\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{2/3}} \, dx\) [55]

Optimal result

Integrand size = 33, antiderivative size = 142 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{2/3}} \, dx=-\frac {3 b (A-2 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \sec (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}} \] Output:

-3/5*b*(A-2*C)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*s 
ec(d*x+c))^(5/3)/(sin(d*x+c)^2)^(1/2)-3/2*B*hypergeom([1/3, 1/2],[4/3],cos 
(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/2)+3*C*tan( 
d*x+c)/d/(b*sec(d*x+c))^(2/3)
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.83 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 \cot (c+d x) \left (-\left ((A-2 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )+2 \left (C \tan ^2(c+d x)+B \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right ) \sec (c+d x) \sqrt {-\tan ^2(c+d x)}\right )\right )}{2 d (b \sec (c+d x))^{2/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

\(\displaystyle (A-2 C) \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {3 b (A-2 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}}-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}\)

Input:

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

Output:

(-3*b*(A - 2*C)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + 
d*x])/(5*d*(b*Sec[c + d*x])^(5/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Hypergeomet 
ric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*d*(b*Sec[c + d*x])^ 
(2/3)*Sqrt[Sin[c + d*x]^2]) + (3*C*Tan[c + d*x])/(d*(b*Sec[c + d*x])^(2/3) 
)
 

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(1/3)/(b 
*sec(d*x + c)), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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