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

Optimal. Leaf size=35 \[ \frac{2 a \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{d}+\frac{2 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

[Out]

(2*a*EllipticE[(c + d*x)/2, 2])/d + (2*a*EllipticF[(c + d*x)/2, 2])/d

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Rubi [A]  time = 0.0587835, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4225, 2748, 2641, 2639} \[ \frac{2 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*EllipticE[(c + d*x)/2, 2])/d + (2*a*EllipticF[(c + d*x)/2, 2])/d

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, 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]

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]

Rubi steps

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

Mathematica [C]  time = 1.80084, size = 155, normalized size = 4.43 \[ \frac{a \sqrt{\cos (c+d x)} (\cos (c+d x)+1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\tan \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )}}-2 \sin (c) \sqrt{\csc ^2(c)} \sqrt{\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+\tan \left (\tan ^{-1}(\tan (c))+d x\right )\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-2*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^
2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + Tan[d*
x + ArcTan[Tan[c]]] - (HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Tan[d*x + ArcTan[Ta
n[c]]])/Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(2*d)

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Maple [A]  time = 1.16, size = 150, normalized size = 4.3 \begin{align*} -2\,{\frac{\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}a\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ({\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) }{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*
c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-Elli
pticE(cos(1/2*d*x+1/2*c),2^(1/2)))/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{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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