\(\int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx\) [631]

Optimal result

Integrand size = 21, antiderivative size = 118 \[ \int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx=-\frac {4 \sqrt {2} \operatorname {AppellF1}\left (\frac {5}{2},-\frac {3}{2},-m,\frac {7}{2},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos ^3(c+d x) (1-\sin (c+d x)) (a+b \sin (c+d x))^m \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-m}}{5 d (1+\sin (c+d x))^{3/2}} \] Output:

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

Mathematica [F]

\[ \int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx=\int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3183, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^4 (a+b \sin (c+d x))^mdx\)

\(\Big \downarrow \) 3183

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

\(\Big \downarrow \) 155

\(\displaystyle \frac {\cos ^3(c+d x) (a+b \sin (c+d x))^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {3}{2},-\frac {3}{2},m+2,\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{3/2} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{3/2}}\)

Input:

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

Output:

(AppellF1[1 + m, -3/2, -3/2, 2 + m, (a + b*Sin[c + d*x])/(a - b), (a + b*S 
in[c + d*x])/(a + b)]*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^(1 + m))/(b*d*(1 
 + m)*(1 - (a + b*Sin[c + d*x])/(a - b))^(3/2)*(1 - (a + b*Sin[c + d*x])/( 
a + b))^(3/2))
 

Defintions of rubi rules used

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 3042

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

rule 3183

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

Maple [F]

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

Input:

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^m,x)
 

Output:

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^m,x)
 

Fricas [F]

\[ \int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{4} \,d x } \] Input:

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

Output:

integral((b*sin(d*x + c) + a)^m*cos(d*x + c)^4, x)
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)**4*(a+b*sin(d*x+c))**m,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{4} \,d x } \] Input:

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

Output:

integrate((b*sin(d*x + c) + a)^m*cos(d*x + c)^4, x)
 

Giac [F]

\[ \int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )^{4} \,d x } \] Input:

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

Output:

integrate((b*sin(d*x + c) + a)^m*cos(d*x + c)^4, x)
 

Mupad [F(-1)]

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

int(cos(c + d*x)^4*(a + b*sin(c + d*x))^m,x)
 

Output:

int(cos(c + d*x)^4*(a + b*sin(c + d*x))^m, x)
 

Reduce [F]

\[ \int \cos ^4(c+d x) (a+b \sin (c+d x))^m \, dx=\int \left (\sin \left (d x +c \right ) b +a \right )^{m} \cos \left (d x +c \right )^{4}d x \] Input:

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^m,x)
 

Output:

int((sin(c + d*x)*b + a)**m*cos(c + d*x)**4,x)