3.161 \(\int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\)

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

[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 = 0.05, 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 = {2673} \[ -\frac {2 a \cos ^3(c+d x)}{3 d (a \sin (c+d x)+a)^{3/2}} \]

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 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[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 = 0.07, size = 30, normalized size = 1.00 \[ -\frac {2 a \cos ^3(c+d x)}{3 d (a (\sin (c+d x)+1))^{3/2}} \]

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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fricas [B]  time = 0.78, size = 71, normalized size = 2.37 \[ \frac {2 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

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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giac [B]  time = 1.10, size = 143, normalized size = 4.77 \[ \frac {2 \, {\left (\frac {2 \, \sqrt {2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a}} + \frac {{\left ({\left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {3 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3 \, a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {a}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}}\right )}}{3 \, d} \]

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]

2/3*(2*sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/sqrt(a) + (((a*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*d*x + 1/2*c) + 1)
- 3*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 3*a/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x +
1/2*c) - a/sgn(tan(1/2*d*x + 1/2*c) + 1))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2))/d

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maple [A]  time = 0.19, size = 44, normalized size = 1.47 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{2}}{3 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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)

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

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

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

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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