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

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

[Out]

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

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Rubi [A]  time = 0.138768, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4401, 2660, 618, 204, 2668, 31} \[ -\frac{2 (b+c) \tan ^{-1}\left (\frac{b-a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{\log (a-b \sin (x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4401

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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+\cos (x)}{a-b \sin (x)} \, dx &=\int \left (\frac{\left (1+\frac{b}{c}\right ) c}{a-b \sin (x)}+\frac{\cos (x)}{a-b \sin (x)}\right ) \, dx\\ &=(b+c) \int \frac{1}{a-b \sin (x)} \, dx+\int \frac{\cos (x)}{a-b \sin (x)} \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{a+x} \, dx,x,-b \sin (x)\right )}{b}+(2 (b+c)) \operatorname{Subst}\left (\int \frac{1}{a-2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=-\frac{\log (a-b \sin (x))}{b}-(4 (b+c)) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,-2 b+2 a \tan \left (\frac{x}{2}\right )\right )\\ &=-\frac{2 (b+c) \tan ^{-1}\left (\frac{b-a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{\log (a-b \sin (x))}{b}\\ \end{align*}

Mathematica [A]  time = 0.109787, size = 59, normalized size = 1.02 \[ -\frac{2 (b+c) \tan ^{-1}\left (\frac{b-a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-\frac{\log (b \sin (x)-a)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.089, size = 116, normalized size = 2. \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-2\,\tan \left ( x/2 \right ) b+a \right ) }+2\,{\frac{b}{\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{c}{\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*ln(tan(1/2*x)^2+1)-1/b*ln(tan(1/2*x)^2*a-2*tan(1/2*x)*b+a)+2*b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)-
2*b)/(a^2-b^2)^(1/2))+2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)-2*b)/(a^2-b^2)^(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+cos(x))/(a-b*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.12698, size = 575, normalized size = 9.91 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}}{\left (b^{2} + b c\right )} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) - b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) +{\left (a^{2} - b^{2}\right )} \log \left (-b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) + a^{2} + b^{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 \sin \left (x\right ) - b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) +{\left (a^{2} - b^{2}\right )} \log \left (-b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) + a^{2} + b^{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+cos(x))/(a-b*sin(x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 65.2471, size = 643, normalized size = 11.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(c*log(tan(x/2)) - log(tan(x/2)**2 + 1) + log(tan(x/2))), Eq(a, 0) & Eq(b, 0)), (-(b*log(tan(x/
2)) + c*log(tan(x/2)) - log(tan(x/2)**2 + 1) + log(tan(x/2)))/b, Eq(a, 0)), (2*b/(b*tan(x/2) + b) + 2*c/(b*tan
(x/2) + b) - 2*log(tan(x/2) + 1)*tan(x/2)/(b*tan(x/2) + b) - 2*log(tan(x/2) + 1)/(b*tan(x/2) + b) + log(tan(x/
2)**2 + 1)*tan(x/2)/(b*tan(x/2) + b) + log(tan(x/2)**2 + 1)/(b*tan(x/2) + b), Eq(a, -b)), (-2*b/(b*tan(x/2) -
b) - 2*c/(b*tan(x/2) - b) - 2*log(tan(x/2) - 1)*tan(x/2)/(b*tan(x/2) - b) + 2*log(tan(x/2) - 1)/(b*tan(x/2) -
b) + log(tan(x/2)**2 + 1)*tan(x/2)/(b*tan(x/2) - b) - log(tan(x/2)**2 + 1)/(b*tan(x/2) - b), Eq(a, b)), ((c*x
+ sin(x))/a, Eq(b, 0)), (a**2*log(tan(x/2)**2 + 1)/(a**2*b - b**3) - a**2*log(tan(x/2) - b/a - sqrt(-a**2 + b*
*2)/a)/(a**2*b - b**3) - a**2*log(tan(x/2) - b/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) - b**2*sqrt(-a**2 + b
**2)*log(tan(x/2) - b/a - sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) + b**2*sqrt(-a**2 + b**2)*log(tan(x/2) - b/a +
sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) - b**2*log(tan(x/2)**2 + 1)/(a**2*b - b**3) + b**2*log(tan(x/2) - b/a -
sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) + b**2*log(tan(x/2) - b/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) - b*c
*sqrt(-a**2 + b**2)*log(tan(x/2) - b/a - sqrt(-a**2 + b**2)/a)/(a**2*b - b**3) + b*c*sqrt(-a**2 + b**2)*log(ta
n(x/2) - b/a + sqrt(-a**2 + b**2)/a)/(a**2*b - b**3), True))

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Giac [A]  time = 1.07213, size = 122, normalized size = 2.1 \begin{align*} \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) - b}{\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} - 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\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+cos(x))/(a-b*sin(x)),x, algorithm="giac")

[Out]

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