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

Optimal. Leaf size=73 \[ \frac{3 \cos (a-c) \tanh ^{-1}(\cos (b x+c))}{2 b}+\frac{\sin (a-c) \csc (b x+c)}{b}-\frac{\cos (a-c) \cot (b x+c) \csc (b x+c)}{2 b}-\frac{\cos (a+b x)}{b} \]

[Out]

(3*ArcTanh[Cos[c + b*x]]*Cos[a - c])/(2*b) - Cos[a + b*x]/b - (Cos[a - c]*Cot[c + b*x]*Csc[c + b*x])/(2*b) + (
Csc[c + b*x]*Sin[a - c])/b

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Rubi [A]  time = 0.07673, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {4577, 4578, 2638, 3770, 2606, 8, 2611} \[ \frac{3 \cos (a-c) \tanh ^{-1}(\cos (b x+c))}{2 b}+\frac{\sin (a-c) \csc (b x+c)}{b}-\frac{\cos (a-c) \cot (b x+c) \csc (b x+c)}{2 b}-\frac{\cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Cot[c + b*x]^3,x]

[Out]

(3*ArcTanh[Cos[c + b*x]]*Cos[a - c])/(2*b) - Cos[a + b*x]/b - (Cos[a - c]*Cot[c + b*x]*Csc[c + b*x])/(2*b) + (
Csc[c + b*x]*Sin[a - c])/b

Rule 4577

Int[Cos[v_]*Cot[w_]^(n_.), x_Symbol] :> -Int[Sin[v]*Cot[w]^(n - 1), x] + Dist[Cos[v - w], Int[Csc[w]*Cot[w]^(n
 - 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]

Rule 4578

Int[Cot[w_]^(n_.)*Sin[v_], x_Symbol] :> Int[Cos[v]*Cot[w]^(n - 1), x] + Dist[Sin[v - w], Int[Csc[w]*Cot[w]^(n
- 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rubi steps

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

Mathematica [A]  time = 0.335982, size = 71, normalized size = 0.97 \[ \frac{\csc ^2(b x+c) (2 \cos (a-b x-2 c)+\cos (a+3 b x+2 c)-5 \cos (a+b x))+12 \cos (a-c) \tanh ^{-1}\left (\cos (c)-\sin (c) \tan \left (\frac{b x}{2}\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*ArcTanh[Cos[c] - Sin[c]*Tan[(b*x)/2]]*Cos[a - c] + (2*Cos[a - 2*c - b*x] - 5*Cos[a + b*x] + Cos[a + 2*c +
3*b*x])*Csc[c + b*x]^2)/(4*b)

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Maple [C]  time = 0.104, size = 179, normalized size = 2.5 \begin{align*} -{\frac{{{\rm e}^{i \left ( bx+a \right ) }}}{2\,b}}-{\frac{{{\rm e}^{-i \left ( bx+a \right ) }}}{2\,b}}-{\frac{-3\,{{\rm e}^{i \left ( 3\,bx+5\,a+2\,c \right ) }}+{{\rm e}^{i \left ( 3\,bx+3\,a+4\,c \right ) }}+{{\rm e}^{i \left ( bx+5\,a \right ) }}-3\,{{\rm e}^{i \left ( bx+3\,a+2\,c \right ) }}}{2\,b \left ( -{{\rm e}^{2\,i \left ( bx+a+c \right ) }}+{{\rm e}^{2\,ia}} \right ) ^{2}}}+{\frac{3\,\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+{{\rm e}^{i \left ( a-c \right ) }} \right ) \cos \left ( a-c \right ) }{2\,b}}-{\frac{3\,\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-{{\rm e}^{i \left ( a-c \right ) }} \right ) \cos \left ( a-c \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*cot(b*x+c)^3,x)

[Out]

-1/2*exp(I*(b*x+a))/b-1/2/b*exp(-I*(b*x+a))-1/2/b/(-exp(2*I*(b*x+a+c))+exp(2*I*a))^2*(-3*exp(I*(3*b*x+5*a+2*c)
)+exp(I*(3*b*x+3*a+4*c))+exp(I*(b*x+5*a))-3*exp(I*(b*x+3*a+2*c)))+3/2/b*ln(exp(I*(b*x+a))+exp(I*(a-c)))*cos(a-
c)-3/2/b*ln(exp(I*(b*x+a))-exp(I*(a-c)))*cos(a-c)

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Maxima [B]  time = 1.34055, size = 1693, normalized size = 23.19 \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)*cot(b*x+c)^3,x, algorithm="maxima")

[Out]

