\(\int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\) [119]

Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {64 a^3 \cos ^3(c+d x)}{105 d (a+a \sin (c+d x))^{3/2}}-\frac {16 a^2 \cos ^3(c+d x)}{35 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{7 d} \] Output:

-64/105*a^3*cos(d*x+c)^3/d/(a+a*sin(d*x+c))^(3/2)-16/35*a^2*cos(d*x+c)^3/d 
/(a+a*sin(d*x+c))^(1/2)-2/7*a*cos(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \cos ^3(c+d x) (a (1+\sin (c+d x)))^{3/2} \left (71+54 \sin (c+d x)+15 \sin ^2(c+d x)\right )}{105 d (1+\sin (c+d x))^3} \] Input:

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

Output:

(-2*Cos[c + d*x]^3*(a*(1 + Sin[c + d*x]))^(3/2)*(71 + 54*Sin[c + d*x] + 15 
*Sin[c + d*x]^2))/(105*d*(1 + Sin[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3153

\(\displaystyle \frac {8}{7} a \int \cos ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8}{7} a \int \cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx-\frac {2 a \cos ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{7 d}\)

\(\Big \downarrow \) 3153

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3152

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

Input:

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

Output:

(-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*S 
qrt[a + a*Sin[c + d*x]])))/7
 

Defintions of rubi rules used

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]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71

method	result	size
default	\(-\frac {2 \left (1+\sin \left (d x +c \right )\right ) a^{2} \left (\sin \left (d x +c \right )-1\right )^{2} \left (15 \sin \left (d x +c \right )^{2}+54 \sin \left (d x +c \right )+71\right )}{105 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(67\)

Input:

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

Output:

-2/105*(1+sin(d*x+c))*a^2*(sin(d*x+c)-1)^2*(15*sin(d*x+c)^2+54*sin(d*x+c)+ 
71)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.28 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (15 \, a \cos \left (d x + c\right )^{4} + 39 \, a \cos \left (d x + c\right )^{3} - 8 \, a \cos \left (d x + c\right )^{2} + 32 \, a \cos \left (d x + c\right ) + {\left (15 \, a \cos \left (d x + c\right )^{3} - 24 \, a \cos \left (d x + c\right )^{2} - 32 \, a \cos \left (d x + c\right ) - 64 \, a\right )} \sin \left (d x + c\right ) + 64 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{105 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \] Input:

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

Output:

-2/105*(15*a*cos(d*x + c)^4 + 39*a*cos(d*x + c)^3 - 8*a*cos(d*x + c)^2 + 3 
2*a*cos(d*x + c) + (15*a*cos(d*x + c)^3 - 24*a*cos(d*x + c)^2 - 32*a*cos(d 
*x + c) - 64*a)*sin(d*x + c) + 64*a)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + 
 c) + d*sin(d*x + c) + d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.07 \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\frac {16 \, \sqrt {2} {\left (15 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 42 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 35 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )} \sqrt {a}}{105 \, d} \] Input:

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

Output:

16/105*sqrt(2)*(15*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 - 42*a*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1 
/2*d*x + 1/2*c)^5 + 35*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. \[ \int \cos ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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