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

Optimal. Leaf size=285 \[ \frac{\sqrt{2} \sin (e+f x) \left (a^2 C+b^2 (A (m+2)+C (m+1))\right ) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\cos (e+f x)+1}}-\frac{\sqrt{2} a C (a+b) \sin (e+f x) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\cos (e+f x)+1}}+\frac{C \sin (e+f x) (a+b \cos (e+f x))^{m+1}}{b f (m+2)} \]

[Out]

(C*(a + b*Cos[e + f*x])^(1 + m)*Sin[e + f*x])/(b*f*(2 + m)) - (Sqrt[2]*a*(a + b)*C*AppellF1[1/2, 1/2, -1 - m,
3/2, (1 - Cos[e + f*x])/2, (b*(1 - Cos[e + f*x]))/(a + b)]*(a + b*Cos[e + f*x])^m*Sin[e + f*x])/(b^2*f*(2 + m)
*Sqrt[1 + Cos[e + f*x]]*((a + b*Cos[e + f*x])/(a + b))^m) + (Sqrt[2]*(a^2*C + b^2*(C*(1 + m) + A*(2 + m)))*App
ellF1[1/2, 1/2, -m, 3/2, (1 - Cos[e + f*x])/2, (b*(1 - Cos[e + f*x]))/(a + b)]*(a + b*Cos[e + f*x])^m*Sin[e +
f*x])/(b^2*f*(2 + m)*Sqrt[1 + Cos[e + f*x]]*((a + b*Cos[e + f*x])/(a + b))^m)

________________________________________________________________________________________

Rubi [A]  time = 0.342942, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3024, 2756, 2665, 139, 138} \[ \frac{\sqrt{2} \sin (e+f x) \left (a^2 C+b^2 (A (m+2)+C (m+1))\right ) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\cos (e+f x)+1}}-\frac{\sqrt{2} a C (a+b) \sin (e+f x) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m-1;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt{\cos (e+f x)+1}}+\frac{C \sin (e+f x) (a+b \cos (e+f x))^{m+1}}{b f (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(a + b*Cos[e + f*x])^(1 + m)*Sin[e + f*x])/(b*f*(2 + m)) - (Sqrt[2]*a*(a + b)*C*AppellF1[1/2, 1/2, -1 - m,
3/2, (1 - Cos[e + f*x])/2, (b*(1 - Cos[e + f*x]))/(a + b)]*(a + b*Cos[e + f*x])^m*Sin[e + f*x])/(b^2*f*(2 + m)
*Sqrt[1 + Cos[e + f*x]]*((a + b*Cos[e + f*x])/(a + b))^m) + (Sqrt[2]*(a^2*C + b^2*(C*(1 + m) + A*(2 + m)))*App
ellF1[1/2, 1/2, -m, 3/2, (1 - Cos[e + f*x])/2, (b*(1 - Cos[e + f*x]))/(a + b)]*(a + b*Cos[e + f*x])^m*Sin[e +
f*x])/(b^2*f*(2 + m)*Sqrt[1 + Cos[e + f*x]]*((a + b*Cos[e + f*x])/(a + b))^m)

Rule 3024

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] + Dist[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]

Rule 2756

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

Rule 2665

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

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rubi steps

\begin{align*} \int (a+b \cos (e+f x))^m \left (A+C \cos ^2(e+f x)\right ) \, dx &=\frac{C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}+\frac{\int (a+b \cos (e+f x))^m (b (C (1+m)+A (2+m))-a C \cos (e+f x)) \, dx}{b (2+m)}\\ &=\frac{C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}-\frac{(a C) \int (a+b \cos (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac{\left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) \int (a+b \cos (e+f x))^m \, dx}{b^2 (2+m)}\\ &=\frac{C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}+\frac{(a C \sin (e+f x)) \operatorname{Subst}\left (\int \frac{(a+b x)^{1+m}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\cos (e+f x)\right )}{b^2 f (2+m) \sqrt{1-\cos (e+f x)} \sqrt{1+\cos (e+f x)}}-\frac{\left (\left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) \sin (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\cos (e+f x)\right )}{b^2 f (2+m) \sqrt{1-\cos (e+f x)} \sqrt{1+\cos (e+f x)}}\\ &=\frac{C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}-\frac{\left (a (-a-b) C (a+b \cos (e+f x))^m \left (-\frac{a+b \cos (e+f x)}{-a-b}\right )^{-m} \sin (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^{1+m}}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\cos (e+f x)\right )}{b^2 f (2+m) \sqrt{1-\cos (e+f x)} \sqrt{1+\cos (e+f x)}}-\frac{\left (\left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) (a+b \cos (e+f x))^m \left (-\frac{a+b \cos (e+f x)}{-a-b}\right )^{-m} \sin (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^m}{\sqrt{1-x} \sqrt{1+x}} \, dx,x,\cos (e+f x)\right )}{b^2 f (2+m) \sqrt{1-\cos (e+f x)} \sqrt{1+\cos (e+f x)}}\\ &=\frac{C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}-\frac{\sqrt{2} a (a+b) C F_1\left (\frac{1}{2};\frac{1}{2},-1-m;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{b^2 f (2+m) \sqrt{1+\cos (e+f x)}}+\frac{\sqrt{2} \left (a^2 C+b^2 (C (1+m)+A (2+m))\right ) F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\cos (e+f x)),\frac{b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac{a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{b^2 f (2+m) \sqrt{1+\cos (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 26.1871, size = 10836, normalized size = 38.02 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

________________________________________________________________________________________

Maple [F]  time = 1.402, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\cos \left ( fx+e \right ) \right ) ^{m} \left ( A+C \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (f x + e\right )^{2} + A\right )}{\left (b \cos \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (f x + e\right )^{2} + A\right )}{\left (b \cos \left (f x + e\right ) + a\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (f x + e\right )^{2} + A\right )}{\left (b \cos \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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