3.251 \(\int \cos (a+b x) \cot ^2(c+b x) \, dx\)
Optimal. Leaf size=46 \[ -\frac{\cos (a-c) \csc (b x+c)}{b}+\frac{\sin (a-c) \tanh ^{-1}(\cos (b x+c))}{b}-\frac{\sin (a+b x)}{b} \]
[Out]
-((Cos[a - c]*Csc[c + b*x])/b) + (ArcTanh[Cos[c + b*x]]*Sin[a - c])/b - Sin[a + b*x]/b
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Rubi [A] time = 0.0413559, 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 =
{4577, 4578, 2637, 3770, 2606, 8} \[ -\frac{\cos (a-c) \csc (b x+c)}{b}+\frac{\sin (a-c) \tanh ^{-1}(\cos (b x+c))}{b}-\frac{\sin (a+b x)}{b} \]
Antiderivative was successfully verified.
[In]
Int[Cos[a + b*x]*Cot[c + b*x]^2,x]
[Out]
-((Cos[a - c]*Csc[c + b*x])/b) + (ArcTanh[Cos[c + b*x]]*Sin[a - c])/b - Sin[a + b*x]/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 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 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 \cos (a+b x) \cot ^2(c+b x) \, dx &=\cos (a-c) \int \cot (c+b x) \csc (c+b x) \, dx-\int \cot (c+b x) \sin (a+b x) \, dx\\ &=-\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{\tanh ^{-1}(\cos (c+b x)) \sin (a-c)}{b}-\frac{\sin (a+b x)}{b}\\ \end{align*}
Mathematica [C] time = 0.0985741, size = 112, normalized size = 2.43 \[ -\frac{\cos (a-c) \csc (b x+c)}{b}+\frac{2 i \sin (a-c) \tan ^{-1}\left (\frac{(\cos (c)-i \sin (c)) \left (\cos (c) \cos \left (\frac{b x}{2}\right )-\sin (c) \sin \left (\frac{b x}{2}\right )\right )}{\sin (c) \cos \left (\frac{b x}{2}\right )+i \cos (c) \cos \left (\frac{b x}{2}\right )}\right )}{b}-\frac{\sin (a) \cos (b x)}{b}-\frac{\cos (a) \sin (b x)}{b} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[a + b*x]*Cot[c + b*x]^2,x]
[Out]
-((Cos[a - c]*Csc[c + b*x])/b) - (Cos[b*x]*Sin[a])/b + ((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])]*Sin[a - c])/b - (Cos[a]*Sin[b*x])/b
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Maple [C] time = 0.095, size = 145, normalized size = 3.2 \begin{align*}{\frac{{\frac{i}{2}}{{\rm e}^{i \left ( bx+a \right ) }}}{b}}-{\frac{{\frac{i}{2}}{{\rm e}^{-i \left ( bx+a \right ) }}}{b}}+{\frac{i \left ({{\rm e}^{i \left ( bx+3\,a \right ) }}+{{\rm e}^{i \left ( bx+a+2\,c \right ) }} \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 ) \sin \left ( a-c \right ) }{b}}+{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+{{\rm e}^{i \left ( a-c \right ) }} \right ) \sin \left ( a-c \right ) }{b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(b*x+a)*cot(b*x+c)^2,x)
[Out]
1/2*I*exp(I*(b*x+a))/b-1/2*I/b*exp(-I*(b*x+a))+I/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)))*sin(a-c)+1/b*ln(exp(I*(b*x+a))+exp(I*(a-c)))*sin(a-c)
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Maxima [B] time = 1.3287, size = 828, normalized size = 18. \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)^2,x, algorithm="maxima")
[Out]
1/2*((sin(3*b*x + a + 2*c) - sin(b*x + a))*cos(4*b*x + 2*a + 2*c) + 3*(sin(2*b*x + 2*a) + sin(2*b*x + 2*c))*co
s(3*b*x + a + 2*c) - (cos(3*b*x + a + 2*c)^2*sin(-a + c) - 2*cos(3*b*x + a + 2*c)*cos(b*x + a)*sin(-a + c) + c
os(b*x + a)^2*sin(-a + c) + sin(3*b*x + a + 2*c)^2*sin(-a + c) - 2*sin(3*b*x + a + 2*c)*sin(b*x + a)*sin(-a +
