Integrand size = 35, antiderivative size = 152 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {(A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A-C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}-\frac {(A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \]
-(A+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))+(A+3*C)*sin(d*x+c)*s ec(d*x+c)^(1/2)/a/d-(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/ a/d+(A-C)*(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 = 8.48 (sec) , antiderivative size = 776, normalized size of antiderivative = 5.11 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\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)}} \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (c+d x) \csc \left (\frac {c}{2}\right ) \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 ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right )}{3 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))}+\frac {\sqrt {2} C 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)}} \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (c+d x) \csc \left (\frac {c}{2}\right ) \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 ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right )}{d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))}+\frac {2 A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{d (A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec (c+d x)} (a+a \sec (c+d x))}-\frac {2 C \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{d (A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec (c+d x)} (a+a \sec (c+d x))}+\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+C \sec ^2(c+d x)\right ) \left (\frac {2 (A+3 C) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{d}-\frac {4 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{d}-\frac {4 (A+C) \tan \left (\frac {c}{2}\right )}{d}\right )}{(A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec (c+d x)} (a+a \sec (c+d x))} \]
(Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2* I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-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))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(3*d*E ^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (Sqrt[2]*C *Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d* x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-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))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (2*A*Cos[c/2 + (d*x)/2 ]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*(A + C* Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x ]]*(a + a*Sec[c + d*x])) - (2*C*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Cs c[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sin[c])/( d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])) + (Cos[c/2 + (d*x)/2]^2*(A + C*Sec[c + d*x]^2)*((2*(A + 3*C)*Cos[d*x]*Csc[ c/2]*Sec[c/2])/d - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[ (d*x)/2]))/d - (4*(A + C)*Tan[c/2])/d))/((A + 2*C + A*Cos[2*c + 2*d*x])*Sq rt[Sec[c + d*x]]*(a + a*Sec[c + d*x]))
Time = 0.77 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4573, 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.
\(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a \sec (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
\(\Big \downarrow \) 4573 |
\(\displaystyle -\frac {\int -\frac {1}{2} \sqrt {\sec (c+d x)} (a (A-C)+a (A+3 C) \sec (c+d x))dx}{a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sqrt {\sec (c+d x)} (a (A-C)+a (A+3 C) \sec (c+d x))dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (A-C)+a (A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {a (A+3 C) \int \sec ^{\frac {3}{2}}(c+d x)dx+a (A-C) \int \sqrt {\sec (c+d x)}dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A-C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+a (A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {a (A-C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+a (A+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A-C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+a (A+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {a (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+a (A+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (A+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {a (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (A+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {2 a (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+a (A+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\) |
-(((A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) + (( 2*a*(A - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x] ])/d + a*(A + 3*C)*((-2*Sqrt[Cos[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))/(2*a^2)
3.3.33.3.1 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[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 1.61 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.08
method | result | size |
default | \(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +3 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +5 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(316\) |
-1/a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-cos(1/2*d *x+1/2*c)*(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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*EllipticF(cos(1/2*d* x+1/2*c),2^(1/2))+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*EllipticF(cos( 1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(-2*s in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+3*C)*sin(1/2*d*x+1/2*c) ^4+(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+5*C)*sin(1/2*d* x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^3/(2*sin(1/2*d*x+1/2*c)^2-1)/cos(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 = 0.11 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {{\left (\sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left ({\left (A + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/2*((sqrt(2)*(-I*A + I*C)*cos(d*x + c) + sqrt(2)*(-I*A + I*C))*weierstras sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - I*C)*cos (d*x + c) + sqrt(2)*(I*A - I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + (sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 3*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I *sin(d*x + c))) + (sqrt(2)*(I*A + 3*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 3*I *C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*si n(d*x + c))) + 2*((A + 3*C)*cos(d*x + c) + 2*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sqrt {\sec {\left (c + d x \right )}}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{\frac {5}{2}}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
(Integral(A*sqrt(sec(c + d*x))/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**(5/2)/(sec(c + d*x) + 1), x))/a
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {\sec \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {\sec \left (d x + c\right )}}{a \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]