3.120 \(\int \cos ^5(a+b x) \cot (a+b x) \, dx\)

Optimal. Leaf size=53 \[ \frac{\cos ^5(a+b x)}{5 b}+\frac{\cos ^3(a+b x)}{3 b}+\frac{\cos (a+b x)}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b} \]

[Out]

-(ArcTanh[Cos[a + b*x]]/b) + Cos[a + b*x]/b + Cos[a + b*x]^3/(3*b) + Cos[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.029208, antiderivative size = 53, 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} \[ \frac{\cos ^5(a+b x)}{5 b}+\frac{\cos ^3(a+b x)}{3 b}+\frac{\cos (a+b x)}{b}-\frac{\tanh ^{-1}(\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*Cot[a + b*x],x]

[Out]

-(ArcTanh[Cos[a + b*x]]/b) + Cos[a + b*x]/b + Cos[a + b*x]^3/(3*b) + Cos[a + b*x]^5/(5*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 ^5(a+b x) \cot (a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{1-x^2} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-1-x^2-x^4+\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{\cos ^5(a+b x)}{5 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}+\frac{\cos ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0316335, size = 75, normalized size = 1.42 \[ \frac{11 \cos (a+b x)}{8 b}+\frac{7 \cos (3 (a+b x))}{48 b}+\frac{\cos (5 (a+b x))}{80 b}+\frac{\log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b}-\frac{\log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^5*Cot[a + b*x],x]

[Out]

(11*Cos[a + b*x])/(8*b) + (7*Cos[3*(a + b*x)])/(48*b) + Cos[5*(a + b*x)]/(80*b) - Log[Cos[(a + b*x)/2]]/b + Lo
g[Sin[(a + b*x)/2]]/b

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Maple [A]  time = 0.016, size = 58, normalized size = 1.1 \begin{align*}{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{5}}{5\,b}}+{\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)^6/sin(b*x+a),x)

[Out]

1/5*cos(b*x+a)^5/b+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 = 1.03192, size = 76, normalized size = 1.43 \begin{align*} \frac{6 \, \cos \left (b x + a\right )^{5} + 10 \, \cos \left (b x + a\right )^{3} + 30 \, \cos \left (b x + a\right ) - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{30 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a),x, algorithm="maxima")

[Out]

1/30*(6*cos(b*x + a)^5 + 10*cos(b*x + a)^3 + 30*cos(b*x + a) - 15*log(cos(b*x + a) + 1) + 15*log(cos(b*x + a)
- 1))/b

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Fricas [A]  time = 1.73491, size = 178, normalized size = 3.36 \begin{align*} \frac{6 \, \cos \left (b x + a\right )^{5} + 10 \, \cos \left (b x + a\right )^{3} + 30 \, \cos \left (b x + a\right ) - 15 \, \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + 15 \, \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right )}{30 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a),x, algorithm="fricas")

[Out]

1/30*(6*cos(b*x + a)^5 + 10*cos(b*x + a)^3 + 30*cos(b*x + a) - 15*log(1/2*cos(b*x + a) + 1/2) + 15*log(-1/2*co
s(b*x + a) + 1/2))/b

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Sympy [A]  time = 11.0172, size = 1085, normalized size = 20.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**6/sin(b*x+a),x)

[Out]

Piecewise((15*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**10/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8
+ 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 75*log(tan(a/2 +
b*x/2))*tan(a/2 + b*x/2)**8/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6
+ 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 150*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**
6/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**
4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 150*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(15*b*tan(a/2 + b*x/2)**1
0 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**
2 + 15*b) + 75*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8
+ 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 15*log(tan(a/2 +
b*x/2))/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b
*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 90*tan(a/2 + b*x/2)**8/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2
+ b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 180*
tan(a/2 + b*x/2)**6/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*
tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 280*tan(a/2 + b*x/2)**4/(15*b*tan(a/2 + b*x/2)**10 +
75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 +
15*b) + 140*tan(a/2 + b*x/2)**2/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)
**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b) + 46/(15*b*tan(a/2 + b*x/2)**10 + 75*b*tan(
a/2 + b*x/2)**8 + 150*b*tan(a/2 + b*x/2)**6 + 150*b*tan(a/2 + b*x/2)**4 + 75*b*tan(a/2 + b*x/2)**2 + 15*b), Ne
(b, 0)), (x*cos(a)**6/sin(a), True))

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Giac [B]  time = 1.17656, size = 196, normalized size = 3.7 \begin{align*} \frac{\frac{4 \,{\left (\frac{70 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{140 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{90 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac{45 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - 23\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{5}} + 15 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{30 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a),x, algorithm="giac")

[Out]

1/30*(4*(70*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 140*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 90*(cos(b*
x + a) - 1)^3/(cos(b*x + a) + 1)^3 - 45*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 - 23)/((cos(b*x + a) - 1)/(c
os(b*x + a) + 1) - 1)^5 + 15*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b