\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx\) [12]

Optimal result

Integrand size = 38, antiderivative size = 92 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {c \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 a f \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2} \sqrt {c-c \sin (e+f x)}}{4 a f} \] Output:

1/6*c*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/a/f/(c-c*sin(f*x+e))^(1/2)+1/4*cos 
(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(1/2)/a/f
 

Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {a \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (-12 \cos (2 (e+f x))-3 \cos (4 (e+f x))+8 (9 \sin (e+f x)+\sin (3 (e+f x))))}{96 f} \] Input:

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x 
]],x]
 

Output:

(a*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(-12*C 
os[2*(e + f*x)] - 3*Cos[4*(e + f*x)] + 8*(9*Sin[e + f*x] + Sin[3*(e + f*x) 
])))/(96*f)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3320, 3042, 3219, 3042, 3217}

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.

\(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{3/2}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{3/2}dx}{a c}\)

\(\Big \downarrow \) 3219

\(\displaystyle \frac {\frac {1}{2} c \int (\sin (e+f x) a+a)^{5/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}{4 f}}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} c \int (\sin (e+f x) a+a)^{5/2} \sqrt {c-c \sin (e+f x)}dx+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}{4 f}}{a c}\)

\(\Big \downarrow \) 3217

\(\displaystyle \frac {\frac {c^2 \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 f \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) (a \sin (e+f x)+a)^{5/2} \sqrt {c-c \sin (e+f x)}}{4 f}}{a c}\)

Input:

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2)*Sqrt[c - c*Sin[e + f*x]],x]
 

Output:

((c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(6*f*Sqrt[c - c*Sin[e + f*x 
]]) + (c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*Sqrt[c - c*Sin[e + f*x]]) 
/(4*f))/(a*c)
 

Defintions of rubi rules used

rule 3042

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

rule 3217

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f 
_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ 
n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n 
}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
 

rule 3219

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n 
))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre 
eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I 
GtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m]) &&  !( 
ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
 

rule 3320

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(a^(p/ 
2)*c^(p/2))   Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(n + 
p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && 
EqQ[a^2 - b^2, 0] && IntegerQ[p/2]
 

Maple [F]

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

Input:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x)
 

Output:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (3 \, a \cos \left (f x + e\right )^{4} - 4 \, {\left (a \cos \left (f x + e\right )^{2} + 2 \, a\right )} \sin \left (f x + e\right ) - 3 \, a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="fricas")
 

Output:

-1/12*(3*a*cos(f*x + e)^4 - 4*(a*cos(f*x + e)^2 + 2*a)*sin(f*x + e) - 3*a) 
*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))
 

Sympy [F]

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

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(1/2),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**(3/2)*sqrt(-c*(sin(e + f*x) - 1))*cos(e + 
 f*x)**2, x)
 

Maxima [F]

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

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^(3/2)*sqrt(-c*sin(f*x + e) + c)*cos(f*x + e 
)^2, x)
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.53 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=\frac {4 \, {\left (3 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 8 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 6 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}\right )} \sqrt {a} \sqrt {c}}{3 \, f} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x, al 
gorithm="giac")
 

Output:

4/3*(3*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1 
/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8 - 8*a*sgn(cos(-1/4*pi + 1/2*f*x + 
1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) 
^6 + 6*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1 
/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4)*sqrt(a)*sqrt(c)/f
 

Mupad [B] (verification not implemented)

Time = 18.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,\cos \left (e+f\,x\right )+15\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )-80\,\sin \left (2\,e+2\,f\,x\right )-8\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:

int(cos(e + f*x)^2*(a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(1/2),x 
)
 

Output:

-(a*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(12*cos(e + 
 f*x) + 15*cos(3*e + 3*f*x) + 3*cos(5*e + 5*f*x) - 80*sin(2*e + 2*f*x) - 8 
*sin(4*e + 4*f*x)))/(96*f*(cos(2*e + 2*f*x) + 1))
 

Reduce [F]

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

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(1/2),x)
 

Output:

sqrt(c)*sqrt(a)*a*(int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*co 
s(e + f*x)**2*sin(e + f*x),x) + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + 
 f*x) + 1)*cos(e + f*x)**2,x))