\(\int \sqrt [3]{a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [788]

Optimal result

Integrand size = 25, antiderivative size = 102 \[ \int \sqrt [3]{a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {3 B \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{4 d}+\frac {(4 A+B) \sqrt [3]{a+a \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{2 \sqrt [6]{2} d (1+\cos (c+d x))^{5/6}} \] Output:

3/4*B*(a+a*cos(d*x+c))^(1/3)*sin(d*x+c)/d+1/4*(4*A+B)*(a+a*cos(d*x+c))^(1/ 
3)*hypergeom([1/6, 1/2],[3/2],1/2-1/2*cos(d*x+c))*sin(d*x+c)*2^(5/6)/d/(1+ 
cos(d*x+c))^(5/6)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(304\) vs. \(2(102)=204\).

Time = 4.05 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.98 \[ \int \sqrt [3]{a+a \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {\sqrt [3]{a (1+\cos (c+d x))} \left (-6 (4 A+B) \cot \left (\frac {c}{2}\right ) \sqrt {\sec ^2\left (\frac {c}{2}\right )}+5 (4 A+B) \cos \left (\frac {1}{2} \left (c-d x-2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )+4 A \cos \left (\frac {1}{2} \left (c+d x+2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )+B \cos \left (\frac {1}{2} \left (c+d x+2 \arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right )+6 B \sqrt {\sec ^2\left (\frac {c}{2}\right )} \sin (c+d x)-\frac {4 (4 A+B) \, _2F_1\left (-\frac {1}{2},-\frac {1}{6};\frac {5}{6};\cos ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}{\sqrt {\sin ^2\left (\frac {d x}{2}+\arctan \left (\tan \left (\frac {c}{2}\right )\right )\right )}}\right )}{8 d \sqrt {\sec ^2\left (\frac {c}{2}\right )}} \] Input:

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

Output:

((a*(1 + Cos[c + d*x]))^(1/3)*(-6*(4*A + B)*Cot[c/2]*Sqrt[Sec[c/2]^2] + 5* 
(4*A + B)*Cos[(c - d*x - 2*ArcTan[Tan[c/2]])/2]*Csc[c/2]*Sec[c/2]*Sec[(c + 
 d*x)/2] + 4*A*Cos[(c + d*x + 2*ArcTan[Tan[c/2]])/2]*Csc[c/2]*Sec[c/2]*Sec 
[(c + d*x)/2] + B*Cos[(c + d*x + 2*ArcTan[Tan[c/2]])/2]*Csc[c/2]*Sec[c/2]* 
Sec[(c + d*x)/2] + 6*B*Sqrt[Sec[c/2]^2]*Sin[c + d*x] - (4*(4*A + B)*Hyperg 
eometricPFQ[{-1/2, -1/6}, {5/6}, Cos[(d*x)/2 + ArcTan[Tan[c/2]]]^2]*Sec[c/ 
2]*Sec[(c + d*x)/2]*Sin[(d*x)/2 + ArcTan[Tan[c/2]]])/Sqrt[Sin[(d*x)/2 + Ar 
cTan[Tan[c/2]]]^2]))/(8*d*Sqrt[Sec[c/2]^2])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 102, 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.

\(\displaystyle \int \sqrt [3]{a \cos (c+d x)+a} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

\(\displaystyle \frac {1}{4} (4 A+B) \int \sqrt [3]{\cos (c+d x) a+a}dx+\frac {3 B \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} (4 A+B) \int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {3 B \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {(4 A+B) \sqrt [3]{a \cos (c+d x)+a} \int \sqrt [3]{\cos (c+d x)+1}dx}{4 \sqrt [3]{\cos (c+d x)+1}}+\frac {3 B \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(4 A+B) \sqrt [3]{a \cos (c+d x)+a} \int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right )+1}dx}{4 \sqrt [3]{\cos (c+d x)+1}}+\frac {3 B \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {(4 A+B) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{2 \sqrt [6]{2} d (\cos (c+d x)+1)^{5/6}}+\frac {3 B \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 d}\)

Input:

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

Output:

(3*B*(a + a*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(4*d) + ((4*A + B)*(a + a*Co 
s[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 3/2, (1 - Cos[c + d*x])/2]*S 
in[c + d*x])/(2*2^(1/6)*d*(1 + Cos[c + d*x])^(5/6))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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