\(\int \frac {\cos ^2(c+d x) (A+C \sec ^2(c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [20]

Optimal result

Integrand size = 33, antiderivative size = 93 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 b (7 A+10 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{70 d (b \sec (c+d x))^{7/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^2 \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}} \] Output:

-3/70*b*(7*A+10*C)*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/d/ 
(b*sec(d*x+c))^(7/3)/(sin(d*x+c)^2)^(1/2)+3/10*A*b^2*tan(d*x+c)/d/(b*sec(d 
*x+c))^(10/3)
 

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\frac {\left (3 \sqrt [6]{\cos ^2(c+d x)} (9 A+10 C+2 A \cos (2 (c+d x)))+(7 A+10 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\sin ^2(c+d x)\right )\right ) \tan (c+d x)}{40 d \sqrt [6]{\cos ^2(c+d x)} (b \sec (c+d x))^{4/3}} \] Input:

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

Output:

((3*(Cos[c + d*x]^2)^(1/6)*(9*A + 10*C + 2*A*Cos[2*(c + d*x)]) + (7*A + 10 
*C)*Hypergeometric2F1[1/2, 5/6, 3/2, Sin[c + d*x]^2])*Tan[c + d*x])/(40*d* 
(Cos[c + d*x]^2)^(1/6)*(b*Sec[c + d*x])^(4/3))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 2030, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b^2 \int \frac {C \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{10/3}}dx\)

\(\Big \downarrow \) 4533

\(\displaystyle b^2 \left (\frac {(7 A+10 C) \int \frac {1}{(b \sec (c+d x))^{4/3}}dx}{10 b^2}+\frac {3 A \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {(7 A+10 C) \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx}{10 b^2}+\frac {3 A \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}\right )\)

\(\Big \downarrow \) 4259

\(\displaystyle b^2 \left (\frac {(7 A+10 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \left (\frac {\cos (c+d x)}{b}\right )^{4/3}dx}{10 b^2}+\frac {3 A \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \left (\frac {(7 A+10 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{4/3}dx}{10 b^2}+\frac {3 A \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}\right )\)

\(\Big \downarrow \) 3122

\(\displaystyle b^2 \left (\frac {3 A \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}-\frac {3 (7 A+10 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{70 b d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}\right )\)

Input:

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

Output:

b^2*((-3*(7*A + 10*C)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Si 
n[c + d*x])/(70*b*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin[c + d*x]^2]) + (3*A*Ta 
n[c + d*x])/(10*d*(b*Sec[c + d*x])^(10/3)))
 

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(int(cos(c + d*x)**2/(sec(c + d*x)**(1/3)*sec(c + d*x)),x)*a + int((cos(c 
+ d*x)**2*sec(c + d*x))/sec(c + d*x)**(1/3),x)*c)/(b**(1/3)*b)