\(\int \cos ^2(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [323]

Optimal result

Integrand size = 31, antiderivative size = 124 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {8 a^2 \cos ^3(c+d x)}{63 d (a+a \sin (c+d x))^{3/2}}-\frac {2 a \cos ^3(c+d x)}{21 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{21 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{9 a d} \] Output:

-8/63*a^2*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-2/21*a*cos(d*x+c)^3/d/(a+a 
*sin(d*x+c))^(1/2)+4/21*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d-2/9*cos(d*x+ 
c)^3*(a+a*sin(d*x+c))^(3/2)/a/d
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.80 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \sqrt {a (1+\sin (c+d x))} (-62+30 \cos (2 (c+d x))-69 \sin (c+d x)+7 \sin (3 (c+d x)))}{126 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

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

Output:

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x])]*(-62 + 
 30*Cos[2*(c + d*x)] - 69*Sin[c + d*x] + 7*Sin[3*(c + d*x)]))/(126*d*(Cos[ 
(c + d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3357, 27, 3042, 3335, 3042, 3153, 3042, 3152}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3357

\(\displaystyle \frac {2 \int \frac {3}{2} \cos ^2(c+d x) (a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{9 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos ^2(c+d x) (a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{3 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \cos (c+d x)^2 (a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{3 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}\)

\(\Big \downarrow \) 3335

\(\displaystyle \frac {\frac {5}{7} a \int \cos ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{3 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{7} a \int \cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{3 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {\frac {5}{7} a \left (\frac {4}{5} a \int \frac {\cos ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {4 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{3 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5}{7} a \left (\frac {4}{5} a \int \frac {\cos (c+d x)^2}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {4 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{3 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {\frac {5}{7} a \left (-\frac {8 a^2 \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {4 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}}{3 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{9 a d}\)

Input:

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

Output:

(-2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(9*a*d) + ((4*a*Cos[c + d*x 
]^3*Sqrt[a + a*Sin[c + d*x]])/(7*d) + (5*a*((-8*a^2*Cos[c + d*x]^3)/(15*d* 
(a + a*Sin[c + d*x])^(3/2)) - (2*a*Cos[c + d*x]^3)/(5*d*Sqrt[a + a*Sin[c + 
 d*x]])))/7)/(3*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3152

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] && EqQ[a^2 
 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
 

rule 3153

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 + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && 
NeQ[m + p, 0]
 

rule 3335

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] + S 
imp[(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] /; FreeQ[{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 3357

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

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.05 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (7 \, \cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{4} - 11 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - {\left (7 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} - 4 \, \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right ) - 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{63 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

2/63*(7*cos(d*x + c)^5 - cos(d*x + c)^4 - 11*cos(d*x + c)^3 + cos(d*x + c) 
^2 - (7*cos(d*x + c)^4 + 8*cos(d*x + c)^3 - 3*cos(d*x + c)^2 - 4*cos(d*x + 
 c) - 8)*sin(d*x + c) - 4*cos(d*x + c) - 8)*sqrt(a*sin(d*x + c) + a)/(d*co 
s(d*x + c) + d*sin(d*x + c) + d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.03 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {8 \, \sqrt {2} {\left (28 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 72 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )} \sqrt {a}}{63 \, d} \] Input:

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

Output:

-8/63*sqrt(2)*(28*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d* 
x + 1/2*c)^9 - 72*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d* 
x + 1/2*c)^7 + 63*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d* 
x + 1/2*c)^5 - 21*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d* 
x + 1/2*c)^3)*sqrt(a)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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