\(\int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m \, dx\) [358]

Optimal result

Integrand size = 27, antiderivative size = 142 \[ \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m \, dx=-\frac {(e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m}{d e (2-m)}+\frac {2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{1+m}}{a d e (2-m) m}-\frac {2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e m \left (4-m^2\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.54 \[ \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-m} \sec ^2(c+d x) (a (1+\sin (c+d x)))^m \left (-2+m^2-2 m \sin (c+d x)+2 \sin ^2(c+d x)\right )}{d e^3 (-2+m) m (2+m)} \] Input:

Integrate[(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^m,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3151

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3151

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3150

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

Input:

Int[(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^m,x]
 

Output:

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

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

Input:

int((e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^m,x)
 

Output:

int((e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^m,x)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.53 \[ \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m \, dx=\frac {{\left (m^{2} \cos \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{3} - 2 \, m \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} - 4 \, d m} \] Input:

integrate((e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")
 

Output:

(m^2*cos(d*x + c) - 2*cos(d*x + c)^3 - 2*m*cos(d*x + c)*sin(d*x + c))*(e*c 
os(d*x + c))^(-m - 3)*(a*sin(d*x + c) + a)^m/(d*m^3 - 4*d*m)
 

Sympy [F]

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

integrate((e*cos(d*x+c))**(-3-m)*(a+a*sin(d*x+c))**m,x)
 

Output:

Integral((a*(sin(c + d*x) + 1))**m*(e*cos(c + d*x))**(-m - 3), x)
 

Maxima [F]

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

integrate((e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")
 

Output:

integrate((e*cos(d*x + c))^(-m - 3)*(a*sin(d*x + c) + a)^m, x)
 

Giac [F]

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

integrate((e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")
 

Output:

integrate((e*cos(d*x + c))^(-m - 3)*(a*sin(d*x + c) + a)^m, x)
 

Mupad [B] (verification not implemented)

Time = 16.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.73 \[ \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m \, dx=-\frac {2\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (-2\,\cos \left (c+d\,x\right )\,m^2+2\,\sin \left (2\,c+2\,d\,x\right )\,m+3\,\cos \left (c+d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\right )}{d\,e^3\,m\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\left (m^2-4\right )\,\left (3\,\cos \left (c+d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\right )} \] Input:

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(m + 3),x)
 

Output:

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

Reduce [F]

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

int((e*cos(d*x+c))^(-3-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)**3),x)/(e**m*e** 
3)