\(\int (a+a \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx\) [787]

Optimal result

Integrand size = 25, antiderivative size = 104 \[ \int (a+a \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx=\frac {3 B (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 d}+\frac {2 \sqrt [6]{2} (5 A+2 B) (a+a \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{5 d (1+\cos (c+d x))^{7/6}} \] Output:

3/5*B*(a+a*cos(d*x+c))^(2/3)*sin(d*x+c)/d+2/5*2^(1/6)*(5*A+2*B)*(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)/d/ 
(1+cos(d*x+c))^(7/6)
 

Mathematica [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.58 \[ \int (a+a \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx=\frac {(a (1+\cos (c+d x)))^{2/3} \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (3\ 2^{5/6} (5 A+4 B+2 B \cos (c+d x)) \sqrt [6]{1-\cos \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )} \sin (c+d x)-2 (5 A+2 B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\cos ^2\left (\frac {d x}{2}-\arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )\right ) \sin \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )\right )}{20\ 2^{5/6} d \sqrt [6]{1-\cos \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )}} \] Input:

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

Output:

((a*(1 + Cos[c + d*x]))^(2/3)*Sec[(c + d*x)/2]^2*(3*2^(5/6)*(5*A + 4*B + 2 
*B*Cos[c + d*x])*(1 - Cos[d*x - 2*ArcTan[Cot[c/2]]])^(1/6)*Sin[c + d*x] - 
2*(5*A + 2*B)*Hypergeometric2F1[1/2, 5/6, 3/2, Cos[(d*x)/2 - ArcTan[Cot[c/ 
2]]]^2]*Sin[d*x - 2*ArcTan[Cot[c/2]]]))/(20*2^(5/6)*d*(1 - Cos[d*x - 2*Arc 
Tan[Cot[c/2]]])^(1/6))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3131

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3130

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

Input:

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

Output:

(3*B*(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*d) + (2*2^(1/6)*(5*A + 2* 
B)*(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))
 

Defintions of rubi rules used

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)]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int (a+a \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx=\int \left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {2}{3}} \left (A + B \cos {\left (c + d x \right )}\right )\, dx \] Input:

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

Output:

Integral((a*(cos(c + d*x) + 1))**(2/3)*(A + B*cos(c + d*x)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+a \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx=a^{\frac {2}{3}} \left (\left (\int \left (\cos \left (d x +c \right )+1\right )^{\frac {2}{3}}d x \right ) a +\left (\int \left (\cos \left (d x +c \right )+1\right )^{\frac {2}{3}} \cos \left (d x +c \right )d x \right ) b \right ) \] Input:

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

Output:

a**(2/3)*(int((cos(c + d*x) + 1)**(2/3),x)*a + int((cos(c + d*x) + 1)**(2/ 
3)*cos(c + d*x),x)*b)