\(\int (a+a \cos (c+d x))^{2/3} (A+C \cos ^2(c+d x)) \, dx\) [199]

Optimal result

Integrand size = 27, antiderivative size = 135 \[ \int (a+a \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {9 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{40 d}+\frac {3 C (a+a \cos (c+d x))^{5/3} \sin (c+d x)}{8 a d}+\frac {(40 A+19 C) (a+a \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{10\ 2^{5/6} d (1+\cos (c+d x))^{7/6}} \]

[Out]

-9/40*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+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 = 0.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3103, 2830, 2731, 2730} \[ \int (a+a \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(40 A+19 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{10\ 2^{5/6} d (\cos (c+d x)+1)^{7/6}}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}-\frac {9 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{40 d} \]

[In]

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

[Out]

(-9*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 + 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 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*} \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)-a C \cos (c+d x)\right ) \, dx}{8 a} \\ & = -\frac {9 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+19 C) \int (a+a \cos (c+d x))^{2/3} \, dx \\ & = -\frac {9 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+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 {9 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+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 [A] (verified)

Time = 0.80 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.30 \[ \int (a+a \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(a (1+\cos (c+d x)))^{2/3} \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (6\ 2^{5/6} (40 A+28 C+14 C \cos (c+d x)+5 C \cos (2 (c+d x))) \sqrt [6]{1-\cos \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )} \sin (c+d x)-4 (40 A+19 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\cos ^2\left (\frac {d x}{2}-\arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )\right ) \sin \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )\right )}{320\ 2^{5/6} d \sqrt [6]{1-\cos \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )}} \]

[In]

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

[Out]

((a*(1 + Cos[c + d*x]))^(2/3)*Sec[(c + d*x)/2]^2*(6*2^(5/6)*(40*A + 28*C + 14*C*Cos[c + d*x] + 5*C*Cos[2*(c +
d*x)])*(1 - Cos[d*x - 2*ArcTan[Cot[c/2]]])^(1/6)*Sin[c + d*x] - 4*(40*A + 19*C)*Hypergeometric2F1[1/2, 5/6, 3/
2, Cos[(d*x)/2 - ArcTan[Cot[c/2]]]^2]*Sin[d*x - 2*ArcTan[Cot[c/2]]]))/(320*2^(5/6)*d*(1 - Cos[d*x - 2*ArcTan[C
ot[c/2]]])^(1/6))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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