\(\int \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [892]

Optimal result

Integrand size = 31, antiderivative size = 117 \[ \int \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {3 A b^2 \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.80 \[ \int \sqrt [3]{b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx=\frac {3 \sqrt [3]{b \cos (c+d x)} \csc (c+d x) \left (2 A \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )+5 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )\right ) \sec ^2(c+d x) \sqrt {\sin ^2(c+d x)}}{10 d} \] Input:

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

Output:

(3*(b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(2*A*Hypergeometric2F1[-5/6, 1/2, 1 
/6, Cos[c + d*x]^2] + 5*B*Cos[c + d*x]*Hypergeometric2F1[-1/3, 1/2, 2/3, C 
os[c + d*x]^2])*Sec[c + d*x]^2*Sqrt[Sin[c + d*x]^2])/(10*d)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b^3 \int \frac {A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{8/3}}dx\)

\(\Big \downarrow \) 3227

\(\displaystyle b^3 \left (A \int \frac {1}{(b \cos (c+d x))^{8/3}}dx+\frac {B \int \frac {1}{(b \cos (c+d x))^{5/3}}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^3 \left (A \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{8/3}}dx+\frac {B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}}dx}{b}\right )\)

\(\Big \downarrow \) 3122

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

Input:

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

Output:

b^3*((3*A*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]*Sin[c + d*x])/ 
(5*b*d*(b*Cos[c + d*x])^(5/3)*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2 
F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*b^2*d*(b*Cos[c + d*x]) 
^(2/3)*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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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