\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx\) [82]

Optimal result

Integrand size = 38, antiderivative size = 114 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-3-m}}{a c f (5+2 m)}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{-2-m}}{a c^2 f \left (15+16 m+4 m^2\right )} \] Output:

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

Mathematica [A] (verified)

Time = 5.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\frac {\cos ^3(e+f x) (4+2 m-\sin (e+f x)) (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-m}}{c^4 f (3+2 m) (5+2 m) (-1+\sin (e+f x))^4} \] Input:

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-4 - 
 m),x]
 

Output:

(Cos[e + f*x]^3*(4 + 2*m - Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^m)/(c^4*f* 
(3 + 2*m)*(5 + 2*m)*(-1 + Sin[e + f*x])^4*(c - c*Sin[e + f*x])^m)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, 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, 3222, 3042, 3221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3222

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3221

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

Input:

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-4 - m),x]
 

Output:

((Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(-3 - m)) 
/(f*(5 + 2*m)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + 
 f*x])^(-2 - m))/(c*f*(3 + 2*m)*(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 3221

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*( 
(c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && Ne 
Q[m, -2^(-1)]
 

rule 3222

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*( 
(c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[(m + n + 1)/(a*(2*m + 1) 
)   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free 
Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && 
 ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || 
!SumSimplerQ[n, 1])
 

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]

\[\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-4-m}d x\]

Input:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-4-m),x)
 

Output:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-4-m),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\frac {{\left (2 \, {\left (m + 2\right )} \cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{3} \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4}}{4 \, f m^{2} + 16 \, f m + 15 \, f} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-4-m),x, algor 
ithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{- m - 4} \cos ^{2}{\left (e + f x \right )}\, dx \] Input:

integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-4-m),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**(-m - 4)*cos(e 
 + f*x)**2, x)
 

Maxima [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4} \cos \left (f x + e\right )^{2} \,d x } \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-4-m),x, algor 
ithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 4)*cos(f*x + 
e)^2, x)
 

Giac [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{-m - 4} \cos \left (f x + e\right )^{2} \,d x } \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-4-m),x, algor 
ithm="giac")
 

Output:

integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(-m - 4)*cos(f*x + 
e)^2, x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 19.36 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.55 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (2\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )+48\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+16\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2+12\,m\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+4\,m\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )-32\right )}{c^4\,f\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^m\,\left (4\,m^2+16\,m+15\right )\,\left (56\,{\sin \left (e+f\,x\right )}^2-56\,\sin \left (e+f\,x\right )-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+8\,\sin \left (3\,e+3\,f\,x\right )+8\right )} \] Input:

int((cos(e + f*x)^2*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(m + 4),x 
)
 

Output:

-((a*(sin(e + f*x) + 1))^m*(2*sin(2*e + 2*f*x) + sin(4*e + 4*f*x) + 48*sin 
(e/2 + (f*x)/2)^2 + 16*sin((3*e)/2 + (3*f*x)/2)^2 + 12*m*(2*sin(e/2 + (f*x 
)/2)^2 - 1) + 4*m*(2*sin((3*e)/2 + (3*f*x)/2)^2 - 1) - 32))/(c^4*f*(-c*(si 
n(e + f*x) - 1))^m*(16*m + 4*m^2 + 15)*(8*sin(3*e + 3*f*x) - 56*sin(e + f* 
x) - 2*sin(2*e + 2*f*x)^2 + 56*sin(e + f*x)^2 + 8))
 

Reduce [F]

\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-4-m} \, dx=\frac {\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{2}}{\left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )^{4}-4 \left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )^{3}+6 \left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )^{2}-4 \left (-\sin \left (f x +e \right ) c +c \right )^{m} \sin \left (f x +e \right )+\left (-\sin \left (f x +e \right ) c +c \right )^{m}}d x}{c^{4}} \] Input:

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-4-m),x)
 

Output:

int(((sin(e + f*x)*a + a)**m*cos(e + f*x)**2)/(( - sin(e + f*x)*c + c)**m* 
sin(e + f*x)**4 - 4*( - sin(e + f*x)*c + c)**m*sin(e + f*x)**3 + 6*( - sin 
(e + f*x)*c + c)**m*sin(e + f*x)**2 - 4*( - sin(e + f*x)*c + c)**m*sin(e + 
 f*x) + ( - sin(e + f*x)*c + c)**m),x)/c**4