3.3.3 \(\int (a+b \cos (c+d x))^{2/3} (A+C \cos ^2(c+d x)) \, dx\) [203]

Optimal. Leaf size=277 \[ \frac {3 C (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b d}-\frac {3 a (a+b) C F_1\left (\frac {1}{2};\frac {1}{2},-\frac {5}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{4 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\left (3 a^2 C+b^2 (8 A+5 C)\right ) F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{4 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \]

[Out]

3/8*C*(a+b*cos(d*x+c))^(5/3)*sin(d*x+c)/b/d-3/8*a*(a+b)*C*AppellF1(1/2,-5/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2
-1/2*cos(d*x+c))*(a+b*cos(d*x+c))^(2/3)*sin(d*x+c)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(2/3)*2^(1/2)/(1+cos(d*x+c))
^(1/2)+1/8*(3*a^2*C+b^2*(8*A+5*C))*AppellF1(1/2,-2/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x+c))*(a+b*c
os(d*x+c))^(2/3)*sin(d*x+c)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(2/3)*2^(1/2)/(1+cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.24, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3103, 2835, 2744, 144, 143} \begin {gather*} \frac {\left (3 a^2 C+b^2 (8 A+5 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{4 \sqrt {2} b^2 d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}-\frac {3 a C (a+b) \sin (c+d x) (a+b \cos (c+d x))^{2/3} F_1\left (\frac {1}{2};\frac {1}{2},-\frac {5}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{4 \sqrt {2} b^2 d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^(2/3)*(A + C*Cos[c + d*x]^2),x]

[Out]

(3*C*(a + b*Cos[c + d*x])^(5/3)*Sin[c + d*x])/(8*b*d) - (3*a*(a + b)*C*AppellF1[1/2, 1/2, -5/3, 3/2, (1 - Cos[
c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(4*Sqrt[2]*b^2*d*Sqrt[1
+ Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3)) + ((3*a^2*C + b^2*(8*A + 5*C))*AppellF1[1/2, 1/2, -2/3,
3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(4*Sqrt[2]
*b^2*d*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3))

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 3103

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] &&  !Lt
Q[m, -1]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {3 C (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b d}+\frac {3 \int (a+b \cos (c+d x))^{2/3} \left (\frac {1}{3} b (8 A+5 C)-a C \cos (c+d x)\right ) \, dx}{8 b}\\ &=\frac {3 C (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b d}-\frac {(3 a C) \int (a+b \cos (c+d x))^{5/3} \, dx}{8 b^2}+\frac {1}{8} \left (8 A+\left (5+\frac {3 a^2}{b^2}\right ) C\right ) \int (a+b \cos (c+d x))^{2/3} \, dx\\ &=\frac {3 C (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b d}+\frac {(3 a C \sin (c+d x)) \text {Subst}\left (\int \frac {(a+b x)^{5/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{8 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}+\frac {\left (\left (-8 A-\left (5+\frac {3 a^2}{b^2}\right ) C\right ) \sin (c+d x)\right ) \text {Subst}\left (\int \frac {(a+b x)^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{8 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)}}\\ &=\frac {3 C (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b d}-\frac {\left (3 a (-a-b) C (a+b \cos (c+d x))^{2/3} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{5/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{8 b^2 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \left (-\frac {a+b \cos (c+d x)}{-a-b}\right )^{2/3}}+\frac {\left (\left (-8 A-\left (5+\frac {3 a^2}{b^2}\right ) C\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{2/3}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{8 d \sqrt {1-\cos (c+d x)} \sqrt {1+\cos (c+d x)} \left (-\frac {a+b \cos (c+d x)}{-a-b}\right )^{2/3}}\\ &=\frac {3 C (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b d}-\frac {3 a (a+b) C F_1\left (\frac {1}{2};\frac {1}{2},-\frac {5}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{4 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\left (8 A+\left (5+\frac {3 a^2}{b^2}\right ) C\right ) F_1\left (\frac {1}{2};\frac {1}{2},-\frac {2}{3};\frac {3}{2};\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{4 \sqrt {2} d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 2.78, size = 279, normalized size = 1.01 \begin {gather*} -\frac {3 (a+b \cos (c+d x))^{2/3} \csc (c+d x) \left (60 a \left (a^2-b^2\right ) C F_1\left (\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}}+4 \left (40 A b^2-6 a^2 C+25 b^2 C\right ) F_1\left (\frac {5}{3};\frac {1}{2},\frac {1}{2};\frac {8}{3};\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))-20 b^2 C (2 a+5 b \cos (c+d x)) \sin ^2(c+d x)\right )}{800 b^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^(2/3)*(A + C*Cos[c + d*x]^2),x]

[Out]

(-3*(a + b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(60*a*(a^2 - b^2)*C*AppellF1[2/3, 1/2, 1/2, 5/3, (a + b*Cos[c + d*
x])/(a - b), (a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Cos[c + d*x
]))/(a - b))] + 4*(40*A*b^2 - 6*a^2*C + 25*b^2*C)*AppellF1[5/3, 1/2, 1/2, 8/3, (a + b*Cos[c + d*x])/(a - b), (
a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*(a
+ b*Cos[c + d*x]) - 20*b^2*C*(2*a + 5*b*Cos[c + d*x])*Sin[c + d*x]^2))/(800*b^3*d)

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \left (a +b \cos \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^(2/3)*(A+C*cos(d*x+c)^2),x)

[Out]

int((a+b*cos(d*x+c))^(2/3)*(A+C*cos(d*x+c)^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(d*x+c))^(2/3)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(2/3), 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(d*x+c))^(2/3)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(2/3), 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(d*x+c))**(2/3)*(A+C*cos(d*x+c)**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(d*x+c))^(2/3)*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(2/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{2/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(2/3),x)

[Out]

int((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(2/3), x)

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