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

Optimal. Leaf size=60 \[ \frac{2 a \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^{3/2}}-\frac{2 \cos ^3(c+d x)}{5 d \sqrt{a \sin (c+d x)+a}} \]

[Out]

(2*a*Cos[c + d*x]^3)/(15*d*(a + a*Sin[c + d*x])^(3/2)) - (2*Cos[c + d*x]^3)/(5*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.126818, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2856, 2673} \[ \frac{2 a \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^{3/2}}-\frac{2 \cos ^3(c+d x)}{5 d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Cos[c + d*x]^3)/(15*d*(a + a*Sin[c + d*x])^(3/2)) - (2*Cos[c + d*x]^3)/(5*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2856

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1
, 0]

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) \sin (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=-\frac{2 \cos ^3(c+d x)}{5 d \sqrt{a+a \sin (c+d x)}}-\frac{1}{5} \int \frac{\cos ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{2 a \cos ^3(c+d x)}{15 d (a+a \sin (c+d x))^{3/2}}-\frac{2 \cos ^3(c+d x)}{5 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.419678, size = 77, normalized size = 1.28 \[ -\frac{2 (3 \sin (c+d x)+2) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{15 d \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(2 + 3*Sin[c + d*x]))/(15*d*
Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.648, size = 54, normalized size = 0.9 \begin{align*} -{\frac{ \left ( 2+2\,\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2} \left ( 3\,\sin \left ( dx+c \right ) +2 \right ) }{15\,d\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(1+sin(d*x+c))*(sin(d*x+c)-1)^2*(3*sin(d*x+c)+2)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{2} \sin \left (d x + c\right )}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58472, size = 251, normalized size = 4.18 \begin{align*} -\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} -{\left (3 \, \cos \left (d x + c\right )^{2} - \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}}{15 \,{\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(3*cos(d*x + c)^3 + 4*cos(d*x + c)^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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{\sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.74696, size = 198, normalized size = 3.3 \begin{align*} -\frac{\frac{{\left ({\left (\frac{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{7}} - \frac{5 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{7}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{5 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{7}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{7}}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{5}{2}}} - \frac{\sqrt{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}{a^{\frac{19}{2}}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((((sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)^2/a^7 - 5*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^7)*tan(
1/2*d*x + 1/2*c) + 5*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^7)*tan(1/2*d*x + 1/2*c)^2 - sgn(tan(1/2*d*x + 1/2*c) + 1)
/a^7)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2) - sqrt(2)*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(19/2))/d