\(\int \frac {\cos ^2(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\) [338]

Optimal result

Integrand size = 29, antiderivative size = 60 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 a \cos ^3(c+d x)}{15 d (a+a \sin (c+d x))^{3/2}}-\frac {2 \cos ^3(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}} \]

[Out]

2/15*a*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-2/5*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2935, 2752} \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\frac {2 a \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}} \]

[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 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]

Rule 2935

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] /; F
reeQ[{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]

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (2+3 \sin (c+d x))}{15 d \sqrt {a (1+\sin (c+d x))}} \]

[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])])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) \left (\sin \left (d x +c \right )-1\right )^{2} \left (3 \sin \left (d x +c \right )+2\right )}{15 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(54\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, {\left (a d \cos \left (d x + c\right ) + a d \sin \left (d x + c\right ) + a d\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]

[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)

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^2(c+d x) \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {4 \, \sqrt {2} {\left (6 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )}}{15 \, a d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]

[In]

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

[Out]

-4/15*sqrt(2)*(6*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 5*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3)/(a*d*s
gn(cos(-1/4*pi + 1/2*d*x + 1/2*c)))

Mupad [F(-1)]

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

[In]

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

[Out]

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