\(\int (a+a \cos (e+f x))^m (A+C \cos ^2(e+f x)) \, dx\) [198]

Optimal result

Integrand size = 25, antiderivative size = 170 \[ \int (a+a \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=-\frac {C (a+a \cos (e+f x))^m \sin (e+f x)}{f \left (2+3 m+m^2\right )}+\frac {C (a+a \cos (e+f x))^{1+m} \sin (e+f x)}{a f (2+m)}+\frac {2^{\frac {1}{2}+m} \left (C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) (1+\cos (e+f x))^{-\frac {1}{2}-m} (a+a \cos (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\cos (e+f x))\right ) \sin (e+f x)}{f (1+m) (2+m)} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.98 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.42 \[ \int (a+a \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx=\frac {i 4^{-1-m} e^{-i (2+m) (e+f x)} \left (1+e^{i (e+f x)}\right ) \left (e^{-\frac {1}{2} i (e+f x)} \left (1+e^{i (e+f x)}\right )\right )^{2 m} \cos ^{-2 m}\left (\frac {1}{2} (e+f x)\right ) (a (1+\cos (e+f x)))^m \left (C e^{i m (e+f x)} (-2+m) m \operatorname {Hypergeometric2F1}\left (1,-1+m,-1-m,-e^{i (e+f x)}\right )+e^{i (2+m) (e+f x)} (2+m) \left (2 (2 A+C) (-2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,1-m,-e^{i (e+f x)}\right )+C e^{2 i (e+f x)} m \operatorname {Hypergeometric2F1}\left (1,3+m,3-m,-e^{i (e+f x)}\right )\right )\right )}{f (-2+m) m (2+m)} \] Input:

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

Output:

(I*4^(-1 - m)*(1 + E^(I*(e + f*x)))*((1 + E^(I*(e + f*x)))/E^((I/2)*(e + f 
*x)))^(2*m)*(a*(1 + Cos[e + f*x]))^m*(C*E^(I*m*(e + f*x))*(-2 + m)*m*Hyper 
geometric2F1[1, -1 + m, -1 - m, -E^(I*(e + f*x))] + E^(I*(2 + m)*(e + f*x) 
)*(2 + m)*(2*(2*A + C)*(-2 + m)*Hypergeometric2F1[1, 1 + m, 1 - m, -E^(I*( 
e + f*x))] + C*E^((2*I)*(e + f*x))*m*Hypergeometric2F1[1, 3 + m, 3 - m, -E 
^(I*(e + f*x))])))/(E^(I*(2 + m)*(e + f*x))*f*(-2 + m)*m*(2 + m)*Cos[(e + 
f*x)/2]^(2*m))
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3503, 3042, 3230, 3042, 3131, 3042, 3130}

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.

\(\displaystyle \int (a \cos (e+f x)+a)^m \left (A+C \cos ^2(e+f x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a \sin \left (e+f x+\frac {\pi }{2}\right )+a\right )^m \left (A+C \sin \left (e+f x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 3503

\(\displaystyle \frac {\int (\cos (e+f x) a+a)^m (a (C (m+1)+A (m+2))-a C \cos (e+f x))dx}{a (m+2)}+\frac {C \sin (e+f x) (a \cos (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (\sin \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^m \left (a (C (m+1)+A (m+2))-a C \sin \left (e+f x+\frac {\pi }{2}\right )\right )dx}{a (m+2)}+\frac {C \sin (e+f x) (a \cos (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {a \left (A \left (m^2+3 m+2\right )+C \left (m^2+m+1\right )\right ) \int (\cos (e+f x) a+a)^mdx}{m+1}-\frac {a C \sin (e+f x) (a \cos (e+f x)+a)^m}{f (m+1)}}{a (m+2)}+\frac {C \sin (e+f x) (a \cos (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a \left (A \left (m^2+3 m+2\right )+C \left (m^2+m+1\right )\right ) \int \left (\sin \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^mdx}{m+1}-\frac {a C \sin (e+f x) (a \cos (e+f x)+a)^m}{f (m+1)}}{a (m+2)}+\frac {C \sin (e+f x) (a \cos (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {a \left (A \left (m^2+3 m+2\right )+C \left (m^2+m+1\right )\right ) (\cos (e+f x)+1)^{-m} (a \cos (e+f x)+a)^m \int (\cos (e+f x)+1)^mdx}{m+1}-\frac {a C \sin (e+f x) (a \cos (e+f x)+a)^m}{f (m+1)}}{a (m+2)}+\frac {C \sin (e+f x) (a \cos (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a \left (A \left (m^2+3 m+2\right )+C \left (m^2+m+1\right )\right ) (\cos (e+f x)+1)^{-m} (a \cos (e+f x)+a)^m \int \left (\sin \left (e+f x+\frac {\pi }{2}\right )+1\right )^mdx}{m+1}-\frac {a C \sin (e+f x) (a \cos (e+f x)+a)^m}{f (m+1)}}{a (m+2)}+\frac {C \sin (e+f x) (a \cos (e+f x)+a)^{m+1}}{a f (m+2)}\)

\(\Big \downarrow \) 3130

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3130

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 
 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 3230

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

rule 3503

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + 
 (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^ 
(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int[(a + b*Sin[e + f*x])^ 
m*Simp[A*b*(m + 2) + b*C*(m + 1) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a 
, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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