\(\int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [1018]

Optimal result

Integrand size = 31, antiderivative size = 93 \[ \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {2 (A-B) (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)}-\frac {(A-3 B) (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)}-\frac {B (a+a \sin (e+f x))^{4+m}}{a^4 f (4+m)} \] Output:

2*(A-B)*(a+a*sin(f*x+e))^(2+m)/a^2/f/(2+m)-(A-3*B)*(a+a*sin(f*x+e))^(3+m)/ 
a^3/f/(3+m)-B*(a+a*sin(f*x+e))^(4+m)/a^4/f/(4+m)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00 \[ \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {2 (A-B) (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)}-\frac {(A-3 B) (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)}-\frac {B (a+a \sin (e+f x))^{4+m}}{a^4 f (4+m)} \] Input:

Integrate[Cos[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]
 

Output:

(2*(A - B)*(a + a*Sin[e + f*x])^(2 + m))/(a^2*f*(2 + m)) - ((A - 3*B)*(a + 
 a*Sin[e + f*x])^(3 + m))/(a^3*f*(3 + m)) - (B*(a + a*Sin[e + f*x])^(4 + m 
))/(a^4*f*(4 + m))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 2009}

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.

\(\displaystyle \int \cos ^3(e+f x) (a \sin (e+f x)+a)^m (A+B \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (e+f x)^3 (a \sin (e+f x)+a)^m (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {\int \frac {(a-a \sin (e+f x)) (\sin (e+f x) a+a)^{m+1} (a A+a B \sin (e+f x))}{a}d(a \sin (e+f x))}{a^3 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a-a \sin (e+f x)) (\sin (e+f x) a+a)^{m+1} (a A+a B \sin (e+f x))d(a \sin (e+f x))}{a^4 f}\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {\int \left (2 a^2 (A-B) (\sin (e+f x) a+a)^{m+1}-a (A-3 B) (\sin (e+f x) a+a)^{m+2}-B (\sin (e+f x) a+a)^{m+3}\right )d(a \sin (e+f x))}{a^4 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {2 a^2 (A-B) (a \sin (e+f x)+a)^{m+2}}{m+2}-\frac {a (A-3 B) (a \sin (e+f x)+a)^{m+3}}{m+3}-\frac {B (a \sin (e+f x)+a)^{m+4}}{m+4}}{a^4 f}\)

Input:

Int[Cos[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]
 

Output:

((2*a^2*(A - B)*(a + a*Sin[e + f*x])^(2 + m))/(2 + m) - (a*(A - 3*B)*(a + 
a*Sin[e + f*x])^(3 + m))/(3 + m) - (B*(a + a*Sin[e + f*x])^(4 + m))/(4 + m 
))/(a^4*f)
 

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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3315

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

Maple [A] (verified)

Time = 3.12 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.59

method	result	size
parallelrisch	\(\frac {\left (\left (\left (A +B \right ) m +4 A \right ) \left (2+m \right ) \sin \left (3 f x +3 e \right )+2 \left (A \,m^{2}+\left (4 A -5 B \right ) m -6 B \right ) \cos \left (2 f x +2 e \right )-\frac {B \left (3+m \right ) \left (2+m \right ) \cos \left (4 f x +4 e \right )}{2}+\left (\left (A +B \right ) m +4 A \right ) \left (m +18\right ) \sin \left (f x +e \right )+\left (2 A +\frac {B}{2}\right ) m^{2}+\left (24 A +\frac {17 B}{2}\right ) m +64 A -9 B \right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{m}}{4 \left (m^{3}+9 m^{2}+26 m +24\right ) f}\)	\(148\)
derivativedivides	\(\frac {\left (A \,m^{2}+8 A m -B m +16 A -6 B \right ) {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (m^{3}+9 m^{2}+26 m +24\right )}+\frac {\left (A \,m^{2}+B \,m^{2}+10 A m +6 B m +24 A \right ) \sin \left (f x +e \right ) {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (m^{3}+9 m^{2}+26 m +24\right )}-\frac {B \sin \left (f x +e \right )^{4} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (4+m \right )}-\frac {\left (A \,m^{2}-B \,m^{2}+4 A m -10 B m -12 B \right ) \sin \left (f x +e \right )^{2} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (m^{3}+9 m^{2}+26 m +24\right )}-\frac {\left (A m +B m +4 A \right ) \sin \left (f x +e \right )^{3} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (m^{2}+7 m +12\right )}\)	\(261\)
default	\(\frac {\left (A \,m^{2}+8 A m -B m +16 A -6 B \right ) {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (m^{3}+9 m^{2}+26 m +24\right )}+\frac {\left (A \,m^{2}+B \,m^{2}+10 A m +6 B m +24 A \right ) \sin \left (f x +e \right ) {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (m^{3}+9 m^{2}+26 m +24\right )}-\frac {B \sin \left (f x +e \right )^{4} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (4+m \right )}-\frac {\left (A \,m^{2}-B \,m^{2}+4 A m -10 B m -12 B \right ) \sin \left (f x +e \right )^{2} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (m^{3}+9 m^{2}+26 m +24\right )}-\frac {\left (A m +B m +4 A \right ) \sin \left (f x +e \right )^{3} {\mathrm e}^{m \ln \left (a \left (1+\sin \left (f x +e \right )\right )\right )}}{f \left (m^{2}+7 m +12\right )}\)	\(261\)

