\(\int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx\) [98]

Optimal result

Integrand size = 21, antiderivative size = 56 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 a \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

2/3*a*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2830, 2725} \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}+\frac {2 a \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

(2*a*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 2725

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x
]])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*S
in[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m
, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {1}{3} \int \sqrt {a+a \cos (c+d x)} \, dx \\ & = \frac {2 a \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (3 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 d} \]

[In]

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

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/(3*d)

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \sqrt {2}}{3 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(58\)

[In]

int(cos(d*x+c)*(a+cos(d*x+c)*a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(1+2*cos(1/2*d*x+1/2*c)^2)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/
d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

2/3*sqrt(a*cos(d*x + c) + a)*(cos(d*x + c) + 2)*sin(d*x + c)/(d*cos(d*x + c) + d)

Sympy [F]

\[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \cos {\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*cos(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.64 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {{\left (\sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{3 \, d} \]

[In]

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

[Out]

1/3*(sqrt(2)*sin(3/2*d*x + 3/2*c) + 3*sqrt(2)*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

Giac [A] (verification not implemented)

none

Time = 0.55 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\frac {\sqrt {2} {\left (\mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{3 \, d} \]

[In]

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

[Out]

1/3*sqrt(2)*(sgn(cos(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c) + 3*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)
)*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) \sqrt {a+a \cos (c+d x)} \, dx=\int \cos \left (c+d\,x\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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