\(\int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx\) [368]

Optimal result

Integrand size = 27, antiderivative size = 70 \[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-2 (1+m)} \operatorname {Hypergeometric2F1}\left (2,-1-m,-m,\frac {1}{2} (1-\sin (c+d x))\right ) (a+a \sin (c+d x))^{1+m}}{4 a d e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.09 \[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-2 m} \operatorname {Hypergeometric2F1}\left (2,-1-m,-m,\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^2(c+d x) (a (1+\sin (c+d x)))^{1+m}}{4 a d e^3 (1+m)} \] Input:

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

Output:

(Hypergeometric2F1[2, -1 - m, -m, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^2*(a* 
(1 + Sin[c + d*x]))^(1 + m))/(4*a*d*e^3*(1 + m)*(e*Cos[c + d*x])^(2*m))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3168, 27, 78}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 78

\(\displaystyle \frac {(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-2 (m+1)} \operatorname {Hypergeometric2F1}\left (2,-m-1,-m,\frac {1}{2} (1-\sin (c+d x))\right )}{4 a d e (m+1)}\)

Input:

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

Output:

(Hypergeometric2F1[2, -1 - m, -m, (1 - Sin[c + d*x])/2]*(a + a*Sin[c + d*x 
])^(1 + m))/(4*a*d*e*(1 + m)*(e*Cos[c + d*x])^(2*(1 + m)))
 

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 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 3042

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

rule 3168

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin 
[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)))   Subst[Int[(a + 
b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre 
eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]
 

Maple [F]

\[\int \left (e \cos \left (d x +c \right )\right )^{-3-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]

Input:

int((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x)
 

Output:

int((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x)
 

Fricas [F]

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

integrate((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas" 
)
 

Output:

integral((e*cos(d*x + c))^(-2*m - 3)*(a*sin(d*x + c) + a)^m, x)
 

Sympy [F]

\[ \int (e \cos (c+d x))^{-3-2 m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{- 2 m - 3}\, dx \] Input:

integrate((e*cos(d*x+c))**(-3-2*m)*(a+a*sin(d*x+c))**m,x)
 

Output:

Integral((a*(sin(c + d*x) + 1))**m*(e*cos(c + d*x))**(-2*m - 3), x)
 

Maxima [F]

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

integrate((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima" 
)
 

Output:

integrate((e*cos(d*x + c))^(-2*m - 3)*(a*sin(d*x + c) + a)^m, x)
 

Giac [F]

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

integrate((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")
 

Output:

integrate((e*cos(d*x + c))^(-2*m - 3)*(a*sin(d*x + c) + a)^m, x)
 

Mupad [F(-1)]

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

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(2*m + 3),x)
 

Output:

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(2*m + 3), x)
 

Reduce [F]

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

int((e*cos(d*x+c))^(-3-2*m)*(a+a*sin(d*x+c))^m,x)
 

Output:

int((sin(c + d*x)*a + a)**m/(cos(c + d*x)**(2*m)*cos(c + d*x)**3),x)/(e**( 
2*m)*e**3)