3.29 \(\int \frac{\csc (x)}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=53 \[ \frac{b \log (a+b \cos (x))}{a^2-b^2}+\frac{\log (1-\cos (x))}{2 (a+b)}-\frac{\log (\cos (x)+1)}{2 (a-b)} \]

[Out]

Log[1 - Cos[x]]/(2*(a + b)) - Log[1 + Cos[x]]/(2*(a - b)) + (b*Log[a + b*Cos[x]])/(a^2 - b^2)

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Rubi [A]  time = 0.0722379, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {2668, 706, 31, 633} \[ \frac{b \log (a+b \cos (x))}{a^2-b^2}+\frac{\log (1-\cos (x))}{2 (a+b)}-\frac{\log (\cos (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a + b*Cos[x]),x]

[Out]

Log[1 - Cos[x]]/(2*(a + b)) - Log[1 + Cos[x]]/(2*(a - b)) + (b*Log[a + b*Cos[x]])/(a^2 - b^2)

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rubi steps

\begin{align*} \int \frac{\csc (x)}{a+b \cos (x)} \, dx &=-\left (b \operatorname{Subst}\left (\int \frac{1}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \cos (x)\right )\right )\\ &=\frac{b \operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,b \cos (x)\right )}{a^2-b^2}+\frac{b \operatorname{Subst}\left (\int \frac{-a+x}{b^2-x^2} \, dx,x,b \cos (x)\right )}{a^2-b^2}\\ &=\frac{b \log (a+b \cos (x))}{a^2-b^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{-b-x} \, dx,x,b \cos (x)\right )}{2 (a-b)}-\frac{\operatorname{Subst}\left (\int \frac{1}{b-x} \, dx,x,b \cos (x)\right )}{2 (a+b)}\\ &=\frac{\log (1-\cos (x))}{2 (a+b)}-\frac{\log (1+\cos (x))}{2 (a-b)}+\frac{b \log (a+b \cos (x))}{a^2-b^2}\\ \end{align*}

Mathematica [A]  time = 0.0411299, size = 50, normalized size = 0.94 \[ \frac{(a-b) \log (1-\cos (x))-(a+b) \log (\cos (x)+1)+2 b \log (a+b \cos (x))}{2 (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(a + b*Cos[x]),x]

[Out]

((a - b)*Log[1 - Cos[x]] - (a + b)*Log[1 + Cos[x]] + 2*b*Log[a + b*Cos[x]])/(2*(a - b)*(a + b))

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Maple [A]  time = 0.04, size = 54, normalized size = 1. \begin{align*}{\frac{b\ln \left ( a+b\cos \left ( x \right ) \right ) }{ \left ( a-b \right ) \left ( a+b \right ) }}+{\frac{\ln \left ( -1+\cos \left ( x \right ) \right ) }{2\,a+2\,b}}-{\frac{\ln \left ( \cos \left ( x \right ) +1 \right ) }{2\,a-2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)/(a+b*cos(x)),x)

[Out]

b/(a-b)/(a+b)*ln(a+b*cos(x))+1/(2*a+2*b)*ln(-1+cos(x))-1/(2*a-2*b)*ln(cos(x)+1)

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Maxima [A]  time = 1.52995, size = 63, normalized size = 1.19 \begin{align*} \frac{b \log \left (b \cos \left (x\right ) + a\right )}{a^{2} - b^{2}} - \frac{\log \left (\cos \left (x\right ) + 1\right )}{2 \,{\left (a - b\right )}} + \frac{\log \left (\cos \left (x\right ) - 1\right )}{2 \,{\left (a + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*cos(x)),x, algorithm="maxima")

[Out]

b*log(b*cos(x) + a)/(a^2 - b^2) - 1/2*log(cos(x) + 1)/(a - b) + 1/2*log(cos(x) - 1)/(a + b)

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Fricas [A]  time = 1.74471, size = 144, normalized size = 2.72 \begin{align*} \frac{2 \, b \log \left (-b \cos \left (x\right ) - a\right ) -{\left (a + b\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (a - b\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{2} - b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*cos(x)),x, algorithm="fricas")

[Out]

1/2*(2*b*log(-b*cos(x) - a) - (a + b)*log(1/2*cos(x) + 1/2) + (a - b)*log(-1/2*cos(x) + 1/2))/(a^2 - b^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc{\left (x \right )}}{a + b \cos{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*cos(x)),x)

[Out]

Integral(csc(x)/(a + b*cos(x)), x)

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Giac [A]  time = 1.15603, size = 73, normalized size = 1.38 \begin{align*} \frac{b^{2} \log \left ({\left | b \cos \left (x\right ) + a \right |}\right )}{a^{2} b - b^{3}} - \frac{\log \left (\cos \left (x\right ) + 1\right )}{2 \,{\left (a - b\right )}} + \frac{\log \left (-\cos \left (x\right ) + 1\right )}{2 \,{\left (a + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*cos(x)),x, algorithm="giac")

[Out]

b^2*log(abs(b*cos(x) + a))/(a^2*b - b^3) - 1/2*log(cos(x) + 1)/(a - b) + 1/2*log(-cos(x) + 1)/(a + b)