3.4.93 \(\int (a+b \cos (e+f x))^m (A+B \cos (e+f x)+C \cos ^2(e+f x)) \, dx\) [393]

Optimal. Leaf size=303 \[ \frac {C (a+b \cos (e+f x))^{1+m} \sin (e+f x)}{b f (2+m)}-\frac {\sqrt {2} (a+b) (a C-b B (2+m)) F_1\left (\frac {1}{2};\frac {1}{2},-1-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{b^2 f (2+m) \sqrt {1+\cos (e+f x)}}+\frac {\sqrt {2} \left (a^2 C+b^2 C (1+m)+A b^2 (2+m)-a b B (2+m)\right ) F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{b^2 f (2+m) \sqrt {1+\cos (e+f x)}} \]

[Out]

C*(a+b*cos(f*x+e))^(1+m)*sin(f*x+e)/b/f/(2+m)-(a+b)*(a*C-b*B*(2+m))*AppellF1(1/2,-1-m,1/2,3/2,b*(1-cos(f*x+e))
/(a+b),1/2-1/2*cos(f*x+e))*(a+b*cos(f*x+e))^m*sin(f*x+e)*2^(1/2)/b^2/f/(2+m)/(((a+b*cos(f*x+e))/(a+b))^m)/(1+c
os(f*x+e))^(1/2)+(a^2*C+b^2*C*(1+m)+A*b^2*(2+m)-a*b*B*(2+m))*AppellF1(1/2,-m,1/2,3/2,b*(1-cos(f*x+e))/(a+b),1/
2-1/2*cos(f*x+e))*(a+b*cos(f*x+e))^m*sin(f*x+e)*2^(1/2)/b^2/f/(2+m)/(((a+b*cos(f*x+e))/(a+b))^m)/(1+cos(f*x+e)
)^(1/2)

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Rubi [A]
time = 0.25, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3102, 2835, 2744, 144, 143} \begin {gather*} \frac {\sqrt {2} \sin (e+f x) \left (a^2 C-a b B (m+2)+A b^2 (m+2)+b^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+b) \sin (e+f x) (a C-b B (m+2)) (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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[e + f*x])^m*(A + B*Cos[e + f*x] + 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 + b)*(a*C - b*B*(2 + m))*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*b^2*(2 + m) - a*b*B*(2 + m))*AppellF1[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 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 144

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 2744

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 2835

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 3102

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+b \cos (e+f x))^m \left (A+B \cos (e+f x)+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-b B (2+m)) \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+b B (2+m)) \int (a+b \cos (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac {\left (b^2 (C (1+m)+A (2+m))-a (-a C+b B (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+b B (2+m)) \sin (e+f x)) \text {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 (b^2 (C (1+m)+A (2+m))-a (-a C+b B (2+m))\right ) \sin (e+f x)\right ) \text {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-b) (-a C+b B (2+m)) (a+b \cos (e+f x))^m \left (-\frac {a+b \cos (e+f x)}{-a-b}\right )^{-m} \sin (e+f x)\right ) \text {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 (b^2 (C (1+m)+A (2+m))-a (-a C+b B (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 ) \text {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+b) (a C-b B (2+m)) 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 b^2 (2+m)-a b B (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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(16142\) vs. \(2(303)=606\).
time = 27.18, size = 16142, normalized size = 53.27 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \left (a +b \cos \left (f x +e \right )\right )^{m} \left (A +B \cos \left (f x +e \right )+C \left (\cos ^{2}\left (f x +e \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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