\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx\) [68]

Optimal result

Integrand size = 32, antiderivative size = 82 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=-\frac {2^{\frac {3}{2}+m} a c \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-\frac {1}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {3}{2}-m} (a+a \sin (e+f x))^{-1+m}}{5 f} \] Output:

-1/5*2^(3/2+m)*a*c*cos(f*x+e)^5*hypergeom([5/2, -1/2-m],[7/2],1/2-1/2*sin( 
f*x+e))*(1+sin(f*x+e))^(-3/2-m)*(a+a*sin(f*x+e))^(-1+m)/f
 

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.20 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x)) \, dx=\frac {c \cos ^3(e+f x) (1+\sin (e+f x))^{-\frac {3}{2}-m} (a (1+\sin (e+f x)))^m \left (-2^{\frac {3}{2}+m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right )+(1+\sin (e+f x))^{\frac {3}{2}+m}\right )}{f (3+m)} \] Input:

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

Output:

(c*Cos[e + f*x]^3*(1 + Sin[e + f*x])^(-3/2 - m)*(a*(1 + Sin[e + f*x]))^m*( 
-(2^(3/2 + m)*Hypergeometric2F1[3/2, -1/2 - m, 5/2, (1 - Sin[e + f*x])/2]) 
 + (1 + Sin[e + f*x])^(3/2 + m)))/(f*(3 + m))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3042, 3319, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (e+f x)^2 (c-c \sin (e+f x)) (a \sin (e+f x)+a)^mdx\)

\(\Big \downarrow \) 3319

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

\(\displaystyle \frac {a^3 c \cos ^5(e+f x) \int (a-a \sin (e+f x))^{3/2} (\sin (e+f x) a+a)^{m+\frac {1}{2}}d\sin (e+f x)}{f (a-a \sin (e+f x))^{5/2} (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {a^3 c 2^{m+\frac {1}{2}} \cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m-2} \int \left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m+\frac {1}{2}} (a-a \sin (e+f x))^{3/2}d\sin (e+f x)}{f (a-a \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {a^2 c 2^{m+\frac {3}{2}} \cos ^5(e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-m-\frac {1}{2},\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{5 f}\)

Input:

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

Output:

-1/5*(2^(3/2 + m)*a^2*c*Cos[e + f*x]^5*Hypergeometric2F1[5/2, -1/2 - m, 7/ 
2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x] 
)^(-2 + m))/f
 

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]
 

rule 3319

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*(c^m/g^(2*m))   Int[(g*Cos[e + f*x])^(2*m + p)*(c + d*Sin[e + f*x])^(n 
- m), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] 
&& EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && LtQ[n^2, m^2])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

-c*(Integral(-(a*sin(e + f*x) + a)**m*cos(e + f*x)**2, x) + Integral((a*si 
n(e + f*x) + a)**m*sin(e + f*x)*cos(e + f*x)**2, x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e)),x, algorithm="g 
iac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

c*( - int((sin(e + f*x)*a + a)**m*cos(e + f*x)**2*sin(e + f*x),x) + int((s 
in(e + f*x)*a + a)**m*cos(e + f*x)**2,x))