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

Optimal. Leaf size=23 \[ -\frac{\sin (a+b x)}{b}-\frac{\csc (a+b x)}{b} \]

[Out]

-(Csc[a + b*x]/b) - Sin[a + b*x]/b

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Rubi [A]  time = 0.0203114, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2590, 14} \[ -\frac{\sin (a+b x)}{b}-\frac{\csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Cot[a + b*x]^2,x]

[Out]

-(Csc[a + b*x]/b) - Sin[a + b*x]/b

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cos (a+b x) \cot ^2(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,-\sin (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,-\sin (a+b x)\right )}{b}\\ &=-\frac{\csc (a+b x)}{b}-\frac{\sin (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0118145, size = 23, normalized size = 1. \[ -\frac{\sin (a+b x)}{b}-\frac{\csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Cot[a + b*x]^2,x]

[Out]

-(Csc[a + b*x]/b) - Sin[a + b*x]/b

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Maple [A]  time = 0.011, size = 42, normalized size = 1.8 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{\sin \left ( bx+a \right ) }}- \left ( 2+ \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) \sin \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3/sin(b*x+a)^2,x)

[Out]

1/b*(-cos(b*x+a)^4/sin(b*x+a)-(2+cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 0.992325, size = 27, normalized size = 1.17 \begin{align*} -\frac{\frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-(1/sin(b*x + a) + sin(b*x + a))/b

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Fricas [A]  time = 1.83447, size = 53, normalized size = 2.3 \begin{align*} \frac{\cos \left (b x + a\right )^{2} - 2}{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)^3/sin(b*x+a)^2,x, algorithm="fricas")

[Out]

(cos(b*x + a)^2 - 2)/(b*sin(b*x + a))

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Sympy [A]  time = 0.983448, size = 39, normalized size = 1.7 \begin{align*} \begin{cases} - \frac{2 \sin{\left (a + b x \right )}}{b} - \frac{\cos ^{2}{\left (a + b x \right )}}{b \sin{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{3}{\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)**3/sin(b*x+a)**2,x)

[Out]

Piecewise((-2*sin(a + b*x)/b - cos(a + b*x)**2/(b*sin(a + b*x)), Ne(b, 0)), (x*cos(a)**3/sin(a)**2, True))

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Giac [A]  time = 1.13176, size = 27, normalized size = 1.17 \begin{align*} -\frac{\frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(b*x+a)^2,x, algorithm="giac")

[Out]

-(1/sin(b*x + a) + sin(b*x + a))/b