3.1.26 \(\int \frac {\sec ^m(c+d x) (A+C \sec ^2(c+d x))}{(b \sec (c+d x))^{4/3}} \, dx\) [26]

3.1.26.1 Optimal result

Integrand size = 33, antiderivative size = 148 \[ \int \frac {\sec ^m(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 C \sec ^m(c+d x) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \sec (c+d x)}}-\frac {3 (A+C (4-3 m)-3 A m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7-3 m),\frac {1}{6} (13-3 m),\cos ^2(c+d x)\right ) \sec ^{-2+m}(c+d x) \sin (c+d x)}{b d (1-3 m) (7-3 m) \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

output

-3*C*sec(d*x+c)^m*sin(d*x+c)/b/d/(1-3*m)/(b*sec(d*x+c))^(1/3)-3*(A+C*(4-3* 
m)-3*A*m)*hypergeom([1/2, 7/6-1/2*m],[13/6-1/2*m],cos(d*x+c)^2)*sec(d*x+c) 
^(-2+m)*sin(d*x+c)/b/d/(9*m^2-24*m+7)/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^ 
(1/2)
 

3.1.26.2 Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.97 \[ \int \frac {\sec ^m(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \left (A (2+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-4+3 m),\frac {1}{6} (2+3 m),\sec ^2(c+d x)\right )+C (-4+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-4+3 m) (2+3 m) (b \sec (c+d x))^{7/3}} \]

input

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

output

(3*b*Csc[c + d*x]*Sec[c + d*x]^m*(A*(2 + 3*m)*Hypergeometric2F1[1/2, (-4 + 
 3*m)/6, (2 + 3*m)/6, Sec[c + d*x]^2] + C*(-4 + 3*m)*Hypergeometric2F1[1/2 
, (2 + 3*m)/6, (8 + 3*m)/6, Sec[c + d*x]^2]*Sec[c + d*x]^2)*Sqrt[-Tan[c + 
d*x]^2])/(d*(-4 + 3*m)*(2 + 3*m)*(b*Sec[c + d*x])^(7/3))
 

3.1.26.3 Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 152, 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 = {2034, 3042, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4534

\(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (\frac {(A (1-3 m)+C (4-3 m)) \int \sec ^{m-\frac {4}{3}}(c+d x)dx}{1-3 m}-\frac {3 C \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x)}{d (1-3 m)}\right )}{b \sqrt [3]{b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\sqrt [3]{\sec (c+d x)} \left (-\frac {3 (A (1-3 m)+C (4-3 m)) \sin (c+d x) \sec ^{m-\frac {7}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7-3 m),\frac {1}{6} (13-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) (7-3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x)}{d (1-3 m)}\right )}{b \sqrt [3]{b \sec (c+d x)}}\)

input

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

output

(Sec[c + d*x]^(1/3)*((-3*C*Sec[c + d*x]^(-1/3 + m)*Sin[c + d*x])/(d*(1 - 3 
*m)) - (3*(A*(1 - 3*m) + C*(4 - 3*m))*Hypergeometric2F1[1/2, (7 - 3*m)/6, 
(13 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-7/3 + m)*Sin[c + d*x])/(d*(1 
- 3*m)*(7 - 3*m)*Sqrt[Sin[c + d*x]^2])))/(b*(b*Sec[c + d*x])^(1/3))
 

3.1.26.3.1 Defintions of rubi rules used

rule 2034

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart 
[n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n]))   Int[(a*v)^(m + n 
)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && 
  !IntegerQ[m + n]
 

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]
 

3.1.26.4 Maple [F]

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

input

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

output

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

3.1.26.5 Fricas [F]

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

input

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

output

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

3.1.26.6 Sympy [F]

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

input

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

output

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

3.1.26.7 Maxima [F]

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

input

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

3.1.26.8 Giac [F]

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

input

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

3.1.26.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^m(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]

input

int(((A + C/cos(c + d*x)^2)*(1/cos(c + d*x))^m)/(b/cos(c + d*x))^(4/3),x)
 

output

int(((A + C/cos(c + d*x)^2)*(1/cos(c + d*x))^m)/(b/cos(c + d*x))^(4/3), x)