3.123 \(\int \frac {1}{a+\cos (x)+b \sin (x)} \, dx\)

Optimal. Leaf size=47 \[ -\frac {2 \tanh ^{-1}\left (\frac {b-(1-a) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+1}}\right )}{\sqrt {-a^2+b^2+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3124, 618, 206} \[ -\frac {2 \tanh ^{-1}\left (\frac {b-(1-a) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2+1}}\right )}{\sqrt {-a^2+b^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + Cos[x] + b*Sin[x])^(-1),x]

[Out]

(-2*ArcTanh[(b - (1 - a)*Tan[x/2])/Sqrt[1 - a^2 + b^2]])/Sqrt[1 - a^2 + b^2]

Rule 206

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 44, normalized size = 0.94 \[ \frac {2 \tan ^{-1}\left (\frac {(a-1) \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2-1}}\right )}{\sqrt {a^2-b^2-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[(b + (-1 + a)*Tan[x/2])/Sqrt[-1 + a^2 - b^2]])/Sqrt[-1 + a^2 - b^2]

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 287, normalized size = 6.11 \[ \left [-\frac {\sqrt {-a^{2} + b^{2} + 1} \log \left (-\frac {b^{4} + {\left (a^{2} + 3\right )} b^{2} - {\left (2 \, a^{2} b^{2} - b^{4} - 2 \, a^{2} + 1\right )} \cos \relax (x)^{2} - a^{2} + 2 \, {\left (a b^{2} + a\right )} \cos \relax (x) + 2 \, {\left (a b^{3} + a b - {\left (b^{3} - {\left (2 \, a^{2} - 1\right )} b\right )} \cos \relax (x)\right )} \sin \relax (x) - 2 \, {\left (2 \, a b \cos \relax (x)^{2} - a b + {\left (b^{3} + b\right )} \cos \relax (x) - {\left (b^{2} - {\left (a b^{2} - a\right )} \cos \relax (x) + 1\right )} \sin \relax (x)\right )} \sqrt {-a^{2} + b^{2} + 1} + 2}{{\left (b^{2} - 1\right )} \cos \relax (x)^{2} - a^{2} - b^{2} - 2 \, a \cos \relax (x) - 2 \, {\left (a b + b \cos \relax (x)\right )} \sin \relax (x)}\right )}{2 \, {\left (a^{2} - b^{2} - 1\right )}}, \frac {\arctan \left (-\frac {{\left (a b \sin \relax (x) + b^{2} + a \cos \relax (x) + 1\right )} \sqrt {a^{2} - b^{2} - 1}}{{\left (b^{3} - {\left (a^{2} - 1\right )} b\right )} \cos \relax (x) + {\left (a^{2} - b^{2} - 1\right )} \sin \relax (x)}\right )}{\sqrt {a^{2} - b^{2} - 1}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.19, size = 60, normalized size = 1.28 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b - \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2} - 1}}\right )\right )}}{\sqrt {a^{2} - b^{2} - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.07, size = 43, normalized size = 0.91 \[ \frac {2 \arctan \left (\frac {2 b +2 \left (a -1\right ) \tan \left (\frac {x}{2}\right )}{2 \sqrt {a^{2}-b^{2}-1}}\right )}{\sqrt {a^{2}-b^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+cos(x)+b*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b^2-a^2+1>0)', see `assume?` f
or more details)Is b^2-a^2+1 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 0.26, size = 58, normalized size = 1.23 \[ \left \{\begin {array}{cl} \frac {\ln \left (b\,\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{b} & \text {\ if\ \ }a=1\\ \frac {2\,\mathrm {atan}\left (\frac {b+\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a-1\right )}{\sqrt {a^2-b^2-1}}\right )}{\sqrt {a^2-b^2-1}} & \text {\ if\ \ }a\neq 1 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + cos(x) + b*sin(x)),x)

[Out]

piecewise(a == 1, log(b*tan(x/2) + 1)/b, a ~= 1, (2*atan((b + tan(x/2)*(a - 1))/(a^2 - b^2 - 1)^(1/2)))/(a^2 -
b^2 - 1)^(1/2))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+cos(x)+b*sin(x)),x)

[Out]

Timed out

________________________________________________________________________________________