c) + sin(b*x + a)^2*sin(-a + 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) + (cos(3*b*x + a + 2*c)^2*sin(-a + c) - 2*cos(3*b*x + a + 2*c)*cos(b*x + a)*sin(-a + c) + cos(b*
x + a)^2*sin(-a + c) + sin(3*b*x + a + 2*c)^2*sin(-a + c) - 2*sin(3*b*x + a + 2*c)*sin(b*x + a)*sin(-a + c) +
sin(b*x + a)^2*sin(-a + 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) + s
in(c)^2) - (cos(3*b*x + a + 2*c) - cos(b*x + a))*sin(4*b*x + 2*a + 2*c) - (3*cos(2*b*x + 2*a) + 3*cos(2*b*x +
2*c) - 1)*sin(3*b*x + a + 2*c) - 3*cos(b*x + a)*sin(2*b*x + 2*a) - 3*cos(b*x + a)*sin(2*b*x + 2*c) + 3*cos(2*b
*x + 2*a)*sin(b*x + a) + 3*cos(2*b*x + 2*c)*sin(b*x + a) - sin(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 = 0.543556, size = 856, normalized size = 18.61 \begin{align*} \frac{4 \,{\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right )^{2} - 4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac{\sqrt{2}{\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) -{\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \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}} - 8 \, \cos \left (-2 \, a + 2 \, c\right ) - 8}{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(cos(b*x+a)*cot(b*x+c)^2,x, algorithm="fricas")
[Out]
1/4*(4*(cos(-2*a + 2*c) + 1)*cos(b*x + a)^2 - 4*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) + sqrt(2)*((cos(-2*a
+ 2*c) + 1)*sin(b*x + a)*sin(-2*a + 2*c) - (cos(-2*a + 2*c)^2 - 1)*cos(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) - 8*cos(-2*a +
2*c) - 8)/(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 \cos{\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(cos(b*x+a)*cot(b*x+c)**2,x)
[Out]
Integral(cos(a + b*x)*cot(b*x + c)**2, x)
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Giac [B] time = 1.23013, size = 846, normalized size = 18.39 \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)^2,x, algorithm="giac")
[Out]
1/2*(4*(tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)*tan(1/2*c)^3 + tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2)*log(abs(ta
n(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))
- 4*(tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c))*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*tan(1/2*a)^2*tan(1/2*c
)^4 - 6*tan(1/2*b*x)^3*tan(1/2*a)^2*tan(1/2*c)^2 + 4*tan(1/2*b*x)^3*tan(1/2*a)*tan(1/2*c)^3 - 6*tan(1/2*b*x)^2
*tan(1/2*a)^2*tan(1/2*c)^3 - tan(1/2*b*x)^3*tan(1/2*c)^4 + tan(1/2*b*x)*tan(1/2*a)^2*tan(1/2*c)^4 + tan(1/2*b*
x)^3*tan(1/2*a)^2 - 4*tan(1/2*b*x)^3*tan(1/2*a)*tan(1/2*c) + 6*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*c) + 6*tan(
1/2*b*x)^3*tan(1/2*c)^2 + 2*tan(1/2*b*x)*tan(1/2*a)^2*tan(1/2*c)^2 + 6*tan(1/2*b*x)^2*tan(1/2*c)^3 + 12*tan(1/
2*b*x)*tan(1/2*a)*tan(1/2*c)^3 - 2*tan(1/2*a)^2*tan(1/2*c)^3 - tan(1/2*b*x)*tan(1/2*c)^4 - tan(1/2*b*x)^3 + ta
n(1/2*b*x)*tan(1/2*a)^2 - 6*tan(1/2*b*x)^2*tan(1/2*c) - 12*tan(1/2*b*x)*tan(1/2*a)*tan(1/2*c) + 2*tan(1/2*a)^2
*tan(1/2*c) - 2*tan(1/2*b*x)*tan(1/2*c)^2 - 16*tan(1/2*a)*tan(1/2*c)^2 + 2*tan(1/2*c)^3 - tan(1/2*b*x) - 2*tan
(1/2*c))/((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