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

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

[Out]

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

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Rubi [A]  time = 0.0573418, 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} \[ \frac{2 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0674208, size = 32, normalized size = 0.91 \[ \frac{2 \left (b \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 1.513, size = 152, 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}}\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 ( b{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -a{\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+b*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)
^2+1)^(1/2)*(b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-a*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+
1/2*c)^4+sin(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 (b \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+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((b*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 (b \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+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((b*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*} \int \left (a + b \sec{\left (c + d x \right )}\right ) \sqrt{\cos{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \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+b*sec(d*x+c)),x, algorithm="giac")

[Out]

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