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

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

[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]])

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Rubi [A]  time = 0.0296477, antiderivative size = 75, normalized size of antiderivative = 1., 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 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])

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]

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*}

Mathematica [C]  time = 0.186312, size = 92, normalized size = 1.23 \[ \frac{2 \tan \left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d x\right ) \sqrt{\cos ^2\left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d x\right )} \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d x\right )\right )}{d \sqrt{b \sqrt{\frac{a^2}{b^2}+1} \sin \left (\tan ^{-1}\left (\frac{a}{b}\right )+c+d x\right )}} \]

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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Maple [A]  time = 1.072, size = 121, normalized size = 1.6 \begin{align*}{\frac{1}{\cos \left ( dx+c-\arctan \left ( -a,b \right ) \right ) d}\sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) +2}\sqrt{-\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }{\it EllipticF} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \sqrt{{a}^{2}+{b}^{2}}}}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}}\,{d x} \end{align*}

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

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

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

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)