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]

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

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, 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 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]]

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]

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*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {\log (\sin (a+b x))}{b}-\frac {\sin ^2(a+b x)}{2 b} \]

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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fricas [A]  time = 0.44, size = 25, normalized size = 0.93 \[ \frac {\cos \left (b x + a\right )^{2} + 2 \, \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right )}{2 \, b} \]

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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giac [A]  time = 0.57, size = 25, normalized size = 0.93 \[ -\frac {\sin \left (b x + a\right )^{2} - \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \]

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

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maple [A]  time = 0.03, size = 26, normalized size = 0.96 \[ \frac {\cos ^{2}\left (b x +a \right )}{2 b}+\frac {\ln \left (\sin \left (b x +a \right )\right )}{b} \]

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 = 0.54, size = 25, normalized size = 0.93 \[ -\frac {\sin \left (b x + a\right )^{2} - \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \]

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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mupad [B]  time = 0.41, size = 35, normalized size = 1.30 \[ \frac {\frac {{\cos \left (a+b\,x\right )}^2}{2}-\frac {\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{2}+\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3/sin(a + b*x),x)

[Out]

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

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sympy [A]  time = 2.35, size = 369, normalized size = 13.67 \[ \begin {cases} - \frac {\log {\left (\tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )} \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {2 \log {\left (\tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )} \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {\log {\left (\tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )}}{b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {\log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {2 \log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {\log {\left (\tan {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )}}{b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {2 \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tan ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 2 b \tan ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} & \text {for}\: b \neq 0 \\\frac {x \cos ^{3}{\relax (a )}}{\sin {\relax (a )}} & \text {otherwise} \end {cases} \]

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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