\(\int \frac {\sqrt [3]{\cos (c+d x)}}{a+b \cos (c+d x)} \, dx\) [677]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int \frac {\sqrt [3]{\cos (c+d x)}}{a+b \cos (c+d x)} \, dx=-\frac {b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{6},1,\frac {3}{2},\sin ^2(c+d x),-\frac {b^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sqrt [3]{\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt [6]{\cos ^2(c+d x)}}-\frac {3 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},\cos ^2(c+d x),\frac {b^2 \cos ^2(c+d x)}{a^2}\right ) \cos ^{\frac {4}{3}}(c+d x) \sin (c+d x)}{4 a d \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4613\) vs. \(2(161)=322\).

Time = 34.55 (sec) , antiderivative size = 4613, normalized size of antiderivative = 28.65 \[ \int \frac {\sqrt [3]{\cos (c+d x)}}{a+b \cos (c+d x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

(9*(a^2 - b^2)*Sin[c + d*x]*((a*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2 
, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2 
)*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 
- b^2))] + (-6*a^2*AppellF1[3/2, 1/6, 2, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[ 
c + d*x]^2)/(a^2 - b^2))] + (-a^2 + b^2)*AppellF1[3/2, 7/6, 1, 5/2, -Tan[c 
 + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c + d*x]^2) + (b*Appe 
llF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2) 
)])/(-9*(a^2 - b^2)*AppellF1[1/2, 2/3, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan 
[c + d*x]^2)/(a^2 - b^2))] + 2*(3*a^2*AppellF1[3/2, 2/3, 2, 5/2, -Tan[c + 
d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + 2*(a^2 - b^2)*AppellF1[3/2, 
 5/3, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])*Tan[c 
 + d*x]^2)))/(d*Cos[c + d*x]^(2/3)*(a + b*Cos[c + d*x])*(Sec[c + d*x]^2)^( 
2/3)*(-b^2 + a^2*Sec[c + d*x]^2)*((9*(a^2 - b^2)*(Sec[c + d*x]^2)^(1/3)*(( 
a*AppellF1[1/2, 1/6, 1, 3/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 
- b^2))]*Sqrt[Sec[c + d*x]^2])/(9*(a^2 - b^2)*AppellF1[1/2, 1/6, 1, 3/2, - 
Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (-6*a^2*AppellF1[3/ 
2, 1/6, 2, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))] + (- 
a^2 + b^2)*AppellF1[3/2, 7/6, 1, 5/2, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x] 
^2)/(a^2 - b^2))])*Tan[c + d*x]^2) + (b*AppellF1[1/2, 2/3, 1, 3/2, -Tan[c 
+ d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))])/(-9*(a^2 - b^2)*AppellF...
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3302, 3042, 3668, 333}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3302

\(\displaystyle a \int \frac {\sqrt [3]{\cos (c+d x)}}{a^2-b^2 \cos ^2(c+d x)}dx-b \int \frac {\cos ^{\frac {4}{3}}(c+d x)}{a^2-b^2 \cos ^2(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3668

\(\displaystyle \frac {a \sqrt [3]{\cos ^2(c+d x)} \int \frac {1}{\sqrt [3]{1-\sin ^2(c+d x)} \left (a^2-b^2+b^2 \sin ^2(c+d x)\right )}d\sin (c+d x)}{d \cos ^{\frac {2}{3}}(c+d x)}-\frac {b \sqrt [3]{\cos (c+d x)} \int \frac {\sqrt [6]{1-\sin ^2(c+d x)}}{a^2-b^2+b^2 \sin ^2(c+d x)}d\sin (c+d x)}{d \sqrt [6]{\cos ^2(c+d x)}}\)

\(\Big \downarrow \) 333

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 3042

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

rule 3302

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

rule 3668

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[( 
-ff)*d^(2*IntPart[(m - 1)/2] + 1)*((d*Sin[e + f*x])^(2*FracPart[(m - 1)/2]) 
/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2]))   Subst[Int[(1 - ff^2*x^2)^((m - 
 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, 
 d, e, f, m, p}, x] &&  !IntegerQ[m]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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