\(\int \sqrt [3]{a+b \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [204]

Optimal result

Integrand size = 27, antiderivative size = 274 \[ \int \sqrt [3]{a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C (a+b \cos (c+d x))^{4/3} \sin (c+d x)}{7 b d}-\frac {3 \sqrt {2} a C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{4/3} \sin (c+d x)}{7 b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{4/3}}+\frac {\sqrt {2} \left (3 a^2 C+b^2 (7 A+4 C)\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{7 b^2 d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}} \] Output:

3/7*C*(a+b*cos(d*x+c))^(4/3)*sin(d*x+c)/b/d-3/7*2^(1/2)*a*C*AppellF1(1/2,- 
4/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x+c))*(a+b*cos(d*x+c))^(4 
/3)*sin(d*x+c)/b^2/d/(1+cos(d*x+c))^(1/2)/((a+b*cos(d*x+c))/(a+b))^(4/3)+1 
/7*2^(1/2)*(3*a^2*C+b^2*(7*A+4*C))*AppellF1(1/2,-1/3,1/2,3/2,b*(1-cos(d*x+ 
c))/(a+b),1/2-1/2*cos(d*x+c))*(a+b*cos(d*x+c))^(1/3)*sin(d*x+c)/b^2/d/(1+c 
os(d*x+c))^(1/2)/((a+b*cos(d*x+c))/(a+b))^(1/3)
 

Mathematica [A] (warning: unable to verify)

Time = 3.20 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.01 \[ \int \sqrt [3]{a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 \sqrt [3]{a+b \cos (c+d x)} \csc (c+d x) \left (12 a \left (a^2-b^2\right ) C \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}}+\left (28 A b^2-3 a^2 C+16 b^2 C\right ) \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))-4 b^2 C (a+4 b \cos (c+d x)) \sin ^2(c+d x)\right )}{112 b^3 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^(1/3)*(A + C*Cos[c + d*x]^2),x]
 

Output:

(-3*(a + b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(12*a*(a^2 - b^2)*C*AppellF1[1 
/3, 1/2, 1/2, 4/3, (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))] + (28*A*b^2 - 3*a^2*C + 16*b^2*C)*AppellF1[4/3, 1/2, 1/2, 7/3, 
 (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*C 
os[c + d*x]) - 4*b^2*C*(a + 4*b*Cos[c + d*x])*Sin[c + d*x]^2))/(112*b^3*d)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3503, 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.

\(\displaystyle \int \sqrt [3]{a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 3503

\(\displaystyle \frac {3 \int \frac {1}{3} \sqrt [3]{a+b \cos (c+d x)} (b (7 A+4 C)-3 a C \cos (c+d x))dx}{7 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sqrt [3]{a+b \cos (c+d x)} (b (7 A+4 C)-3 a C \cos (c+d x))dx}{7 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (b (7 A+4 C)-3 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d}\)

\(\Big \downarrow \) 3235

\(\displaystyle \frac {\frac {\left (3 a^2 C+b^2 (7 A+4 C)\right ) \int \sqrt [3]{a+b \cos (c+d x)}dx}{b}-\frac {3 a C \int (a+b \cos (c+d x))^{4/3}dx}{b}}{7 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (3 a^2 C+b^2 (7 A+4 C)\right ) \int \sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {3 a C \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx}{b}}{7 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d}\)

\(\Big \downarrow \) 3144

\(\displaystyle \frac {\frac {3 a C \sin (c+d x) \int \frac {(a+b \cos (c+d x))^{4/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {\left (3 a^2 C+b^2 (7 A+4 C)\right ) \sin (c+d x) \int \frac {\sqrt [3]{a+b \cos (c+d x)}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}}{7 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {\frac {3 a C (a+b) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \int \frac {\left (\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}\right )^{4/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (3 a^2 C+b^2 (7 A+4 C)\right ) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \int \frac {\sqrt [3]{\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}}{7 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\frac {\sqrt {2} \left (3 a^2 C+b^2 (7 A+4 C)\right ) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}-\frac {3 \sqrt {2} a C (a+b) \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {4}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}}{7 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{4/3}}{7 b d}\)

Input:

Int[(a + b*Cos[c + d*x])^(1/3)*(A + C*Cos[c + d*x]^2),x]
 

Output:

(3*C*(a + b*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*b*d) + ((-3*Sqrt[2]*a*(a 
+ b)*C*AppellF1[1/2, 1/2, -4/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c + 
 d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(b*d*Sqrt[1 + Co 
s[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(1/3)) + (Sqrt[2]*(3*a^2*C + b^ 
2*(7*A + 4*C))*AppellF1[1/2, 1/2, -1/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - 
 Cos[c + d*x]))/(a + b)]*(a + b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(b*d*Sqr 
t[1 + Cos[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(1/3)))/(7*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 3503

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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] + Simp[1/(b*(m + 2))   Int[(a + b*Sin[e + f*x])^ 
m*Simp[A*b*(m + 2) + b*C*(m + 1) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a 
, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

Maple [F]

\[\int \left (a +b \cos \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \cos \left (d x +c \right )^{2}\right )d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((a+b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(1/3), x)
 

Sympy [F]

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

integrate((a+b*cos(d*x+c))**(1/3)*(A+C*cos(d*x+c)**2),x)
 

Output:

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

Maxima [F]

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

integrate((a+b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(1/3), x)
 

Giac [F]

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

integrate((a+b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)^(1/3), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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