\(\int \frac {\cos (c+d x) (A+C \sec ^2(c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [19]

Optimal result

Integrand size = 31, antiderivative size = 90 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 (4 A+7 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{28 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 A b \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}} \] Output:

-3/28*(4*A+7*C)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*s 
ec(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)+3/7*A*b*tan(d*x+c)/d/(b*sec(d*x+c))^ 
(7/3)
 

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 \cot (c+d x) \left (A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\sec ^2(c+d x)\right )+7 C \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{7 b d \sqrt [3]{b \sec (c+d x)}} \] Input:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 4533

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b \left (\frac {3 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}-\frac {3 (4 A+7 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{28 b d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}\right )\)

Input:

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

Output:

b*((-3*(4*A + 7*C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c 
+ d*x])/(28*b*d*(b*Sec[c + d*x])^(4/3)*Sqrt[Sin[c + d*x]^2]) + (3*A*Tan[c 
+ d*x])/(7*d*(b*Sec[c + d*x])^(7/3)))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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