\(\int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [330]

Optimal result

Integrand size = 31, antiderivative size = 156 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {64 a^3 \cos ^3(c+d x)}{385 d (a+a \sin (c+d x))^{3/2}}-\frac {48 a^2 \cos ^3(c+d x)}{385 d \sqrt {a+a \sin (c+d x)}}-\frac {6 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{77 d}+\frac {4 \cos ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{33 d}-\frac {2 \cos ^3(c+d x) (a+a \sin (c+d x))^{5/2}}{11 a d} \] Output:

-64/385*a^3*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-48/385*a^2*cos(d*x+c)^3/ 
d/(a+a*sin(d*x+c))^(1/2)-6/77*a*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d+4/33 
*cos(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d-2/11*cos(d*x+c)^3*(a+a*sin(d*x+c))^ 
(5/2)/a/d
 

Mathematica [A] (verified)

Time = 7.60 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \sqrt {a (1+\sin (c+d x))} (4159-2280 \cos (2 (c+d x))+105 \cos (4 (c+d x))+5076 \sin (c+d x)-700 \sin (3 (c+d x)))}{4620 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

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

Output:

-1/4620*(a*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*Sqrt[a*(1 + Sin[c + d*x 
])]*(4159 - 2280*Cos[2*(c + d*x)] + 105*Cos[4*(c + d*x)] + 5076*Sin[c + d* 
x] - 700*Sin[3*(c + d*x)]))/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3357, 27, 3042, 3335, 3042, 3153, 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.

\(\displaystyle \int \sin ^2(c+d x) \cos ^2(c+d x) (a \sin (c+d x)+a)^{3/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3357

\(\displaystyle \frac {2 \int \frac {1}{2} \cos ^2(c+d x) (5 a-6 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \cos ^2(c+d x) (5 a-6 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \cos (c+d x)^2 (5 a-6 a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 3335

\(\displaystyle \frac {3 a \int \cos ^2(c+d x) (\sin (c+d x) a+a)^{3/2}dx+\frac {4 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \int \cos (c+d x)^2 (\sin (c+d x) a+a)^{3/2}dx+\frac {4 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {3 a \left (\frac {8}{7} a \int \cos ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )+\frac {4 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \left (\frac {8}{7} a \int \cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )+\frac {4 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {3 a \left (\frac {8}{7} a \left (\frac {4}{5} a \int \frac {\cos ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )+\frac {4 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \left (\frac {8}{7} a \left (\frac {4}{5} a \int \frac {\cos (c+d x)^2}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )+\frac {4 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {3 a \left (\frac {8}{7} a \left (-\frac {8 a^2 \cos ^3(c+d x)}{15 d (a \sin (c+d x)+a)^{3/2}}-\frac {2 a \cos ^3(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}\right )-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\right )+\frac {4 a \cos ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}}{11 a}-\frac {2 \cos ^3(c+d x) (a \sin (c+d x)+a)^{5/2}}{11 a d}\)

Input:

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

Output:

(-2*Cos[c + d*x]^3*(a + a*Sin[c + d*x])^(5/2))/(11*a*d) + ((4*a*Cos[c + d* 
x]^3*(a + a*Sin[c + d*x])^(3/2))/(3*d) + 3*a*((-2*a*Cos[c + d*x]^3*Sqrt[a 
+ a*Sin[c + d*x]])/(7*d) + (8*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 
))/(11*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]

\[\int \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (105 \, a \cos \left (d x + c\right )^{6} + 245 \, a \cos \left (d x + c\right )^{5} - 185 \, a \cos \left (d x + c\right )^{4} - 397 \, a \cos \left (d x + c\right )^{3} + 24 \, a \cos \left (d x + c\right )^{2} - 96 \, a \cos \left (d x + c\right ) + {\left (105 \, a \cos \left (d x + c\right )^{5} - 140 \, a \cos \left (d x + c\right )^{4} - 325 \, a \cos \left (d x + c\right )^{3} + 72 \, a \cos \left (d x + c\right )^{2} + 96 \, a \cos \left (d x + c\right ) + 192 \, a\right )} \sin \left (d x + c\right ) - 192 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{1155 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="f 
ricas")
 

Output:

2/1155*(105*a*cos(d*x + c)^6 + 245*a*cos(d*x + c)^5 - 185*a*cos(d*x + c)^4 
 - 397*a*cos(d*x + c)^3 + 24*a*cos(d*x + c)^2 - 96*a*cos(d*x + c) + (105*a 
*cos(d*x + c)^5 - 140*a*cos(d*x + c)^4 - 325*a*cos(d*x + c)^3 + 72*a*cos(d 
*x + c)^2 + 96*a*cos(d*x + c) + 192*a)*sin(d*x + c) - 192*a)*sqrt(a*sin(d* 
x + c) + a)/(d*cos(d*x + c) + d*sin(d*x + c) + d)
 

Sympy [F]

\[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}\, dx \] Input:

integrate(cos(d*x+c)**2*sin(d*x+c)**2*(a+a*sin(d*x+c))**(3/2),x)
 

Output:

Integral((a*(sin(c + d*x) + 1))**(3/2)*sin(c + d*x)**2*cos(c + d*x)**2, x)
 

Maxima [F]

\[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )^{2} \,d x } \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="m 
axima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {16 \, \sqrt {2} {\left (420 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1540 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2145 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1386 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )} \sqrt {a}}{1155 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c))^(3/2),x, algorithm="g 
iac")
 

Output:

16/1155*sqrt(2)*(420*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1 
/2*d*x + 1/2*c)^11 - 1540*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*p 
i + 1/2*d*x + 1/2*c)^9 + 2145*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1 
/4*pi + 1/2*d*x + 1/2*c)^7 - 1386*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*si 
n(-1/4*pi + 1/2*d*x + 1/2*c)^5 + 385*a*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. \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3}d x +\int \sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2}d x \right ) \] Input:

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

Output:

sqrt(a)*a*(int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x)**3,x) + 
 int(sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*sin(c + d*x)**2,x))