\(\int (a \cos (c+d x)+b \sin (c+d x))^n \, dx\) [140]

Optimal result

Integrand size = 19, antiderivative size = 136 \[ \int (a \cos (c+d x)+b \sin (c+d x))^n \, dx=-\frac {\cos ^{1+n}\left (c+d x-\tan ^{-1}(a,b)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\cos ^2\left (c+d x-\tan ^{-1}(a,b)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^n \left (\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}\right )^{-n} \sin \left (c+d x-\tan ^{-1}(a,b)\right )}{d (1+n) \sqrt {\sin ^2\left (c+d x-\tan ^{-1}(a,b)\right )}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.69 \[ \int (a \cos (c+d x)+b \sin (c+d x))^n \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3}{2},\cos ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^n \sin ^2\left (c+d x+\arctan \left (\frac {a}{b}\right )\right )^{-\frac {1}{2}-\frac {n}{2}} \sin \left (2 \left (c+d x+\arctan \left (\frac {a}{b}\right )\right )\right )}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 136, 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, 3557, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3557

\(\displaystyle (a \cos (c+d x)+b \sin (c+d x))^n \left (\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}\right )^{-n} \int \cos ^n\left (c+d x-\tan ^{-1}(a,b)\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (a \cos (c+d x)+b \sin (c+d x))^n \left (\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}\right )^{-n} \int \sin \left (c+d x-\tan ^{-1}(a,b)+\frac {\pi }{2}\right )^ndx\)

\(\Big \downarrow \) 3122

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

Input:

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

Output:

-((Cos[c + d*x - ArcTan[a, b]]^(1 + n)*Hypergeometric2F1[1/2, (1 + n)/2, ( 
3 + n)/2, Cos[c + d*x - ArcTan[a, b]]^2]*(a*Cos[c + d*x] + b*Sin[c + d*x]) 
^n*Sin[c + d*x - ArcTan[a, b]])/(d*(1 + n)*((a*Cos[c + d*x] + b*Sin[c + d* 
x])/Sqrt[a^2 + b^2])^n*Sqrt[Sin[c + d*x - ArcTan[a, b]]^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 3557

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x 
_Symbol] :> Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*S 
in[c + d*x])/Sqrt[a^2 + b^2])^n   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] || EqQ[a^2 + b^2, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((a*cos(c + d*x) + b*sin(c + d*x))**n, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - (cos(c + d*x)*a + sin(c + d*x)*b)**n*b + int(((cos(c + d*x)*a + sin(c 
+ d*x)*b)**n*cos(c + d*x))/(cos(c + d*x)*a + sin(c + d*x)*b),x)*a**2*d*n + 
 int(((cos(c + d*x)*a + sin(c + d*x)*b)**n*cos(c + d*x))/(cos(c + d*x)*a + 
 sin(c + d*x)*b),x)*b**2*d*n)/(a*d*n)