Integrand size = 32, antiderivative size = 176 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\frac {6 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {10 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{21 b^5 d}+\frac {2 B \sin (c+d x)}{7 b^2 d (b \sec (c+d x))^{5/2}}+\frac {2 C \sin (c+d x)}{5 b^3 d (b \sec (c+d x))^{3/2}}+\frac {10 B \sin (c+d x)}{21 b^4 d \sqrt {b \sec (c+d x)}} \] Output:
6/5*C*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^4/d/cos(d*x+c)^(1/2)/(b*sec( d*x+c))^(1/2)+10/21*B*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^5/d+2/7*B*sin(d*x+c)/b^2/d/(b*sec(d*x+c))^(5/2) +2/5*C*sin(d*x+c)/b^3/d/(b*sec(d*x+c))^(3/2)+10/21*B*sin(d*x+c)/b^4/d/(b*s ec(d*x+c))^(1/2)
Time = 1.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.59 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\frac {\sqrt {b \sec (c+d x)} \left (252 C \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+100 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(65 B+42 C \cos (c+d x)+15 B \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{210 b^5 d} \] Input:
Integrate[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(9/2),x]
Output:
(Sqrt[b*Sec[c + d*x]]*(252*C*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 100*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (65*B + 42*C*Cos[c + d*x] + 15*B*Cos[2*(c + d*x)])*Sin[2*(c + d*x)]))/(210*b^5*d)
Time = 0.78 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.406, Rules used = {3042, 4535, 27, 2030, 3042, 4256, 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.
| \(\displaystyle \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {B \int \frac {1}{(b \sec (c+d x))^{7/2}}dx}{b}+\int \frac {C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B \int \frac {1}{(b \sec (c+d x))^{7/2}}dx}{b}+C \int \frac {\sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}}dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle \frac {C \int \frac {1}{(b \sec (c+d x))^{5/2}}dx}{b^2}+\frac {B \int \frac {1}{(b \sec (c+d x))^{7/2}}dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {C \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{b^2}+\frac {B \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{b}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {B \left (\frac {5 \int \frac {1}{(b \sec (c+d x))^{3/2}}dx}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )}{b}+\frac {C \left (\frac {3 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \left (\frac {5 \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )}{b}+\frac {C \left (\frac {3 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b^2}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {B \left (\frac {5 \left (\frac {\int \sqrt {b \sec (c+d x)}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )}{b}+\frac {C \left (\frac {3 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \left (\frac {5 \left (\frac {\int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )}{b}+\frac {C \left (\frac {3 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {B \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )}{b}+\frac {C \left (\frac {3 \int \sqrt {\cos (c+d x)}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )}{b}+\frac {C \left (\frac {3 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {B \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )}{b}+\frac {C \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {B \left (\frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 \sin (c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\right )}{7 b^2}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}}\right )}{b}+\frac {C \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x)}{5 b d (b \sec (c+d x))^{3/2}}\right )}{b^2}\) |
Input:
Int[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(b*Sec[c + d*x])^(9/2),x]
Output:
(C*((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^2 + (B*((2 *Sin[c + d*x])/(7*b*d*(b*Sec[c + d*x])^(5/2)) + (5*((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)))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
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]
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]
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]
Result contains complex when optimal does not.
Time = 3.92 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.69
| method | result | size |
| parts | \(\frac {2 B \left (\sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{2}+5\right )+i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-5-5 \sec \left (d x +c \right )\right )\right )}{21 d \sqrt {b \sec \left (d x +c \right )}\, b^{4}}+\frac {2 C \left (\sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right )-3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+3 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b^{4}}\) | \(297\) |
| default | \(\frac {\frac {6 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) C \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{5}-\frac {10 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right ) B \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )}{21}-\frac {6 i C \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (\cos \left (d x +c \right )+2+\sec \left (d x +c \right )\right )}{5}+\frac {2 \sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+5 \cos \left (d x +c \right )+5\right ) B}{21}+\frac {2 \sin \left (d x +c \right ) \left (\cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+3\right ) C}{5}}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {b \sec \left (d x +c \right )}\, b^{4}}\) | \(304\) |
Input:
int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x,method=_RETURNVER BOSE)
Output:
2/21*B/d/(b*sec(d*x+c))^(1/2)/b^4*(sin(d*x+c)*(3*cos(d*x+c)^2+5)+I*(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)))+2/5*C/d/(cos(d*x+c)+1)/(b*sec(d*x+c))^ (1/2)/b^4*(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)*(cos(d*x+c)+2+sec(d*x+c))*EllipticF (I*(csc(d*x+c)-cot(d*x+c)),I)+3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(co s(d*x+c)+1))^(1/2)*EllipticE(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)+2+se c(d*x+c)))
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.99 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\frac {-25 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 63 i \, \sqrt {2} C \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 i \, \sqrt {2} C \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (15 \, B \cos \left (d x + c\right )^{3} + 21 \, C \cos \left (d x + c\right )^{2} + 25 \, B \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, b^{5} d} \] Input:
integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm= "fricas")
Output:
1/105*(-25*I*sqrt(2)*B*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I *sin(d*x + c)) + 25*I*sqrt(2)*B*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 63*I*sqrt(2)*C*sqrt(b)*weierstrassZeta(-4, 0, we ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 63*I*sqrt(2)*C* sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I *sin(d*x + c))) + 2*(15*B*cos(d*x + c)^3 + 21*C*cos(d*x + c)^2 + 25*B*cos( d*x + c))*sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^5*d)
Timed out. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\text {Timed out} \] Input:
integrate((B*sec(d*x+c)+C*sec(d*x+c)**2)/(b*sec(d*x+c))**(9/2),x)
Output:
Timed out
\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm= "maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))/(b*sec(d*x + c))^(9/2), x)
\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \] Input:
integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x, algorithm= "giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))/(b*sec(d*x + c))^(9/2), x)
Timed out. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{9/2}} \,d x \] Input:
int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(9/2),x)
Output:
int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(b/cos(c + d*x))^(9/2), x)
\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{9/2}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}}d x \right ) c \right )}{b^{5}} \] Input:
int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(9/2),x)
Output:
(sqrt(b)*(int(sqrt(sec(c + d*x))/sec(c + d*x)**4,x)*b + int(sqrt(sec(c + d *x))/sec(c + d*x)**3,x)*c))/b**5