3.122 \(\int \cos ^3(a+b x) \cot (a+b x) \, dx\)

Optimal. Leaf size=38 \[ \frac{\cos ^3(a+b x)}{3 b}+\frac{\cos (a+b x)}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b} \]

[Out]

-(ArcTanh[Cos[a + b*x]]/b) + Cos[a + b*x]/b + Cos[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.025659, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2592, 302, 206} \[ \frac{\cos ^3(a+b x)}{3 b}+\frac{\cos (a+b x)}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Cot[a + b*x],x]

[Out]

-(ArcTanh[Cos[a + b*x]]/b) + Cos[a + b*x]/b + Cos[a + b*x]^3/(3*b)

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \cos ^3(a+b x) \cot (a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=\frac{\cos (a+b x)}{b}+\frac{\cos ^3(a+b x)}{3 b}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\tanh ^{-1}(\cos (a+b x))}{b}+\frac{\cos (a+b x)}{b}+\frac{\cos ^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.024518, size = 60, normalized size = 1.58 \[ \frac{5 \cos (a+b x)}{4 b}+\frac{\cos (3 (a+b x))}{12 b}+\frac{\log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b}-\frac{\log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Cot[a + b*x],x]

[Out]

(5*Cos[a + b*x])/(4*b) + Cos[3*(a + b*x)]/(12*b) - Log[Cos[(a + b*x)/2]]/b + Log[Sin[(a + b*x)/2]]/b

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Maple [A]  time = 0.011, size = 45, normalized size = 1.2 \begin{align*}{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{3\,b}}+{\frac{\cos \left ( bx+a \right ) }{b}}+{\frac{\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^4/sin(b*x+a),x)

[Out]

1/3*cos(b*x+a)^3/b+cos(b*x+a)/b+1/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [A]  time = 0.95757, size = 62, normalized size = 1.63 \begin{align*} \frac{2 \, \cos \left (b x + a\right )^{3} + 6 \, \cos \left (b x + a\right ) - 3 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^4/sin(b*x+a),x, algorithm="maxima")

[Out]

1/6*(2*cos(b*x + a)^3 + 6*cos(b*x + a) - 3*log(cos(b*x + a) + 1) + 3*log(cos(b*x + a) - 1))/b

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Fricas [A]  time = 1.67045, size = 146, normalized size = 3.84 \begin{align*} \frac{2 \, \cos \left (b x + a\right )^{3} + 6 \, \cos \left (b x + a\right ) - 3 \, \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + 3 \, \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^4/sin(b*x+a),x, algorithm="fricas")

[Out]

1/6*(2*cos(b*x + a)^3 + 6*cos(b*x + a) - 3*log(1/2*cos(b*x + a) + 1/2) + 3*log(-1/2*cos(b*x + a) + 1/2))/b

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Sympy [A]  time = 3.41468, size = 410, normalized size = 10.79 \begin{align*} \begin{cases} \frac{3 \log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )} \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{3 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 3 b} + \frac{9 \log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )} \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{3 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 3 b} + \frac{9 \log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )} \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{3 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 3 b} + \frac{3 \log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )}}{3 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 3 b} - \frac{4 \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{3 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 3 b} + \frac{4}{3 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 9 b \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 3 b} & \text{for}\: b \neq 0 \\\frac{x \cos ^{4}{\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)**4/sin(b*x+a),x)

[Out]

Piecewise((3*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(3*b*tan(a/2 + b*x/2)**6 + 9*b*tan(a/2 + b*x/2)**4 + 9*
b*tan(a/2 + b*x/2)**2 + 3*b) + 9*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(3*b*tan(a/2 + b*x/2)**6 + 9*b*tan(
a/2 + b*x/2)**4 + 9*b*tan(a/2 + b*x/2)**2 + 3*b) + 9*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(3*b*tan(a/2 +
b*x/2)**6 + 9*b*tan(a/2 + b*x/2)**4 + 9*b*tan(a/2 + b*x/2)**2 + 3*b) + 3*log(tan(a/2 + b*x/2))/(3*b*tan(a/2 +
b*x/2)**6 + 9*b*tan(a/2 + b*x/2)**4 + 9*b*tan(a/2 + b*x/2)**2 + 3*b) - 4*tan(a/2 + b*x/2)**6/(3*b*tan(a/2 + b*
x/2)**6 + 9*b*tan(a/2 + b*x/2)**4 + 9*b*tan(a/2 + b*x/2)**2 + 3*b) + 4/(3*b*tan(a/2 + b*x/2)**6 + 9*b*tan(a/2
+ b*x/2)**4 + 9*b*tan(a/2 + b*x/2)**2 + 3*b), Ne(b, 0)), (x*cos(a)**4/sin(a), True))

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Giac [B]  time = 1.16256, size = 136, normalized size = 3.58 \begin{align*} \frac{\frac{8 \,{\left (\frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 2\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{3}} + 3 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^4/sin(b*x+a),x, algorithm="giac")

[Out]

1/6*(8*(3*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 2)/((cos(b*x +
a) - 1)/(cos(b*x + a) + 1) - 1)^3 + 3*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b