\(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [21]

Optimal result

Integrand size = 25, antiderivative size = 77 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}} \] Output:

2/5*(3*A+5*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/d/cos(d*x+c)^(1/2) 
/(b*sec(d*x+c))^(1/2)+2/5*A*tan(d*x+c)/d/(b*sec(d*x+c))^(5/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.02 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.73 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {e^{-i d x} \sec ^2(c+d x) (\cos (d x)+i \sin (d x)) \left (12 i (3 A+5 C)-\frac {8 i (3 A+5 C) e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+6 A \sin (2 (c+d x))\right )}{30 d (b \sec (c+d x))^{5/2}} \] Input:

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(5/2),x]
 

Output:

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])*((12*I)*(3*A + 5*C) - ((8*I)*(3*A 
+ 5*C)*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + 
 d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + 6*A*Sin[2*(c + d*x)]))/(30*d*E^(I 
*d*x)*(b*Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4533, 3042, 4258, 3042, 3119}

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.

\(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4533

\(\displaystyle \frac {(3 A+5 C) \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{5 b^2}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(3 A+5 C) \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {(3 A+5 C) \int \sqrt {\cos (c+d x)}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(3 A+5 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 (3 A+5 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/2}}\)

Input:

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(5/2),x]
 

Output:

(2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt 
[b*Sec[c + d*x]]) + (2*A*Tan[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/2))
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

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 [C] (verified)

Result contains complex when optimal does not.

Time = 6.30 (sec) , antiderivative size = 343, normalized size of antiderivative = 4.45

method	result	size
default	\(\frac {\frac {6 i A \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )}{5}+2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) C \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )-\frac {6 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) A \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{5}-2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+\frac {2 \sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right ) A}{5}+2 C \sin \left (d x +c \right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b^{2}}\)	\(343\)
parts	\(\frac {2 A \left (\sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )-3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b^{2}}+\frac {2 C \left (i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+\sin \left (d x +c \right )\right )}{b^{2} d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}\)	\(371\)

Input:

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/5/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)/b^2*(3*I*A*(1/(cos(d*x+c)+1))^(1 
/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)), 
I)*(cos(d*x+c)+2+sec(d*x+c))+5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos 
(d*x+c)+1))^(1/2)*(cos(d*x+c)+2+sec(d*x+c))*C*EllipticE(I*(csc(d*x+c)-cot( 
d*x+c)),I)-3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* 
(cos(d*x+c)+2+sec(d*x+c))*A*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)-5*I*C*( 
1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(csc 
(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)+2+sec(d*x+c))+sin(d*x+c)*(cos(d*x+c)^2+ 
cos(d*x+c)+3)*A+5*C*sin(d*x+c))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.40 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\frac {2 \, A \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{5 \, b^{3} d} \] Input:

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/5*(2*A*sqrt(b/cos(d*x + c))*cos(d*x + c)^2*sin(d*x + c) + sqrt(2)*(3*I*A 
 + 5*I*C)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* 
x + c) + I*sin(d*x + c))) + sqrt(2)*(-3*I*A - 5*I*C)*sqrt(b)*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(b^3 
*d)
 

Sympy [F]

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

integrate((A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(5/2),x)
 

Output:

Integral((A + C*sec(c + d*x)**2)/(b*sec(c + d*x))**(5/2), x)
 

Maxima [F]

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

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(5/2), x)
 

Giac [F]

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

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(5/2),x)
 

Output:

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(5/2), x)
 

Reduce [F]

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

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

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