\(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [62]

Optimal result

Integrand size = 39, antiderivative size = 149 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 (2 A-C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x)}{2 b d \sqrt [3]{b \sec (c+d x)}} \] Output:

-3/8*(2*A-C)*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)+3/2*C*tan(d 
*x+c)/b/d/(b*sec(d*x+c))^(1/3)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.81 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 C \tan (c+d x)-3 (2 A-C) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}+3 B \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{2 b d \sqrt [3]{b \sec (c+d x)}} \] Input:

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

Output:

(3*C*Tan[c + d*x] - 3*(2*A - C)*Cot[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 
5/6, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2] + 3*B*Csc[c + d*x]*Hypergeometr 
ic2F1[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))
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2030, 3042, 4535, 3042, 4259, 3042, 3122, 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.

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \frac {-\frac {3 b (2 A-C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}}{b}\)

Input:

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

Output:

((-3*b*(2*A - C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + 
d*x])/(8*d*(b*Sec[c + d*x])^(4/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Hypergeomet 
ric2F1[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]) + (3*C*Tan[c + d*x])/(2*d*(b*Sec[c + d*x])^(1/3) 
))/b
 

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 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]
 

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\frac {\left (\int \sec \left (d x +c \right )^{\frac {5}{3}}d x \right ) c +\left (\int \sec \left (d x +c \right )^{\frac {2}{3}}d x \right ) b +\left (\int \frac {1}{\sec \left (d x +c \right )^{\frac {1}{3}}}d x \right ) a}{b^{\frac {4}{3}}} \] Input:

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

Output:

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