3.3.7 \(\int (a+b \cos (e+f x))^m (A-A \cos ^2(e+f x)) \, dx\) [207]

Optimal. Leaf size=211 \[ -\frac {4 \sqrt {2} A F_1\left (\frac {1}{2};-\frac {3}{2},-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{f \sqrt {1+\cos (e+f x)}}+\frac {4 \sqrt {2} A F_1\left (\frac {1}{2};-\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x)}{f \sqrt {1+\cos (e+f x)}} \]

[Out]

-4*A*AppellF1(1/2,-m,-3/2,3/2,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)
/f/(((a+b*cos(f*x+e))/(a+b))^m)/(1+cos(f*x+e))^(1/2)+4*A*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)/f/(((a+b*cos(f*x+e))/(a+b))^m)/(1+cos(f*x+e))^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3097, 2834, 144, 143, 2863} \begin {gather*} \frac {4 \sqrt {2} A \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;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right )}{f \sqrt {\cos (e+f x)+1}}-\frac {4 \sqrt {2} A \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 {3}{2},-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right )}{f \sqrt {\cos (e+f x)+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[2]*A*AppellF1[1/2, -3/2, -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])/(f*Sqrt[1 + Cos[e + f*x]]*((a + b*Cos[e + f*x])/(a + b))^m) + (4*Sqrt[2]*A*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])/(
f*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 2834

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

Rule 2863

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

Rule 3097

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A - C, Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e + f*x]), x], x] + Dist[C, Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e +
f*x])^2, x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A + C, 0] &&  !IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+b \cos (e+f x))^m \left (A-A \cos ^2(e+f x)\right ) \, dx &=-\left (A \int (1+\cos (e+f x))^2 (a+b \cos (e+f x))^m \, dx\right )+(2 A) \int (1+\cos (e+f x)) (a+b \cos (e+f x))^m \, dx\\ &=\frac {(A \sin (e+f x)) \text {Subst}\left (\int \frac {(1+x)^{3/2} (a+b x)^m}{\sqrt {1-x}} \, dx,x,\cos (e+f x)\right )}{f \sqrt {1-\cos (e+f x)} \sqrt {1+\cos (e+f x)}}-\frac {(2 A \sin (e+f x)) \text {Subst}\left (\int \frac {\sqrt {1+x} (a+b x)^m}{\sqrt {1-x}} \, dx,x,\cos (e+f x)\right )}{f \sqrt {1-\cos (e+f x)} \sqrt {1+\cos (e+f x)}}\\ &=\frac {\left (A (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 {(1+x)^{3/2} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x}} \, dx,x,\cos (e+f x)\right )}{f \sqrt {1-\cos (e+f x)} \sqrt {1+\cos (e+f x)}}-\frac {\left (2 A (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 {\sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x}} \, dx,x,\cos (e+f x)\right )}{f \sqrt {1-\cos (e+f x)} \sqrt {1+\cos (e+f x)}}\\ &=-\frac {4 \sqrt {2} A F_1\left (\frac {1}{2};-\frac {3}{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)}{f \sqrt {1+\cos (e+f x)}}+\frac {4 \sqrt {2} A 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)}{f \sqrt {1+\cos (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 119, normalized size = 0.56 \begin {gather*} \frac {4 A F_1\left (\frac {3}{2};-\frac {1}{2},-m;\frac {5}{2};\sin ^2\left (\frac {1}{2} (e+f x)\right ),\frac {2 b \sin ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right ) \sqrt {\cos ^2\left (\frac {1}{2} (e+f x)\right )} (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} \sin (e+f x) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*A*AppellF1[3/2, -1/2, -m, 5/2, Sin[(e + f*x)/2]^2, (2*b*Sin[(e + f*x)/2]^2)/(a + b)]*Sqrt[Cos[(e + f*x)/2]^
2]*(a + b*Cos[e + f*x])^m*Sin[e + f*x]*Tan[(e + f*x)/2]^2)/(3*f*((a + b*Cos[e + f*x])/(a + b))^m)

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Maple [F]
time = 0.21, size = 0, normalized size = 0.00 \[\int \left (a +b \cos \left (f x +e \right )\right )^{m} \left (A -A \left (\cos ^{2}\left (f x +e \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(-(A*cos(f*x + e)^2 - 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-A*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-A*cos(f*x+e)^2),x, algorithm="giac")

[Out]

integrate(-(A*cos(f*x + e)^2 - 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-A\,{\cos \left (e+f\,x\right )}^2\right )\,{\left (a+b\,\cos \left (e+f\,x\right )\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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