3.1.23 \(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [23]

3.1.23.1 Optimal result

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

output

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

3.1.23.2 Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.80 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 \csc (c+d x) \left (2 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right )-B \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{2 b d \sqrt [3]{b \sec (c+d x)}} \]

input

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(b*Sec[c + d*x])^(4/3),x]
 

output

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

3.1.23.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2030, 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.

\(\displaystyle \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\)

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

input

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(b*Sec[c + d*x])^(4/3),x]
 

output

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

3.1.23.3.1 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 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]
 

3.1.23.4 Maple [F]

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

input

int(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x)
 

output

int(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x)
 

3.1.23.5 Fricas [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

input

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="f 
ricas")
 

output

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

3.1.23.6 Sympy [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]

input

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))**(4/3),x)
 

output

Integral((A + B*sec(c + d*x))*sec(c + d*x)/(b*sec(c + d*x))**(4/3), x)
 

3.1.23.7 Maxima [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

input

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="m 
axima")
 

output

integrate((B*sec(d*x + c) + A)*sec(d*x + c)/(b*sec(d*x + c))^(4/3), x)
 

3.1.23.8 Giac [F]

\[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]

input

integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(b*sec(d*x+c))^(4/3),x, algorithm="g 
iac")
 

output

integrate((B*sec(d*x + c) + A)*sec(d*x + c)/(b*sec(d*x + c))^(4/3), x)
 

3.1.23.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\cos \left (c+d\,x\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]

input

int((A + B/cos(c + d*x))/(cos(c + d*x)*(b/cos(c + d*x))^(4/3)),x)
 

output

int((A + B/cos(c + d*x))/(cos(c + d*x)*(b/cos(c + d*x))^(4/3)), x)