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

Optimal. Leaf size=108 \[ \frac{2 \sqrt{\frac{a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt{b^2+c^2}}} \text{EllipticF}\left (\frac{1}{2} \left (-\tan ^{-1}(b,c)+d+e x\right ),\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right )}{e \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}} \]

[Out]

(2*EllipticF[(d + e*x - ArcTan[b, c])/2, (2*Sqrt[b^2 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[(a + b*Cos[d + e*x] +
c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])])/(e*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.0702036, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3127, 2661} \[ \frac{2 \sqrt{\frac{a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt{b^2+c^2}}} F\left (\frac{1}{2} \left (d+e x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right )}{e \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]],x]

[Out]

(2*EllipticF[(d + e*x - ArcTan[b, c])/2, (2*Sqrt[b^2 + c^2])/(a + Sqrt[b^2 + c^2])]*Sqrt[(a + b*Cos[d + e*x] +
c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])])/(e*Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]])

Rule 3127

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a +
b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])]/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], Int[1/Sqrt[
a/(a + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; Free
Q[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+b \cos (d+e x)+c \sin (d+e x)}} \, dx &=\frac{\sqrt{\frac{a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt{b^2+c^2}}} \int \frac{1}{\sqrt{\frac{a}{a+\sqrt{b^2+c^2}}+\frac{\sqrt{b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}{a+\sqrt{b^2+c^2}}}} \, dx}{\sqrt{a+b \cos (d+e x)+c \sin (d+e x)}}\\ &=\frac{2 F\left (\frac{1}{2} \left (d+e x-\tan ^{-1}(b,c)\right )|\frac{2 \sqrt{b^2+c^2}}{a+\sqrt{b^2+c^2}}\right ) \sqrt{\frac{a+b \cos (d+e x)+c \sin (d+e x)}{a+\sqrt{b^2+c^2}}}}{e \sqrt{a+b \cos (d+e x)+c \sin (d+e x)}}\\ \end{align*}

Mathematica [C]  time = 0.647812, size = 285, normalized size = 2.64 \[ \frac{2 \sec \left (\tan ^{-1}\left (\frac{b}{c}\right )+d+e x\right ) \sqrt{-\frac{c \sqrt{\frac{b^2}{c^2}+1} \left (\sin \left (\tan ^{-1}\left (\frac{b}{c}\right )+d+e x\right )-1\right )}{a+c \sqrt{\frac{b^2}{c^2}+1}}} \sqrt{\frac{c \sqrt{\frac{b^2}{c^2}+1} \left (\sin \left (\tan ^{-1}\left (\frac{b}{c}\right )+d+e x\right )+1\right )}{c \sqrt{\frac{b^2}{c^2}+1}-a}} \sqrt{a+c \sqrt{\frac{b^2}{c^2}+1} \sin \left (\tan ^{-1}\left (\frac{b}{c}\right )+d+e x\right )} F_1\left (\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{3}{2};\frac{a+\sqrt{\frac{b^2}{c^2}+1} c \sin \left (d+e x+\tan ^{-1}\left (\frac{b}{c}\right )\right )}{a-\sqrt{\frac{b^2}{c^2}+1} c},\frac{a+\sqrt{\frac{b^2}{c^2}+1} c \sin \left (d+e x+\tan ^{-1}\left (\frac{b}{c}\right )\right )}{a+\sqrt{\frac{b^2}{c^2}+1} c}\right )}{c e \sqrt{\frac{b^2}{c^2}+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]],x]

[Out]

(2*AppellF1[1/2, 1/2, 1/2, 3/2, (a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x + ArcTan[b/c]])/(a - Sqrt[1 + b^2/c^2]*c)
, (a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x + ArcTan[b/c]])/(a + Sqrt[1 + b^2/c^2]*c)]*Sec[d + e*x + ArcTan[b/c]]*S
qrt[-((Sqrt[1 + b^2/c^2]*c*(-1 + Sin[d + e*x + ArcTan[b/c]]))/(a + Sqrt[1 + b^2/c^2]*c))]*Sqrt[(Sqrt[1 + b^2/c
^2]*c*(1 + Sin[d + e*x + ArcTan[b/c]]))/(-a + Sqrt[1 + b^2/c^2]*c)]*Sqrt[a + Sqrt[1 + b^2/c^2]*c*Sin[d + e*x +
ArcTan[b/c]]])/(Sqrt[1 + b^2/c^2]*c*e)

________________________________________________________________________________________

Maple [B]  time = 2.115, size = 303, normalized size = 2.8 \begin{align*} -2\,{\frac{-a+\sqrt{{b}^{2}+{c}^{2}}}{\sqrt{{b}^{2}+{c}^{2}}\cos \left ( ex+d-\arctan \left ( -b,c \right ) \right ) e}\sqrt{-{\frac{\sqrt{{b}^{2}+{c}^{2}}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) +a}{-a+\sqrt{{b}^{2}+{c}^{2}}}}}\sqrt{-{\frac{ \left ( \sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) -1 \right ) \sqrt{{b}^{2}+{c}^{2}}}{a+\sqrt{{b}^{2}+{c}^{2}}}}}\sqrt{{\frac{ \left ( 1+\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) \right ) \sqrt{{b}^{2}+{c}^{2}}}{-a+\sqrt{{b}^{2}+{c}^{2}}}}}{\it EllipticF} \left ( \sqrt{-{\frac{\sqrt{{b}^{2}+{c}^{2}}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) +a}{-a+\sqrt{{b}^{2}+{c}^{2}}}}},\sqrt{-{\frac{-a+\sqrt{{b}^{2}+{c}^{2}}}{a+\sqrt{{b}^{2}+{c}^{2}}}}} \right ){\frac{1}{\sqrt{{\frac{{b}^{2}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) +{c}^{2}\sin \left ( ex+d-\arctan \left ( -b,c \right ) \right ) +a\sqrt{{b}^{2}+{c}^{2}}}{\sqrt{{b}^{2}+{c}^{2}}}}}}}} \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))^(1/2),x)

[Out]

-2*(-a+(b^2+c^2)^(1/2))*(-((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)/(-a+(b^2+c^2)^(1/2)))^(1/2)*(-(sin(e*x+d
-arctan(-b,c))-1)*(b^2+c^2)^(1/2)/(a+(b^2+c^2)^(1/2)))^(1/2)*((1+sin(e*x+d-arctan(-b,c)))*(b^2+c^2)^(1/2)/(-a+
(b^2+c^2)^(1/2)))^(1/2)*EllipticF((-((b^2+c^2)^(1/2)*sin(e*x+d-arctan(-b,c))+a)/(-a+(b^2+c^2)^(1/2)))^(1/2),(-
(-a+(b^2+c^2)^(1/2))/(a+(b^2+c^2)^(1/2)))^(1/2))/(b^2+c^2)^(1/2)/cos(e*x+d-arctan(-b,c))/((b^2*sin(e*x+d-arcta
n(-b,c))+c^2*sin(e*x+d-arctan(-b,c))+a*(b^2+c^2)^(1/2))/(b^2+c^2)^(1/2))^(1/2)/e

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a}}\,{d x} \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))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\sqrt{b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a}}, x\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))^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \cos{\left (d + e x \right )} + c \sin{\left (d + e x \right )}}}\, dx \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))**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + a}}\,{d x} \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))^(1/2),x, algorithm="giac")

[Out]

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