3.138 \(\int \cot ^2(a+b x) \, dx\)

Optimal. Leaf size=15 \[ -\frac{\cot (a+b x)}{b}-x \]

[Out]

-x - Cot[a + b*x]/b

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Rubi [A]  time = 0.0078847, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3473, 8} \[ -\frac{\cot (a+b x)}{b}-x \]

Antiderivative was successfully verified.

[In]

Int[Cot[a + b*x]^2,x]

[Out]

-x - Cot[a + b*x]/b

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [C]  time = 0.0132666, size = 29, normalized size = 1.93 \[ -\frac{\cot (a+b x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\tan ^2(a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[a + b*x]^2,x]

[Out]

-((Cot[a + b*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[a + b*x]^2])/b)

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Maple [A]  time = 0.01, size = 21, normalized size = 1.4 \begin{align*}{\frac{-\cot \left ( bx+a \right ) -bx-a}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/sin(b*x+a)^2,x)

[Out]

1/b*(-cot(b*x+a)-b*x-a)

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Maxima [A]  time = 1.47918, size = 24, normalized size = 1.6 \begin{align*} -\frac{b x + a + \frac{1}{\tan \left (b x + a\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-(b*x + a + 1/tan(b*x + a))/b

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Fricas [A]  time = 1.93702, size = 72, normalized size = 4.8 \begin{align*} -\frac{b x \sin \left (b x + a\right ) + \cos \left (b x + a\right )}{b \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^2,x, algorithm="fricas")

[Out]

-(b*x*sin(b*x + a) + cos(b*x + a))/(b*sin(b*x + a))

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Sympy [A]  time = 0.693617, size = 29, normalized size = 1.93 \begin{align*} \begin{cases} - x - \frac{\cos{\left (a + b x \right )}}{b \sin{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{2}{\left (a \right )}}{\sin ^{2}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2/sin(b*x+a)**2,x)

[Out]

Piecewise((-x - cos(a + b*x)/(b*sin(a + b*x)), Ne(b, 0)), (x*cos(a)**2/sin(a)**2, True))

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Giac [B]  time = 1.16921, size = 47, normalized size = 3.13 \begin{align*} -\frac{2 \, b x + 2 \, a + \frac{1}{\tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )} - \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/2*(2*b*x + 2*a + 1/tan(1/2*b*x + 1/2*a) - tan(1/2*b*x + 1/2*a))/b