3.227 \(\int \cos (a+b x) \csc (c+b x) \, dx\)
Optimal. Leaf size=27 \[ \frac{\cos (a-c) \log (\sin (b x+c))}{b}-x \sin (a-c) \]
[Out]
(Cos[a - c]*Log[Sin[c + b*x]])/b - x*Sin[a - c]
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Rubi [A] time = 0.0168558, antiderivative size = 27, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used =
{4581, 3475, 8} \[ \frac{\cos (a-c) \log (\sin (b x+c))}{b}-x \sin (a-c) \]
Antiderivative was successfully verified.
[In]
Int[Cos[a + b*x]*Csc[c + b*x],x]
[Out]
(Cos[a - c]*Log[Sin[c + b*x]])/b - x*Sin[a - c]
Rule 4581
Int[Cos[v_]*Csc[w_]^(n_.), x_Symbol] :> Dist[Cos[v - w], Int[Cot[w]*Csc[w]^(n - 1), x], x] - Dist[Sin[v - w],
Int[Csc[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]
Rule 3475
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rubi steps
\begin{align*} \int \cos (a+b x) \csc (c+b x) \, dx &=\cos (a-c) \int \cot (c+b x) \, dx-\sin (a-c) \int 1 \, dx\\ &=\frac{\cos (a-c) \log (\sin (c+b x))}{b}-x \sin (a-c)\\ \end{align*}
Mathematica [C] time = 0.173198, size = 58, normalized size = 2.15 \[ \frac{-2 b x \sin (a-c)-2 i \cos (a-c) \tan ^{-1}(\tan (b x+c))+\cos (a-c) \left (\log \left (\sin ^2(b x+c)\right )+2 i b x\right )}{2 b} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[a + b*x]*Csc[c + b*x],x]
[Out]
((-2*I)*ArcTan[Tan[c + b*x]]*Cos[a - c] + Cos[a - c]*((2*I)*b*x + Log[Sin[c + b*x]^2]) - 2*b*x*Sin[a - c])/(2*
b)
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Maple [B] time = 0.184, size = 325, normalized size = 12. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(b*x+a)/sin(b*x+c),x)
[Out]
-1/2/b/(cos(c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*ln(1+tan(b*x+a)^2)*cos(a)*cos(c)-1/2/b/(cos(c)^2+sin(c)^2)/(cos
(a)^2+sin(a)^2)*ln(1+tan(b*x+a)^2)*sin(a)*sin(c)+1/b/(cos(c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*cos(a)*sin(c)*arc
tan(tan(b*x+a))-1/b/(cos(c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*cos(c)*sin(a)*arctan(tan(b*x+a))+1/b/(cos(a)^2*cos
(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)*ln(tan(b*x+a)*cos(a)*cos(c)+tan(b*x+a)*sin(a)*sin
(c)+cos(a)*sin(c)-sin(a)*cos(c))*cos(a)*cos(c)+1/b/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(
a)^2*sin(c)^2)*ln(tan(b*x+a)*cos(a)*cos(c)+tan(b*x+a)*sin(a)*sin(c)+cos(a)*sin(c)-sin(a)*cos(c))*sin(a)*sin(c)
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Maxima [B] time = 1.113, size = 143, normalized size = 5.3 \begin{align*} \frac{2 \, b x \sin \left (-a + c\right ) + \cos \left (-a + c\right ) \log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \left (c\right ) + \cos \left (c\right )^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}\right ) + \cos \left (-a + c\right ) \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \left (c\right ) + \cos \left (c\right )^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}\right )}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)/sin(b*x+c),x, algorithm="maxima")
