3.235 \(\int \cot ^3(c+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=74 \[ -\frac {\cos (a-c) \csc (b x+c)}{b}+\frac {3 \sin (a-c) \tanh ^{-1}(\cos (b x+c))}{2 b}-\frac {\sin (a-c) \cot (b x+c) \csc (b x+c)}{2 b}-\frac {\sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 74, 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 = {4578, 4577, 2637, 3770, 2606, 8, 2611} \[ -\frac {\cos (a-c) \csc (b x+c)}{b}+\frac {3 \sin (a-c) \tanh ^{-1}(\cos (b x+c))}{2 b}-\frac {\sin (a-c) \cot (b x+c) \csc (b x+c)}{2 b}-\frac {\sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + b*x]^3*Sin[a + b*x],x]

[Out]

-((Cos[a - c]*Csc[c + b*x])/b) + (3*ArcTanh[Cos[c + b*x]]*Sin[a - c])/(2*b) - (Cot[c + b*x]*Csc[c + b*x]*Sin[a
- c])/(2*b) - Sin[a + b*x]/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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 \cot ^3(c+b x) \sin (a+b x) \, dx &=\sin (a-c) \int \cot ^2(c+b x) \csc (c+b x) \, dx+\int \cos (a+b x) \cot ^2(c+b x) \, dx\\ &=-\frac {\cot (c+b x) \csc (c+b x) \sin (a-c)}{2 b}+\cos (a-c) \int \cot (c+b x) \csc (c+b x) \, dx-\frac {1}{2} \sin (a-c) \int \csc (c+b x) \, dx-\int \cot (c+b x) \sin (a+b x) \, dx\\ &=\frac {\tanh ^{-1}(\cos (c+b x)) \sin (a-c)}{2 b}-\frac {\cot (c+b x) \csc (c+b x) \sin (a-c)}{2 b}-\frac {\cos (a-c) \operatorname {Subst}(\int 1 \, dx,x,\csc (c+b x))}{b}-\sin (a-c) \int \csc (c+b x) \, dx-\int \cos (a+b x) \, dx\\ &=-\frac {\cos (a-c) \csc (c+b x)}{b}+\frac {3 \tanh ^{-1}(\cos (c+b x)) \sin (a-c)}{2 b}-\frac {\cot (c+b x) \csc (c+b x) \sin (a-c)}{2 b}-\frac {\sin (a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 71, normalized size = 0.96 \[ \frac {\csc ^2(b x+c) (2 \sin (a-b x-2 c)+\sin (a+3 b x+2 c)-5 \sin (a+b x))+12 \sin (a-c) \tanh ^{-1}\left (\cos (c)-\sin (c) \tan \left (\frac {b x}{2}\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.51, size = 372, normalized size = 5.03 \[ \frac {\frac {3 \, \sqrt {2} {\left (2 \, {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + {\left (2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) - 1\right )} \sin \left (-2 \, a + 2 \, c\right )\right )} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) + 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) - 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - 4 \, {\left (4 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 