3.490 \(\int \sqrt{a+b \cos (c+d x)} \sec (c+d x) \, dx\)

Optimal. Leaf size=118 \[ \frac{2 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}} \]

[Out]

(2*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) +
(2*a*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]
)

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Rubi [A]  time = 0.226215, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2803, 2663, 2661, 2807, 2805} \[ \frac{2 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x],x]

[Out]

(2*b*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) +
(2*a*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]
)

Rule 2803

Int[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[d/b
, Int[1/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[(b*c - a*d)/b, Int[1/((a + b*Sin[e + f*x])*Sqrt[c + d*Sin[e +
f*x]]), x], 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]

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]

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]

Rubi steps

\begin{align*} \int \sqrt{a+b \cos (c+d x)} \sec (c+d x) \, dx &=a \int \frac{\sec (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx+b \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{\left (a \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{\sqrt{a+b \cos (c+d x)}}+\frac{\left (b \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{\sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 b \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.18378, size = 81, normalized size = 0.69 \[ \frac{2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+a \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )}{d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x],x]

[Out]

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + a*EllipticPi[2, (c + d*x)/2,
(2*b)/(a + b)]))/(d*Sqrt[a + b*Cos[c + d*x]])

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Maple [A]  time = 2.727, size = 194, normalized size = 1.6 \begin{align*} -2\,{\frac{\sqrt{ \left ( 2\,b \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+a-b \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{\sqrt{-2\,b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b}d}\sqrt{{\frac{2\,b \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+a-b}{a-b}}} \left ({\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ) b-{\it EllipticPi} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,2,\sqrt{-2\,{\frac{b}{a-b}}} \right ) a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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