\(\int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx\) [789]

Optimal result

Integrand size = 25, antiderivative size = 101 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {3 B \sin (c+d x)}{2 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {(2 A-B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{2^{5/6} d \sqrt [6]{1+\cos (c+d x)} \sqrt [3]{a+a \cos (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.41 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32 \[ \int \frac {A+B \cos (c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {3\ 2^{5/6} B \sqrt [6]{1-\cos \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )} \sin (c+d x)-2 (2 A-B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\cos ^2\left (\frac {d x}{2}-\arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )\right ) \sin \left (d x-2 \arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )}{4 d \sqrt [3]{a (1+\cos (c+d x))} \sqrt [6]{\sin ^2\left (\frac {d x}{2}-\arctan \left (\cot \left (\frac {c}{2}\right )\right )\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 101, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3131

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3130

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

Input:

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

Output:

(3*B*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(1/3)) + ((2*A - B)*Hypergeom 
etric2F1[1/2, 5/6, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x])/(2^(5/6)*d*(1 
+ Cos[c + d*x])^(1/6)*(a + a*Cos[c + d*x])^(1/3))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

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

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

Output:

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