\(\int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx\) [288]

Optimal result

Integrand size = 25, antiderivative size = 78 \[ \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx=\frac {2^{5/6} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {4}{3},\frac {1}{6},\frac {3}{2},1-\cos (c+d x),\frac {1}{2} (1-\cos (c+d x))\right ) \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{d (1+\cos (c+d x))^{5/6}} \] Output:

2^(5/6)*AppellF1(1/2,-4/3,1/6,3/2,1-cos(d*x+c),1/2-1/2*cos(d*x+c))*(a+a*co 
s(d*x+c))^(1/3)*sin(d*x+c)/d/(1+cos(d*x+c))^(5/6)
 

Mathematica [F]

\[ \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx=\int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx \] Input:

Integrate[Cos[c + d*x]^(4/3)*(a + a*Cos[c + d*x])^(1/3),x]
 

Output:

Integrate[Cos[c + d*x]^(4/3)*(a + a*Cos[c + d*x])^(1/3), x]
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3266, 3042, 3264, 150}

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.

\(\displaystyle \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a \cos (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3266

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3264

\(\displaystyle \frac {\sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \int \frac {\cos ^{\frac {4}{3}}(c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt [6]{\cos (c+d x)+1}}d(1-\cos (c+d x))}{d \sqrt {1-\cos (c+d x)} (\cos (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {2^{5/6} \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {4}{3},\frac {1}{6},\frac {3}{2},1-\cos (c+d x),\frac {1}{2} (1-\cos (c+d x))\right )}{d (\cos (c+d x)+1)^{5/6}}\)

Input:

Int[Cos[c + d*x]^(4/3)*(a + a*Cos[c + d*x])^(1/3),x]
 

Output:

(2^(5/6)*AppellF1[1/2, -4/3, 1/6, 3/2, 1 - Cos[c + d*x], (1 - Cos[c + d*x] 
)/2]*(a + a*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])^(5/6))
 

Defintions of rubi rules used

rule 150

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

rule 3042

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

rule 3264

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(d/b)^n*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e 
 + f*x]]*Sqrt[a - b*Sin[e + f*x]]))   Subst[Int[(a - x)^n*((2*a - x)^(m - 1 
/2)/Sqrt[x]), x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n} 
, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]
 

rule 3266

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Sin[e + f*x])^FracPart[m 
]/(1 + (b/a)*Sin[e + f*x])^FracPart[m])   Int[(1 + (b/a)*Sin[e + f*x])^m*(d 
*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 
2, 0] &&  !IntegerQ[m] &&  !GtQ[a, 0]
 

Maple [F]

\[\int \cos \left (d x +c \right )^{\frac {4}{3}} \left (a +a \cos \left (d x +c \right )\right )^{\frac {1}{3}}d x\]

Input:

int(cos(d*x+c)^(4/3)*(a+a*cos(d*x+c))^(1/3),x)
 

Output:

int(cos(d*x+c)^(4/3)*(a+a*cos(d*x+c))^(1/3),x)
 

Fricas [F]

\[ \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \cos \left (d x + c\right )^{\frac {4}{3}} \,d x } \] Input:

integrate(cos(d*x+c)^(4/3)*(a+a*cos(d*x+c))^(1/3),x, algorithm="fricas")
 

Output:

integral((a*cos(d*x + c) + a)^(1/3)*cos(d*x + c)^(4/3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(4/3)*(a+a*cos(d*x+c))**(1/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \cos \left (d x + c\right )^{\frac {4}{3}} \,d x } \] Input:

integrate(cos(d*x+c)^(4/3)*(a+a*cos(d*x+c))^(1/3),x, algorithm="maxima")
 

Output:

integrate((a*cos(d*x + c) + a)^(1/3)*cos(d*x + c)^(4/3), x)
 

Giac [F]

\[ \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}} \cos \left (d x + c\right )^{\frac {4}{3}} \,d x } \] Input:

integrate(cos(d*x+c)^(4/3)*(a+a*cos(d*x+c))^(1/3),x, algorithm="giac")
 

Output:

integrate((a*cos(d*x + c) + a)^(1/3)*cos(d*x + c)^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx=\int {\cos \left (c+d\,x\right )}^{4/3}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{1/3} \,d x \] Input:

int(cos(c + d*x)^(4/3)*(a + a*cos(c + d*x))^(1/3),x)
 

Output:

int(cos(c + d*x)^(4/3)*(a + a*cos(c + d*x))^(1/3), x)
 

Reduce [F]

\[ \int \cos ^{\frac {4}{3}}(c+d x) \sqrt [3]{a+a \cos (c+d x)} \, dx=a^{\frac {1}{3}} \left (\int \left (\cos \left (d x +c \right )+1\right )^{\frac {1}{3}} \cos \left (d x +c \right )^{\frac {4}{3}}d x \right ) \] Input:

int(cos(d*x+c)^(4/3)*(a+a*cos(d*x+c))^(1/3),x)
 

Output:

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