3.383 \(\int (a+a \cos (e+f x))^m (A+B \cos (e+f x)+C \cos ^2(e+f x)) \, dx\)

Optimal. Leaf size=183 \[ \frac {2^{m+\frac {1}{2}} \left (A \left (m^2+3 m+2\right )+B m (m+2)+C \left (m^2+m+1\right )\right ) \sin (e+f x) (\cos (e+f x)+1)^{-m-\frac {1}{2}} (a \cos (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x))\right )}{f (m+1) (m+2)}-\frac {(C-B (m+2)) \sin (e+f x) (a \cos (e+f x)+a)^m}{f (m+1) (m+2)}+\frac {C \sin (e+f x) (a \cos (e+f x)+a)^{m+1}}{a f (m+2)} \]

[Out]

-(C-B*(2+m))*(a+a*cos(f*x+e))^m*sin(f*x+e)/f/(1+m)/(2+m)+C*(a+a*cos(f*x+e))^(1+m)*sin(f*x+e)/a/f/(2+m)+2^(1/2+
m)*(B*m*(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/(m^2+3*m+2)

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Rubi [A]  time = 0.25, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3023, 2751, 2652, 2651} \[ \frac {2^{m+\frac {1}{2}} \left (A \left (m^2+3 m+2\right )+B m (m+2)+C \left (m^2+m+1\right )\right ) \sin (e+f x) (\cos (e+f x)+1)^{-m-\frac {1}{2}} (a \cos (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x))\right )}{f (m+1) (m+2)}-\frac {(C-B (m+2)) \sin (e+f x) (a \cos (e+f x)+a)^m}{f (m+1) (m+2)}+\frac {C \sin (e+f x) (a \cos (e+f x)+a)^{m+1}}{a f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((C - B*(2 + m))*(a + a*Cos[e + f*x])^m*Sin[e + f*x])/(f*(1 + m)*(2 + m))) + (C*(a + a*Cos[e + f*x])^(1 + m)
*Sin[e + f*x])/(a*f*(2 + m)) + (2^(1/2 + m)*(B*m*(2 + m) + 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)*(2 + m))

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2652

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

Rule 2751

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] + Dist[(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 3023

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

Rubi steps

\begin {align*} \int (a+a \cos (e+f x))^m \left (A+B \cos (e+f x)+C \cos ^2(e+f x)\right ) \, dx &=\frac {C (a+a \cos (e+f x))^{1+m} \sin (e+f x)}{a f (2+m)}+\frac {\int (a+a \cos (e+f x))^m (a (C (1+m)+A (2+m))-a (C-B (2+m)) \cos (e+f x)) \, dx}{a (2+m)}\\ &=-\frac {(C-B (2+m)) (a+a \cos (e+f x))^m \sin (e+f x)}{f (1+m) (2+m)}+\frac {C (a+a \cos (e+f x))^{1+m} \sin (e+f x)}{a f (2+m)}+\frac {\left (B m (2+m)+C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) \int (a+a \cos (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=-\frac {(C-B (2+m)) (a+a \cos (e+f x))^m \sin (e+f x)}{f (1+m) (2+m)}+\frac {C (a+a \cos (e+f x))^{1+m} \sin (e+f x)}{a f (2+m)}+\frac {\left (\left (B m (2+m)+C \left (1+m+m^2\right )+A \left (2+3 m+m^2\right )\right ) (1+\cos (e+f x))^{-m} (a+a \cos (e+f x))^m\right ) \int (1+\cos (e+f x))^m \, dx}{(1+m) (2+m)}\\ &=-\frac {(C-B (2+m)) (a+a \cos (e+f x))^m \sin (e+f x)}{f (1+m) (2+m)}+\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 (B m (2+m)+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 \, _2F_1\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)}\\ \end {align*}

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Mathematica [C]  time = 3.82, size = 557, normalized size = 3.04 \[ \frac {\cos ^{-2 m}\left (\frac {1}{2} (e+f x)\right ) (a (\cos (e+f x)+1))^m \left (\frac {i A 4^{1-m} \left (1+e^{i (e+f x)}\right ) \left (e^{-\frac {1}{2} i (e+f x)} \left (1+e^{i (e+f x)}\right )\right )^{2 m} \, _2F_1\left (1,m+1;1-m;-e^{i (e+f x)}\right )}{m}+\frac {2 i B e^{-i f x} (\cos (f x)+i \sin (f x)) \cos ^{2 m}\left (\frac {1}{2} (e+f x)\right ) \left (i \sin (e) e^{i f x}+\cos (e) e^{i f x}+1\right )^{-2 m} \left ((m-1) (\cos (e+f x)-i \sin (e+f x)) \, _2F_1\left (-m-1,-2 m;-m;-e^{i f x} (\cos (e)+i \sin (e))\right )+(m+1) (\cos (e+f x)+i \sin (e+f x)) \, _2F_1\left (1-m,-2 m;2-m;-e^{i f x} (\cos (e)+i \sin (e))\right )\right )}{m^2-1}+\frac {C e^{-2 i f x} \cos ^{2 m}\left (\frac {1}{2} (e+f x)\right ) \left (i \sin (e) e^{i f x}+\cos (e) e^{i f x}+1\right )^{-2 m} \left (i (m+2) e^{4 i f x} (\cos (2 e)+i \sin (2 e)) \, _2F_1\left (2-m,-2 m;3-m;-e^{i f x} (\cos (e)+i \sin (e))\right )+(m-2) (\sin (2 e)+i \cos (2 e)) \, _2F_1\left (-m-2,-2 m;-m-1;-e^{i f x} (\cos (e)+i \sin (e))\right )\right )}{m^2-4}+\frac {i C 2^{1-2 m} \left (1+e^{i (e+f x)}\right ) \left (e^{-\frac {1}{2} i (e+f x)} \left (1+e^{i (e+f x)}\right )\right )^{2 m} \, _2F_1\left (1,m+1;1-m;-e^{i (e+f x)}\right )}{m}\right )}{4 f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 2.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (f x + e\right )^{2} + B \cos \left (f x + e\right ) + A\right )} {\left (a \cos \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (f x + e\right )^{2} + B \cos \left (f x + e\right ) + A\right )} {\left (a \cos \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.66, size = 0, normalized size = 0.00 \[ \int \left (a +a \cos \left (f x +e \right )\right )^{m} \left (A +B \cos \left (f x +e \right )+C \left (\cos ^{2}\left (f x +e \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (f x + e\right )^{2} + B \cos \left (f x + e\right ) + A\right )} {\left (a \cos \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\cos \left (e+f\,x\right )\right )}^m\,\left (C\,{\cos \left (e+f\,x\right )}^2+B\,\cos \left (e+f\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\cos {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \cos {\left (e + f x \right )} + C \cos ^{2}{\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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