \right ) }}}{1+m}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*Pi*x^(1+m)/(1+m)-b/(2+m)*x^2*exp(m*ln(x))-(arccot(cot(b*x+a))-b*x)/(1+m)*x*exp(m*ln(x))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07195, size = 101, normalized size = 2.81 \begin{align*} -\frac{{\left (2 \,{\left (b m + b\right )} x^{2} -{\left (\pi{\left (m + 2\right )} - 2 \, a m - 4 \, a\right )} x\right )} x^{m}}{2 \,{\left (m^{2} + 3 \, m + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(b*m + b)*x^2 - (pi*(m + 2) - 2*a*m - 4*a)*x)*x^m/(m^2 + 3*m + 2)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.09397, size = 84, normalized size = 2.33 \begin{align*} -\frac{2 \, b m x^{2} x^{m} - \pi m x x^{m} + 2 \, a m x x^{m} + 2 \, b x^{2} x^{m} - 2 \, \pi x x^{m} + 4 \, a x x^{m}}{2 \,{\left (m^{2} + 3 \, m + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b*m*x^2*x^m - pi*m*x*x^m + 2*a*m*x*x^m + 2*b*x^2*x^m - 2*pi*x*x^m + 4*a*x*x^m)/(m^2 + 3*m + 2)