3 \, \cos \left (-2 \, a + 2 \, c\right ) - 5\right )} \sin \left (b x + a\right ) - 4 \, {\left (4 \, \cos \left (b x + a\right )^{3} - 5 \, \cos \left (b x + a\right )\right )} \sin \left (-2 \, a + 2 \, c\right )}{8 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*sqrt(2)*(2*(cos(-2*a + 2*c)^2 - 1)*cos(b*x + a)*sin(b*x + a) + (2*cos(b*x + a)^2*cos(-2*a + 2*c) - cos(
-2*a + 2*c) - 1)*sin(-2*a + 2*c))*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) - 4*(4*cos(b*x + a)^2*cos(-2*a + 2*c) - 3*cos(-2*a + 2*c) - 5)
*sin(b*x + a) - 4*(4*cos(b*x + a)^3 - 5*cos(b*x + a))*sin(-2*a + 2*c))/(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(cot(b*x+c)^3*sin(b*x+a),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*((tan(b*x/2)*tan(a/2)^2-tan(b*x/2)-2*tan(a/2))/(tan(a/2)^2
+1)/(tan(b*x/2)^2+1)+(2*tan(b*x/2)^3*tan(c/2)^7*tan(a/2)+6*tan(b*x/2)^3*tan(c/2)^5*tan(a/2)-4*tan(b*x/2)^3*tan
(c/2)^4*tan(a/2)^2+4*tan(b*x/2)^3*tan(c/2)^4-6*tan(b*x/2)^3*tan(c/2)^3*tan(a/2)-2*tan(b*x/2)^3*tan(c/2)*tan(a/
2)+tan(b*x/2)^2*tan(c/2)^8*tan(a/2)+tan(b*x/2)^2*tan(c/2)^7*tan(a/2)^2-tan(b*x/2)^2*tan(c/2)^7+2*tan(b*x/2)^2*
tan(c/2)^6*tan(a/2)-5*tan(b*x/2)^2*tan(c/2)^5*tan(a/2)^2+5*tan(b*x/2)^2*tan(c/2)^5-22*tan(b*x/2)^2*tan(c/2)^4*
tan(a/2)+5*tan(b*x/2)^2*tan(c/2)^3*tan(a/2)^2-5*tan(b*x/2)^2*tan(c/2)^3+2*tan(b*x/2)^2*tan(c/2)^2*tan(a/2)-tan
(b*x/2)^2*tan(c/2)*tan(a/2)^2+tan(b*x/2)^2*tan(c/2)+tan(b*x/2)^2*tan(a/2)-2*tan(b*x/2)*tan(c/2)^7*tan(a/2)-4*t
an(b*x/2)*tan(c/2)^6*tan(a/2)^2+4*tan(b*x/2)*tan(c/2)^6-14*tan(b*x/2)*tan(c/2)^5*tan(a/2)+4*tan(b*x/2)*tan(c/2
)^4*tan(a/2)^2-4*tan(b*x/2)*tan(c/2)^4+14*tan(b*x/2)*tan(c/2)^3*tan(a/2)-4*tan(b*x/2)*tan(c/2)^2*tan(a/2)^2+4*
tan(b*x/2)*tan(c/2)^2+2*tan(b*x/2)*tan(c/2)*tan(a/2)+2*tan(c/2)^6*tan(a/2)+2*tan(c/2)^5*tan(a/2)^2-2*tan(c/2)^
5+12*tan(c/2)^4*tan(a/2)-2*tan(c/2)^3*tan(a/2)^2+2*tan(c/2)^3+2*tan(c/2)^2*tan(a/2))/(-8*tan(c/2)^2*tan(a/2)^2
-8*tan(c/2)^2)/(tan(b*x/2)^2*tan(c/2)+tan(b*x/2)*tan(c/2)^2-tan(b*x/2)-tan(c/2))^2+(-3*tan(c/2)^2*tan(a/2)+3*t
an(c/2)*tan(a/2)^2-3*tan(c/2)+3*tan(a/2))/(-2*tan(c/2)^2*tan(a/2)^2-2*tan(c/2)^2-2*tan(a/2)^2-2)*ln(abs(tan(b*
x/2)+tan(c/2)))+(-3*tan(c/2)^3*tan(a/2)+3*tan(c/2)^2*tan(a/2)^2-3*tan(c/2)^2+3*tan(c/2)*tan(a/2))/(2*tan(c/2)^
3*tan(a/2)^2+2*tan(c/2)^3+2*tan(c/2)*tan(a/2)^2+2*tan(c/2))*ln(abs(tan(b*x/2)*tan(c/2)-1)))

