3.384 \(\int (a+a \cos (c+d x))^{2/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=144 \[ \frac{(40 A+16 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right )}{10\ 2^{5/6} d (\cos (c+d x)+1)^{7/6}}+\frac{3 (8 B-3 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{40 d}+\frac{3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d} \]

[Out]

(3*(8*B - 3*C)*(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(40*d) + (3*C*(a + a*Cos[c + d*x])^(5/3)*Sin[c + d*x])
/(8*a*d) + ((40*A + 16*B + 19*C)*(a + a*Cos[c + d*x])^(2/3)*Hypergeometric2F1[-1/6, 1/2, 3/2, (1 - Cos[c + d*x
])/2]*Sin[c + d*x])/(10*2^(5/6)*d*(1 + Cos[c + d*x])^(7/6))

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Rubi [A]  time = 0.18872, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {3023, 2751, 2652, 2651} \[ \frac{(40 A+16 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right )}{10\ 2^{5/6} d (\cos (c+d x)+1)^{7/6}}+\frac{3 (8 B-3 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{40 d}+\frac{3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(8*B - 3*C)*(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(40*d) + (3*C*(a + a*Cos[c + d*x])^(5/3)*Sin[c + d*x])
/(8*a*d) + ((40*A + 16*B + 19*C)*(a + a*Cos[c + d*x])^(2/3)*Hypergeometric2F1[-1/6, 1/2, 3/2, (1 - Cos[c + d*x
])/2]*Sin[c + d*x])/(10*2^(5/6)*d*(1 + Cos[c + d*x])^(7/6))

Rule 3023

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]

Rule 2751

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

Rule 2652

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

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{3 C (a+a \cos (c+d x))^{5/3} \sin (c+d x)}{8 a d}+\frac{3 \int (a+a \cos (c+d x))^{2/3} \left (\frac{1}{3} a (8 A+5 C)+\frac{1}{3} a (8 B-3 C) \cos (c+d x)\right ) \, dx}{8 a}\\ &=\frac{3 (8 B-3 C) (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{40 d}+\frac{3 C (a+a \cos (c+d x))^{5/3} \sin (c+d x)}{8 a d}+\frac{1}{40} (40 A+16 B+19 C) \int (a+a \cos (c+d x))^{2/3} \, dx\\ &=\frac{3 (8 B-3 C) (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{40 d}+\frac{3 C (a+a \cos (c+d x))^{5/3} \sin (c+d x)}{8 a d}+\frac{\left ((40 A+16 B+19 C) (a+a \cos (c+d x))^{2/3}\right ) \int (1+\cos (c+d x))^{2/3} \, dx}{40 (1+\cos (c+d x))^{2/3}}\\ &=\frac{3 (8 B-3 C) (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{40 d}+\frac{3 C (a+a \cos (c+d x))^{5/3} \sin (c+d x)}{8 a d}+\frac{(40 A+16 B+19 C) (a+a \cos (c+d x))^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{3}{2};\frac{1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{10\ 2^{5/6} d (1+\cos (c+d x))^{7/6}}\\ \end{align*}

Mathematica [C]  time = 0.833666, size = 137, normalized size = 0.95 \[ \frac{3 \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a (\cos (c+d x)+1))^{2/3} \left (2 \sin (c+d x) (40 A+2 (8 B+7 C) \cos (c+d x)+32 B+5 C \cos (2 (c+d x))+28 C)-2 i (40 A+16 B+19 C) \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-e^{i (c+d x)}\right ) (i \sin (c+d x)+\cos (c+d x)+1)^{2/3}\right )}{320 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a*(1 + Cos[c + d*x]))^(2/3)*Sec[(c + d*x)/2]^2*((-2*I)*(40*A + 16*B + 19*C)*Hypergeometric2F1[1/3, 2/3, 4/
3, -E^(I*(c + d*x))]*(1 + Cos[c + d*x] + I*Sin[c + d*x])^(2/3) + 2*(40*A + 32*B + 28*C + 2*(8*B + 7*C)*Cos[c +
d*x] + 5*C*Cos[2*(c + d*x)])*Sin[c + d*x]))/(320*d)

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Maple [F]  time = 0.341, size = 0, normalized size = 0. \begin{align*} \int \left ( a+\cos \left ( dx+c \right ) a \right ) ^{{\frac{2}{3}}} \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{2}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+a*cos(d*x+c))**(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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