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]

3/2*arctanh(cos(b*x+c))*cos(a-c)/b-cos(b*x+a)/b-1/2*cos(a-c)*cot(b*x+c)*csc(b*x+c)/b+csc(b*x+c)*sin(a-c)/b

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Rubi [A]  time = 0.08, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 8

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

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 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]

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*}

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Mathematica [A]  time = 0.34, 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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fricas [B]  time = 0.51, size = 385, normalized size = 5.27 \[ -\frac {16 \, \cos \left (b x + a\right )^{3} \cos \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) + 5\right )} \cos \left (b x + a\right ) + \frac {3 \, \sqrt {2} {\left (2 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 2 \, {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} + \cos \left (-2 \, a + 2 \, c\right )\right )} \cos \left (b x + a\right )^{2} + \cos \left (-2 \, a + 2 \, c\right )^{2} + 2 \, \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \log \left (\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) + 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) - 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}}}{8 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) - b\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/b*((2*tan(b*x/2)*tan(a/2)+tan(a/2)^2-1)/(tan(a/2)^2+1)/(tan(
b*x/2)^2+1)+(-2*tan(b*x/2)^3*tan(c/2)^7*tan(a/2)^2+2*tan(b*x/2)^3*tan(c/2)^7-6*tan(b*x/2)^3*tan(c/2)^5*tan(a/2
)^2+6*tan(b*x/2)^3*tan(c/2)^5-16*tan(b*x/2)^3*tan(c/2)^4*tan(a/2)+6*tan(b*x/2)^3*tan(c/2)^3*tan(a/2)^2-6*tan(b
*x/2)^3*tan(c/2)^3+2*tan(b*x/2)^3*tan(c/2)*tan(a/2)^2-2*tan(b*x/2)^3*tan(c/2)-tan(b*x/2)^2*tan(c/2)^8*tan(a/2)
^2+tan(b*x/2)^2*tan(c/2)^8+4*tan(b*x/2)^2*tan(c/2)^7*tan(a/2)-2*tan(b*x/2)^2*tan(c/2)^6*tan(a/2)^2+2*tan(b*x/2
)^2*tan(c/2)^6-20*tan(b*x/2)^2*tan(c/2)^5*tan(a/2)+22*tan(b*x/2)^2*tan(c/2)^4*tan(a/2)^2-22*tan(b*x/2)^2*tan(c
/2)^4+20*tan(b*x/2)^2*tan(c/2)^3*tan(a/2)-2*tan(b*x/2)^2*tan(c/2)^2*tan(a/2)^2+2*tan(b*x/2)^2*tan(c/2)^2-4*tan
(b*x/2)^2*tan(c/2)*tan(a/2)-tan(b*x/2)^2*tan(a/2)^2+tan(b*x/2)^2+2*tan(b*x/2)*tan(c/2)^7*tan(a/2)^2-2*tan(b*x/
2)*tan(c/2)^7-16*tan(b*x/2)*tan(c/2)^6*tan(a/2)+14*tan(b*x/2)*tan(c/2)^5*tan(a/2)^2-14*tan(b*x/2)*tan(c/2)^5+1
6*tan(b*x/2)*tan(c/2)^4*tan(a/2)-14*tan(b*x/2)*tan(c/2)^3*tan(a/2)^2+14*tan(b*x/2)*tan(c/2)^3-16*tan(b*x/2)*ta
n(c/2)^2*tan(a/2)-2*tan(b*x/2)*tan(c/2)*tan(a/2)^2+2*tan(b*x/2)*tan(c/2)-2*tan(c/2)^6*tan(a/2)^2+2*tan(c/2)^6+
8*tan(c/2)^5*tan(a/2)-12*tan(c/2)^4*tan(a/2)^2+12*tan(c/2)^4-8*tan(c/2)^3*tan(a/2)-2*tan(c/2)^2*tan(a/2)^2+2*t
an(c/2)^2)/(-16*tan(c/2)^2*tan(a/2)^2-16*tan(c/2)^2)/(tan(b*x/2)^2*tan(c/2)+tan(b*x/2)*tan(c/2)^2-tan(b*x/2)-t
an(c/2))^2+(3*tan(c/2)^2*tan(a/2)^2-3*tan(c/2)^2+12*tan(c/2)*tan(a/2)-3*tan(a/2)^2+3)/(-4*tan(c/2)^2*tan(a/2)^
2-4*tan(c/2)^2-4*tan(a/2)^2-4)*ln(abs(tan(b*x/2)+tan(c/2)))+(3*tan(c/2)^3*tan(a/2)^2-3*tan(c/2)^3+12*tan(c/2)^
2*tan(a/2)-3*tan(c/2)*tan(a/2)^2+3*tan(c/2))/(4*tan(c/2)^3*tan(a/2)^2+4*tan(c/2)^3+4*tan(c/2)*tan(a/2)^2+4*tan
(c/2))*ln(abs(tan(b*x/2)*tan(c/2)-1)))

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maple [C]  time = 1.01, size = 179, normalized size = 2.45 \[ -\frac {{\mathrm e}^{i \left (b x +a \right )}}{2 b}-\frac {{\mathrm e}^{-i \left (b x +a \right )}}{2 b}-\frac {-3 \,{\mathrm e}^{i \left (3 b x +5 a +2 c \right )}+{\mathrm e}^{i \left (3 b x +3 a +4 c \right )}+{\mathrm e}^{i \left (b x +5 a \right )}-3 \,{\mathrm e}^{i \left (b x +3 a +2 c \right )}}{2 b \left (-{\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-{\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{2 b} \]

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

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maxima [B]  time = 0.40, size = 1254, normalized size = 17.18 \[ \text {result too large to display} \]

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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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*cot(c + b*x)^3,x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos {\left (a + b x \right )} \cot ^{3}{\left (b x + c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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