\(\int \frac {\cos ^2(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [689]

Optimal result

Integrand size = 33, antiderivative size = 256 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (6 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}-\frac {2 b \left (4 a^2 A b^2-3 A b^4+2 a^4 C-a^2 b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]

[Out]

1/2*(6*A*b^2+a^2*(A+2*C))*x/a^4-2*b*(4*A*a^2*b^2-3*A*b^4+2*C*a^4-C*a^2*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/
2*c)/(a+b)^(1/2))/a^4/(a-b)^(3/2)/(a+b)^(3/2)/d+b*(3*A*b^2-a^2*(2*A-C))*sin(d*x+c)/a^3/(a^2-b^2)/d-1/2*(3*A*b^
2-a^2*(A-2*C))*cos(d*x+c)*sin(d*x+c)/a^2/(a^2-b^2)/d+(A*b^2+C*a^2)*cos(d*x+c)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*se
c(d*x+c))

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4186, 4189, 4004, 3916, 2738, 214} \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \sin (c+d x) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \cos (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {2 b \left (2 a^4 C+4 a^2 A b^2-a^2 b^2 C-3 A b^4\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {x \left (a^2 (A+2 C)+6 A b^2\right )}{2 a^4}+\frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right )} \]

[In]

Int[(Cos[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]

[Out]

((6*A*b^2 + a^2*(A + 2*C))*x)/(2*a^4) - (2*b*(4*a^2*A*b^2 - 3*A*b^4 + 2*a^4*C - a^2*b^2*C)*ArcTanh[(Sqrt[a - b
]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(a - b)^(3/2)*(a + b)^(3/2)*d) + (b*(3*A*b^2 - a^2*(2*A - C))*Sin[c + d
*x])/(a^3*(a^2 - b^2)*d) - ((3*A*b^2 - a^2*(A - 2*C))*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*(a^2 - b^2)*d) + ((A*b
^2 + a^2*C)*Cos[c + d*x]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4186

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*b^2 + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x]
)^n/(a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e
+ f*x])^n*Simp[a^2*(A + C)*(m + 1) - (A*b^2 + a^2*C)*(m + n + 1) - a*b*(A + C)*(m + 1)*Csc[e + f*x] + (A*b^2
+ a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && NeQ[a^2 - b^2, 0] &&
LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])

Rule 4189

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (A b^2+a^2 C\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\cos ^2(c+d x) \left (3 A b^2-a^2 (A-2 C)+a b (A+C) \sec (c+d x)-2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )} \\ & = -\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {\cos (c+d x) \left (2 b \left (3 A b^2-a^2 (2 A-C)\right )+a \left (A b^2+a^2 (A+2 C)\right ) \sec (c+d x)-b \left (3 A b^2-a^2 (A-2 C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )} \\ & = \frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {-\left (\left (a^2-b^2\right ) \left (6 A b^2+a^2 (A+2 C)\right )\right )+a b \left (3 A b^2-a^2 (A-2 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )} \\ & = \frac {\left (6 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}+\frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (b \left (3 A b^4-a^2 b^2 (4 A-C)-2 a^4 C\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )} \\ & = \frac {\left (6 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}+\frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (3 A b^4-a^2 b^2 (4 A-C)-2 a^4 C\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^4 \left (a^2-b^2\right )} \\ & = \frac {\left (6 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}+\frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (2 \left (3 A b^4-a^2 b^2 (4 A-C)-2 a^4 C\right )\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 \left (a^2-b^2\right ) d} \\ & = \frac {\left (6 A b^2+a^2 (A+2 C)\right ) x}{2 a^4}-\frac {2 b \left (4 a^2 A b^2-3 A b^4+2 a^4 C-a^2 b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b \left (3 A b^2-a^2 (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^2-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^2+a^2 C\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.48 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {2 \left (6 A b^2+a^2 (A+2 C)\right ) (c+d x)-\frac {8 b \left (3 A b^4-2 a^4 C+a^2 b^2 (-4 A+C)\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-8 a A b \sin (c+d x)+\frac {4 a b^2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))}+a^2 A \sin (2 (c+d x))}{4 a^4 d} \]

[In]

Integrate[(Cos[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]

[Out]

