3.399 \(\int \frac{1}{a+b \cos (d+e x)+c \sin (d+e x)} \, dx\)

Optimal. Leaf size=61 \[ \frac{2 \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{e \sqrt{a^2-b^2-c^2}} \]

[Out]

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

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Rubi [A]  time = 0.0837496, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3124, 618, 204} \[ \frac{2 \tan ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{a^2-b^2-c^2}}\right )}{e \sqrt{a^2-b^2-c^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(-1),x]

[Out]

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

Rule 3124

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Module[{f = Free
Factors[Tan[(d + e*x)/2], x]}, Dist[(2*f)/e, Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d +
e*x)/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^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{1}{a+b \cos (d+e x)+c \sin (d+e x)} \, dx &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{a+b+2 c x+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{e}\\ &=-\frac{4 \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2-c^2\right )-x^2} \, dx,x,2 c+2 (a-b) \tan \left (\frac{1}{2} (d+e x)\right )\right )}{e}\\ &=\frac{2 \tan ^{-1}\left (\frac{c+(a-b) \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a^2-b^2-c^2}}\right )}{\sqrt{a^2-b^2-c^2} e}\\ \end{align*}

Mathematica [A]  time = 0.117129, size = 57, normalized size = 0.93 \[ -\frac{2 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (d+e x)\right )+c}{\sqrt{-a^2+b^2+c^2}}\right )}{e \sqrt{-a^2+b^2+c^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(-1),x]

[Out]

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

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Maple [A]  time = 0.079, size = 61, normalized size = 1. \begin{align*} 2\,{\frac{1}{e\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( a-b \right ) \tan \left ( d/2+1/2\,ex \right ) +2\,c}{\sqrt{{a}^{2}-{b}^{2}-{c}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(e*x+d)+c*sin(e*x+d)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.33889, size = 953, normalized size = 15.62 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2} + c^{2}} \log \left (-\frac{a^{2} b^{2} - 2 \, b^{4} - c^{4} -{\left (a^{2} + 3 \, b^{2}\right )} c^{2} -{\left (2 \, a^{2} b^{2} - b^{4} - 2 \, a^{2} c^{2} + c^{4}\right )} \cos \left (e x + d\right )^{2} - 2 \,{\left (a b^{3} + a b c^{2}\right )} \cos \left (e x + d\right ) - 2 \,{\left (a b^{2} c + a c^{3} -{\left (b c^{3} -{\left (2 \, a^{2} b - b^{3}\right )} c\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right ) + 2 \,{\left (2 \, a b c \cos \left (e x + d\right )^{2} - a b c +{\left (b^{2} c + c^{3}\right )} \cos \left (e x + d\right ) -{\left (b^{3} + b c^{2} +{\left (a b^{2} - a c^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )} \sqrt{-a^{2} + b^{2} + c^{2}}}{2 \, a b \cos \left (e x + d\right ) +{\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} + a^{2} + c^{2} + 2 \,{\left (b c \cos \left (e x + d\right ) + a c\right )} \sin \left (e x + d\right )}\right )}{2 \,{\left (a^{2} - b^{2} - c^{2}\right )} e}, \frac{\arctan \left (-\frac{{\left (a b \cos \left (e x + d\right ) + a c \sin \left (e x + d\right ) + b^{2} + c^{2}\right )} \sqrt{a^{2} - b^{2} - c^{2}}}{{\left (c^{3} -{\left (a^{2} - b^{2}\right )} c\right )} \cos \left (e x + d\right ) +{\left (a^{2} b - b^{3} - b c^{2}\right )} \sin \left (e x + d\right )}\right )}{\sqrt{a^{2} - b^{2} - c^{2}} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d)),x, algorithm="fricas")

[Out]

[-1/2*sqrt(-a^2 + b^2 + c^2)*log(-(a^2*b^2 - 2*b^4 - c^4 - (a^2 + 3*b^2)*c^2 - (2*a^2*b^2 - b^4 - 2*a^2*c^2 +
c^4)*cos(e*x + d)^2 - 2*(a*b^3 + a*b*c^2)*cos(e*x + d) - 2*(a*b^2*c + a*c^3 - (b*c^3 - (2*a^2*b - b^3)*c)*cos(
e*x + d))*sin(e*x + d) + 2*(2*a*b*c*cos(e*x + d)^2 - a*b*c + (b^2*c + c^3)*cos(e*x + d) - (b^3 + b*c^2 + (a*b^
2 - a*c^2)*cos(e*x + d))*sin(e*x + d))*sqrt(-a^2 + b^2 + c^2))/(2*a*b*cos(e*x + d) + (b^2 - c^2)*cos(e*x + d)^
2 + a^2 + c^2 + 2*(b*c*cos(e*x + d) + a*c)*sin(e*x + d)))/((a^2 - b^2 - c^2)*e), arctan(-(a*b*cos(e*x + d) + a
*c*sin(e*x + d) + b^2 + c^2)*sqrt(a^2 - b^2 - c^2)/((c^3 - (a^2 - b^2)*c)*cos(e*x + d) + (a^2*b - b^3 - b*c^2)
*sin(e*x + d)))/(sqrt(a^2 - b^2 - c^2)*e)]

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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(1/(a+b*cos(e*x+d)+c*sin(e*x+d)),x)

[Out]

Timed out

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Giac [A]  time = 1.11689, size = 123, normalized size = 2.02 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x e + d}{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 e + \frac{1}{2} \, d\right ) - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) + c}{\sqrt{a^{2} - b^{2} - c^{2}}}\right )\right )} e^{\left (-1\right )}}{\sqrt{a^{2} - b^{2} - c^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d)),x, algorithm="giac")

[Out]

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