Input:

int(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x,method=_RETURNVERBO 
SE)
 

Output:

1/4*(((A+B)*m+4*A)*(2+m)*sin(3*f*x+3*e)+2*(A*m^2+(4*A-5*B)*m-6*B)*cos(2*f* 
x+2*e)-1/2*B*(3+m)*(2+m)*cos(4*f*x+4*e)+((A+B)*m+4*A)*(m+18)*sin(f*x+e)+(2 
*A+1/2*B)*m^2+(24*A+17/2*B)*m+64*A-9*B)*(a*(1+sin(f*x+e)))^m/(m^3+9*m^2+26 
*m+24)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.45 \[ \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {{\left ({\left (B m^{2} + 5 \, B m + 6 \, B\right )} \cos \left (f x + e\right )^{4} - {\left ({\left (A + B\right )} m^{2} + 4 \, A m\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (A + B\right )} m - {\left ({\left ({\left (A + B\right )} m^{2} + 2 \, {\left (3 \, A + B\right )} m + 8 \, A\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (A + B\right )} m + 16 \, A\right )} \sin \left (f x + e\right ) - 16 \, A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{f m^{3} + 9 \, f m^{2} + 26 \, f m + 24 \, f} \] Input:

integrate(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="f 
ricas")
 

Output:

-((B*m^2 + 5*B*m + 6*B)*cos(f*x + e)^4 - ((A + B)*m^2 + 4*A*m)*cos(f*x + e 
)^2 - 4*(A + B)*m - (((A + B)*m^2 + 2*(3*A + B)*m + 8*A)*cos(f*x + e)^2 + 
4*(A + B)*m + 16*A)*sin(f*x + e) - 16*A)*(a*sin(f*x + e) + a)^m/(f*m^3 + 9 
*f*m^2 + 26*f*m + 24*f)
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5243 vs. \(2 (78) = 156\).

Time = 11.77 (sec) , antiderivative size = 5243, normalized size of antiderivative = 56.38 \[ \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Too large to display} \] Input:

integrate(cos(f*x+e)**3*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)
 

Output:

Piecewise((x*(A + B*sin(e))*(a*sin(e) + a)**m*cos(e)**3, Eq(f, 0)), (4*A*s 
in(e + f*x)**2/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18* 
a**4*f*sin(e + f*x) + 6*a**4*f) + 6*A*sin(e + f*x)/(6*a**4*f*sin(e + f*x)* 
*3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 2*A* 
cos(e + f*x)**2/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18 
*a**4*f*sin(e + f*x) + 6*a**4*f) + 2*A/(6*a**4*f*sin(e + f*x)**3 + 18*a**4 
*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 6*B*log(sin(e + 
f*x) + 1)*sin(e + f*x)**3/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f* 
x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 18*B*log(sin(e + f*x) + 1)*si 
n(e + f*x)**2/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a 
**4*f*sin(e + f*x) + 6*a**4*f) - 18*B*log(sin(e + f*x) + 1)*sin(e + f*x)/( 
6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f 
*x) + 6*a**4*f) - 6*B*log(sin(e + f*x) + 1)/(6*a**4*f*sin(e + f*x)**3 + 18 
*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - 10*B*sin(e 
+ f*x)**2/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4* 
f*sin(e + f*x) + 6*a**4*f) - 3*B*sin(e + f*x)*cos(e + f*x)**2/(6*a**4*f*si 
n(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*a** 
4*f) - 18*B*sin(e + f*x)/(6*a**4*f*sin(e + f*x)**3 + 18*a**4*f*sin(e + f*x 
)**2 + 18*a**4*f*sin(e + f*x) + 6*a**4*f) - B*cos(e + f*x)**2/(6*a**4*f*si 
n(e + f*x)**3 + 18*a**4*f*sin(e + f*x)**2 + 18*a**4*f*sin(e + f*x) + 6*...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (93) = 186\).

