\(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx\) [52]

Optimal result

Integrand size = 32, antiderivative size = 147 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^4 d}+\frac {2 B \sin (c+d x)}{5 b^2 d (b \sec (c+d x))^{3/2}}+\frac {2 C \sin (c+d x)}{3 b^3 d \sqrt {b \sec (c+d x)}} \] Output:

6/5*B*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^3/d/cos(d*x+c)^(1/2)/(b*sec( 
d*x+c))^(1/2)+2/3*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+2/5*B*sin(d*x+c)/b^2/d/(b*sec(d*x+c))^(3/2)+2 
/3*C*sin(d*x+c)/b^3/d/(b*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.62 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\frac {2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \left (9 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 C \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (5 C+3 B \cos (c+d x)) \sin (c+d x)\right )}{15 b^4 d} \] Input:

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

Output:

(2*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]*(9*B*EllipticE[(c + d*x)/2, 2] 
+ 5*C*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*C + 3*B*Cos[c + d* 
x])*Sin[c + d*x]))/(15*b^4*d)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {3042, 4535, 27, 2030, 3042, 4256, 3042, 4258, 3042, 3119, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {C \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b^2}+\frac {B \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {B \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b}+\frac {C \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{b^2}\)

Input:

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

Output:

(B*((6*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c 
 + d*x]]) + (2*Sin[c + d*x])/(5*b*d*(b*Sec[c + d*x])^(3/2))))/b + (C*((2*S 
qrt[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]])))/b^2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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 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 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 [C] (verified)

Result contains complex when optimal does not.

Time = 3.82 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.95

method	result	size
parts	\(-\frac {2 B \left (\sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )-3\right )-3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b^{3}}+\frac {C \left (-\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (1+\sec \left (d x +c \right )\right )}{3}+\frac {2 \sin \left (d x +c \right )}{3}\right )}{d \sqrt {b \sec \left (d x +c \right )}\, b^{3}}\)	\(287\)
default	\(-\frac {2 \left (9 i B \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-\cos \left (d x +c \right )-2-\sec \left (d x +c \right )\right )+9 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) B \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+5 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )+3 \sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )-3\right ) B +5 \sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )-1\right ) C \right )}{15 d \,b^{3} \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}}\)	\(292\)

Input:

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(7/2),x,method=_RETURNVER 
BOSE)
 

Output:

-2/5*B/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^(1/2)/b^3*(sin(d*x+c)*(-cos(d*x+c)^ 
2-cos(d*x+c)-3)-3*I*(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))+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+s 
ec(d*x+c))*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I))+C/d*(-2/3*I*(1/(cos(d*x 
+c)+1))^(1/2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*(1+sec(d*x+c))+2/3*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 = 164, normalized size of antiderivative = 1.12 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\frac {-5 i \, \sqrt {2} C \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} C \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, B \cos \left (d x + c\right )^{2} + 5 \, C \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, b^{4} d} \] Input:

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

Output:

1/15*(-5*I*sqrt(2)*C*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s 
in(d*x + c)) + 5*I*sqrt(2)*C*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + 
c) - I*sin(d*x + c)) + 9*I*sqrt(2)*B*sqrt(b)*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*I*sqrt(2)*B*sqrt( 
b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin( 
d*x + c))) + 2*(3*B*cos(d*x + c)^2 + 5*C*cos(d*x + c))*sqrt(b/cos(d*x + c) 
)*sin(d*x + c))/(b^4*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}}d x \right ) c \right )}{b^{4}} \] Input:

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