\(\int (a+b \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx\) [795]

Optimal result

Integrand size = 25, antiderivative size = 226 \[ \int (a+b \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx=\frac {\sqrt {2} B \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{5/3}}+\frac {\sqrt {2} (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{b d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \] Output:

2^(1/2)*B*AppellF1(1/2,-5/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x 
+c))*(a+b*cos(d*x+c))^(5/3)*sin(d*x+c)/b/d/(1+cos(d*x+c))^(1/2)/((a+b*cos( 
d*x+c))/(a+b))^(5/3)+2^(1/2)*(A*b-B*a)*AppellF1(1/2,-2/3,1/2,3/2,b*(1-cos( 
d*x+c))/(a+b),1/2-1/2*cos(d*x+c))*(a+b*cos(d*x+c))^(2/3)*sin(d*x+c)/b/d/(1 
+cos(d*x+c))^(1/2)/((a+b*cos(d*x+c))/(a+b))^(2/3)
 

Mathematica [A] (warning: unable to verify)

Time = 1.58 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.15 \[ \int (a+b \cos (c+d x))^{2/3} (A+B \cos (c+d x)) \, dx=\frac {3 (a+b \cos (c+d x))^{2/3} \left (5 \left (a^2-b^2\right ) B \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} \csc (c+d x)-(5 A b+2 a B) \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x)) \csc (c+d x)+5 b^2 B \sin (c+d x)\right )}{25 b^2 d} \] Input:

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

Output:

(3*(a + b*Cos[c + d*x])^(2/3)*(5*(a^2 - b^2)*B*AppellF1[2/3, 1/2, 1/2, 5/3 
, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(- 
1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Cos[c + d*x]))/(a - b))]*Csc[c 
 + d*x] - (5*A*b + 2*a*B)*AppellF1[5/3, 1/2, 1/2, 8/3, (a + b*Cos[c + d*x] 
)/(a - b), (a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a 
 + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*(a + b*Cos[c + d*x])*Csc[c + 
 d*x] + 5*b^2*B*Sin[c + d*x]))/(25*b^2*d)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3235, 3042, 3144, 156, 155}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3235

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3144

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle \frac {\sqrt {2} (A b-a B) \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} B (a+b) \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}\)

Input:

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

Output:

(Sqrt[2]*(a + b)*B*AppellF1[1/2, 1/2, -5/3, 3/2, (1 - Cos[c + d*x])/2, (b* 
(1 - Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(b*d 
*Sqrt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3)) + (Sqrt[2]*( 
A*b - a*B)*AppellF1[1/2, 1/2, -2/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos 
[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(b*d*Sqrt[1 
+ Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3))
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 3042

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

rule 3144

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

rule 3235

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)/b   Int[(a + b*Sin[e + f*x])^m, 
 x], x] + Simp[d/b   Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, 
b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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