(2*(6*A*b^2 + a^2*(A + 2*C))*(c + d*x) - (8*b*(3*A*b^4 - 2*a^4*C + a^2*b^2*(-4*A + C))*ArcTanh[((-a + b)*Tan[(
c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - 8*a*A*b*Sin[c + d*x] + (4*a*b^2*(A*b^2 + a^2*C)*Sin[c + d*x
])/((a - b)*(a + b)*(b + a*Cos[c + d*x])) + a^2*A*Sin[2*(c + d*x)])/(4*a^4*d)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2} A -2 a A b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} a^{2} A -2 a A b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (a^{2} A +6 A \,b^{2}+2 C \,a^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}+\frac {2 b \left (-\frac {a b \left (A \,b^{2}+C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (4 A \,a^{2} b^{2}-3 A \,b^{4}+2 a^{4} C -C \,a^{2} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}}{d}\)	\(266\)
default	\(\frac {\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2} A -2 a A b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {1}{2} a^{2} A -2 a A b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (a^{2} A +6 A \,b^{2}+2 C \,a^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{4}}+\frac {2 b \left (-\frac {a b \left (A \,b^{2}+C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (4 A \,a^{2} b^{2}-3 A \,b^{4}+2 a^{4} C -C \,a^{2} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4}}}{d}\)	\(266\)
risch	\(\frac {A x}{2 a^{2}}+\frac {3 x A \,b^{2}}{a^{4}}+\frac {x C}{a^{2}}-\frac {i A b \,{\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{3}}+\frac {i A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}+\frac {i A b \,{\mathrm e}^{i \left (d x +c \right )}}{d \,a^{3}}-\frac {i A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}+\frac {2 i b^{2} \left (A \,b^{2}+C \,a^{2}\right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{4} \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {3 b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{4}}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {4 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {3 b^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{4}}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}\)	\(870\)

[In]

int(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(2/a^4*(((-1/2*a^2*A-2*a*A*b)*tan(1/2*d*x+1/2*c)^3+(1/2*a^2*A-2*a*A*b)*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+
1/2*c)^2)^2+1/2*(A*a^2+6*A*b^2+2*C*a^2)*arctan(tan(1/2*d*x+1/2*c)))+2*b/a^4*(-a*b*(A*b^2+C*a^2)/(a^2-b^2)*tan(
1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)-(4*A*a^2*b^2-3*A*b^4+2*C*a^4-C*a^2*b^2)/(a+
b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 838, normalized size of antiderivative = 3.27 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {{\left ({\left (A + 2 \, C\right )} a^{7} + 4 \, {\left (A - C\right )} a^{5} b^{2} - {\left (11 \, A - 2 \, C\right )} a^{3} b^{4} + 6 \, A a b^{6}\right )} d x \cos \left (d x + c\right ) + {\left ({\left (A + 2 \, C\right )} a^{6} b + 4 \, {\left (A - C\right )} a^{4} b^{3} - {\left (11 \, A - 2 \, C\right )} a^{2} b^{5} + 6 \, A b^{7}\right )} d x + {\left (2 \, C a^{4} b^{2} + {\left (4 \, A - C\right )} a^{2} b^{4} - 3 \, A b^{6} + {\left (2 \, C a^{5} b + {\left (4 \, A - C\right )} a^{3} b^{3} - 3 \, A a b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (2 \, {\left (2 \, A - C\right )} a^{5} b^{2} - 2 \, {\left (5 \, A - C\right )} a^{3} b^{4} + 6 \, A a b^{6} - {\left (A a^{7} - 2 \, A a^{5} b^{2} + A a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (A a^{6} b - 2 \, A a^{4} b^{3} + A a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d\right )}}, \frac {{\left ({\left (A + 2 \, C\right )} a^{7} + 4 \, {\left (A - C\right )} a^{5} b^{2} - {\left (11 \, A - 2 \, C\right )} a^{3} b^{4} + 6 \, A a b^{6}\right )} d x \cos \left (d x + c\right ) + {\left ({\left (A + 2 \, C\right )} a^{6} b + 4 \, {\left (A - C\right )} a^{4} b^{3} - {\left (11 \, A - 2 \, C\right )} a^{2} b^{5} + 6 \, A b^{7}\right )} d x - 2 \, {\left (2 \, C a^{4} b^{2} + {\left (4 \, A - C\right )} a^{2} b^{4} - 3 \, A b^{6} + {\left (2 \, C a^{5} b + {\left (4 \, A - C\right )} a^{3} b^{3} - 3 \, A a b^{5}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (2 \, {\left (2 \, A - C\right )} a^{5} b^{2} - 2 \, {\left (5 \, A - C\right )} a^{3} b^{4} + 6 \, A a b^{6} - {\left (A a^{7} - 2 \, A a^{5} b^{2} + A a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (A a^{6} b - 2 \, A a^{4} b^{3} + A a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d\right )}}\right ] \]

