\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx\) [75]

Optimal result

Integrand size = 36, antiderivative size = 107 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=\frac {8 c \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f \left (15+16 m+4 m^2\right ) \sqrt {c-c \sin (e+f x)}}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)}}{a f (5+2 m)} \] Output:

8*c*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(4*m^2+16*m+15)/(c-c*sin(f*x+e)) 
^(1/2)+2*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(1/2)/a/f/(5+2 
*m)
 

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.04 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (a (1+\sin (e+f x)))^m \sqrt {c-c \sin (e+f x)} (-7-2 m+(3+2 m) \sin (e+f x))}{f (3+2 m) (5+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(-2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(a*(1 + Sin[e + f*x]))^m*Sqrt[ 
c - c*Sin[e + f*x]]*(-7 - 2*m + (3 + 2*m)*Sin[e + f*x]))/(f*(3 + 2*m)*(5 + 
 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{3/2}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{3/2}dx}{a c}\)

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3217

\(\displaystyle \frac {\frac {8 c^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{f (2 m+3) (2 m+5) \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^{m+1}}{f (2 m+5)}}{a c}\)

Input:

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

Output:

((8*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(f*(3 + 2*m)*(5 + 2*m)* 
Sqrt[c - c*Sin[e + f*x]]) + (2*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m) 
*Sqrt[c - c*Sin[e + f*x]])/(f*(5 + 2*m)))/(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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.45 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 \, {\left ({\left (2 \, m + 3\right )} \cos \left (f x + e\right )^{3} + {\left (2 \, m - 1\right )} \cos \left (f x + e\right )^{2} + {\left ({\left (2 \, m + 3\right )} \cos \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) + 8\right )} \sin \left (f x + e\right ) + 4 \, \cos \left (f x + e\right ) + 8\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{4 \, f m^{2} + 16 \, f m + {\left (4 \, f m^{2} + 16 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) - {\left (4 \, f m^{2} + 16 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2),x, algori 
thm="fricas")
 

Output:

2*((2*m + 3)*cos(f*x + e)^3 + (2*m - 1)*cos(f*x + e)^2 + ((2*m + 3)*cos(f* 
x + e)^2 + 4*cos(f*x + e) + 8)*sin(f*x + e) + 4*cos(f*x + e) + 8)*sqrt(-c* 
sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(4*f*m^2 + 16*f*m + (4*f*m^2 + 16 
*f*m + 15*f)*cos(f*x + e) - (4*f*m^2 + 16*f*m + 15*f)*sin(f*x + e) + 15*f)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (103) = 206\).

Time = 0.19 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.92 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (a^{m} \sqrt {c} {\left (2 \, m + 7\right )} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 15\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{m} \sqrt {c} {\left (2 \, m - 5\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{m} \sqrt {c} {\left (2 \, m - 5\right )} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{m} \sqrt {c} {\left (2 \, m + 7\right )} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (4 \, m^{2} + 16 \, m + \frac {2 \, {\left (4 \, m^{2} + 16 \, m + 15\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (4 \, m^{2} + 16 \, m + 15\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 15\right )} f \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2),x, algori 
thm="maxima")
 

Output:

-2*(a^m*sqrt(c)*(2*m + 7) + a^m*sqrt(c)*(2*m + 15)*sin(f*x + e)/(cos(f*x + 
 e) + 1) - 2*a^m*sqrt(c)*(2*m - 5)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2 
*a^m*sqrt(c)*(2*m - 5)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a^m*sqrt(c)*( 
2*m + 15)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^m*sqrt(c)*(2*m + 7)*sin( 
f*x + e)^5/(cos(f*x + e) + 1)^5)*e^(2*m*log(sin(f*x + e)/(cos(f*x + e) + 1 
) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((4*m^2 + 16*m + 
2*(4*m^2 + 16*m + 15)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (4*m^2 + 16*m 
+ 15)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 15)*f*sqrt(sin(f*x + e)^2/(cos 
(f*x + e) + 1)^2 + 1))
 

Giac [F]

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

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1/2),x, algori 
thm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.97 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \sqrt {c-c \sin (e+f x)} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (25\,\cos \left (e+f\,x\right )+3\,\cos \left (3\,e+3\,f\,x\right )+8\,\sin \left (2\,e+2\,f\,x\right )+6\,m\,\cos \left (e+f\,x\right )+2\,m\,\cos \left (3\,e+3\,f\,x\right )\right )}{2\,f\,\left (\sin \left (e+f\,x\right )-1\right )\,\left (4\,m^2+16\,m+15\right )} \] Input:

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

Output:

-((a*(sin(e + f*x) + 1))^m*(-c*(sin(e + f*x) - 1))^(1/2)*(25*cos(e + f*x) 
+ 3*cos(3*e + 3*f*x) + 8*sin(2*e + 2*f*x) + 6*m*cos(e + f*x) + 2*m*cos(3*e 
 + 3*f*x)))/(2*f*(sin(e + f*x) - 1)*(16*m + 4*m^2 + 15))
 

Reduce [F]

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

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

Output:

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