Integrand size = 33, antiderivative size = 229 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {3 (5 A-7 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}+\frac {5 (A-B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {3 (5 A-7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}+\frac {5 (A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac {(5 A-7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {(A-B) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \]
5/3*(A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d-1/5*(5*A-7*B)*sec(d*x+c)^(5/2)*s in(d*x+c)/a/d+(A-B)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))-3/5*(5* A-7*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d+3/5*(5*A-7*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)/a/d+5/3*(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)/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.74 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.01 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) (A+B \sec (c+d x)) \left (-30 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+42 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+100 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-100 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+2 \sqrt {\sec (c+d x)} \left (-18 (5 A-7 B) \cos (d x) \csc (c)+(25 A-13 B+4 (5 A-2 B) \cos (c+d x)+25 (A-B) \cos (2 (c+d x))) \sec ^2(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )\right )}{30 a d (B+A \cos (c+d x)) (1+\sec (c+d x))} \]
(Cos[(c + d*x)/2]^2*(A + B*Sec[c + d*x])*((-30*Sqrt[2]*A*Sqrt[E^(I*(c + d* x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sq rt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeomet ric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + (42*Sqrt[2]*B*Sq rt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x)) ]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)* c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + 1 00*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 100 *B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + 2*Sqr t[Sec[c + d*x]]*(-18*(5*A - 7*B)*Cos[d*x]*Csc[c] + (25*A - 13*B + 4*(5*A - 2*B)*Cos[c + d*x] + 25*(A - B)*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Tan[(c + d*x)/2])))/(30*a*d*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x]))
Time = 1.04 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4507, 27, 3042, 4274, 3042, 4255, 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.
\(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle \frac {\int \frac {1}{2} \sec ^{\frac {5}{2}}(c+d x) (5 a (A-B)-a (5 A-7 B) \sec (c+d x))dx}{a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec ^{\frac {5}{2}}(c+d x) (5 a (A-B)-a (5 A-7 B) \sec (c+d x))dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (5 a (A-B)-a (5 A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {5 a (A-B) \int \sec ^{\frac {5}{2}}(c+d x)dx-a (5 A-7 B) \int \sec ^{\frac {7}{2}}(c+d x)dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 a (A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx-a (5 A-7 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}dx}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \int \sec ^{\frac {3}{2}}(c+d x)dx+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {5 a (A-B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )-a (5 A-7 B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {5 a (A-B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )-a (5 A-7 B) \left (\frac {2 \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {3}{5} \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )}{2 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
((A - B)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + (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)) - a*(5*A - 7*B)*((2* Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (3*((-2*Sqrt[Cos[c + d*x]]*Ellipt icE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/5))/(2*a^2)
3.3.1.3.1 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_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(778\) vs. \(2(255)=510\).
Time = 22.64 (sec) , antiderivative size = 779, normalized size of antiderivative = 3.40
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((A-B)*(cos(1 /2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2 )*(EllipticF(cos(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)^4+sin(1/2*d*x+1/2*c)^2)/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)+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+2*B)/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))+(2*A-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)*Ellip...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.48 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 \, {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-5 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-5 i \, A + 7 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (5 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (5 i \, A - 7 i \, B\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (9 \, {\left (5 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (10 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right ) - 6 \, B\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]
-1/30*(25*(sqrt(2)*(I*A - I*B)*cos(d*x + c)^3 + sqrt(2)*(I*A - I*B)*cos(d* x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*( sqrt(2)*(-I*A + I*B)*cos(d*x + c)^3 + sqrt(2)*(-I*A + I*B)*cos(d*x + c)^2) *weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*(sqrt(2)*(- 5*I*A + 7*I*B)*cos(d*x + c)^3 + sqrt(2)*(-5*I*A + 7*I*B)*cos(d*x + c)^2)*w eierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*(sqrt(2)*(5*I*A - 7*I*B)*cos(d*x + c)^3 + sqrt(2)*(5*I*A - 7*I* B)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d *x + c) - I*sin(d*x + c))) + 2*(9*(5*A - 7*B)*cos(d*x + c)^3 + 2*(10*A - 1 9*B)*cos(d*x + c)^2 - 2*(5*A - 2*B)*cos(d*x + c) - 6*B)*sin(d*x + c)/sqrt( cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{a \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x))}{a+a \sec (c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]