\(\int \frac {\cos (c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx\) [197]

Optimal result

Integrand size = 19, antiderivative size = 58 \[ \int \frac {\cos (c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 b^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d (b \sec (c+d x))^{7/3} \sqrt {\sin ^2(c+d x)}} \] Output:

-3/7*b^2*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x 
+c))^(7/3)/(sin(d*x+c)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 b \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{4 d (b \sec (c+d x))^{4/3}} \] Input:

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

Output:

(-3*b*Cot[c + d*x]*Hypergeometric2F1[-2/3, 1/2, 1/3, Sec[c + d*x]^2]*Sqrt[ 
-Tan[c + d*x]^2])/(4*d*(b*Sec[c + d*x])^(4/3))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 2030, 4259, 3042, 3122}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b \int \frac {1}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{4/3}}dx\)

\(\Big \downarrow \) 4259

\(\displaystyle b \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \left (\frac {\cos (c+d x)}{b}\right )^{4/3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{4/3}dx\)

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {3 b^2 \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}\)

Input:

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

Output:

(-3*b^2*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7 
*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin[c + d*x]^2])
 

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*sec(d*x + c))^(2/3)*cos(d*x + c)/(b*sec(d*x + c)), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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