\(\int (b \sec (c+d x))^{4/3} (A+C \sec ^2(c+d x)) \, dx\) [8]

Optimal result

Integrand size = 25, antiderivative size = 90 \[ \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b (7 A+4 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86 \[ \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \cot (c+d x) (b \sec (c+d x))^{4/3} \left (4 C \tan ^2(c+d x)+(7 A+4 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )}{28 d} \] Input:

Integrate[(b*Sec[c + d*x])^(4/3)*(A + C*Sec[c + d*x]^2),x]
 

Output:

(3*Cot[c + d*x]*(b*Sec[c + d*x])^(4/3)*(4*C*Tan[c + d*x]^2 + (7*A + 4*C)*H 
ypergeometric2F1[1/2, 2/3, 5/3, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2]))/(2 
8*d)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {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.

\(\displaystyle \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 4534

\(\displaystyle \frac {1}{7} (7 A+4 C) \int (b \sec (c+d x))^{4/3}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+4 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {1}{7} (7 A+4 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{4/3}}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 A+4 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{4/3}}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}\)

\(\Big \downarrow \) 3122

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

Input:

Int[(b*Sec[c + d*x])^(4/3)*(A + C*Sec[c + d*x]^2),x]
 

Output:

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

Defintions of rubi rules used

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]

\[\int \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]

Input:

int((b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x)
 

Output:

int((b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x)
 

Fricas [F]

\[ \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \] Input:

integrate((b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \] Input:

integrate((b*sec(d*x+c))**(4/3)*(A+C*sec(d*x+c)**2),x)
 

Output:

Integral((b*sec(c + d*x))**(4/3)*(A + C*sec(c + d*x)**2), x)
 

Maxima [F]

\[ \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \] Input:

integrate((b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^(4/3), x)
 

Giac [F]

\[ \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \] Input:

integrate((b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \] Input:

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(4/3),x)
 

Output:

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^(4/3), x)
 

Reduce [F]

\[ \int (b \sec (c+d x))^{4/3} \left (A+C \sec ^2(c+d x)\right ) \, dx=b^{\frac {4}{3}} \left (\left (\int \sec \left (d x +c \right )^{\frac {10}{3}}d x \right ) c +\left (\int \sec \left (d x +c \right )^{\frac {4}{3}}d x \right ) a \right ) \] Input:

int((b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x)
 

Output:

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