Integrand size = 33, antiderivative size = 178 \[ \int (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 b^2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 b B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b (5 A+3 C) \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 B (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {2 C (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
2/3*B*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/d-2/5*b^2*(5*A+3*C)*(cos(1/2*d*x+1/2 *c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/co s(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+2/5*b*(5*A+3*C)*sin(d*x+c)*(b*sec(d*x+ c))^(1/2)/d+2/3*b*B*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt icF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/d+2/ 5*C*(b*sec(d*x+c))^(3/2)*tan(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.71 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.73 \[ \int (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 e^{-i c} \left (-1+e^{2 i c}\right ) \cos ^3(c+d x) \csc (c) \left (5 B-15 A e^{i (c+d x)}-3 C e^{i (c+d x)}-30 A e^{3 i (c+d x)}-24 C e^{3 i (c+d x)}-5 B e^{4 i (c+d x)}-15 A e^{5 i (c+d x)}-9 C e^{5 i (c+d x)}-5 i B \left (1+e^{2 i (c+d x)}\right )^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(5 A+3 C) e^{i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{15 d \left (1+e^{2 i (c+d x)}\right )^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \]
(2*(-1 + E^((2*I)*c))*Cos[c + d*x]^3*Csc[c]*(5*B - 15*A*E^(I*(c + d*x)) - 3*C*E^(I*(c + d*x)) - 30*A*E^((3*I)*(c + d*x)) - 24*C*E^((3*I)*(c + d*x)) - 5*B*E^((4*I)*(c + d*x)) - 15*A*E^((5*I)*(c + d*x)) - 9*C*E^((5*I)*(c + d *x)) - (5*I)*B*(1 + E^((2*I)*(c + d*x)))^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (5*A + 3*C)*E^(I*(c + d*x))*(1 + E^((2*I)*(c + d*x)))^(5/2 )*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*(b*Sec[c + d*x]) ^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(15*d*E^(I*c)*(1 + E^((2*I )*(c + d*x)))^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))
Time = 0.97 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.01, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4535, 3042, 4255, 3042, 4258, 3042, 3120, 4534, 3042, 4255, 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.
| \(\displaystyle \int (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \int (b \sec (c+d x))^{3/2} \left (C \sec ^2(c+d x)+A\right )dx+\frac {B \int (b \sec (c+d x))^{5/2}dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{b}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \left (\frac {1}{3} b^2 \int \sqrt {b \sec (c+d x)}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \left (\frac {1}{3} b^2 \int \sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \left (\frac {1}{3} b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {1}{5} (5 A+3 C) \int (b \sec (c+d x))^{3/2}dx+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+3 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} (5 A+3 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \sec (c+d x)}}dx\right )+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+3 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-b^2 \int \frac {1}{\sqrt {b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {1}{5} (5 A+3 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} (5 A+3 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {b^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} (5 A+3 C) \left (\frac {2 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{d}-\frac {2 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\right )+\frac {B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {b \sec (c+d x)}}{3 d}+\frac {2 b \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 d}\right )}{b}+\frac {2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d}\) |
((5*A + 3*C)*((-2*b^2*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqr t[b*Sec[c + d*x]]) + (2*b*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/d))/5 + (B*(( 2*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/( 3*d) + (2*b*(b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d)))/b + (2*C*(b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)
3.1.65.3.1 Defintions of rubi rules used
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[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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_.)*(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]
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 = 2.95 (sec) , antiderivative size = 949, normalized size of antiderivative = 5.33
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(949\) |
| default | \(\text {Expression too large to display}\) | \(998\) |
-2*A/d*(I*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),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-I*EllipticE(I*(cot(d*x+c)-c sc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*c os(d*x+c)^2+2*I*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),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*I*EllipticE(I*(cot(d* x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^( 1/2)*cos(d*x+c)+I*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*(1/(cos(d*x+c)+1) )^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-I*(1/(cos(d*x+c)+1))^(1/2)*(cos( d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)-sin(d* x+c))*(b*sec(d*x+c))^(1/2)*b/(cos(d*x+c)+1)+2/3*B/d*(b*sec(d*x+c))^(1/2)*b *(-I*EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos (d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d *x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-cot(d*x+c)+csc(d*x+c)),I)+tan(d* x+c))+2/5*C/d*(b*sec(d*x+c))^(1/2)*b/(cos(d*x+c)+1)*(3*I*(1/(cos(d*x+c)+1) )^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)-csc(d*x+ c)),I)*cos(d*x+c)^2-3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*cos(d*x+c)^2+6*I*(1/(cos(d *x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cot(d*x+c)- csc(d*x+c)),I)*cos(d*x+c)-6*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d* x+c)+1))^(1/2)*EllipticF(I*(cot(d*x+c)-csc(d*x+c)),I)*cos(d*x+c)+3*I*(1...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.26 \[ \int (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-5 i \, \sqrt {2} B b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, {\left (5 \, A + 3 \, C\right )} b \cos \left (d x + c\right )^{2} + 5 \, B b \cos \left (d x + c\right ) + 3 \, C b\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{2}} \]
1/15*(-5*I*sqrt(2)*B*b^(3/2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos (d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*B*b^(3/2)*cos(d*x + c)^2*weierst rassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(5*A + 3* C)*b^(3/2)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0 , cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(5*A + 3*C)*b^(3/2)*cos(d* x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(5*A + 3*C)*b*cos(d*x + c)^2 + 5*B*b*cos(d*x + c) + 3*C*b)*sqrt(b/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \]
\[ \int (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]
\[ \int (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]