\(\int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx\) [372]

Optimal result

Integrand size = 27, antiderivative size = 93 \[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-1-2 m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2} (-1-2 m),\frac {1}{2} (1-2 m),\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {1+\sin (c+d x)} (a+a \sin (c+d x))^m}{\sqrt {2} d e (1+2 m)} \] Output:

1/2*(e*cos(d*x+c))^(-1-2*m)*hypergeom([3/2, -1/2-m],[1/2-m],1/2-1/2*sin(d* 
x+c))*(1+sin(d*x+c))^(1/2)*(a+a*sin(d*x+c))^m*2^(1/2)/d/e/(1+2*m)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94 \[ \int (e \cos (c+d x))^{-2-2 m} (a+a \sin (c+d x))^m \, dx=\frac {(e \cos (c+d x))^{-1-2 m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {1}{2}-m,\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt {1+\sin (c+d x)} (a (1+\sin (c+d x)))^m}{\sqrt {2} e (d+2 d m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.37, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3168, 80, 79}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle -\frac {2^{-m-\frac {1}{2}} (1-\sin (c+d x))^{m+\frac {1}{2}} (a-a \sin (c+d x))^{-m+\frac {1}{2} (2 m+1)-\frac {1}{2}} (a \sin (c+d x)+a)^{\frac {1}{2} (2 m+1)-\frac {1}{2}} (e \cos (c+d x))^{-2 m-1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (2 m+3),\frac {1}{2},\frac {1}{2} (\sin (c+d x)+1)\right )}{d e}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((e*cos(d*x+c))**(-2-2*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 - 2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((e*cos(d*x+c))^(-2-2*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)**2),x)/(e**( 
2*m)*e**2)