[Out]
1/2*(2*b*x*sin(-a + c) + cos(-a + c)*log(cos(b*x)^2 + 2*cos(b*x)*cos(c) + cos(c)^2 + sin(b*x)^2 - 2*sin(b*x)*s
in(c) + sin(c)^2) + cos(-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))/b
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Fricas [A] time = 0.50052, size = 77, normalized size = 2.85 \begin{align*} \frac{b x \sin \left (-a + c\right ) + \cos \left (-a + c\right ) \log \left (\frac{1}{2} \, \sin \left (b x + c\right )\right )}{b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)/sin(b*x+c),x, algorithm="fricas")
[Out]
(b*x*sin(-a + c) + cos(-a + c)*log(1/2*sin(b*x + c)))/b
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Sympy [B] time = 10.3423, size = 335, normalized size = 12.41 \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)/sin(b*x+c),x)
[Out]
-Piecewise((0, Eq(b, 0) & (Eq(b, 0) | Eq(c, 0))), (x, Eq(c, 0)), (-b*x*tan(c/2)**2/(b*tan(c/2)**2 + b) + b*x/(
b*tan(c/2)**2 + b) - 2*log(tan(c/2) + tan(b*x/2))*tan(c/2)/(b*tan(c/2)**2 + b) - 2*log(tan(b*x/2) - 1/tan(c/2)
)*tan(c/2)/(b*tan(c/2)**2 + b) + 2*log(tan(b*x/2)**2 + 1)*tan(c/2)/(b*tan(c/2)**2 + b), True))*sin(a) + Piecew
ise((zoo*x, Eq(b, 0) & Eq(c, 0)), (x/sin(c), Eq(b, 0)), (log(sin(b*x))/b, Eq(c, 0)), (2*b*x*tan(c/2)/(b*tan(c/
2)**2 + b) - log(tan(c/2) + tan(b*x/2))*tan(c/2)**2/(b*tan(c/2)**2 + b) + log(tan(c/2) + tan(b*x/2))/(b*tan(c/
2)**2 + b) - log(tan(b*x/2) - 1/tan(c/2))*tan(c/2)**2/(b*tan(c/2)**2 + b) + log(tan(b*x/2) - 1/tan(c/2))/(b*ta
n(c/2)**2 + b) + log(tan(b*x/2)**2 + 1)*tan(c/2)**2/(b*tan(c/2)**2 + b) - log(tan(b*x/2)**2 + 1)/(b*tan(c/2)**
2 + b), True))*cos(a)
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Giac [B] time = 1.228, size = 651, normalized size = 24.11 \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)/sin(b*x+c),x, algorithm="giac")
[Out]
-1/2*(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))*(b*x + a)/(tan(1/2*a)^2*
tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) + (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(tan(b*x + a)^2 + 1)/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 +
1) - 2*(tan(1/2*a)^4*tan(1/2*c)^4 - 2*tan(1/2*a)^4*tan(1/2*c)^2 + 8*tan(1/2*a)^3*tan(1/2*c)^3 - 2*tan(1/2*a)^
2*tan(1/2*c)^4 + tan(1/2*a)^4 - 8*tan(1/2*a)^3*tan(1/2*c) + 20*tan(1/2*a)^2*tan(1/2*c)^2 - 8*tan(1/2*a)*tan(1/
2*c)^3 + tan(1/2*c)^4 - 2*tan(1/2*a)^2 + 8*tan(1/2*a)*tan(1/2*c) - 2*tan(1/2*c)^2 + 1)*log(abs(tan(b*x + a)*ta
n(1/2*a)^2*tan(1/2*c)^2 - tan(b*x + a)*tan(1/2*a)^2 + 4*tan(b*x + a)*tan(1/2*a)*tan(1/2*c) - 2*tan(1/2*a)^2*ta
n(1/2*c) - tan(b*x + a)*tan(1/2*c)^2 + 2*tan(1/2*a)*tan(1/2*c)^2 + tan(b*x + a) - 2*tan(1/2*a) + 2*tan(1/2*c))
)/(tan(1/2*a)^4*tan(1/2*c)^4 + 4*tan(1/2*a)^3*tan(1/2*c)^3 - tan(1/2*a)^4 + 4*tan(1/2*a)^3*tan(1/2*c) + 4*tan(
1/2*a)*tan(1/2*c)^3 - tan(1/2*c)^4 + 4*tan(1/2*a)*tan(1/2*c) + 1))/b