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

Optimal. Leaf size=76 \[ \frac{\sqrt{2} \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}} \text{EllipticF}\left (c+d x-\frac{\pi }{4},\frac{2 b}{2 a+b}\right )}{d \sqrt{2 a+b \sin (2 c+2 d x)}} \]

[Out]

(Sqrt[2]*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])/(d*Sqrt[2*a +
b*Sin[2*c + 2*d*x]])

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Rubi [A]  time = 0.0665995, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2666, 2663, 2661} \[ \frac{\sqrt{2} \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}} F\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )}{d \sqrt{2 a+b \sin (2 c+2 d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])/(d*Sqrt[2*a +
b*Sin[2*c + 2*d*x]])

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 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 (c+d x) \sin (c+d x)}} \, dx &=\int \frac{1}{\sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}} \, dx\\ &=\frac{\sqrt{\frac{a+\frac{1}{2} b \sin (2 c+2 d x)}{a+\frac{b}{2}}} \int \frac{1}{\sqrt{\frac{a}{a+\frac{b}{2}}+\frac{b \sin (2 c+2 d x)}{2 \left (a+\frac{b}{2}\right )}}} \, dx}{\sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}}\\ &=\frac{\sqrt{2} F\left (c-\frac{\pi }{4}+d x|\frac{2 b}{2 a+b}\right ) \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}}{d \sqrt{2 a+b \sin (2 c+2 d x)}}\\ \end{align*}

Mathematica [A]  time = 0.140003, size = 70, normalized size = 0.92 \[ \frac{\sqrt{\frac{2 a+b \sin (2 (c+d x))}{2 a+b}} \text{EllipticF}\left (c+d x-\frac{\pi }{4},\frac{2 b}{2 a+b}\right )}{d \sqrt{a+\frac{1}{2} b \sin (2 (c+d x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*(c + d*x)])/(2*a + b)])/(d*Sqrt[a + (b*Sin[2*(
c + d*x)])/2])

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Maple [A]  time = 2.135, size = 165, normalized size = 2.2 \begin{align*} 2\,{\frac{2\,a-b}{b\cos \left ( 2\,dx+2\,c \right ) \sqrt{4\,a+2\,b\sin \left ( 2\,dx+2\,c \right ) }d}\sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( 2\,dx+2\,c \right ) -1 \right ) b}{2\,a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( 2\,dx+2\,c \right ) \right ) b}{2\,a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}},\sqrt{{\frac{2\,a-b}{2\,a+b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*a-b)*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/
(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))/b/cos(2*d*x+2*c)/(4*a
+2*b*sin(2*d*x+2*c))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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