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

3.4.84.1 Optimal result
3.4.84.2 Mathematica [C] (verified)
3.4.84.3 Rubi [A] (verified)
3.4.84.4 Maple [F]
3.4.84.5 Fricas [F]
3.4.84.6 Sympy [F(-1)]
3.4.84.7 Maxima [F]
3.4.84.8 Giac [F]
3.4.84.9 Mupad [F(-1)]

3.4.84.1 Optimal result

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

output
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)*hypergeo 
m([-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)
 
3.4.84.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

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

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

Time = 0.64 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3502, 27, 3042, 3230, 3042, 3131, 3042, 3130}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{2/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {3 \int \frac {1}{3} (\cos (c+d x) a+a)^{2/3} (a (8 A+5 C)+a (8 B-3 C) \cos (c+d x))dx}{8 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (\cos (c+d x) a+a)^{2/3} (a (8 A+5 C)+a (8 B-3 C) \cos (c+d x))dx}{8 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3} \left (a (8 A+5 C)+a (8 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{8 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {1}{5} a (40 A+16 B+19 C) \int (\cos (c+d x) a+a)^{2/3}dx+\frac {3 a (8 B-3 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 d}}{8 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} a (40 A+16 B+19 C) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{2/3}dx+\frac {3 a (8 B-3 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 d}}{8 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {a (40 A+16 B+19 C) (a \cos (c+d x)+a)^{2/3} \int (\cos (c+d x)+1)^{2/3}dx}{5 (\cos (c+d x)+1)^{2/3}}+\frac {3 a (8 B-3 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 d}}{8 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a (40 A+16 B+19 C) (a \cos (c+d x)+a)^{2/3} \int \left (\sin \left (c+d x+\frac {\pi }{2}\right )+1\right )^{2/3}dx}{5 (\cos (c+d x)+1)^{2/3}}+\frac {3 a (8 B-3 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 d}}{8 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {2 \sqrt [6]{2} a (40 A+16 B+19 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{5 d (\cos (c+d x)+1)^{7/6}}+\frac {3 a (8 B-3 C) \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 d}}{8 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{5/3}}{8 a d}\)

input
Int[(a + a*Cos[c + d*x])^(2/3)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
 
output
(3*C*(a + a*Cos[c + d*x])^(5/3)*Sin[c + d*x])/(8*a*d) + ((3*a*(8*B - 3*C)* 
(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*d) + (2*2^(1/6)*a*(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])/(5*d*(1 + Cos[c + d*x])^(7/6)))/(8*a)
 

3.4.84.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3130
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]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[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 3230
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] + Simp[(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 3502
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
 
3.4.84.4 Maple [F]

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

input
int((a+cos(d*x+c)*a)^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 
output
int((a+cos(d*x+c)*a)^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 
3.4.84.5 Fricas [F]

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

input
integrate((a+a*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algori 
thm="fricas")
 
output
integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(2/3 
), x)
 
3.4.84.6 Sympy [F(-1)]

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

input
integrate((a+a*cos(d*x+c))**(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)
 
output
Timed out
 
3.4.84.7 Maxima [F]

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

input
integrate((a+a*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algori 
thm="maxima")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(2/ 
3), x)
 
3.4.84.8 Giac [F]

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

input
integrate((a+a*cos(d*x+c))^(2/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algori 
thm="giac")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^(2/ 
3), x)
 
3.4.84.9 Mupad [F(-1)]

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

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