3.8 \(\int \frac{A+B \csc (x)}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=99 \[ \frac{b B \log (a+b \cos (x))}{a^2-b^2}+\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (1-\cos (x))}{2 (a+b)}-\frac{B \log (\cos (x)+1)}{2 (a-b)} \]

[Out]

(2*A*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) + (B*Log[1 - Cos[x]])/(2*(a + b)) -
(B*Log[1 + Cos[x]])/(2*(a - b)) + (b*B*Log[a + b*Cos[x]])/(a^2 - b^2)

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Rubi [A]  time = 0.250172, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {4225, 4401, 2659, 205, 2668, 706, 31, 633} \[ \frac{b B \log (a+b \cos (x))}{a^2-b^2}+\frac{2 A \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{B \log (1-\cos (x))}{2 (a+b)}-\frac{B \log (\cos (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Csc[x])/(a + b*Cos[x]),x]

[Out]

(2*A*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]) + (B*Log[1 - Cos[x]])/(2*(a + b)) -
(B*Log[1 + Cos[x]])/(2*(a - b)) + (b*B*Log[a + b*Cos[x]])/(a^2 - b^2)

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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

Mathematica [A]  time = 0.190406, size = 116, normalized size = 1.17 \[ \frac{-2 A \left (a^2-b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )-B \sqrt{b^2-a^2} \left ((b-a) \log \left (\sin \left (\frac{x}{2}\right )\right )+(a+b) \log \left (\cos \left (\frac{x}{2}\right )\right )-b \log (a+b \cos (x))\right )}{(a-b) (a+b) \sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Csc[x])/(a + b*Cos[x]),x]

[Out]

(-2*A*(a^2 - b^2)*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]] - Sqrt[-a^2 + b^2]*B*((a + b)*Log[Cos[x/2]] - b
*Log[a + b*Cos[x]] + (-a + b)*Log[Sin[x/2]]))/((a - b)*(a + b)*Sqrt[-a^2 + b^2])

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Maple [A]  time = 0.027, size = 134, normalized size = 1.4 \begin{align*}{\frac{B}{a+b}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{Bb}{ \left ( a+b \right ) \left ( a-b \right ) }\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}a- \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}b+a+b \right ) }+2\,{\frac{Aa}{ \left ( a+b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{Ab}{ \left ( a+b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B/(a+b)*ln(tan(1/2*x))+1/(a+b)*B*b/(a-b)*ln(tan(1/2*x)^2*a-tan(1/2*x)^2*b+a+b)+2/(a+b)/((a+b)*(a-b))^(1/2)*arc
tan((a-b)*tan(1/2*x)/((a+b)*(a-b))^(1/2))*A*a+2/(a+b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tan(1/2*x)/((a+b)*(a-b)
)^(1/2))*A*b

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 48.048, size = 680, normalized size = 6.87 \begin{align*} \left [\frac{B b \log \left (b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}\right ) - \sqrt{-a^{2} + b^{2}} A \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) -{\left (B a + B b\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (B a - B b\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{2} - b^{2}\right )}}, \frac{B b \log \left (b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}\right ) + 2 \, \sqrt{a^{2} - b^{2}} A \arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right ) -{\left (B a + B b\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (B a - B b\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{2} - b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(B*b*log(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2) - sqrt(-a^2 + b^2)*A*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)
^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2)) - (B*a + B*b
)*log(1/2*cos(x) + 1/2) + (B*a - B*b)*log(-1/2*cos(x) + 1/2))/(a^2 - b^2), 1/2*(B*b*log(b^2*cos(x)^2 + 2*a*b*c
os(x) + a^2) + 2*sqrt(a^2 - b^2)*A*arctan(-(a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x))) - (B*a + B*b)*log(1/2*cos(
x) + 1/2) + (B*a - B*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{A + B \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((A+B*csc(x))/(a+b*cos(x)),x)

[Out]

Integral((A + B*csc(x))/(a + b*cos(x)), x)

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Giac [A]  time = 1.16498, size = 155, normalized size = 1.57 \begin{align*} \frac{B b \log \left (-a \tan \left (\frac{1}{2} \, x\right )^{2} + b \tan \left (\frac{1}{2} \, x\right )^{2} - a - b\right )}{a^{2} - b^{2}} - \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )} A}{\sqrt{a^{2} - b^{2}}} + \frac{B \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{a + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b*log(-a*tan(1/2*x)^2 + b*tan(1/2*x)^2 - a - b)/(a^2 - b^2) - 2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) +
arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*A/sqrt(a^2 - b^2) + B*log(abs(tan(1/2*x)))/(a + b)