3.2.61 \(\int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [161]

Optimal. Leaf size=30 \[ -\frac {2 a \cos ^3(c+d x)}{3 d (a+a \sin (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2752} \begin {gather*} -\frac {2 a \cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

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

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=-\frac {2 a \cos ^3(c+d x)}{3 d (a+a \sin (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 30, normalized size = 1.00 \begin {gather*} -\frac {2 a \cos ^3(c+d x)}{3 d (a (1+\sin (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

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

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Maple [A]
time = 0.34, size = 44, normalized size = 1.47

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{2}}{3 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (26) = 52\).
time = 0.33, size = 71, normalized size = 2.37 \begin {gather*} \frac {2 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 6.33, size = 40, normalized size = 1.33 \begin {gather*} \frac {4 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{3 \, \sqrt {a} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3/(sqrt(a)*d*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^2}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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