\(\int \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [890]

Optimal result

Integrand size = 29, antiderivative size = 114 \[ \int \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \sec (c+d x) \, dx=-\frac {3 A \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 b d \sqrt {\sin ^2(c+d x)}} \] Output:

-3*A*(b*cos(d*x+c))^(1/3)*hypergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x 
+c)/d/(sin(d*x+c)^2)^(1/2)-3/4*B*(b*cos(d*x+c))^(4/3)*hypergeom([1/2, 2/3] 
,[5/3],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.75 \[ \int \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \sec (c+d x) \, dx=-\frac {3 b \cot (c+d x) \left (4 A \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )+B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{4 d (b \cos (c+d x))^{2/3}} \] Input:

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

Output:

(-3*b*Cot[c + d*x]*(4*A*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2] + 
 B*Cos[c + d*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2])*Sqrt[Sin 
[c + d*x]^2])/(4*d*(b*Cos[c + d*x])^(2/3))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 2030, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3227

\(\displaystyle b \left (A \int \frac {1}{(b \cos (c+d x))^{2/3}}dx+\frac {B \int \sqrt [3]{b \cos (c+d x)}dx}{b}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b \left (-\frac {3 A \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 b^2 d \sqrt {\sin ^2(c+d x)}}\right )\)

Input:

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

Output:

b*((-3*A*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d 
*x]^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(4 
/3)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*b^2* 
d*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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((b*cos(d*x+c))^(1/3)*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="f 
ricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((b*cos(d*x+c))^(1/3)*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="m 
axima")
 

Output:

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

Giac [F]

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

integrate((b*cos(d*x+c))^(1/3)*(A+B*cos(d*x+c))*sec(d*x+c),x, algorithm="g 
iac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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