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

Optimal. Leaf size=75 \[ \frac {2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{d \sqrt {a \cos (c+d x)+b \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3078, 2641} \[ \frac {2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{d \sqrt {a \cos (c+d x)+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3078

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[c + d*x] +
b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2])^n, Int[Cos[c + d*x - ArcTan[a, b]]^n, x]
, x] /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) &&  !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])

Rubi steps

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

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Mathematica [C]  time = 0.19, size = 92, normalized size = 1.23 \[ \frac {2 \tan \left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right ) \sqrt {\cos ^2\left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\tan ^{-1}\left (\frac {a}{b}\right )\right )\right )}{d \sqrt {b \sqrt {\frac {a^2}{b^2}+1} \sin \left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Cos[c + d*x + ArcTan[a/b]]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[a/b]]^2]*Tan[c
+ d*x + ArcTan[a/b]])/(d*Sqrt[Sqrt[1 + a^2/b^2]*b*Sin[c + d*x + ArcTan[a/b]]])

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.25, size = 121, normalized size = 1.61 \[ \frac {\sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \sqrt {-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}, \frac {\sqrt {2}}{2}\right )}{\cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2),x)

[Out]

(1+sin(d*x+c-arctan(-a,b)))^(1/2)*(-2*sin(d*x+c-arctan(-a,b))+2)^(1/2)*(-sin(d*x+c-arctan(-a,b)))^(1/2)*Ellipt
icF((1+sin(d*x+c-arctan(-a,b)))^(1/2),1/2*2^(1/2))/cos(d*x+c-arctan(-a,b))/(sin(d*x+c-arctan(-a,b))*(a^2+b^2)^
(1/2))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(c + d*x) + b*sin(c + d*x))^(1/2),x)

[Out]

int(1/(a*cos(c + d*x) + b*sin(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))**(1/2),x)

[Out]

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

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