3.7.49 \(\int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx\) [649]

3.7.49.1 Optimal result

Integrand size = 27, antiderivative size = 201 \[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=-\frac {(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (1+m)}+\frac {2^{\frac {1}{2}-\frac {m}{2}} a (e \cos (c+d x))^{-1-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},\frac {(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m)} \]

output

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

3.7.49.2 Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.84 \[ \int (e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^m \, dx=-\frac {2^{\frac {1}{2} (-1-m)} (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^{1+m} \left (2^{\frac {1+m}{2}} (a+b)-2 a \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1-m),\frac {1+m}{2},\frac {1-m}{2},-\frac {(a-b) (-1+\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) \left (\frac {(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac {1+m}{2}}\right )}{(a-b) (a+b) d e (1+m)} \]

input

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

output

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

3.7.49.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3178, 3042, 3399, 142}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3178

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3399

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

\(\Big \downarrow \) 142

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

input

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

output

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

3.7.49.3.1 Defintions of rubi rules used

rule 142

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

rule 3042

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

rule 3178

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

rule 3399

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

3.7.49.4 Maple [F]

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

input

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

output

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

3.7.49.5 Fricas [F]

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

input

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

output

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

3.7.49.6 Sympy [F]

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

input

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

output

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

3.7.49.7 Maxima [F]

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

input

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

output

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

3.7.49.8 Giac [F]

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

input

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

output

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

3.7.49.9 Mupad [F(-1)]

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

input

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

output

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