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[x/2])/Sqrt[p^2 - q^2 - r^2]])/Sqrt[p^2 - q^2 - r^2]

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Rubi [A]  time = 0.0681303, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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}{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*}

Mathematica [A]  time = 0.0812852, 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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Maple [A]  time = 0.044, size = 53, normalized size = 1. \begin{align*} 2\,{\frac{1}{\sqrt{{p}^{2}-{q}^{2}-{r}^{2}}}\arctan \left ( 1/2\,{\frac{2\, \left ( p-q \right ) \tan \left ( x/2 \right ) +2\,r}{\sqrt{{p}^{2}-{q}^{2}-{r}^{2}}}} \right ) } \end{align*}

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., 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/(p+q*cos(x)+r*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89938, size = 813, normalized size = 15.06 \begin{align*} \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 \left (x\right )^{2} - 2 \,{\left (p q^{3} + p q r^{2}\right )} \cos \left (x\right ) - 2 \,{\left (p q^{2} r + p r^{3} -{\left (q r^{3} -{\left (2 \, p^{2} q - q^{3}\right )} r\right )} \cos \left (x\right )\right )} \sin \left (x\right ) + 2 \,{\left (2 \, p q r \cos \left (x\right )^{2} - p q r +{\left (q^{2} r + r^{3}\right )} \cos \left (x\right ) -{\left (q^{3} + q r^{2} +{\left (p q^{2} - p r^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )\right )} \sqrt{-p^{2} + q^{2} + r^{2}}}{2 \, p q \cos \left (x\right ) +{\left (q^{2} - r^{2}\right )} \cos \left (x\right )^{2} + p^{2} + r^{2} + 2 \,{\left (q r \cos \left (x\right ) + p r\right )} \sin \left (x\right )}\right )}{2 \,{\left (p^{2} - q^{2} - r^{2}\right )}}, \frac{\arctan \left (-\frac{{\left (p q \cos \left (x\right ) + p r \sin \left (x\right ) + q^{2} + r^{2}\right )} \sqrt{p^{2} - q^{2} - r^{2}}}{{\left (r^{3} -{\left (p^{2} - q^{2}\right )} r\right )} \cos \left (x\right ) +{\left (p^{2} q - q^{3} - q r^{2}\right )} \sin \left (x\right )}\right )}{\sqrt{p^{2} - q^{2} - r^{2}}}\right ] \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.08755, size = 97, normalized size = 1.8 \begin{align*} -\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}}} \end{align*}

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)