3.170 \(\int \frac{\sqrt{a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=76 \[ \frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 \sqrt{a} B \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d} \]

[Out]

(2*Sqrt[a]*B*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a*A*Sin[c + d*x])/(d*Sqrt[Cos[c +
d*x]]*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.164648, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {2980, 2774, 216} \[ \frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}+\frac{2 \sqrt{a} B \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(2*Sqrt[a]*B*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a*A*Sin[c + d*x])/(d*Sqrt[Cos[c +
d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}+B \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}-\frac{(2 B) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}\\ &=\frac{2 \sqrt{a} B \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}+\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.162091, size = 86, normalized size = 1.13 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} \left (2 A \sin \left (\frac{1}{2} (c+d x)\right )+\sqrt{2} B \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{\cos (c+d x)}\right )}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(Sqrt[2]*B*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] +
2*A*Sin[(c + d*x)/2]))/(d*Sqrt[Cos[c + d*x]])

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Maple [A]  time = 0.657, size = 109, normalized size = 1.4 \begin{align*} -2\,{\frac{\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) }}{d\sin \left ( dx+c \right ) \sqrt{\cos \left ( dx+c \right ) }} \left ( -B\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \sin \left ( dx+c \right ) +A\cos \left ( dx+c \right ) -A \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x)

[Out]

-2/d*(a*(1+cos(d*x+c)))^(1/2)*(-B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)/cos(d*x+c))*sin(d*x+c)+A*cos(d*x+c)-A)/sin(d*x+c)/cos(d*x+c)^(1/2)

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Maxima [B]  time = 2.30896, size = 331, normalized size = 4.36 \begin{align*} \frac{B \sqrt{a} \arctan \left ({\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac{1}{4}} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \sin \left (d x + c\right ),{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac{1}{4}} \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \cos \left (d x + c\right )\right ) + \frac{2 \, A{\left (\frac{\sqrt{2} \sqrt{a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sqrt{2} \sqrt{a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{3}{2}}{\left (-\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{3}{2}}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

(B*sqrt(a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c)) + 2*A*(sqrt(2)*sqrt
(a)*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/((sin(d*x + c)/(cos
(d*x + c) + 1) + 1)^(3/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)))/d

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Fricas [A]  time = 1.62316, size = 298, normalized size = 3.92 \begin{align*} \frac{2 \,{\left (\sqrt{a \cos \left (d x + c\right ) + a} A \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) -{\left (B \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt{a} \arctan \left (\frac{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right )\right )}}{d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

2*(sqrt(a*cos(d*x + c) + a)*A*sqrt(cos(d*x + c))*sin(d*x + c) - (B*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(a)*ar
ctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(d*cos(d*x + c)^2 + d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \left (\cos{\left (c + d x \right )} + 1\right )} \left (A + B \cos{\left (c + d x \right )}\right )}{\cos ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(3/2),x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + B*cos(c + d*x))/cos(c + d*x)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{a \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)/cos(d*x + c)^(3/2), x)