3.272 \(\int \frac{\sqrt{g \sec (e+f x)} \sqrt{c+d \sec (e+f x)}}{a+b \cos (e+f x)} \, dx\)

Optimal. Leaf size=168 \[ \frac{2 (a c-b d) \sqrt{g \sec (e+f x)} \sqrt{\frac{c \cos (e+f x)+d}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} (e+f x)|\frac{2 c}{c+d}\right )}{a f (a+b) \sqrt{c+d \sec (e+f x)}}+\frac{2 d \sqrt{g \sec (e+f x)} \sqrt{\frac{c \cos (e+f x)+d}{c+d}} \Pi \left (2;\frac{1}{2} (e+f x)|\frac{2 c}{c+d}\right )}{a f \sqrt{c+d \sec (e+f x)}} \]

[Out]

(2*d*Sqrt[(d + c*Cos[e + f*x])/(c + d)]*EllipticPi[2, (e + f*x)/2, (2*c)/(c + d)]*Sqrt[g*Sec[e + f*x]])/(a*f*S
qrt[c + d*Sec[e + f*x]]) + (2*(a*c - b*d)*Sqrt[(d + c*Cos[e + f*x])/(c + d)]*EllipticPi[(2*b)/(a + b), (e + f*
x)/2, (2*c)/(c + d)]*Sqrt[g*Sec[e + f*x]])/(a*(a + b)*f*Sqrt[c + d*Sec[e + f*x]])

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Rubi [A]  time = 1.06961, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2962, 3971, 3859, 2807, 2805, 3975} \[ \frac{2 (a c-b d) \sqrt{g \sec (e+f x)} \sqrt{\frac{c \cos (e+f x)+d}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} (e+f x)|\frac{2 c}{c+d}\right )}{a f (a+b) \sqrt{c+d \sec (e+f x)}}+\frac{2 d \sqrt{g \sec (e+f x)} \sqrt{\frac{c \cos (e+f x)+d}{c+d}} \Pi \left (2;\frac{1}{2} (e+f x)|\frac{2 c}{c+d}\right )}{a f \sqrt{c+d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[g*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]])/(a + b*Cos[e + f*x]),x]

[Out]

(2*d*Sqrt[(d + c*Cos[e + f*x])/(c + d)]*EllipticPi[2, (e + f*x)/2, (2*c)/(c + d)]*Sqrt[g*Sec[e + f*x]])/(a*f*S
qrt[c + d*Sec[e + f*x]]) + (2*(a*c - b*d)*Sqrt[(d + c*Cos[e + f*x])/(c + d)]*EllipticPi[(2*b)/(a + b), (e + f*
x)/2, (2*c)/(c + d)]*Sqrt[g*Sec[e + f*x]])/(a*(a + b)*f*Sqrt[c + d*Sec[e + f*x]])

Rule 2962

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.)
+ (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[g^m, Int[(g*Csc[e + f*x])^(p - m)*(b + a*Csc[e + f*x])^m*(c + d*Csc[e
+ f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[m]

Rule 3971

Int[((csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(csc[(e_.) + (f_.)*(x_)
]*(d_.) + (c_)), x_Symbol] :> Dist[b/d, Int[(g*Csc[e + f*x])^(3/2)/Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*
c - a*d)/d, Int[(g*Csc[e + f*x])^(3/2)/(Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])), x], x] /; FreeQ[{a, b,
c, d, e, f, g}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2807

