\(\int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^m \, dx\) [357]

Optimal result

Integrand size = 27, antiderivative size = 201 \[ \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^m \, dx=-\frac {(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}-\frac {3 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m}}{a d e (1-m) (3-m)}+\frac {6 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e (3-m) \left (1-m^2\right )}-\frac {6 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{3+m}}{a^3 d e \left (9-10 m^2+m^4\right )} \] Output:

-(e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^m/d/e/(3-m)-3*(e*cos(d*x+c))^(-3-m 
)*(a+a*sin(d*x+c))^(1+m)/a/d/e/(1-m)/(3-m)+6*(e*cos(d*x+c))^(-3-m)*(a+a*si 
n(d*x+c))^(2+m)/a^2/d/e/(3-m)/(-m^2+1)-6*(e*cos(d*x+c))^(-3-m)*(a+a*sin(d* 
x+c))^(3+m)/a^3/d/e/(m^4-10*m^2+9)
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.50 \[ \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-m} \sec ^3(c+d x) (a (1+\sin (c+d x)))^m \left (m \left (-7+m^2\right )-3 \left (-3+m^2\right ) \sin (c+d x)+6 m \sin ^2(c+d x)-6 \sin ^3(c+d x)\right )}{d e^4 (-3+m) (-1+m) (1+m) (3+m)} \] Input:

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

Output:

(Sec[c + d*x]^3*(a*(1 + Sin[c + d*x]))^m*(m*(-7 + m^2) - 3*(-3 + m^2)*Sin[ 
c + d*x] + 6*m*Sin[c + d*x]^2 - 6*Sin[c + d*x]^3))/(d*e^4*(-3 + m)*(-1 + m 
)*(1 + m)*(3 + m)*(e*Cos[c + d*x])^m)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3151, 3042, 3151, 3042, 3151, 3042, 3150}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3151

\(\displaystyle \frac {3 \int (e \cos (c+d x))^{-m-4} (\sin (c+d x) a+a)^{m+1}dx}{a (3-m)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int (e \cos (c+d x))^{-m-4} (\sin (c+d x) a+a)^{m+1}dx}{a (3-m)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {3 \left (\frac {2 \int (e \cos (c+d x))^{-m-4} (\sin (c+d x) a+a)^{m+2}dx}{a (1-m)}-\frac {(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-3}}{d e (1-m)}\right )}{a (3-m)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {2 \int (e \cos (c+d x))^{-m-4} (\sin (c+d x) a+a)^{m+2}dx}{a (1-m)}-\frac {(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-3}}{d e (1-m)}\right )}{a (3-m)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {3 \left (\frac {2 \left (\frac {(a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-3}}{d e (m+1)}-\frac {\int (e \cos (c+d x))^{-m-4} (\sin (c+d x) a+a)^{m+3}dx}{a (m+1)}\right )}{a (1-m)}-\frac {(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-3}}{d e (1-m)}\right )}{a (3-m)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {2 \left (\frac {(a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-3}}{d e (m+1)}-\frac {\int (e \cos (c+d x))^{-m-4} (\sin (c+d x) a+a)^{m+3}dx}{a (m+1)}\right )}{a (1-m)}-\frac {(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-3}}{d e (1-m)}\right )}{a (3-m)}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}\)

\(\Big \downarrow \) 3150

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3150

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/(a*f*g*m)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] 
 && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]
 

rule 3151

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/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl 
ify[2*m + p + 1])   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] && ILtQ[Simpli 
fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.52 \[ \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^m \, dx=-\frac {{\left (6 \, m \cos \left (d x + c\right )^{3} - {\left (m^{3} - m\right )} \cos \left (d x + c\right ) - 3 \, {\left (2 \, \cos \left (d x + c\right )^{3} - {\left (m^{2} - 1\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{4} - 10 \, d m^{2} + 9 \, d} \] Input:

integrate((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")
 

Output:

-(6*m*cos(d*x + c)^3 - (m^3 - m)*cos(d*x + c) - 3*(2*cos(d*x + c)^3 - (m^2 
 - 1)*cos(d*x + c))*sin(d*x + c))*(e*cos(d*x + c))^(-m - 4)*(a*sin(d*x + c 
) + a)^m/(d*m^4 - 10*d*m^2 + 9*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 16.74 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.68 \[ \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^m \, dx=\frac {2\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (12\,\sin \left (2\,c+2\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )-22\,m\,\cos \left (c+d\,x\right )-6\,m\,\cos \left (3\,c+3\,d\,x\right )+4\,m^3\,\cos \left (c+d\,x\right )-6\,m^2\,\sin \left (2\,c+2\,d\,x\right )\right )}{d\,e^4\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\left (4\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+3\right )\,\left (m^4-10\,m^2+9\right )} \] Input:

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

Output:

(2*(a*(sin(c + d*x) + 1))^m*(12*sin(2*c + 2*d*x) + 3*sin(4*c + 4*d*x) - 22 
*m*cos(c + d*x) - 6*m*cos(3*c + 3*d*x) + 4*m^3*cos(c + d*x) - 6*m^2*sin(2* 
c + 2*d*x)))/(d*e^4*(e*cos(c + d*x))^m*(4*cos(2*c + 2*d*x) + cos(4*c + 4*d 
*x) + 3)*(m^4 - 10*m^2 + 9))
 

Reduce [F]

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

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

Output:

int((sin(c + d*x)*a + a)**m/(cos(c + d*x)**m*cos(c + d*x)**4),x)/(e**m*e** 
4)