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

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

[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 = 0.25, 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 = {3018, 2755, 139, 138, 2784} \[ \frac {4 \sqrt {2} A \sin (e+f x) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};-\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right )}{f \sqrt {\cos (e+f x)+1}}-\frac {4 \sqrt {2} A \sin (e+f x) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};-\frac {3}{2},-m;\frac {3}{2};\frac {1}{2} (1-\cos (e+f x)),\frac {b (1-\cos (e+f x))}{a+b}\right )}{f \sqrt {\cos (e+f x)+1}} \]

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 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])

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 2755

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*x)/c])/Sqrt
[1 - (d*x)/c], 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 2784

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*x)/a)^(m - 1/2)*(c
+ d*x)^n)/Sqrt[1 - (b*x)/a], 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 3018

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)) \operatorname {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)) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.37, size = 119, normalized size = 0.56 \[ \frac {4 A \sin (e+f x) \sqrt {\cos ^2\left (\frac {1}{2} (e+f x)\right )} \tan ^2\left (\frac {1}{2} (e+f x)\right ) (a+b \cos (e+f x))^m \left (\frac {a+b \cos (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {3}{2};-\frac {1}{2},-m;\frac {5}{2};\sin ^2\left (\frac {1}{2} (e+f x)\right ),\frac {2 b \sin ^2\left (\frac {1}{2} (e+f x)\right )}{a+b}\right )}{3 f} \]

Warning: Unable to verify antiderivative.

[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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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (A \cos \left (f x + e\right )^{2} - A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

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

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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maple [F]  time = 1.27, 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 \[ -\int {\left (A \cos \left (f x + e\right )^{2} - A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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