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

Rule 2805

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

Rule 3975

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))), x_Symbol] :> Dist[(g*Sqrt[g*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]],
Int[1/(Sqrt[b + a*Sin[e + f*x]]*(d + c*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{g \sec (e+f x)} \sqrt{c+d \sec (e+f x)}}{a+b \cos (e+f x)} \, dx &=\frac{\int \frac{(g \sec (e+f x))^{3/2} \sqrt{c+d \sec (e+f x)}}{b+a \sec (e+f x)} \, dx}{g}\\ &=\frac{d \int \frac{(g \sec (e+f x))^{3/2}}{\sqrt{c+d \sec (e+f x)}} \, dx}{a g}+\frac{(a c-b d) \int \frac{(g \sec (e+f x))^{3/2}}{(b+a \sec (e+f x)) \sqrt{c+d \sec (e+f x)}} \, dx}{a g}\\ &=\frac{\left (d \sqrt{d+c \cos (e+f x)} \sqrt{g \sec (e+f x)}\right ) \int \frac{\sec (e+f x)}{\sqrt{d+c \cos (e+f x)}} \, dx}{a \sqrt{c+d \sec (e+f x)}}+\frac{\left ((a c-b d) \sqrt{d+c \cos (e+f x)} \sqrt{g \sec (e+f x)}\right ) \int \frac{1}{(a+b \cos (e+f x)) \sqrt{d+c \cos (e+f x)}} \, dx}{a \sqrt{c+d \sec (e+f x)}}\\ &=\frac{\left (d \sqrt{\frac{d+c \cos (e+f x)}{c+d}} \sqrt{g \sec (e+f x)}\right ) \int \frac{\sec (e+f x)}{\sqrt{\frac{d}{c+d}+\frac{c \cos (e+f x)}{c+d}}} \, dx}{a \sqrt{c+d \sec (e+f x)}}+\frac{\left ((a c-b d) \sqrt{\frac{d+c \cos (e+f x)}{c+d}} \sqrt{g \sec (e+f x)}\right ) \int \frac{1}{(a+b \cos (e+f x)) \sqrt{\frac{d}{c+d}+\frac{c \cos (e+f x)}{c+d}}} \, dx}{a \sqrt{c+d \sec (e+f x)}}\\ &=\frac{2 d \sqrt{\frac{d+c \cos (e+f x)}{c+d}} \Pi \left (2;\frac{1}{2} (e+f x)|\frac{2 c}{c+d}\right ) \sqrt{g \sec (e+f x)}}{a f \sqrt{c+d \sec (e+f x)}}+\frac{2 (a c-b d) \sqrt{\frac{d+c \cos (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} (e+f x)|\frac{2 c}{c+d}\right ) \sqrt{g \sec (e+f x)}}{a (a+b) f \sqrt{c+d \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 3.94151, size = 222, normalized size = 1.32 \[ -\frac{2 i \cot (e+f x) \sqrt{g \sec (e+f x)} \sqrt{-\frac{c (\cos (e+f x)-1)}{c+d}} \sqrt{\frac{c (\cos (e+f x)+1)}{c-d}} \sqrt{c+d \sec (e+f x)} \left (\Pi \left (1-\frac{c}{d};i \sinh ^{-1}\left (\sqrt{\frac{1}{c-d}} \sqrt{d+c \cos (e+f x)}\right )|\frac{d-c}{c+d}\right )-\Pi \left (\frac{b (d-c)}{b d-a c};i \sinh ^{-1}\left (\sqrt{\frac{1}{c-d}} \sqrt{d+c \cos (e+f x)}\right )|\frac{d-c}{c+d}\right )\right )}{a f \sqrt{\frac{1}{c-d}} \sqrt{c \cos (e+f x)+d}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[g*Sec[e + f*x]]*Sqrt[c + d*Sec[e + f*x]])/(a + b*Cos[e + f*x]),x]

[Out]

((-2*I)*Sqrt[-((c*(-1 + Cos[e + f*x]))/(c + d))]*Sqrt[(c*(1 + Cos[e + f*x]))/(c - d)]*Cot[e + f*x]*(EllipticPi
[1 - c/d, I*ArcSinh[Sqrt[(c - d)^(-1)]*Sqrt[d + c*Cos[e + f*x]]], (-c + d)/(c + d)] - EllipticPi[(b*(-c + d))/
(-(a*c) + b*d), I*ArcSinh[Sqrt[(c - d)^(-1)]*Sqrt[d + c*Cos[e + f*x]]], (-c + d)/(c + d)])*Sqrt[g*Sec[e + f*x]
]*Sqrt[c + d*Sec[e + f*x]])/(a*Sqrt[(c - d)^(-1)]*f*Sqrt[d + c*Cos[e + f*x]])

