3.162 \(\int \frac {\cos (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=22 \[ \frac {2 \sqrt {a \sin (c+d x)+a}}{a d} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2667, 32} \[ \frac {2 \sqrt {a \sin (c+d x)+a}}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + a*Sin[c + d*x]])/(a*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 2667

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 22, normalized size = 1.00 \[ \frac {2 \sqrt {a \sin (c+d x)+a}}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + a*Sin[c + d*x]])/(a*d)

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fricas [A]  time = 0.67, size = 20, normalized size = 0.91 \[ \frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(a*sin(d*x + c) + a)/(a*d)

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giac [A]  time = 0.84, size = 20, normalized size = 0.91 \[ \frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(a*sin(d*x + c) + a)/(a*d)

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maple [A]  time = 0.03, size = 21, normalized size = 0.95 \[ \frac {2 \sqrt {a +a \sin \left (d x +c \right )}}{d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 20, normalized size = 0.91 \[ \frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(a*sin(d*x + c) + a)/(a*d)

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mupad [B]  time = 4.82, size = 20, normalized size = 0.91 \[ \frac {2\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.21, size = 32, normalized size = 1.45 \[ \begin {cases} \frac {2 \sqrt {a \sin {\left (c + d x \right )} + a}}{a d} & \text {for}\: d \neq 0 \\\frac {x \cos {\relax (c )}}{\sqrt {a \sin {\relax (c )} + a}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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