\(\int \frac {\cos (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx\) [158]

Optimal result

Integrand size = 21, antiderivative size = 75 \[ \int \frac {\cos (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=-\frac {3 \sqrt {2} \operatorname {AppellF1}\left (\frac {1}{6},\frac {1}{2},2,\frac {7}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \] Output:

-3*2^(1/2)*AppellF1(1/6,2,1/2,7/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*tan(d*x 
+c)/d/(1-sec(d*x+c))^(1/2)/(a+a*sec(d*x+c))^(1/3)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(240\) vs. \(2(75)=150\).

Time = 1.27 (sec) , antiderivative size = 240, normalized size of antiderivative = 3.20 \[ \int \frac {\cos (c+d x)}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {(a (1+\sec (c+d x)))^{2/3} \left (\frac {20 \operatorname {AppellF1}\left (\frac {3}{2},\frac {2}{3},1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sin ^3\left (\frac {1}{2} (c+d x)\right )}{6 \left (3 \operatorname {AppellF1}\left (\frac {5}{2},\frac {2}{3},2,\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 \operatorname {AppellF1}\left (\frac {5}{2},\frac {5}{3},1,\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+45 \operatorname {AppellF1}\left (\frac {3}{2},\frac {2}{3},1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))}+\sin (c+d x)-\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a d} \] Input:

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

Output:

((a*(1 + Sec[c + d*x]))^(2/3)*((20*AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x 
)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]*Sin[(c + d*x)/2]^3)/(6*(3*Ap 
pellF1[5/2, 2/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*Appe 
llF1[5/2, 5/3, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 + Cos 
[c + d*x]) + 45*AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d 
*x)/2]^2]*(1 + Cos[c + d*x])) + Sin[c + d*x] - Tan[(c + d*x)/2]))/(a*d)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4315, 3042, 4314, 149, 936}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 4315

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{\sec (c+d x)+1} \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{\sqrt [3]{a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4314

\(\displaystyle -\frac {\tan (c+d x) \int \frac {\cos ^2(c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 149

\(\displaystyle -\frac {6 \tan (c+d x) \int \frac {\cos ^2(c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 936

\(\displaystyle -\frac {3 \sqrt {2} \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{6},2,\frac {1}{2},\frac {7}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}\)

Input:

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

Output:

(-3*Sqrt[2]*AppellF1[1/6, 2, 1/2, 7/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x] 
)/2]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(1/3))
 

Defintions of rubi rules used

rule 149

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) 
)^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b   Subst[Int[x^(k*(m + 1 
) - 1)*(c - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + 
 b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && 
IntegerQ[2*n] && IntegerQ[p]
 

rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (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 4314

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

rule 4315

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Csc[e + f*x])^FracPart[m 
]/(1 + (b/a)*Csc[e + f*x])^FracPart[m])   Int[(1 + (b/a)*Csc[e + f*x])^m*(d 
*Csc[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]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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