3.694 \(\int \frac{\frac{b B}{a}+B \sin (x)}{a+b \sin (x)} \, dx\)

Optimal. Leaf size=54 \[ \frac{B x}{b}-\frac{2 B \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a b} \]

[Out]

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

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Rubi [A]  time = 0.0816141, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2735, 2660, 618, 204} \[ \frac{B x}{b}-\frac{2 B \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a b} \]

Antiderivative was successfully verified.

[In]

Int[((b*B)/a + B*Sin[x])/(a + b*Sin[x]),x]

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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])

Rubi steps

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

Mathematica [A]  time = 0.0748122, size = 52, normalized size = 0.96 \[ \frac{B \left (a x-2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )\right )}{a b} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*B)/a + B*Sin[x])/(a + b*Sin[x]),x]

[Out]

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

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Maple [B]  time = 0.037, size = 99, normalized size = 1.8 \begin{align*} 2\,{\frac{B\arctan \left ( \tan \left ( x/2 \right ) \right ) }{b}}-2\,{\frac{Ba}{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{bB}{a\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*B/a+B*sin(x))/(a+b*sin(x)),x)

[Out]

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

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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*B/a+B*sin(x))/(a+b*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57714, size = 377, normalized size = 6.98 \begin{align*} \left [\frac{2 \, B a x + \sqrt{-a^{2} + b^{2}} B \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 )}{2 \, a b}, \frac{B a x + \sqrt{a^{2} - b^{2}} B \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right )}{a b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*B/a+B*sin(x))/(a+b*sin(x)),x, algorithm="fricas")

[Out]

[1/2*(2*B*a*x + sqrt(-a^2 + b^2)*B*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*b), (B*a*x + sqrt(a^2 - b^2)*B*a
rctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))))/(a*b)]

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Sympy [A]  time = 126.164, size = 87, normalized size = 1.61 \begin{align*} \begin{cases} \frac{B x}{b} + \frac{B \sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{x}{2} \right )} + \frac{b}{a} - \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a b} - \frac{B \sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{x}{2} \right )} + \frac{b}{a} + \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a b} & \text{for}\: b \neq 0 \\- \frac{B \cos{\left (x \right )}}{a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*B/a+B*sin(x))/(a+b*sin(x)),x)

[Out]

Piecewise((B*x/b + B*sqrt(-a**2 + b**2)*log(tan(x/2) + b/a - sqrt(-a**2 + b**2)/a)/(a*b) - B*sqrt(-a**2 + b**2
)*log(tan(x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a*b), Ne(b, 0)), (-B*cos(x)/a, True))

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Giac [A]  time = 1.27009, size = 99, normalized size = 1.83 \begin{align*} \frac{B x}{b} - \frac{2 \,{\left (B a^{2} - B b^{2}\right )}{\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 )}}{\sqrt{a^{2} - b^{2}} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*B/a+B*sin(x))/(a+b*sin(x)),x, algorithm="giac")

[Out]

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