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]

3/40*(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)

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Rubi [A]  time = 0.19, antiderivative size = 144, normalized size of antiderivative = 1.00, 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 0.86, 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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fricas [F]  time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}, x\right ) \]

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

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)

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maple [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \int \left (a +a \cos \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\cos \left (c+d\,x\right )\right )}^{2/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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+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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