[In]

integrate(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*(((A + 2*C)*a^7 + 4*(A - C)*a^5*b^2 - (11*A - 2*C)*a^3*b^4 + 6*A*a*b^6)*d*x*cos(d*x + c) + ((A + 2*C)*a^6
*b + 4*(A - C)*a^4*b^3 - (11*A - 2*C)*a^2*b^5 + 6*A*b^7)*d*x + (2*C*a^4*b^2 + (4*A - C)*a^2*b^4 - 3*A*b^6 + (2
*C*a^5*b + (4*A - C)*a^3*b^3 - 3*A*a*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2
)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*
a*b*cos(d*x + c) + b^2)) - (2*(2*A - C)*a^5*b^2 - 2*(5*A - C)*a^3*b^4 + 6*A*a*b^6 - (A*a^7 - 2*A*a^5*b^2 + A*a
^3*b^4)*cos(d*x + c)^2 + 3*(A*a^6*b - 2*A*a^4*b^3 + A*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9 - 2*a^7*b^2 +
a^5*b^4)*d*cos(d*x + c) + (a^8*b - 2*a^6*b^3 + a^4*b^5)*d), 1/2*(((A + 2*C)*a^7 + 4*(A - C)*a^5*b^2 - (11*A -
2*C)*a^3*b^4 + 6*A*a*b^6)*d*x*cos(d*x + c) + ((A + 2*C)*a^6*b + 4*(A - C)*a^4*b^3 - (11*A - 2*C)*a^2*b^5 + 6*
A*b^7)*d*x - 2*(2*C*a^4*b^2 + (4*A - C)*a^2*b^4 - 3*A*b^6 + (2*C*a^5*b + (4*A - C)*a^3*b^3 - 3*A*a*b^5)*cos(d*
x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (2*(2*A -
C)*a^5*b^2 - 2*(5*A - C)*a^3*b^4 + 6*A*a*b^6 - (A*a^7 - 2*A*a^5*b^2 + A*a^3*b^4)*cos(d*x + c)^2 + 3*(A*a^6*b
- 2*A*a^4*b^3 + A*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9 - 2*a^7*b^2 + a^5*b^4)*d*cos(d*x + c) + (a^8*b -
2*a^6*b^3 + a^4*b^5)*d)]

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

integrate(cos(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**2,x)

[Out]

Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**2/(a + b*sec(c + d*x))**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (2 \, C a^{4} b + 4 \, A a^{2} b^{3} - C a^{2} b^{3} - 3 \, A b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {4 \, {\left (C a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} - \frac {{\left (A a^{2} + 2 \, C a^{2} + 6 \, A b^{2}\right )} {\left (d x + c\right )}}{a^{4}} + \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}}}{2 \, d} \]

[In]

integrate(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(4*(2*C*a^4*b + 4*A*a^2*b^3 - C*a^2*b^3 - 3*A*b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + ar
ctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^6 - a^4*b^2)*sqrt(-a^2 + b^2))
+ 4*(C*a^2*b^2*tan(1/2*d*x + 1/2*c) + A*b^4*tan(1/2*d*x + 1/2*c))/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 -
b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - (A*a^2 + 2*C*a^2 + 6*A*b^2)*(d*x + c)/a^4 + 2*(A*a*tan(1/2*d*x + 1/2*c)^
3 + 4*A*b*tan(1/2*d*x + 1/2*c)^3 - A*a*tan(1/2*d*x + 1/2*c) + 4*A*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*
c)^2 + 1)^2*a^3))/d

Mupad [B] (verification not implemented)

Time = 26.06 (sec) , antiderivative size = 6519, normalized size of antiderivative = 25.46 \[ \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

int((cos(c + d*x)^2*(A + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^2,x)

[Out]

