\(\int (a+a \cos (c+d x))^{2/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [384]

Optimal result

Integrand size = 35, antiderivative size = 144 \[ \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 (8 B-3 C) (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{40 d}+\frac {3 C (a+a \cos (c+d x))^{5/3} \sin (c+d x)}{8 a d}+\frac {(40 A+16 B+19 C) (a+a \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{10\ 2^{5/6} d (1+\cos (c+d x))^{7/6}} \]

[Out]

3/40*(8*B-3*C)*(a+a*cos(d*x+c))^(2/3)*sin(d*x+c)/d+3/8*C*(a+a*cos(d*x+c))^(5/3)*sin(d*x+c)/a/d+1/20*(40*A+16*B
+19*C)*(a+a*cos(d*x+c))^(2/3)*hypergeom([-1/6, 1/2],[3/2],1/2-1/2*cos(d*x+c))*sin(d*x+c)*2^(1/6)/d/(1+cos(d*x+
c))^(7/6)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3102, 2830, 2731, 2730} \[ \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {(40 A+16 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{10\ 2^{5/6} d (\cos (c+d x)+1)^{7/6}}+\frac {3 (8 B-3 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{40 d}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d} \]

[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 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

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

Rule 2830

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*S
in[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 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*} \text {integral}& = \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} \operatorname {Hypergeometric2F1}\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] (verified)

Result contains complex when optimal does not.

Time = 0.99 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.95 \[ \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 (a (1+\cos (c+d x)))^{2/3} \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-2 i (40 A+16 B+19 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-e^{i (c+d x)}\right ) (1+\cos (c+d x)+i \sin (c+d x))^{2/3}+2 (40 A+32 B+28 C+2 (8 B+7 C) \cos (c+d x)+5 C \cos (2 (c+d x))) \sin (c+d x)\right )}{320 d} \]

[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)

Maple [F]

\[\int \left (a +\cos \left (d x +c \right ) a \right )^{\frac {2}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]

[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)

Fricas [F]

\[ \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[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

Maxima [F]

\[ \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]

[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)

Giac [F]

\[ \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^{2/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int {\left (a+a\,\cos \left (c+d\,x\right )\right )}^{2/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

[In]

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

[Out]

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