3.5 \(\int \frac{b+c+\sin (x)}{a-b \cos (x)} \, dx\)

Optimal. Leaf size=58 \[ \frac{2 (b+c) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan \left (\frac{x}{2}\right )}{\sqrt{a-b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{\log (a-b \cos (x))}{b} \]

[Out]

(2*(b + c)*ArcTan[(Sqrt[a + b]*Tan[x/2])/Sqrt[a - b]])/(Sqrt[a - b]*Sqrt[a + b]) + Log[a - b*Cos[x]]/b

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Rubi [A]  time = 0.130403, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4401, 2659, 205, 2668, 31} \[ \frac{2 (b+c) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan \left (\frac{x}{2}\right )}{\sqrt{a-b}}\right )}{\sqrt{a-b} \sqrt{a+b}}+\frac{\log (a-b \cos (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[(b + c + Sin[x])/(a - b*Cos[x]),x]

[Out]

(2*(b + c)*ArcTan[(Sqrt[a + b]*Tan[x/2])/Sqrt[a - b]])/(Sqrt[a - b]*Sqrt[a + b]) + Log[a - b*Cos[x]]/b

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 31

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

Rubi steps

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

Mathematica [A]  time = 0.0913181, size = 55, normalized size = 0.95 \[ \frac{\log (a-b \cos (x))}{b}-\frac{2 (b+c) \tanh ^{-1}\left (\frac{(a+b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b + c + Sin[x])/(a - b*Cos[x]),x]

[Out]

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

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Maple [B]  time = 0.023, size = 110, normalized size = 1.9 \begin{align*} -{\frac{1}{b}\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+{\frac{1}{b}\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{b}{\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{c}{\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((b+c+sin(x))/(a-b*cos(x)),x)

[Out]

-1/b*ln(tan(1/2*x)^2+1)+1/b*ln(tan(1/2*x)^2*a+tan(1/2*x)^2*b+a-b)+2*b/((a+b)*(a-b))^(1/2)*arctan((a+b)*tan(1/2
*x)/((a+b)*(a-b))^(1/2))+2/((a+b)*(a-b))^(1/2)*arctan((a+b)*tan(1/2*x)/((a+b)*(a-b))^(1/2))*c

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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((b+c+sin(x))/(a-b*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10972, size = 540, normalized size = 9.31 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}}{\left (b^{2} + b c\right )} \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 (a^{2} - b^{2}\right )} \log \left (b^{2} \cos \left (x\right )^{2} - 2 \, a b \cos \left (x\right ) + a^{2}\right )}{2 \,{\left (a^{2} b - b^{3}\right )}}, \frac{2 \, \sqrt{a^{2} - b^{2}}{\left (b^{2} + b c\right )} \arctan \left (-\frac{a \cos \left (x\right ) - b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right ) +{\left (a^{2} - b^{2}\right )} \log \left (b^{2} \cos \left (x\right )^{2} - 2 \, a b \cos \left (x\right ) + a^{2}\right )}{2 \,{\left (a^{2} b - b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b+c+sin(x))/(a-b*cos(x)),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(-a^2 + b^2)*(b^2 + b*c)*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)) - (a^2 - b^2)*log(b^2*cos(x)^2 - 2*a*b*cos(x)
+ a^2))/(a^2*b - b^3), 1/2*(2*sqrt(a^2 - b^2)*(b^2 + b*c)*arctan(-(a*cos(x) - b)/(sqrt(a^2 - b^2)*sin(x))) +
(a^2 - b^2)*log(b^2*cos(x)^2 - 2*a*b*cos(x) + a^2))/(a^2*b - b^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b+c+sin(x))/(a-b*cos(x)),x)

[Out]

Timed out

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Giac [B]  time = 1.15525, size = 139, normalized size = 2.4 \begin{align*} \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 )}{\left (b + c\right )}}{\sqrt{a^{2} - b^{2}}} + \frac{\log \left (a \tan \left (\frac{1}{2} \, x\right )^{2} + b \tan \left (\frac{1}{2} \, x\right )^{2} + a - b\right )}{b} - \frac{\log \left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b+c+sin(x))/(a-b*cos(x)),x, algorithm="giac")

[Out]

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)))*(b + c)/sq
rt(a^2 - b^2) + log(a*tan(1/2*x)^2 + b*tan(1/2*x)^2 + a - b)/b - log(tan(1/2*x)^2 + 1)/b