Time = 0.06 (sec) , antiderivative size = 285, normalized size of antiderivative = 3.06 \[ \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {\frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{m} \sin \left (f x + e\right )^{3} + {\left (m^{2} + m\right )} a^{m} \sin \left (f x + e\right )^{2} - 2 \, a^{m} m \sin \left (f x + e\right ) + 2 \, a^{m}\right )} A {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} + \frac {{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} a^{m} \sin \left (f x + e\right )^{4} + {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{m} \sin \left (f x + e\right )^{3} - 3 \, {\left (m^{2} + m\right )} a^{m} \sin \left (f x + e\right )^{2} + 6 \, a^{m} m \sin \left (f x + e\right ) - 6 \, a^{m}\right )} B {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} - \frac {{\left (a^{m} {\left (m + 1\right )} \sin \left (f x + e\right )^{2} + a^{m} m \sin \left (f x + e\right ) - a^{m}\right )} B {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{2} + 3 \, m + 2} - \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m + 1} A}{a {\left (m + 1\right )}}}{f} \] Input:

integrate(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="m 
axima")
 

Output:

-(((m^2 + 3*m + 2)*a^m*sin(f*x + e)^3 + (m^2 + m)*a^m*sin(f*x + e)^2 - 2*a 
^m*m*sin(f*x + e) + 2*a^m)*A*(sin(f*x + e) + 1)^m/(m^3 + 6*m^2 + 11*m + 6) 
 + ((m^3 + 6*m^2 + 11*m + 6)*a^m*sin(f*x + e)^4 + (m^3 + 3*m^2 + 2*m)*a^m* 
sin(f*x + e)^3 - 3*(m^2 + m)*a^m*sin(f*x + e)^2 + 6*a^m*m*sin(f*x + e) - 6 
*a^m)*B*(sin(f*x + e) + 1)^m/(m^4 + 10*m^3 + 35*m^2 + 50*m + 24) - (a^m*(m 
 + 1)*sin(f*x + e)^2 + a^m*m*sin(f*x + e) - a^m)*B*(sin(f*x + e) + 1)^m/(m 
^2 + 3*m + 2) - (a*sin(f*x + e) + a)^(m + 1)*A/(a*(m + 1)))/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 562 vs. \(2 (93) = 186\).

Time = 0.13 (sec) , antiderivative size = 562, normalized size of antiderivative = 6.04 \[ \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx =\text {Too large to display} \] Input:

integrate(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="g 
iac")
 

Output:

