\(\int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx\) [141]

Optimal result

Integrand size = 19, antiderivative size = 95 \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx=-\frac {13^{n/2} \cos ^{1+n}\left (c+d x-\arctan \left (\frac {3}{2}\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\cos ^2\left (c+d x-\arctan \left (\frac {3}{2}\right )\right )\right ) \sin \left (c+d x-\arctan \left (\frac {3}{2}\right )\right )}{d (1+n) \sqrt {\sin ^2\left (c+d x-\arctan \left (\frac {3}{2}\right )\right )}} \] Output:

-13^(1/2*n)*cos(c+d*x-arctan(3/2))^(1+n)*hypergeom([1/2, 1/2+1/2*n],[3/2+1 
/2*n],cos(c+d*x-arctan(3/2))^2)*sin(c+d*x-arctan(3/2))/d/(1+n)/(sin(c+d*x- 
arctan(3/2))^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3}{2},\cos ^2\left (c+d x+\arctan \left (\frac {2}{3}\right )\right )\right ) (2 \cos (c+d x)+3 \sin (c+d x))^n \sin ^2\left (c+d x+\arctan \left (\frac {2}{3}\right )\right )^{-\frac {1}{2}-\frac {n}{2}} \sin \left (2 \left (c+d x+\arctan \left (\frac {2}{3}\right )\right )\right )}{2 d} \] Input:

Integrate[(2*Cos[c + d*x] + 3*Sin[c + d*x])^n,x]
 

Output:

-1/2*(Hypergeometric2F1[1/2, (1 - n)/2, 3/2, Cos[c + d*x + ArcTan[2/3]]^2] 
*(2*Cos[c + d*x] + 3*Sin[c + d*x])^n*(Sin[c + d*x + ArcTan[2/3]]^2)^(-1/2 
- n/2)*Sin[2*(c + d*x + ArcTan[2/3])])/d
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3556, 3042, 3122}

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.

\(\displaystyle \int (3 \sin (c+d x)+2 \cos (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (3 \sin (c+d x)+2 \cos (c+d x))^ndx\)

\(\Big \downarrow \) 3556

\(\displaystyle 13^{n/2} \int \cos ^n\left (c+d x-\arctan \left (\frac {3}{2}\right )\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 13^{n/2} \int \sin \left (c+d x-\arctan \left (\frac {3}{2}\right )+\frac {\pi }{2}\right )^ndx\)

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {13^{n/2} \sin \left (-\arctan \left (\frac {3}{2}\right )+c+d x\right ) \cos ^{n+1}\left (-\arctan \left (\frac {3}{2}\right )+c+d x\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\cos ^2\left (c+d x-\arctan \left (\frac {3}{2}\right )\right )\right )}{d (n+1) \sqrt {\sin ^2\left (-\arctan \left (\frac {3}{2}\right )+c+d x\right )}}\)

Input:

Int[(2*Cos[c + d*x] + 3*Sin[c + d*x])^n,x]
 

Output:

-((13^(n/2)*Cos[c + d*x - ArcTan[3/2]]^(1 + n)*Hypergeometric2F1[1/2, (1 + 
 n)/2, (3 + n)/2, Cos[c + d*x - ArcTan[3/2]]^2]*Sin[c + d*x - ArcTan[3/2]] 
)/(d*(1 + n)*Sqrt[Sin[c + d*x - ArcTan[3/2]]^2]))
 

Defintions of rubi rules used

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3556

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x 
_Symbol] :> Simp[(a^2 + b^2)^(n/2)   Int[Cos[c + d*x - ArcTan[a, b]]^n, x], 
 x] /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) && GtQ[a^2 
+ b^2, 0]
 

Maple [F]

\[\int \left (2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )\right )^{n}d x\]

Input:

int((2*cos(d*x+c)+3*sin(d*x+c))^n,x)
 

Output:

int((2*cos(d*x+c)+3*sin(d*x+c))^n,x)
 

Fricas [F]

\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx=\int { {\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{n} \,d x } \] Input:

integrate((2*cos(d*x+c)+3*sin(d*x+c))^n,x, algorithm="fricas")
 

Output:

integral((2*cos(d*x + c) + 3*sin(d*x + c))^n, x)
 

Sympy [F]

\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx=\int \left (3 \sin {\left (c + d x \right )} + 2 \cos {\left (c + d x \right )}\right )^{n}\, dx \] Input:

integrate((2*cos(d*x+c)+3*sin(d*x+c))**n,x)
 

Output:

Integral((3*sin(c + d*x) + 2*cos(c + d*x))**n, x)
 

Maxima [F]

\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx=\int { {\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{n} \,d x } \] Input:

integrate((2*cos(d*x+c)+3*sin(d*x+c))^n,x, algorithm="maxima")
 

Output:

integrate((2*cos(d*x + c) + 3*sin(d*x + c))^n, x)
 

Giac [F]

\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx=\int { {\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{n} \,d x } \] Input:

integrate((2*cos(d*x+c)+3*sin(d*x+c))^n,x, algorithm="giac")
 

Output:

integrate((2*cos(d*x + c) + 3*sin(d*x + c))^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx=\int {\left (2\,\cos \left (c+d\,x\right )+3\,\sin \left (c+d\,x\right )\right )}^n \,d x \] Input:

int((2*cos(c + d*x) + 3*sin(c + d*x))^n,x)
 

Output:

int((2*cos(c + d*x) + 3*sin(c + d*x))^n, x)
 

Reduce [F]

\[ \int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx=\frac {-3 \left (2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )\right )^{n}+13 \left (\int \frac {\left (2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )\right )^{n} \cos \left (d x +c \right )}{2 \cos \left (d x +c \right )+3 \sin \left (d x +c \right )}d x \right ) d n}{2 d n} \] Input:

int((2*cos(d*x+c)+3*sin(d*x+c))^n,x)
 

Output:

( - 3*(2*cos(c + d*x) + 3*sin(c + d*x))**n + 13*int(((2*cos(c + d*x) + 3*s 
in(c + d*x))**n*cos(c + d*x))/(2*cos(c + d*x) + 3*sin(c + d*x)),x)*d*n)/(2 
*d*n)