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

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 = 0.114686, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, 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 = 0.0729495, size = 57, normalized size = 0.98 \[ -\frac{2 (b+c) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-\frac{\log (a+b \cos (x))}{b} \]

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 = 0.02, size = 150, normalized size = 2.6 \begin{align*}{\frac{1}{b}\ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{a}{b \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 ) }+{\frac{1}{a-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/(a-b)*ln(tan(1/2*x)^2*a-tan(1/2*x)^2*b+a+b)*a+1/(a-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 = 2.18665, size = 539, normalized size = 9.29 \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 = 1.18613, size = 146, normalized size = 2.52 \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)
/sqrt(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