3.9 \(\int \frac {1}{p+q \cos (x)+r \sin (x)} \, dx\)

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

[Out]

2*arctan((r+(p-q)*tan(1/2*x))/(p^2-q^2-r^2)^(1/2))/(p^2-q^2-r^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3124, 618, 204} \[ \frac {2 \tan ^{-1}\left (\frac {(p-q) \tan \left (\frac {x}{2}\right )+r}{\sqrt {p^2-q^2-r^2}}\right )}{\sqrt {p^2-q^2-r^2}} \]

Antiderivative was successfully verified.

[In]

Int[(p + q*Cos[x] + r*Sin[x])^(-1),x]

[Out]

(2*ArcTan[(r + (p - q)*Tan[x/2])/Sqrt[p^2 - q^2 - r^2]])/Sqrt[p^2 - q^2 - r^2]

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

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Mathematica [A]  time = 0.09, size = 50, normalized size = 0.93 \[ -\frac {2 \tanh ^{-1}\left (\frac {(p-q) \tan \left (\frac {x}{2}\right )+r}{\sqrt {-p^2+q^2+r^2}}\right )}{\sqrt {-p^2+q^2+r^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(p + q*Cos[x] + r*Sin[x])^(-1),x]

[Out]

(-2*ArcTanh[(r + (p - q)*Tan[x/2])/Sqrt[-p^2 + q^2 + r^2]])/Sqrt[-p^2 + q^2 + r^2]

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fricas [A]  time = 0.50, size = 364, normalized size = 6.74 \[ \left [-\frac {\sqrt {-p^{2} + q^{2} + r^{2}} \log \left (-\frac {p^{2} q^{2} - 2 \, q^{4} - r^{4} - {\left (p^{2} + 3 \, q^{2}\right )} r^{2} - {\left (2 \, p^{2} q^{2} - q^{4} - 2 \, p^{2} r^{2} + r^{4}\right )} \cos \relax (x)^{2} - 2 \, {\left (p q^{3} + p q r^{2}\right )} \cos \relax (x) - 2 \, {\left (p q^{2} r + p r^{3} - {\left (q r^{3} - {\left (2 \, p^{2} q - q^{3}\right )} r\right )} \cos \relax (x)\right )} \sin \relax (x) + 2 \, {\left (2 \, p q r \cos \relax (x)^{2} - p q r + {\left (q^{2} r + r^{3}\right )} \cos \relax (x) - {\left (q^{3} + q r^{2} + {\left (p q^{2} - p r^{2}\right )} \cos \relax (x)\right )} \sin \relax (x)\right )} \sqrt {-p^{2} + q^{2} + r^{2}}}{2 \, p q \cos \relax (x) + {\left (q^{2} - r^{2}\right )} \cos \relax (x)^{2} + p^{2} + r^{2} + 2 \, {\left (q r \cos \relax (x) + p r\right )} \sin \relax (x)}\right )}{2 \, {\left (p^{2} - q^{2} - r^{2}\right )}}, \frac {\arctan \left (-\frac {{\left (p q \cos \relax (x) + p r \sin \relax (x) + q^{2} + r^{2}\right )} \sqrt {p^{2} - q^{2} - r^{2}}}{{\left (r^{3} - {\left (p^{2} - q^{2}\right )} r\right )} \cos \relax (x) + {\left (p^{2} q - q^{3} - q r^{2}\right )} \sin \relax (x)}\right )}{\sqrt {p^{2} - q^{2} - r^{2}}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(p+q*cos(x)+r*sin(x)),x, algorithm="fricas")

[Out]

[-1/2*sqrt(-p^2 + q^2 + r^2)*log(-(p^2*q^2 - 2*q^4 - r^4 - (p^2 + 3*q^2)*r^2 - (2*p^2*q^2 - q^4 - 2*p^2*r^2 +
r^4)*cos(x)^2 - 2*(p*q^3 + p*q*r^2)*cos(x) - 2*(p*q^2*r + p*r^3 - (q*r^3 - (2*p^2*q - q^3)*r)*cos(x))*sin(x) +
2*(2*p*q*r*cos(x)^2 - p*q*r + (q^2*r + r^3)*cos(x) - (q^3 + q*r^2 + (p*q^2 - p*r^2)*cos(x))*sin(x))*sqrt(-p^2
+ q^2 + r^2))/(2*p*q*cos(x) + (q^2 - r^2)*cos(x)^2 + p^2 + r^2 + 2*(q*r*cos(x) + p*r)*sin(x)))/(p^2 - q^2 - r
^2), arctan(-(p*q*cos(x) + p*r*sin(x) + q^2 + r^2)*sqrt(p^2 - q^2 - r^2)/((r^3 - (p^2 - q^2)*r)*cos(x) + (p^2*
q - q^3 - q*r^2)*sin(x)))/sqrt(p^2 - q^2 - r^2)]

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giac [A]  time = 0.91, size = 72, normalized size = 1.33 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, p + 2 \, q\right ) + \arctan \left (-\frac {p \tan \left (\frac {1}{2} \, x\right ) - q \tan \left (\frac {1}{2} \, x\right ) + r}{\sqrt {p^{2} - q^{2} - r^{2}}}\right )\right )}}{\sqrt {p^{2} - q^{2} - r^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(p+q*cos(x)+r*sin(x)),x, algorithm="giac")

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*p + 2*q) + arctan(-(p*tan(1/2*x) - q*tan(1/2*x) + r)/sqrt(p^2 - q^2 - r^2)
))/sqrt(p^2 - q^2 - r^2)

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maple [A]  time = 0.08, size = 53, normalized size = 0.98 \[ \frac {2 \arctan \left (\frac {2 r +2 \left (p -q \right ) \tan \left (\frac {x}{2}\right )}{2 \sqrt {p^{2}-q^{2}-r^{2}}}\right )}{\sqrt {p^{2}-q^{2}-r^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(p+q*cos(x)+r*sin(x)),x)

[Out]

2/(p^2-q^2-r^2)^(1/2)*arctan(1/2*(2*(p-q)*tan(1/2*x)+2*r)/(p^2-q^2-r^2)^(1/2))

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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/(p+q*cos(x)+r*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(r^2+q^2-p^2>0)', see `assume?`
for more details)Is r^2+q^2-p^2 positive or negative?

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mupad [B]  time = 0.32, size = 68, normalized size = 1.26 \[ \left \{\begin {array}{cl} \frac {\ln \left (q+r\,\mathrm {tan}\left (\frac {x}{2}\right )\right )}{r} & \text {\ if\ \ }p=q\\ \frac {2\,\mathrm {atan}\left (\frac {r+\mathrm {tan}\left (\frac {x}{2}\right )\,\left (p-q\right )}{\sqrt {p^2-q^2-r^2}}\right )}{\sqrt {p^2-q^2-r^2}} & \text {\ if\ \ }p\neq q \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(p + q*cos(x) + r*sin(x)),x)

[Out]

piecewise(p == q, log(q + r*tan(x/2))/r, p ~= q, (2*atan((r + tan(x/2)*(p - q))/(p^2 - q^2 - r^2)^(1/2)))/(p^2
- q^2 - r^2)^(1/2))

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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/(p+q*cos(x)+r*sin(x)),x)

[Out]

Timed out

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