3.234 \(\int \cot ^2(c+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=46 \[ -\frac{\cos (a-c) \tanh ^{-1}(\cos (b x+c))}{b}-\frac{\sin (a-c) \csc (b x+c)}{b}+\frac{\cos (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0404453, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4578, 4577, 2638, 3770, 2606, 8} \[ -\frac{\cos (a-c) \tanh ^{-1}(\cos (b x+c))}{b}-\frac{\sin (a-c) \csc (b x+c)}{b}+\frac{\cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + b*x]^2*Sin[a + b*x],x]

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.0972564, size = 111, normalized size = 2.41 \[ -\frac{\sin (a-c) \csc (b x+c)}{b}-\frac{2 i \cos (a-c) \tan ^{-1}\left (\frac{(\cos (c)-i \sin (c)) \left (\cos (c) \cos \left (\frac{b x}{2}\right )-\sin (c) \sin \left (\frac{b x}{2}\right )\right )}{\sin (c) \cos \left (\frac{b x}{2}\right )+i \cos (c) \cos \left (\frac{b x}{2}\right )}\right )}{b}-\frac{\sin (a) \sin (b x)}{b}+\frac{\cos (a) \cos (b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + b*x]^2*Sin[a + b*x],x]

[Out]

((-2*I)*ArcTan[((Cos[c] - I*Sin[c])*(Cos[c]*Cos[(b*x)/2] - Sin[c]*Sin[(b*x)/2]))/(I*Cos[c]*Cos[(b*x)/2] + Cos[
(b*x)/2]*Sin[c])]*Cos[a - c])/b + (Cos[a]*Cos[b*x])/b - (Csc[c + b*x]*Sin[a - c])/b - (Sin[a]*Sin[b*x])/b

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Maple [C]  time = 0.089, size = 143, normalized size = 3.1 \begin{align*}{\frac{{{\rm e}^{i \left ( bx+a \right ) }}}{2\,b}}+{\frac{{{\rm e}^{-i \left ( bx+a \right ) }}}{2\,b}}+{\frac{{{\rm e}^{i \left ( bx+3\,a \right ) }}-{{\rm e}^{i \left ( bx+a+2\,c \right ) }}}{b \left ( -{{\rm e}^{2\,i \left ( bx+a+c \right ) }}+{{\rm e}^{2\,ia}} \right ) }}-{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+{{\rm e}^{i \left ( a-c \right ) }} \right ) \cos \left ( a-c \right ) }{b}}+{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-{{\rm e}^{i \left ( a-c \right ) }} \right ) \cos \left ( a-c \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(b*x+c)^2*sin(b*x+a),x)

[Out]

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

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Maxima [B]  time = 1.33538, size = 826, normalized size = 17.96 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(b*x+c)^2*sin(b*x+a),x, algorithm="maxima")

[Out]

1/2*((cos(3*b*x + a + 2*c) - cos(b*x + a))*cos(4*b*x + 2*a + 2*c) - (3*cos(2*b*x + 2*a) - 3*cos(2*b*x + 2*c) +
1)*cos(3*b*x + a + 2*c) + 3*cos(2*b*x + 2*a)*cos(b*x + a) - 3*cos(2*b*x + 2*c)*cos(b*x + a) - (cos(3*b*x + a
+ 2*c)^2*cos(-a + c) - 2*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(3*b*x + a + 2*c)^2 - 2*cos(-a + c)*sin(3*b*x + a + 2*c)*sin(b*x + a) + cos(-a + c)*sin(b*x + a)^2)*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) + (cos(3*b*x + a + 2*c
)^2*cos(-a + c) - 2*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)*s
in(3*b*x + a + 2*c)^2 - 2*cos(-a + c)*sin(3*b*x + a + 2*c)*sin(b*x + a) + cos(-a + c)*sin(b*x + a)^2)*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) + (sin(3*b*x + a + 2*c) - s
in(b*x + a))*sin(4*b*x + 2*a + 2*c) - 3*(sin(2*b*x + 2*a) - sin(2*b*x + 2*c))*sin(3*b*x + a + 2*c) + 3*sin(2*b
*x + 2*a)*sin(b*x + a) - 3*sin(2*b*x + 2*c)*sin(b*x + a) + cos(b*x + a))/(b*cos(3*b*x + a + 2*c)^2 - 2*b*cos(3
*b*x + a + 2*c)*cos(b*x + a) + b*cos(b*x + a)^2 + b*sin(3*b*x + a + 2*c)^2 - 2*b*sin(3*b*x + a + 2*c)*sin(b*x
+ a) + b*sin(b*x + a)^2)

