\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx\) [69]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 78 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {3}{2}+m} \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {3}{2}-m} (a+a \sin (e+f x))^m}{3 f} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {3}{2}+m} \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {3}{2}-m} (a (1+\sin (e+f x)))^m}{3 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((a*(sin(e + f*x) + 1))**m*cos(e + f*x)**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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