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maple [C]  time = 0.88, size = 184, normalized size = 2.49 \[ \frac {i {\mathrm e}^{i \left (b x +a \right )}}{2 b}-\frac {i {\mathrm e}^{-i \left (b x +a \right )}}{2 b}+\frac {i \left (-3 \,{\mathrm e}^{i \left (3 b x +5 a +2 c \right )}-{\mathrm e}^{i \left (3 b x +3 a +4 c \right )}+{\mathrm e}^{i \left (b x +5 a \right )}+3 \,{\mathrm e}^{i \left (b x +3 a +2 c \right )}\right )}{2 b \left (-{\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+{\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-{\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*exp(I*(b*x+a))/b-1/2*I/b*exp(-I*(b*x+a))+1/2*I/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*s
in(a-c)-3/2*ln(exp(I*(b*x+a))-exp(I*(a-c)))/b*sin(a-c)

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maxima [B]  time = 0.40, size = 1254, normalized size = 16.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(sin(5*b*x + a + 4*c) - 2*sin(3*b*x + a + 2*c) + sin(b*x + a))*cos(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))*cos(5*b*x + a + 4*c) + 10*(2*sin(
3*b*x + a + 2*c) - sin(b*x + a))*cos(4*b*x + 2*a + 2*c) + 4*(2*sin(3*b*x + a + 2*c) - sin(b*x + a))*cos(4*b*x
+ 4*c) + 4*(2*sin(2*b*x + 2*a) + 5*sin(2*b*x + 2*c))*cos(3*b*x + a + 2*c) - 3*(cos(5*b*x + a + 4*c)^2*sin(-a +
c) + 4*cos(3*b*x + a + 2*c)^2*sin(-a + c) - 4*cos(3*b*x + a + 2*c)*cos(b*x + a)*sin(-a + c) + cos(b*x + a)^2*
sin(-a + c) + sin(5*b*x + a + 4*c)^2*sin(-a + c) + 4*sin(3*b*x + a + 2*c)^2*sin(-a + c) - 4*sin(3*b*x + a + 2*
c)*sin(b*x + a)*sin(-a + c) + sin(b*x + a)^2*sin(-a + c) - 2*(2*cos(3*b*x + a + 2*c)*sin(-a + c) - cos(b*x + a
)*sin(-a + c))*cos(5*b*x + a + 4*c) - 2*(2*sin(3*b*x + a + 2*c)*sin(-a + c) - sin(b*x + a)*sin(-a + c))*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*sin(-a + c) + 4*cos(3*b*x + a + 2*c)^2*sin(-a + c) - 4*cos(3*b*x + a + 2*c)*cos(b*x +
a)*sin(-a + c) + cos(b*x + a)^2*sin(-a + c) + sin(5*b*x + a + 4*c)^2*sin(-a + c) + 4*sin(3*b*x + a + 2*c)^2*s
in(-a + c) - 4*sin(3*b*x + a + 2*c)*sin(b*x + a)*sin(-a + c) + sin(b*x + a)^2*sin(-a + c) - 2*(2*cos(3*b*x + a
+ 2*c)*sin(-a + c) - cos(b*x + a)*sin(-a + c))*cos(5*b*x + a + 4*c) - 2*(2*sin(3*b*x + a + 2*c)*sin(-a + c) -
sin(b*x + a)*sin(-a + c))*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*(cos(5*b*x + a + 4*c) - 2*cos(3*b*x + a + 2*c) + cos(b*x + a))*sin(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)*si
n(5*b*x + a + 4*c) - 10*(2*cos(3*b*x + a + 2*c) - cos(b*x + a))*sin(4*b*x + 2*a + 2*c) - 4*(2*cos(3*b*x + a +
2*c) - cos(b*x + a))*sin(4*b*x + 4*c) - 4*(2*cos(2*b*x + 2*a) + 5*cos(2*b*x + 2*c) - 1)*sin(3*b*x + a + 2*c) -
4*cos(b*x + a)*sin(2*b*x + 2*a) - 10*cos(b*x + a)*sin(2*b*x + 2*c) + 4*cos(2*b*x + 2*a)*sin(b*x + a) + 10*cos
(2*b*x + 2*c)*sin(b*x + a) - 2*sin(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(cot(c + b*x)^3*sin(a + b*x),x)

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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