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Fricas [B]  time = 0.553371, size = 859, normalized size = 18.67 \begin{align*} \frac{4 \,{\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \frac{\sqrt{2}{\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) +{\left (\cos \left (-2 \, a + 2 \, c\right )^{2} + 2 \, \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\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 (\cos \left (b x + a\right )^{2} + 1\right )} \sin \left (-2 \, a + 2 \, c\right )}{4 \,{\left (b \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) +{\left (b \cos \left (-2 \, a + 2 \, c\right ) + b\right )} \sin \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(b*x+c)^2*sin(b*x+a),x, algorithm="fricas")

[Out]

1/4*(4*(cos(-2*a + 2*c) + 1)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*((cos(-2*a + 2*c) + 1)*cos(b*x + a)*sin(-2*a
+ 2*c) + (cos(-2*a + 2*c)^2 + 2*cos(-2*a + 2*c) + 1)*sin(b*x + a))*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*(cos(b*x + a)^2 + 1)*sin(
-2*a + 2*c))/(b*cos(b*x + a)*sin(-2*a + 2*c) + (b*cos(-2*a + 2*c) + b)*sin(b*x + a))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (a + b x \right )} \cot ^{2}{\left (b x + c \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(b*x+c)**2*sin(b*x+a),x)

[Out]

Integral(sin(a + b*x)*cot(b*x + c)**2, x)

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Giac [B]  time = 1.20986, size = 779, normalized size = 16.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(b*x+c)^2*sin(b*x+a),x, algorithm="giac")

[Out]

-((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)) - (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) + (tan(1/2*b*x)^3*ta
n(1/2*a)^2*tan(1/2*c)^3 - tan(1/2*b*x)^3*tan(1/2*a)*tan(1/2*c)^4 - tan(1/2*b*x)^3*tan(1/2*a)^2*tan(1/2*c) + 6*
tan(1/2*b*x)^3*tan(1/2*a)*tan(1/2*c)^2 - tan(1/2*b*x)^3*tan(1/2*c)^3 + 6*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c)^
3 + 3*tan(1/2*b*x)*tan(1/2*a)^2*tan(1/2*c)^3 - tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c)^4 - tan(1/2*b*x)^3*tan(1/2*a
) + tan(1/2*b*x)^3*tan(1/2*c) - 6*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*c) - 3*tan(1/2*b*x)*tan(1/2*a)^2*tan(1/2*c
) - 2*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c)^2 - 4*tan(1/2*a)^2*tan(1/2*c)^2 - 3*tan(1/2*b*x)*tan(1/2*c)^3 + 2*tan
(1/2*a)*tan(1/2*c)^3 - tan(1/2*b*x)*tan(1/2*a) + 3*tan(1/2*b*x)*tan(1/2*c) - 2*tan(1/2*a)*tan(1/2*c) + 4*tan(1
/2*c)^2)/((tan(1/2*b*x)^4*tan(1/2*c) + tan(1/2*b*x)^3*tan(1/2*c)^2 - tan(1/2*b*x)^3 + tan(1/2*b*x)*tan(1/2*c)^
2 - tan(1/2*b*x) - tan(1/2*c))*(tan(1/2*a)^2*tan(1/2*c) + tan(1/2*c))))/b