\(\int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx\) [30]

Optimal result

Integrand size = 31, antiderivative size = 165 \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (4-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d (1-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^m(c+d x) (A+B \sec (c+d x))}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 \csc (c+d x) \left (A (2+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\sec ^2(c+d x)\right )+B (-1+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\sec ^2(c+d x)\right )\right ) \sec ^m(c+d x) \sqrt {-\tan ^2(c+d x)}}{d (-1+3 m) (2+3 m) \sqrt [3]{b \sec (c+d x)}} \] Input:

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

Output:

(3*Csc[c + d*x]*(A*(2 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (-1 + 3*m 
)/6, (5 + 3*m)/6, Sec[c + d*x]^2] + B*(-1 + 3*m)*Hypergeometric2F1[1/2, (2 
 + 3*m)/6, (8 + 3*m)/6, Sec[c + d*x]^2])*Sec[c + d*x]^m*Sqrt[-Tan[c + d*x] 
^2])/(d*(-1 + 3*m)*(2 + 3*m)*(b*Sec[c + d*x])^(1/3))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2034, 3042, 4274, 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.

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

\(\Big \downarrow \) 2034

\(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \int \sec ^{m-\frac {1}{3}}(c+d x) (A+B \sec (c+d x))dx}{\sqrt [3]{b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (-\frac {3 A \sin (c+d x) \sec ^{m-\frac {4}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4-3 m),\frac {1}{6} (10-3 m),\cos ^2(c+d x)\right )}{d (4-3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) \sqrt {\sin ^2(c+d x)}}\right )}{\sqrt [3]{b \sec (c+d x)}}\)

Input:

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

Output:

(Sec[c + d*x]^(1/3)*((-3*A*Hypergeometric2F1[1/2, (4 - 3*m)/6, (10 - 3*m)/ 
6, Cos[c + d*x]^2]*Sec[c + d*x]^(-4/3 + m)*Sin[c + d*x])/(d*(4 - 3*m)*Sqrt 
[Sin[c + d*x]^2]) - (3*B*Hypergeometric2F1[1/2, (1 - 3*m)/6, (7 - 3*m)/6, 
Cos[c + d*x]^2]*Sec[c + d*x]^(-1/3 + m)*Sin[c + d*x])/(d*(1 - 3*m)*Sqrt[Si 
n[c + d*x]^2])))/(b*Sec[c + d*x])^(1/3)
 

Defintions of rubi rules used

rule 2034

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

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 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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