Integrand size = 31, antiderivative size = 172 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {2 a (5 A+3 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (A+B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a (5 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (A+B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]
2/3*a*(A+B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*B*sec(d*x+c)^(5/2)*sin(d*x +c)/d+2/5*a*(5*A+3*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*a*(5*A+3*B)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/3*a*(A+B)*(cos(1/2*d*x+1/2*c)^ 2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+ c)^(1/2)*sec(d*x+c)^(1/2)/d
Time = 1.98 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.98 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a \sec ^2\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x)) (A+B \sec (c+d x)) \left (-3 (5 A+3 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 (A+B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+15 A \sin (c+d x)+9 B \sin (c+d x)+5 A \tan (c+d x)+5 B \tan (c+d x)+3 B \sec (c+d x) \tan (c+d x)\right )}{15 d (B+A \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \]
(a*Sec[(c + d*x)/2]^2*(1 + Sec[c + d*x])*(A + B*Sec[c + d*x])*(-3*(5*A + 3 *B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 5*(A + B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 15*A*Sin[c + d*x] + 9*B*Sin[c + d*x] + 5 *A*Tan[c + d*x] + 5*B*Tan[c + d*x] + 3*B*Sec[c + d*x]*Tan[c + d*x]))/(15*d *(B + A*Cos[c + d*x])*Sec[c + d*x]^(3/2))
Time = 0.77 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4485, 27, 3042, 4274, 3042, 4255, 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.
\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a) (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4485 |
\(\displaystyle \frac {2}{5} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (a (5 A+3 B)+5 a (A+B) \sec (c+d x))dx+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \sec ^{\frac {3}{2}}(c+d x) (a (5 A+3 B)+5 a (A+B) \sec (c+d x))dx+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a (5 A+3 B)+5 a (A+B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {1}{5} \left (a (5 A+3 B) \int \sec ^{\frac {3}{2}}(c+d x)dx+5 a (A+B) \int \sec ^{\frac {5}{2}}(c+d x)dx\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (a (5 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+5 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+a (5 A+3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+a (5 A+3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+a (5 A+3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+a (5 A+3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )+a (5 A+3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{5} \left (5 a (A+B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )+a (5 A+3 B) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {2 a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\) |
(2*a*B*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (a*(5*A + 3*B)*((-2*Sqrt[C os[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec [c + d*x]]*Sin[c + d*x])/d) + 5*a*(A + B)*((2*Sqrt[Cos[c + d*x]]*EllipticF [(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)))/5
3.2.80.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(634\) vs. \(2(200)=400\).
Time = 32.79 (sec) , antiderivative size = 635, normalized size of antiderivative = 3.69
method | result | size |
default | \(-\frac {a \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2 B \left (24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{5 \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2 A \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+4 \left (\frac {A}{2}+\frac {B}{2}\right ) \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(635\) |
parts | \(\text {Expression too large to display}\) | \(763\) |
-a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*B/(8*sin (1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d*x +1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2* c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2 *c)+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))* (2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c) *sin(1/2*d*x+1/2*c)^2-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x +1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))*(-2*sin(1/2*d*x+1/2*c)^ 4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*A/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2* c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d* x+1/2*c)*sin(1/2*d*x+1/2*c)^2-(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1 /2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))+4*(1/2*A+1/2*B)*( -1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/ 2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1 /2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1 /2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2* d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.27 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {-5 i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (A + B\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right )^{2} + 5 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 3 \, B a\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \]
1/15*(-5*I*sqrt(2)*(A + B)*a*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos (d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(A + B)*a*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* B)*a*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*B)*a*cos(d*x + c)^2*wei erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(5*A + 3*B)*a*cos(d*x + c)^2 + 5*(A + B)*a*cos(d*x + c) + 3*B* a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]