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

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Rubi [A]  time = 0.02, antiderivative size = 36, 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} \[ \frac {x^{m+1} \tan ^{-1}(\cot (a+b x))}{m+1}+\frac {b x^{m+2}}{m^2+3 m+2} \]

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 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^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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.55, size = 42, normalized size = 1.17 \[ -\frac {{\left (2 \, {\left (b m + b\right )} x^{2} - {\left (\pi {\left (m + 2\right )} - 2 \, a m - 4 \, a\right )} x\right )} x^{m}}{2 \, {\left (m^{2} + 3 \, m + 2\right )}} \]

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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giac [A]  time = 0.14, size = 62, normalized size = 1.72 \[ -\frac {2 \, b m x^{2} x^{m} - \pi m x x^{m} + 2 \, a m x x^{m} + 2 \, b x^{2} x^{m} - 2 \, \pi x x^{m} + 4 \, a x x^{m}}{2 \, {\left (m^{2} + 3 \, m + 2\right )}} \]

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)

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maple [A]  time = 0.31, size = 56, normalized size = 1.56 \[ \frac {\pi \,x^{1+m}}{2+2 m}-\frac {b \,x^{2} {\mathrm e}^{m \ln \relax (x )}}{2+m}-\frac {\left (\mathrm {arccot}\left (\cot \left (b x +a \right )\right )-b x \right ) x \,{\mathrm e}^{m \ln \relax (x )}}{1+m} \]

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 [A]  time = 0.32, size = 40, normalized size = 1.11 \[ -\frac {b x^{m + 2}}{m + 2} + \frac {\pi x^{m + 1}}{2 \, {\left (m + 1\right )}} - \frac {a x^{m + 1}}{m + 1} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int x^m\,\left (\frac {\Pi }{2}-\mathrm {acot}\left (\mathrm {cot}\left (a+b\,x\right )\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.61, size = 160, normalized size = 4.44 \[ \begin {cases} - b \log {\relax (x )} + \frac {\operatorname {acot}{\left (\cot {\left (a + b x \right )} \right )}}{x} - \frac {\pi }{2 x} & \text {for}\: m = -2 \\b x \log {\relax (x )} - b x - \log {\relax (x )} \operatorname {acot}{\left (\cot {\left (a + b x \right )} \right )} + \frac {\pi \log {\relax (x )}}{2} & \text {for}\: m = -1 \\\frac {2 b x^{2} x^{m}}{2 m^{2} + 6 m + 4} - \frac {2 m x x^{m} \operatorname {acot}{\left (\cot {\left (a + b x \right )} \right )}}{2 m^{2} + 6 m + 4} + \frac {\pi m x x^{m}}{2 m^{2} + 6 m + 4} - \frac {4 x x^{m} \operatorname {acot}{\left (\cot {\left (a + b x \right )} \right )}}{2 m^{2} + 6 m + 4} + \frac {2 \pi x x^{m}}{2 m^{2} + 6 m + 4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-b*log(x) + acot(cot(a + b*x))/x - pi/(2*x), Eq(m, -2)), (b*x*log(x) - b*x - log(x)*acot(cot(a + b*
x)) + pi*log(x)/2, Eq(m, -1)), (2*b*x**2*x**m/(2*m**2 + 6*m + 4) - 2*m*x*x**m*acot(cot(a + b*x))/(2*m**2 + 6*m
+ 4) + pi*m*x*x**m/(2*m**2 + 6*m + 4) - 4*x*x**m*acot(cot(a + b*x))/(2*m**2 + 6*m + 4) + 2*pi*x*x**m/(2*m**2
+ 6*m + 4), True))

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