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

Optimal. Leaf size=27 \[ \frac{\log (\sin (a+b x))}{b}-\frac{\sin ^2(a+b x)}{2 b} \]

[Out]

Log[Sin[a + b*x]]/b - Sin[a + b*x]^2/(2*b)

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Rubi [A]  time = 0.0211221, antiderivative size = 27, 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} \[ \frac{\log (\sin (a+b x))}{b}-\frac{\sin ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Cot[a + b*x],x]

[Out]

Log[Sin[a + b*x]]/b - Sin[a + b*x]^2/(2*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 ^2(a+b x) \cot (a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{x} \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{x}-x\right ) \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac{\log (\sin (a+b x))}{b}-\frac{\sin ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0142919, size = 27, normalized size = 1. \[ \frac{\log (\sin (a+b x))}{b}-\frac{\sin ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Sin[a + b*x]]/b - Sin[a + b*x]^2/(2*b)

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Maple [A]  time = 0.016, size = 26, normalized size = 1. \begin{align*}{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2\,b}}+{\frac{\ln \left ( \sin \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.960211, size = 34, normalized size = 1.26 \begin{align*} -\frac{\sin \left (b x + a\right )^{2} - \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67852, size = 68, normalized size = 2.52 \begin{align*} \frac{\cos \left (b x + a\right )^{2} + 2 \, \log \left (\frac{1}{2} \, \sin \left (b x + a\right )\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.93497, size = 369, normalized size = 13.67 \begin{align*} \begin{cases} - \frac{\log{\left (\tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 1 \right )} \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + b} - \frac{2 \log{\left (\tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + b} - \frac{\log{\left (\tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 1 \right )}}{b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + b} + \frac{\log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )} \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + b} + \frac{2 \log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )} \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + b} + \frac{\log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )}}{b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + b} - \frac{2 \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + b} & \text{for}\: b \neq 0 \\\frac{x \cos ^{3}{\left (a \right )}}{\sin{\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),x)

[Out]

Piecewise((-log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2
+ b) - 2*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b
) - log(tan(a/2 + b*x/2)**2 + 1)/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b) + log(tan(a/2 + b*x/2))
*tan(a/2 + b*x/2)**4/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b) + 2*log(tan(a/2 + b*x/2))*tan(a/2 +
b*x/2)**2/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b) + log(tan(a/2 + b*x/2))/(b*tan(a/2 + b*x/2)**
4 + 2*b*tan(a/2 + b*x/2)**2 + b) - 2*tan(a/2 + b*x/2)**2/(b*tan(a/2 + b*x/2)**4 + 2*b*tan(a/2 + b*x/2)**2 + b)
, Ne(b, 0)), (x*cos(a)**3/sin(a), True))

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Giac [A]  time = 1.17104, size = 34, normalized size = 1.26 \begin{align*} -\frac{\sin \left (b x + a\right )^{2} - \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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