\(\int \frac {\sec (c+d x) (A+C \sec ^2(c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx\) [12]

Optimal result

Integrand size = 31, antiderivative size = 92 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 (5 A+2 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.87 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 \cot (c+d x) (b \sec (c+d x))^{2/3} \left (2 C \tan ^2(c+d x)+(5 A+2 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )}{10 b d} \] Input:

Integrate[(Sec[c + d*x]*(A + C*Sec[c + d*x]^2))/(b*Sec[c + d*x])^(1/3),x]
 

Output:

(3*Cot[c + d*x]*(b*Sec[c + d*x])^(2/3)*(2*C*Tan[c + d*x]^2 + (5*A + 2*C)*H 
ypergeometric2F1[1/3, 1/2, 4/3, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2]))/(1 
0*b*d)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2030, 3042, 4534, 3042, 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.

\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4534

\(\displaystyle \frac {\frac {1}{5} (5 A+2 C) \int (b \sec (c+d x))^{2/3}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} (5 A+2 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 d}}{b}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {\frac {1}{5} (5 A+2 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{2/3}}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 d}}{b}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

Int[(Sec[c + d*x]*(A + C*Sec[c + d*x]^2))/(b*Sec[c + d*x])^(1/3),x]
 

Output:

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

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]
 

rule 4534

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

Maple [F]

\[\int \frac {\sec \left (d x +c \right ) \left (A +C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]

Input:

int(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x)
 

Output:

int(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x)
 

Fricas [F]

\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x, algorithm= 
"fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\sqrt [3]{b \sec {\left (c + d x \right )}}}\, dx \] Input:

integrate(sec(d*x+c)*(A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(1/3),x)
 

Output:

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)/(b*sec(c + d*x))**(1/3), x)
 

Maxima [F]

\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x, algorithm= 
"maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)/(b*sec(d*x + c))^(1/3), x)
 

Giac [F]

\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:

integrate(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x, algorithm= 
"giac")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)/(b*sec(d*x + c))^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\cos \left (c+d\,x\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \] Input:

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)*(b/cos(c + d*x))^(1/3)),x)
 

Output:

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)*(b/cos(c + d*x))^(1/3)), x)
 

Reduce [F]

\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {\left (\int \sec \left (d x +c \right )^{\frac {8}{3}}d x \right ) c +\left (\int \sec \left (d x +c \right )^{\frac {2}{3}}d x \right ) a}{b^{\frac {1}{3}}} \] Input:

int(sec(d*x+c)*(A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(1/3),x)
 

Output:

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