\(\int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) [508]

Optimal result

Integrand size = 21, antiderivative size = 22 \[ \int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \sqrt {a+b \sin (c+d x)}}{b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, 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 = {2747, 32} \[ \int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \sqrt {a+b \sin (c+d x)}}{b d} \]

[In]

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

[Out]

(2*Sqrt[a + b*Sin[c + d*x]])/(b*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \sqrt {a+b \sin (c+d x)}}{b d} \]

[In]

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

[Out]

(2*Sqrt[a + b*Sin[c + d*x]])/(b*d)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {2 \sqrt {a +b \sin \left (d x +c \right )}}{b d}\)	\(21\)
default	\(\frac {2 \sqrt {a +b \sin \left (d x +c \right )}}{b d}\)	\(21\)
risch	\(-\frac {i \sqrt {2}\, \left (2 i a +2 i b \sin \left (d x +c \right )\right )}{\sqrt {2 b \sin \left (d x +c \right )+2 a}\, d b}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \, \sqrt {b \sin \left (d x + c\right ) + a}}{b d} \]

[In]

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

[Out]

2*sqrt(b*sin(d*x + c) + a)/(b*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (17) = 34\).

Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45 \[ \int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\begin {cases} \frac {x \cos {\left (c \right )}}{\sqrt {a}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin {\left (c + d x \right )}}{\sqrt {a} d} & \text {for}\: b = 0 \\\frac {x \cos {\left (c \right )}}{\sqrt {a + b \sin {\left (c \right )}}} & \text {for}\: d = 0 \\\frac {2 \sqrt {a + b \sin {\left (c + d x \right )}}}{b d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*cos(c)/sqrt(a), Eq(b, 0) & Eq(d, 0)), (sin(c + d*x)/(sqrt(a)*d), Eq(b, 0)), (x*cos(c)/sqrt(a + b*
sin(c)), Eq(d, 0)), (2*sqrt(a + b*sin(c + d*x))/(b*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \, \sqrt {b \sin \left (d x + c\right ) + a}}{b d} \]

[In]

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

[Out]

2*sqrt(b*sin(d*x + c) + a)/(b*d)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \, \sqrt {b \sin \left (d x + c\right ) + a}}{b d} \]

[In]

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

[Out]

2*sqrt(b*sin(d*x + c) + a)/(b*d)

Mupad [B] (verification not implemented)

Time = 5.56 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2\,\sqrt {a+b\,\sin \left (c+d\,x\right )}}{b\,d} \]

[In]

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

[Out]

(2*(a + b*sin(c + d*x))^(1/2))/(b*d)