3.10.45 \(\int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx\) [945]

3.10.45.1 Optimal result

Integrand size = 33, antiderivative size = 135 \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\frac {4 \sqrt {2} \operatorname {AppellF1}\left (\frac {5}{2}+m,-\frac {3}{2},-n,\frac {7}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{2+m} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{a^2 f (5+2 m) \sqrt {1-\sin (e+f x)}} \]

output

4*AppellF1(5/2+m,-n,-3/2,7/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e)) 
*cos(f*x+e)*(a+a*sin(f*x+e))^(2+m)*(c+d*sin(f*x+e))^n*2^(1/2)/a^2/f/(5+2*m 
)/(((c+d*sin(f*x+e))/(c-d))^n)/(1-sin(f*x+e))^(1/2)
 

3.10.45.2 Mathematica [F]

\[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx \]

input

Integrate[Cos[e + f*x]^4*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n,x]
 

output

Integrate[Cos[e + f*x]^4*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x]
 

3.10.45.3 Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 3397, 157, 27, 156, 155}

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.

\(\displaystyle \int \cos ^4(e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (e+f x)^4 (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^ndx\)

\(\Big \downarrow \) 3397

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

\(\Big \downarrow \) 157

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle \frac {4 \sqrt {2} \cos (e+f x) (a \sin (e+f x)+a)^{m+2} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} \operatorname {AppellF1}\left (m+\frac {5}{2},-\frac {3}{2},-n,m+\frac {7}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a^2 f (2 m+5) \sqrt {1-\sin (e+f x)}}\)

input

Int[Cos[e + f*x]^4*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n,x]
 

output

(4*Sqrt[2]*AppellF1[5/2 + m, -3/2, -n, 7/2 + m, (1 + Sin[e + f*x])/2, -((d 
*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^(2 + m)*( 
c + d*Sin[e + f*x])^n)/(a^2*f*(5 + 2*m)*Sqrt[1 - Sin[e + f*x]]*((c + d*Sin 
[e + f*x])/(c - d))^n)
 

3.10.45.3.1 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 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

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

rule 157

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

rule 3042

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

rule 3397

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

3.10.45.4 Maple [F]

\[\int \left (\cos ^{4}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]

input

int(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)
 

output

int(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x)
 

3.10.45.5 Fricas [F]

\[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4} \,d x } \]

input

integrate(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x, algorithm= 
"fricas")
 

output

integral((a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n*cos(f*x + e)^4, x)
 

3.10.45.6 Sympy [F(-1)]

Timed out. \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\text {Timed out} \]

input

integrate(cos(f*x+e)**4*(a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**n,x)
 

output

Timed out
 

3.10.45.7 Maxima [F(-1)]

Timed out. \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\text {Timed out} \]

input

integrate(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x, algorithm= 
"maxima")
 

output

Timed out
 

3.10.45.8 Giac [F]

\[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \cos \left (f x + e\right )^{4} \,d x } \]

input

integrate(cos(f*x+e)^4*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n,x, algorithm= 
"giac")
 

output

integrate((a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n*cos(f*x + e)^4, x)
 

3.10.45.9 Mupad [F(-1)]

Timed out. \[ \int \cos ^4(e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx=\int {\cos \left (e+f\,x\right )}^4\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]

input

int(cos(e + f*x)^4*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^n,x)
 

output

int(cos(e + f*x)^4*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^n, x)