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

Optimal. Leaf size=80 \[ \frac{2 A b \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{b \cos (c+d x)}}+\frac{2 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}} \]

[Out]

(2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]) + (2*A*b*Sqrt[Cos[c + d*x]]*Ellipt
icF[(c + d*x)/2, 2])/(d*Sqrt[b*Cos[c + d*x]])

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Rubi [A]  time = 0.0867523, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {16, 2748, 2642, 2641, 2640, 2639} \[ \frac{2 A b \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{b \cos (c+d x)}}+\frac{2 B E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]) + (2*A*b*Sqrt[Cos[c + d*x]]*Ellipt
icF[(c + d*x)/2, 2])/(d*Sqrt[b*Cos[c + d*x]])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, 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]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, 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{b \cos (c+d x)} (A+B \cos (c+d x)) \sec (c+d x) \, dx &=b \int \frac{A+B \cos (c+d x)}{\sqrt{b \cos (c+d x)}} \, dx\\ &=(A b) \int \frac{1}{\sqrt{b \cos (c+d x)}} \, dx+B \int \sqrt{b \cos (c+d x)} \, dx\\ &=\frac{\left (A b \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{\sqrt{b \cos (c+d x)}}+\frac{\left (B \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{\sqrt{\cos (c+d x)}}\\ &=\frac{2 B \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{\cos (c+d x)}}+\frac{2 A b \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0759697, size = 55, normalized size = 0.69 \[ \frac{2 b \sqrt{\cos (c+d x)} \left (A F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 2.906, size = 161, normalized size = 2. \begin{align*} -2\,{\frac{\sqrt{b \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}}b\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 ( A{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -B{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) }{\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/
2*c)^2+1)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-b*(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)/(b*(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{b \cos \left (d x + c\right )} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sec(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{b \cos \left (d x + c\right )} \sec \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sec(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((b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))*sec(d*x+c),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{b \cos \left (d x + c\right )} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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