\(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx\) [18]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 \cot (c+d x) \left (A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\sec ^2(c+d x)\right )-2 C \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{2/3} \sqrt {-\tan ^2(c+d x)}}{4 b^2 d} \] Input:

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

Output:

(-3*Cot[c + d*x]*(A*Cos[c + d*x]^2*Hypergeometric2F1[-2/3, 1/2, 1/3, Sec[c 
 + d*x]^2] - 2*C*Hypergeometric2F1[1/3, 1/2, 4/3, Sec[c + d*x]^2])*(b*Sec[ 
c + d*x])^(2/3)*Sqrt[-Tan[c + d*x]^2])/(4*b^2*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, 4533, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4533

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

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

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 4533

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{4/3}} \, dx=\frac {\left (\int \sec \left (d x +c \right )^{\frac {2}{3}}d x \right ) c +\left (\int \frac {1}{\sec \left (d x +c \right )^{\frac {4}{3}}}d x \right ) a}{b^{\frac {4}{3}}} \] Input:

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

Output:

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