-1/4*(2*(cos(5*b*x + a + 4*c) - 2*cos(3*b*x + a + 2*c) + cos(b*x + a))*cos(6*b*x + 2*a + 4*c) - 2*(5*cos(4*b*x
 + 2*a + 2*c) - 2*cos(4*b*x + 4*c) - 2*cos(2*b*x + 2*a) + 5*cos(2*b*x + 2*c) - 1)*cos(5*b*x + a + 4*c) + 10*(2
*cos(3*b*x + a + 2*c) - cos(b*x + a))*cos(4*b*x + 2*a + 2*c) - 4*(2*cos(3*b*x + a + 2*c) - cos(b*x + a))*cos(4
*b*x + 4*c) - 4*(2*cos(2*b*x + 2*a) - 5*cos(2*b*x + 2*c) + 1)*cos(3*b*x + a + 2*c) + 4*cos(2*b*x + 2*a)*cos(b*
x + a) - 10*cos(2*b*x + 2*c)*cos(b*x + a) - 3*(cos(5*b*x + a + 4*c)^2*cos(-a + c) + 4*cos(3*b*x + a + 2*c)^2*c
os(-a + c) - 4*cos(3*b*x + a + 2*c)*cos(b*x + a)*cos(-a + c) + cos(b*x + a)^2*cos(-a + c) + cos(-a + c)*sin(5*
b*x + a + 4*c)^2 + 4*cos(-a + c)*sin(3*b*x + a + 2*c)^2 - 4*cos(-a + c)*sin(3*b*x + a + 2*c)*sin(b*x + a) + co
s(-a + c)*sin(b*x + a)^2 - 2*(2*cos(3*b*x + a + 2*c)*cos(-a + c) - cos(b*x + a)*cos(-a + c))*cos(5*b*x + a + 4
*c) - 2*(2*cos(-a + c)*sin(3*b*x + a + 2*c) - cos(-a + c)*sin(b*x + a))*sin(5*b*x + a + 4*c))*log(cos(b*x)^2 +
 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(c) + sin(c)^2) + 3*(cos(5*b*x + a + 4*c)^2*cos(-a
+ c) + 4*cos(3*b*x + a + 2*c)^2*cos(-a + c) - 4*cos(3*b*x + a + 2*c)*cos(b*x + a)*cos(-a + c) + cos(b*x + a)^2
*cos(-a + c) + cos(-a + c)*sin(5*b*x + a + 4*c)^2 + 4*cos(-a + c)*sin(3*b*x + a + 2*c)^2 - 4*cos(-a + c)*sin(3
*b*x + a + 2*c)*sin(b*x + a) + cos(-a + c)*sin(b*x + a)^2 - 2*(2*cos(3*b*x + a + 2*c)*cos(-a + c) - cos(b*x +
a)*cos(-a + c))*cos(5*b*x + a + 4*c) - 2*(2*cos(-a + c)*sin(3*b*x + a + 2*c) - cos(-a + c)*sin(b*x + a))*sin(5
*b*x + a + 4*c))*log(cos(b*x)^2 - 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(c) + sin(c)^2) +
2*(sin(5*b*x + a + 4*c) - 2*sin(3*b*x + a + 2*c) + sin(b*x + a))*sin(6*b*x + 2*a + 4*c) - 2*(5*sin(4*b*x + 2*a
 + 2*c) - 2*sin(4*b*x + 4*c) - 2*sin(2*b*x + 2*a) + 5*sin(2*b*x + 2*c))*sin(5*b*x + a + 4*c) + 10*(2*sin(3*b*x
 + a + 2*c) - sin(b*x + a))*sin(4*b*x + 2*a + 2*c) - 4*(2*sin(3*b*x + a + 2*c) - sin(b*x + a))*sin(4*b*x + 4*c
) - 4*(2*sin(2*b*x + 2*a) - 5*sin(2*b*x + 2*c))*sin(3*b*x + a + 2*c) + 4*sin(2*b*x + 2*a)*sin(b*x + a) - 10*si
n(2*b*x + 2*c)*sin(b*x + a) + 2*cos(b*x + a))/(b*cos(5*b*x + a + 4*c)^2 + 4*b*cos(3*b*x + a + 2*c)^2 - 4*b*cos
(3*b*x + a + 2*c)*cos(b*x + a) + b*cos(b*x + a)^2 + b*sin(5*b*x + a + 4*c)^2 + 4*b*sin(3*b*x + a + 2*c)^2 - 4*
b*sin(3*b*x + a + 2*c)*sin(b*x + a) + b*sin(b*x + a)^2 - 2*(2*b*cos(3*b*x + a + 2*c) - b*cos(b*x + a))*cos(5*b
*x + a + 4*c) - 2*(2*b*sin(3*b*x + a + 2*c) - b*sin(b*x + a))*sin(5*b*x + a + 4*c))

