3.320 \(\int \frac{1}{(a+a \cos (c+d x)) \sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=110 \[ \frac{\sin (c+d x) \sqrt{\sec (c+d x)}}{d (a \sec (c+d x)+a)}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

[Out]

-((Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d)) + (Sqrt[Cos[c + d*x]]*EllipticF[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + (Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

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Rubi [A]  time = 0.138834, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3238, 3820, 3787, 3771, 2639, 2641} \[ \frac{\sin (c+d x) \sqrt{\sec (c+d x)}}{d (a \sec (c+d x)+a)}+\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d}-\frac{\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]]),x]

[Out]

-((Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d)) + (Sqrt[Cos[c + d*x]]*EllipticF[(c
+ d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a*d) + (Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

Rule 3238

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

Rule 3820

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[(b*d*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 1))/(a*f*(a + b*Csc[e + f*x])), x] + Dist[(d*(n - 1))/(a*b), Int[(d*Csc[e + f*x]
)^(n - 1)*(a - b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

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

Rubi steps

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

Mathematica [C]  time = 0.952056, size = 181, normalized size = 1.65 \[ -\frac{4 i \left (\left (1+e^{i (c+d x)}\right ) \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 i (c+d x)}\right )+e^{i (c+d x)} \left (1+e^{i (c+d x)}\right ) \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-e^{2 i (c+d x)}\right )-e^{2 i (c+d x)}-1\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec (c+d x)}}{a d \left (1+e^{i (c+d x)}\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]]),x]

[Out]

((-4*I)*Cos[(c + d*x)/2]^2*(-1 - E^((2*I)*(c + d*x)) + (1 + E^(I*(c + d*x)))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hyp
ergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + E^(I*(c + d*x))*(1 + E^(I*(c + d*x)))*Sqrt[1 + E^((2*I)
*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sqrt[Sec[c + d*x]])/(a*d*(1 + E^(I*(c + d
*x)))^3)

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Maple [A]  time = 2.007, size = 198, normalized size = 1.8 \begin{align*} -{\frac{1}{da}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ({\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +{\it EllipticE} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) \right ) +2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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