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Maple [C]  time = 0.397, size = 479, normalized size = 2.9 \begin{align*}{\frac{-2\,i\cos \left ( fx+e \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{af \left ( a+b \right ) \left ( a-b \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) \left ( d+c\cos \left ( fx+e \right ) \right ) } \left ({\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},\sqrt{-{\frac{c-d}{c+d}}} \right ){a}^{2}c-{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},\sqrt{-{\frac{c-d}{c+d}}} \right ){a}^{2}d+{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},\sqrt{-{\frac{c-d}{c+d}}} \right ) abc-{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},\sqrt{-{\frac{c-d}{c+d}}} \right ) abd+2\,{\it EllipticPi} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},-1,i\sqrt{{\frac{c-d}{c+d}}} \right ){a}^{2}d-2\,{\it EllipticPi} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},-1,i\sqrt{{\frac{c-d}{c+d}}} \right ){b}^{2}d-2\,{\it EllipticPi} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},{\frac{a-b}{a+b}},i\sqrt{{\frac{c-d}{c+d}}} \right ) abc+2\,{\it EllipticPi} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},{\frac{a-b}{a+b}},i\sqrt{{\frac{c-d}{c+d}}} \right ){b}^{2}d \right ) \sqrt{{\frac{g}{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{d+c\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}\sqrt{{\frac{d+c\cos \left ( fx+e \right ) }{ \left ( c+d \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) }}}\sqrt{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*sec(f*x+e))^(1/2)*(c+d*sec(f*x+e))^(1/2)/(a+b*cos(f*x+e)),x)

[Out]

-2*I/f/a/(a+b)/(a-b)*(EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),(-(c-d)/(c+d))^(1/2))*a^2*c-EllipticF(I*(-1+cos(f
*x+e))/sin(f*x+e),(-(c-d)/(c+d))^(1/2))*a^2*d+EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),(-(c-d)/(c+d))^(1/2))*a*b
*c-EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),(-(c-d)/(c+d))^(1/2))*a*b*d+2*EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e
),-1,I*((c-d)/(c+d))^(1/2))*a^2*d-2*EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e),-1,I*((c-d)/(c+d))^(1/2))*b^2*d-2*
EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e),(a-b)/(a+b),I*((c-d)/(c+d))^(1/2))*a*b*c+2*EllipticPi(I*(-1+cos(f*x+e)
)/sin(f*x+e),(a-b)/(a+b),I*((c-d)/(c+d))^(1/2))*b^2*d)*cos(f*x+e)*sin(f*x+e)^2*(g/cos(f*x+e))^(1/2)*((d+c*cos(
f*x+e))/cos(f*x+e))^(1/2)*(1/(c+d)*(d+c*cos(f*x+e))/(1+cos(f*x+e)))^(1/2)*(1/(1+cos(f*x+e)))^(1/2)/(-1+cos(f*x
+e))/(d+c*cos(f*x+e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*sec(f*x + e) + c)*sqrt(g*sec(f*x + e))/(b*cos(f*x + e) + a), x)

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Fricas [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((g*sec(f*x+e))^(1/2)*(c+d*sec(f*x+e))^(1/2)/(a+b*cos(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))**(1/2)*(c+d*sec(f*x+e))**(1/2)/(a+b*cos(f*x+e)),x)

[Out]

Integral(sqrt(g*sec(e + f*x))*sqrt(c + d*sec(e + f*x))/(a + b*cos(e + f*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*sec(f*x + e) + c)*sqrt(g*sec(f*x + e))/(b*cos(f*x + e) + a), x)