\(\int (a+b \cos (c+d x))^{2/3} (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [234]

Optimal result

Integrand size = 34, antiderivative size = 281 \[ \int (a+b \cos (c+d x))^{2/3} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b d}+\frac {(8 b B-3 a C) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{5/3} \sin (c+d x)}{4 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{5/3}}-\frac {\left (8 a b B-3 a^2 C-5 b^2 C\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{4 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}} \] Output:

3/8*C*(a+b*cos(d*x+c))^(5/3)*sin(d*x+c)/b/d+1/8*(8*B*b-3*C*a)*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)*2^(1/2)/b^2/d/(1+cos(d*x+c))^(1/2)/((a+b*cos(d*x+c))/(a+b 
))^(5/3)-1/8*(8*B*a*b-3*C*a^2-5*C*b^2)*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)*2^(1/2 
)/b^2/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 = 5.84 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.03 \[ \int (a+b \cos (c+d x))^{2/3} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 (a+b \cos (c+d x))^{2/3} \csc (c+d x) \left (5 \left (-a^2+b^2\right ) (8 b B-3 a C) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}}+\left (16 a b B-6 a^2 C+25 b^2 C\right ) \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}} (a+b \cos (c+d x))-5 b^2 (8 b B+2 a C+5 b C \cos (c+d x)) \sin ^2(c+d x)\right )}{200 b^3 d} \] Input:

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

Output:

(-3*(a + b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(5*(-a^2 + b^2)*(8*b*B - 3*a*C 
)*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 + C 
os[c + d*x]))/(a - b))] + (16*a*b*B - 6*a^2*C + 25*b^2*C)*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)]*S 
qrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Cos[c + d*x]))/(a - 
 b))]*(a + b*Cos[c + d*x]) - 5*b^2*(8*b*B + 2*a*C + 5*b*C*Cos[c + d*x])*Si 
n[c + d*x]^2))/(200*b^3*d)
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 3502, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3235

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3144

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

\(\displaystyle \frac {\frac {\sqrt {2} (a+b) (8 b B-3 a C) \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {5}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}-\frac {\sqrt {2} \left (-3 a^2 C+8 a b B-5 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}}{8 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{5/3}}{8 b d}\)

Input:

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

Output:

(3*C*(a + b*Cos[c + d*x])^(5/3)*Sin[c + d*x])/(8*b*d) + ((Sqrt[2]*(a + b)* 
(8*b*B - 3*a*C)*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*Sq 
rt[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3)) - (Sqrt[2]*(8*a 
*b*B - 3*a^2*C - 5*b^2*C)*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)))/(8*b 
)
 

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 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]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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