-(((a*sin(f*x + e) + a)^m*m^2*sin(f*x + e)^3 + (a*sin(f*x + e) + a)^m*m^2* 
sin(f*x + e)^2 + 3*(a*sin(f*x + e) + a)^m*m*sin(f*x + e)^3 + (a*sin(f*x + 
e) + a)^m*m*sin(f*x + e)^2 + 2*(a*sin(f*x + e) + a)^m*sin(f*x + e)^3 - 2*( 
a*sin(f*x + e) + a)^m*m*sin(f*x + e) + 2*(a*sin(f*x + e) + a)^m)*A/(m^3 + 
6*m^2 + 11*m + 6) + ((a*sin(f*x + e) + a)^m*m^3*sin(f*x + e)^4 + (a*sin(f* 
x + e) + a)^m*m^3*sin(f*x + e)^3 + 6*(a*sin(f*x + e) + a)^m*m^2*sin(f*x + 
e)^4 + 3*(a*sin(f*x + e) + a)^m*m^2*sin(f*x + e)^3 + 11*(a*sin(f*x + e) + 
a)^m*m*sin(f*x + e)^4 - 3*(a*sin(f*x + e) + a)^m*m^2*sin(f*x + e)^2 + 2*(a 
*sin(f*x + e) + a)^m*m*sin(f*x + e)^3 + 6*(a*sin(f*x + e) + a)^m*sin(f*x + 
 e)^4 - 3*(a*sin(f*x + e) + a)^m*m*sin(f*x + e)^2 + 6*(a*sin(f*x + e) + a) 
^m*m*sin(f*x + e) - 6*(a*sin(f*x + e) + a)^m)*B/(m^4 + 10*m^3 + 35*m^2 + 5 
0*m + 24) - ((a*sin(f*x + e) + a)^m*m*sin(f*x + e)^2 + (a*sin(f*x + e) + a 
)^m*m*sin(f*x + e) + (a*sin(f*x + e) + a)^m*sin(f*x + e)^2 - (a*sin(f*x + 
e) + a)^m)*B/(m^2 + 3*m + 2) - (a*sin(f*x + e) + a)^(m + 1)*A/(a*(m + 1))) 
/f
 

Mupad [B] (verification not implemented)

Time = 36.01 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.92 \[ \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (128\,A-18\,B+48\,A\,m+17\,B\,m+144\,A\,\sin \left (e+f\,x\right )-24\,B\,\cos \left (2\,e+2\,f\,x\right )-6\,B\,\cos \left (4\,e+4\,f\,x\right )+4\,A\,m^2+B\,m^2+16\,A\,\sin \left (3\,e+3\,f\,x\right )+4\,A\,m^2\,\cos \left (2\,e+2\,f\,x\right )-B\,m^2\,\cos \left (4\,e+4\,f\,x\right )+2\,A\,m^2\,\sin \left (3\,e+3\,f\,x\right )+2\,B\,m^2\,\sin \left (3\,e+3\,f\,x\right )+44\,A\,m\,\sin \left (e+f\,x\right )+36\,B\,m\,\sin \left (e+f\,x\right )+16\,A\,m\,\cos \left (2\,e+2\,f\,x\right )-20\,B\,m\,\cos \left (2\,e+2\,f\,x\right )-5\,B\,m\,\cos \left (4\,e+4\,f\,x\right )+12\,A\,m\,\sin \left (3\,e+3\,f\,x\right )+2\,A\,m^2\,\sin \left (e+f\,x\right )+4\,B\,m\,\sin \left (3\,e+3\,f\,x\right )+2\,B\,m^2\,\sin \left (e+f\,x\right )\right )}{8\,f\,\left (m^3+9\,m^2+26\,m+24\right )} \] Input:

int(cos(e + f*x)^3*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m,x)
 

Output:

((a*(sin(e + f*x) + 1))^m*(128*A - 18*B + 48*A*m + 17*B*m + 144*A*sin(e + 
f*x) - 24*B*cos(2*e + 2*f*x) - 6*B*cos(4*e + 4*f*x) + 4*A*m^2 + B*m^2 + 16 
*A*sin(3*e + 3*f*x) + 4*A*m^2*cos(2*e + 2*f*x) - B*m^2*cos(4*e + 4*f*x) + 
2*A*m^2*sin(3*e + 3*f*x) + 2*B*m^2*sin(3*e + 3*f*x) + 44*A*m*sin(e + f*x) 
+ 36*B*m*sin(e + f*x) + 16*A*m*cos(2*e + 2*f*x) - 20*B*m*cos(2*e + 2*f*x) 
- 5*B*m*cos(4*e + 4*f*x) + 12*A*m*sin(3*e + 3*f*x) + 2*A*m^2*sin(e + f*x) 
+ 4*B*m*sin(3*e + 3*f*x) + 2*B*m^2*sin(e + f*x)))/(8*f*(26*m + 9*m^2 + m^3 
 + 24))
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \cos ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{3} \sin \left (f x +e \right )d x \right ) b +\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{3}d x \right ) a \] Input:

int(cos(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)
 

Output:

int((sin(e + f*x)*a + a)**m*cos(e + f*x)**3*sin(e + f*x),x)*b + int((sin(e 
 + f*x)*a + a)**m*cos(e + f*x)**3,x)*a