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Fricas [B]  time = 0.560044, size = 1040, normalized size = 14.25 \begin{align*} -\frac{16 \, \cos \left (b x + a\right )^{3} \cos \left (-2 \, a + 2 \, c\right ) - 4 \,{\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 4 \,{\left (\cos \left (-2 \, a + 2 \, c\right ) + 5\right )} \cos \left (b x + a\right ) + \frac{3 \, \sqrt{2}{\left (2 \,{\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 2 \,{\left (\cos \left (-2 \, a + 2 \, c\right )^{2} + \cos \left (-2 \, a + 2 \, c\right )\right )} \cos \left (b x + a\right )^{2} + \cos \left (-2 \, a + 2 \, c\right )^{2} + 2 \, \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \log \left (\frac{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac{2 \, \sqrt{2}{\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt{\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) + 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) - 1}\right )}{\sqrt{\cos \left (-2 \, a + 2 \, c\right ) + 1}}}{8 \,{\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(16*cos(b*x + a)^3*cos(-2*a + 2*c) - 4*(4*cos(b*x + a)^2 + 1)*sin(b*x + a)*sin(-2*a + 2*c) - 4*(cos(-2*a
+ 2*c) + 5)*cos(b*x + a) + 3*sqrt(2)*(2*(cos(-2*a + 2*c) + 1)*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - 2*(c
os(-2*a + 2*c)^2 + cos(-2*a + 2*c))*cos(b*x + a)^2 + cos(-2*a + 2*c)^2 + 2*cos(-2*a + 2*c) + 1)*log((2*cos(b*x
 + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) + 2*sqrt(2)*((cos(-2*a + 2*c) + 1)*cos(b
*x + a) - sin(b*x + a)*sin(-2*a + 2*c))/sqrt(cos(-2*a + 2*c) + 1) - cos(-2*a + 2*c) + 3)/(2*cos(b*x + a)^2*cos
(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - cos(-2*a + 2*c) - 1))/sqrt(cos(-2*a + 2*c) + 1))/
(2*b*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*b*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - b*cos(-2*a + 2*c) - b)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*cot(b*x+c)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.38402, size = 1300, normalized size = 17.81 \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)*cot(b*x+c)^3,x, algorithm="giac")

[Out]

1/8*(12*(tan(1/2*a)^2*tan(1/2*c)^3 - tan(1/2*a)^2*tan(1/2*c) + 4*tan(1/2*a)*tan(1/2*c)^2 - tan(1/2*c)^3 + tan(
1/2*c))*log(abs(tan(1/2*b*x)*tan(1/2*c) - 1))/(tan(1/2*a)^2*tan(1/2*c)^3 + tan(1/2*a)^2*tan(1/2*c) + tan(1/2*c
)^3 + tan(1/2*c)) - 12*(tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)^2 + 4*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + 1)
*log(abs(tan(1/2*b*x) + tan(1/2*c)))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) + 16*(2*tan
(1/2*b*x)*tan(1/2*a) + tan(1/2*a)^2 - 1)/((tan(1/2*b*x)^2 + 1)*(tan(1/2*a)^2 + 1)) + (2*tan(1/2*b*x)^3*tan(1/2
*a)^2*tan(1/2*c)^7 + tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*c)^8 + 6*tan(1/2*b*x)^3*tan(1/2*a)^2*tan(1/2*c)^5 + 2
*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*c)^6 - 2*tan(1/2*b*x)^3*tan(1/2*c)^7 - 4*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/
2*c)^7 - 2*tan(1/2*b*x)*tan(1/2*a)^2*tan(1/2*c)^7 - tan(1/2*b*x)^2*tan(1/2*c)^8 - 6*tan(1/2*b*x)^3*tan(1/2*a)^
2*tan(1/2*c)^3 + 16*tan(1/2*b*x)^3*tan(1/2*a)*tan(1/2*c)^4 - 22*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*c)^4 - 6*t
an(1/2*b*x)^3*tan(1/2*c)^5 + 20*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c)^5 - 14*tan(1/2*b*x)*tan(1/2*a)^2*tan(1/2*
c)^5 - 2*tan(1/2*b*x)^2*tan(1/2*c)^6 + 16*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c)^6 + 2*tan(1/2*a)^2*tan(1/2*c)^6 +
 2*tan(1/2*b*x)*tan(1/2*c)^7 - 2*tan(1/2*b*x)^3*tan(1/2*a)^2*tan(1/2*c) + 2*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/
2*c)^2 + 6*tan(1/2*b*x)^3*tan(1/2*c)^3 - 20*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c)^3 + 14*tan(1/2*b*x)*tan(1/2*a
)^2*tan(1/2*c)^3 + 22*tan(1/2*b*x)^2*tan(1/2*c)^4 - 16*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c)^4 + 12*tan(1/2*a)^2*
tan(1/2*c)^4 + 14*tan(1/2*b*x)*tan(1/2*c)^5 - 8*tan(1/2*a)*tan(1/2*c)^5 - 2*tan(1/2*c)^6 + tan(1/2*b*x)^2*tan(
1/2*a)^2 + 2*tan(1/2*b*x)^3*tan(1/2*c) + 4*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c) + 2*tan(1/2*b*x)*tan(1/2*a)^2*
tan(1/2*c) - 2*tan(1/2*b*x)^2*tan(1/2*c)^2 + 16*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c)^2 + 2*tan(1/2*a)^2*tan(1/2*
c)^2 - 14*tan(1/2*b*x)*tan(1/2*c)^3 + 8*tan(1/2*a)*tan(1/2*c)^3 - 12*tan(1/2*c)^4 - tan(1/2*b*x)^2 - 2*tan(1/2
*b*x)*tan(1/2*c) - 2*tan(1/2*c)^2)/((tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*c)^2)*(tan(1/2*b*x)^2*tan(1/2*c) + ta
n(1/2*b*x)*tan(1/2*c)^2 - tan(1/2*b*x) - tan(1/2*c))^2))/b

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