3.701 \(\int \frac{c+d \sin (e+f x)}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=65 \[ \frac{2 (b c-a d) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b f \sqrt{a^2-b^2}}+\frac{d x}{b} \]

[Out]

(d*x)/b + (2*(b*c - a*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2]*f)

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Rubi [A]  time = 0.0719839, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2735, 2660, 618, 204} \[ \frac{2 (b c-a d) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b f \sqrt{a^2-b^2}}+\frac{d x}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])/(a + b*Sin[e + f*x]),x]

[Out]

(d*x)/b + (2*(b*c - a*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(b*Sqrt[a^2 - b^2]*f)

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

Mathematica [A]  time = 0.0980203, size = 67, normalized size = 1.03 \[ \frac{\frac{2 (b c-a d) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+d (e+f x)}{b f} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sin[e + f*x])/(a + b*Sin[e + f*x]),x]

[Out]

(d*(e + f*x) + (2*(b*c - a*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2])/(b*f)

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Maple [A]  time = 0.042, size = 119, normalized size = 1.8 \begin{align*} 2\,{\frac{d\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{bf}}-2\,{\frac{da}{bf\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{c}{f\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x)

[Out]

2/f*d/b*arctan(tan(1/2*f*x+1/2*e))-2/f/b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/
2))*d*a+2/f/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+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((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.42105, size = 543, normalized size = 8.35 \begin{align*} \left [\frac{2 \,{\left (a^{2} - b^{2}\right )} d f x + \sqrt{-a^{2} + b^{2}}{\left (b c - a d\right )} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right )}{2 \,{\left (a^{2} b - b^{3}\right )} f}, \frac{{\left (a^{2} - b^{2}\right )} d f x - \sqrt{a^{2} - b^{2}}{\left (b c - a d\right )} \arctan \left (-\frac{a \sin \left (f x + e\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (f x + e\right )}\right )}{{\left (a^{2} b - b^{3}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 142.884, size = 502, normalized size = 7.72 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \left (c + d \sin{\left (e \right )}\right )}{\sin{\left (e \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac{2 c}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - b f} + \frac{d f x \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - b f} - \frac{d f x}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - b f} + \frac{2 d}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} - b f} & \text{for}\: a = - b \\- \frac{2 c}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + b f} + \frac{d f x \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + b f} + \frac{d f x}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + b f} + \frac{2 d}{b f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + b f} & \text{for}\: a = b \\\frac{c x - \frac{d \cos{\left (e + f x \right )}}{f}}{a} & \text{for}\: b = 0 \\\frac{x \left (c + d \sin{\left (e \right )}\right )}{a + b \sin{\left (e \right )}} & \text{for}\: f = 0 \\\frac{\frac{c \log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} \right )}}{f} + d x}{b} & \text{for}\: a = 0 \\\frac{a^{2} d f x}{a^{2} b f - b^{3} f} + \frac{a d \sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + \frac{b}{a} - \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a^{2} b f - b^{3} f} - \frac{a d \sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + \frac{b}{a} + \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a^{2} b f - b^{3} f} - \frac{b^{2} d f x}{a^{2} b f - b^{3} f} - \frac{b c \sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + \frac{b}{a} - \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a^{2} b f - b^{3} f} + \frac{b c \sqrt{- a^{2} + b^{2}} \log{\left (\tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + \frac{b}{a} + \frac{\sqrt{- a^{2} + b^{2}}}{a} \right )}}{a^{2} b f - b^{3} f} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x)

[Out]

Piecewise((zoo*x*(c + d*sin(e))/sin(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (2*c/(b*f*tan(e/2 + f*x/2) - b*f) + d
*f*x*tan(e/2 + f*x/2)/(b*f*tan(e/2 + f*x/2) - b*f) - d*f*x/(b*f*tan(e/2 + f*x/2) - b*f) + 2*d/(b*f*tan(e/2 + f
*x/2) - b*f), Eq(a, -b)), (-2*c/(b*f*tan(e/2 + f*x/2) + b*f) + d*f*x*tan(e/2 + f*x/2)/(b*f*tan(e/2 + f*x/2) +
b*f) + d*f*x/(b*f*tan(e/2 + f*x/2) + b*f) + 2*d/(b*f*tan(e/2 + f*x/2) + b*f), Eq(a, b)), ((c*x - d*cos(e + f*x
)/f)/a, Eq(b, 0)), (x*(c + d*sin(e))/(a + b*sin(e)), Eq(f, 0)), ((c*log(tan(e/2 + f*x/2))/f + d*x)/b, Eq(a, 0)
), (a**2*d*f*x/(a**2*b*f - b**3*f) + a*d*sqrt(-a**2 + b**2)*log(tan(e/2 + f*x/2) + b/a - sqrt(-a**2 + b**2)/a)
/(a**2*b*f - b**3*f) - a*d*sqrt(-a**2 + b**2)*log(tan(e/2 + f*x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**2*b*f - b
**3*f) - b**2*d*f*x/(a**2*b*f - b**3*f) - b*c*sqrt(-a**2 + b**2)*log(tan(e/2 + f*x/2) + b/a - sqrt(-a**2 + b**
2)/a)/(a**2*b*f - b**3*f) + b*c*sqrt(-a**2 + b**2)*log(tan(e/2 + f*x/2) + b/a + sqrt(-a**2 + b**2)/a)/(a**2*b*
f - b**3*f), True))

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Giac [A]  time = 1.34768, size = 116, normalized size = 1.78 \begin{align*} \frac{\frac{{\left (f x + e\right )} d}{b} + \frac{2 \,{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}{\left (b c - a d\right )}}{\sqrt{a^{2} - b^{2}} b}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

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