3.1.29 \(\int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [29]

3.1.29.1 Optimal result

Integrand size = 31, antiderivative size = 165 \[ \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} (A+B \sec (c+d x)) \, dx=-\frac {3 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (2-3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1-3 m),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) \sqrt {\sin ^2(c+d x)}} \]

output

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

3.1.29.2 Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.85 \[ \int \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {3 \csc (c+d x) \left (A (4+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1+3 m),\frac {1}{6} (7+3 m),\sec ^2(c+d x)\right )+B (1+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\sec ^2(c+d x)\right )\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sqrt {-\tan ^2(c+d x)}}{d (1+3 m) (4+3 m)} \]

input

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

output

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

3.1.29.3 Rubi [A] (verified)

Time = 0.55 (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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\sqrt [3]{b \sec (c+d x)} \left (A \int \sec ^{m+\frac {1}{3}}(c+d x)dx+B \int \sec ^{m+\frac {4}{3}}(c+d x)dx\right )}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{b \sec (c+d x)} \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {1}{3}}dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}}dx\right )}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {\sqrt [3]{b \sec (c+d x)} \left (A \cos ^{m+\frac {1}{3}}(c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \int \cos ^{-m-\frac {1}{3}}(c+d x)dx+B \cos ^{m+\frac {1}{3}}(c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \int \cos ^{-m-\frac {4}{3}}(c+d x)dx\right )}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{b \sec (c+d x)} \left (A \cos ^{m+\frac {1}{3}}(c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-\frac {1}{3}}dx+B \cos ^{m+\frac {1}{3}}(c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-\frac {4}{3}}dx\right )}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\sqrt [3]{b \sec (c+d x)} \left (\frac {3 B \sin (c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-1),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right )}{d (3 m+1) \sqrt {\sin ^2(c+d x)}}-\frac {3 A \sin (c+d x) \sec ^{m-\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2-3 m),\frac {1}{6} (8-3 m),\cos ^2(c+d x)\right )}{d (2-3 m) \sqrt {\sin ^2(c+d x)}}\right )}{\sqrt [3]{\sec (c+d x)}}\)

input

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

output

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

3.1.29.3.1 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]
 

3.1.29.4 Maple [F]

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

input

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

output

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

3.1.29.5 Fricas [F]

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

input

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

output

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

3.1.29.6 Sympy [F]

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

input

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

output

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

3.1.29.7 Maxima [F]

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

input

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

output

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

3.1.29.8 Giac [F]

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

input

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

output

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

3.1.29.9 Mupad [F(-1)]

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

input

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

output

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