3.563 \(\int \sqrt{\cos (c+d x)} (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=61 \[ \frac{2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 B \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

[Out]

(2*A*EllipticE[(c + d*x)/2, 2])/d + (2*B*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*B*Sqrt[Cos[c + d*x]]*Sin[c + d*
x])/(3*d)

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Rubi [A]  time = 0.0505281, antiderivative size = 61, 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 = {2748, 2639, 2635, 2641} \[ \frac{2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 B \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]

[Out]

(2*A*EllipticE[(c + d*x)/2, 2])/d + (2*B*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*B*Sqrt[Cos[c + d*x]]*Sin[c + d*
x])/(3*d)

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 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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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 \sqrt{\cos (c+d x)} (A+B \cos (c+d x)) \, dx &=A \int \sqrt{\cos (c+d x)} \, dx+B \int \cos ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{2 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 B \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{3} 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 )}{3 d}+\frac{2 B \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.102872, size = 53, normalized size = 0.87 \[ \frac{2 \left (3 A E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+B \left (F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt{\cos (c+d x)}\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]

[Out]

(2*(3*A*EllipticE[(c + d*x)/2, 2] + B*(EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*Sin[c + d*x])))/(3*d)

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Maple [B]  time = 2.297, size = 229, normalized size = 3.8 \begin{align*}{\frac{2}{3\,d}\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 ( -4\,B\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+3\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -B\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) +2\,B \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \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 ( \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((A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x)

[Out]

2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*A*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*B*sin(1/2*d*x+1/
2*c)^2*cos(1/2*d*x+1/2*c))/(-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 \cos \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((A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*cos(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 \cos \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((A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c) + A)*sqrt(cos(d*x + c)), x)

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Sympy [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((A+B*cos(d*x+c))*cos(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \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((A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(cos(d*x + c)), x)