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

Optimal. Leaf size=24 \[ \frac {2 (a+b \sin (c+d x))^{3/2}}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 24, 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 = {2668, 32} \[ \frac {2 (a+b \sin (c+d x))^{3/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*Sin[c + d*x])^(3/2))/(3*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 2668

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

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.00 \[ \frac {2 (a+b \sin (c+d x))^{3/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*Sin[c + d*x])^(3/2))/(3*b*d)

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fricas [A]  time = 1.02, size = 20, normalized size = 0.83 \[ \frac {2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*sin(d*x + c) + a)^(3/2)/(b*d)

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giac [A]  time = 1.45, size = 20, normalized size = 0.83 \[ \frac {2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*sin(d*x + c) + a)^(3/2)/(b*d)

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maple [A]  time = 0.04, size = 21, normalized size = 0.88 \[ \frac {2 \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{3 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 20, normalized size = 0.83 \[ \frac {2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{3 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*sin(d*x + c) + a)^(3/2)/(b*d)

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mupad [B]  time = 5.20, size = 20, normalized size = 0.83 \[ \frac {2\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}}{3\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.55, size = 83, normalized size = 3.46 \[ \begin {cases} \sqrt {a} x \cos {\relax (c )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sqrt {a} \sin {\left (c + d x \right )}}{d} & \text {for}\: b = 0 \\x \sqrt {a + b \sin {\relax (c )}} \cos {\relax (c )} & \text {for}\: d = 0 \\\frac {2 a \sqrt {a + b \sin {\left (c + d x \right )}}}{3 b d} + \frac {2 \sqrt {a + b \sin {\left (c + d x \right )}} \sin {\left (c + d x \right )}}{3 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sqrt(a)*x*cos(c), Eq(b, 0) & Eq(d, 0)), (sqrt(a)*sin(c + d*x)/d, Eq(b, 0)), (x*sqrt(a + b*sin(c))*c
os(c), Eq(d, 0)), (2*a*sqrt(a + b*sin(c + d*x))/(3*b*d) + 2*sqrt(a + b*sin(c + d*x))*sin(c + d*x)/(3*d), True)
)

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