\(\int \sqrt {a+a \cos (c+d x)} \, dx\) [4]

Optimal result

Integrand size = 14, antiderivative size = 26 \[ \int \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 a \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2725} \[ \int \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}} \]

[In]

Int[Sqrt[a + a*Cos[c + d*x]],x]

[Out]

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

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 \sqrt {a (1+\cos (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{d} \]

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(2*Sqrt[a*(1 + Cos[c + d*x])]*Tan[(c + d*x)/2])/d

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65

method	result	size
default	\(\frac {2 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}}{\sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(43\)
risch	\(-\frac {i \sqrt {2}\, \sqrt {a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{2} {\mathrm e}^{-i \left (d x +c \right )}}\, \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) d}\)	\(60\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(a*cos(c + d*x) + a), x)

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 \, \sqrt {2} \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \sqrt {a+a \cos (c+d x)} \, dx=\frac {2 \, \sqrt {2} \sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \sqrt {a+a \cos (c+d x)} \, dx=\frac {2\,\sin \left (c+d\,x\right )\,\sqrt {a\,\left (\cos \left (c+d\,x\right )+1\right )}}{d\,\left (\cos \left (c+d\,x\right )+1\right )} \]

[In]

int((a + a*cos(c + d*x))^(1/2),x)

[Out]

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