\(\int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx\) [22]

Optimal result

Integrand size = 25, antiderivative size = 112 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{21 b^4 d}+\frac {2 (5 A+7 C) \sin (c+d x)}{21 b^3 d \sqrt {b \sec (c+d x)}}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}} \] Output:

2/21*(5*A+7*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(b* 
sec(d*x+c))^(1/2)/b^4/d+2/21*(5*A+7*C)*sin(d*x+c)/b^3/d/(b*sec(d*x+c))^(1/ 
2)+2/7*A*tan(d*x+c)/d/(b*sec(d*x+c))^(7/2)
 

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.71 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\frac {\frac {4 (5 A+7 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{\sqrt {\cos (c+d x)}}+2 (13 A+14 C+3 A \cos (2 (c+d x))) \sin (c+d x)}{42 b^3 d \sqrt {b \sec (c+d x)}} \] Input:

Integrate[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(7/2),x]
 

Output:

((4*(5*A + 7*C)*EllipticF[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*(13*A + 
14*C + 3*A*Cos[2*(c + d*x)])*Sin[c + d*x])/(42*b^3*d*Sqrt[b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4533, 3042, 4256, 3042, 4258, 3042, 3120}

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.

\(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4533

\(\displaystyle \frac {(5 A+7 C) \int \frac {1}{(b \sec (c+d x))^{3/2}}dx}{7 b^2}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(5 A+7 C) \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 b^2}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {(5 A+7 C) \left (\frac {\int \sqrt {b \sec (c+d x)}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(5 A+7 C) \left (\frac {\int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {(5 A+7 C) \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(5 A+7 C) \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 A \tan (c+d x)}{7 d (b \sec (c+d x))^{7/2}}\)

\(\Big \downarrow \) 3120

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

Input:

Int[(A + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(7/2),x]
 

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3120

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

rule 4256

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

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 = 6.57 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.62

method	result	size
default	\(\frac {-\frac {2 i A \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )}{21}-\frac {2 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (7+7 \sec \left (d x +c \right )\right )}{21}-\frac {2 \sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )^{2}-5\right ) A}{21}+\frac {2 C \sin \left (d x +c \right )}{3}}{d \sqrt {b \sec \left (d x +c \right )}\, b^{3}}\)	\(181\)
parts	\(-\frac {2 A \left (\sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )^{2}-5\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (5+5 \sec \left (d x +c \right )\right )\right )}{21 d \sqrt {b \sec \left (d x +c \right )}\, b^{3}}+\frac {C \left (-\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (1+\sec \left (d x +c \right )\right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {b \sec \left (d x +c \right )}\, b^{3}}\)	\(196\)

Input:

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-2/21*I*A*(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)*(5+5*sec(d*x+c))-2/21*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)*(7+7*sec(d*x+c))-2/21*sin(d*x+c)*(-3*cos(d*x+c)^2-5)*A+2/3* 
C*sin(d*x+c))/(b*sec(d*x+c))^(1/2)/b^3
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.06 \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, A \cos \left (d x + c\right )^{3} + {\left (5 \, A + 7 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{21 \, b^{4} d} \] Input:

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

1/21*(sqrt(2)*(-5*I*A - 7*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c)) + sqrt(2)*(5*I*A + 7*I*C)*sqrt(b)*weierstrassPInver 
se(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(3*A*cos(d*x + c)^3 + (5*A + 
7*C)*cos(d*x + c))*sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^4*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((A+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(7/2), x)
 

Giac [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + A)/(b*sec(d*x + c))^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \] Input:

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(7/2),x)
 

Output:

int((A + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) c \right )}{b^{4}} \] Input:

int((A+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x)
 

Output:

(sqrt(b)*(int(sqrt(sec(c + d*x))/sec(c + d*x)**4,x)*a + int(sqrt(sec(c + d 
*x))/sec(c + d*x)**2,x)*c))/b**4