(atan((((A*b^2*3i + a^2*((A*1i)/2 + C*1i))*((8*tan(c/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^
2*a*b^9 - 2*A^2*a^9*b - 8*C^2*a^9*b - 120*A^2*a^2*b^8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23
*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 11*A^2*a^8*b^2 + 8*C^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^7*b
^3 + 12*C^2*a^8*b^2 + 4*A*C*a^10 - 8*A*C*a^9*b + 48*A*C*a^2*b^8 - 48*A*C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^
5*b^5 + 36*A*C*a^6*b^4 - 32*A*C*a^7*b^3 + 20*A*C*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (((8*(2*A*a^15
+ 4*C*a^15 - 12*A*a^8*b^7 + 6*A*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11*b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a
^10*b^5 + 12*C*a^12*b^3 - 4*C*a^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (8*tan(c/2 + (d*x
)/2)*(A*b^2*3i + a^2*((A*1i)/2 + C*1i))*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12
*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2)))*(A*b^2*3i + a^2*((A*1i)/2 + C*1i)))/a^4)*1i)/a^4 + ((A*b^2*3i
+ a^2*((A*1i)/2 + C*1i))*((8*tan(c/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^2*a*b^9 - 2*A^2*a^
9*b - 8*C^2*a^9*b - 120*A^2*a^2*b^8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*
A^2*a^7*b^3 + 11*A^2*a^8*b^2 + 8*C^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^7*b^3 + 12*C^2*a^8*b^
2 + 4*A*C*a^10 - 8*A*C*a^9*b + 48*A*C*a^2*b^8 - 48*A*C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^5*b^5 + 36*A*C*a^6
*b^4 - 32*A*C*a^7*b^3 + 20*A*C*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (((8*(2*A*a^15 + 4*C*a^15 - 12*A*
a^8*b^7 + 6*A*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11*b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a^10*b^5 + 12*C*a^1
2*b^3 - 4*C*a^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (8*tan(c/2 + (d*x)/2)*(A*b^2*3i + a
^2*((A*1i)/2 + C*1i))*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b
+ a^9 - a^6*b^3 - a^7*b^2)))*(A*b^2*3i + a^2*((A*1i)/2 + C*1i)))/a^4)*1i)/a^4)/((16*(108*A^3*b^11 - 54*A^3*a*
b^10 + 8*C^3*a^10*b - 216*A^3*a^2*b^9 + 81*A^3*a^3*b^8 + 63*A^3*a^4*b^7 - 9*A^3*a^5*b^6 + 41*A^3*a^6*b^5 - 4*A
^3*a^7*b^4 + 4*A^3*a^8*b^3 + 4*C^3*a^6*b^5 - 4*C^3*a^7*b^4 - 12*C^3*a^8*b^3 + 8*C^3*a^9*b^2 + 8*A*C^2*a^10*b +
2*A^2*C*a^10*b + 36*A*C^2*a^4*b^7 - 30*A*C^2*a^5*b^6 - 96*A*C^2*a^6*b^5 + 52*A*C^2*a^7*b^4 + 52*A*C^2*a^8*b^3
+ 108*A^2*C*a^2*b^9 - 72*A^2*C*a^3*b^8 - 252*A^2*C*a^4*b^7 + 111*A^2*C*a^5*b^6 + 105*A^2*C*a^6*b^5 - 5*A^2*C*
a^7*b^4 + 37*A^2*C*a^8*b^3 - 2*A^2*C*a^9*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - ((A*b^2*3i + a^2*((A*1i)
/2 + C*1i))*((8*tan(c/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^2*a*b^9 - 2*A^2*a^9*b - 8*C^2*a
^9*b - 120*A^2*a^2*b^8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 +
11*A^2*a^8*b^2 + 8*C^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^7*b^3 + 12*C^2*a^8*b^2 + 4*A*C*a^1
0 - 8*A*C*a^9*b + 48*A*C*a^2*b^8 - 48*A*C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^5*b^5 + 36*A*C*a^6*b^4 - 32*A*C
*a^7*b^3 + 20*A*C*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (((8*(2*A*a^15 + 4*C*a^15 - 12*A*a^8*b^7 + 6*A
*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11*b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a^10*b^5 + 12*C*a^12*b^3 - 4*C*a
^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (8*tan(c/2 + (d*x)/2)*(A*b^2*3i + a^2*((A*1i)/2
+ C*1i))*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*
b^3 - a^7*b^2)))*(A*b^2*3i + a^2*((A*1i)/2 + C*1i)))/a^4))/a^4 + ((A*b^2*3i + a^2*((A*1i)/2 + C*1i))*((8*tan(c
/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^2*a*b^9 - 2*A^2*a^9*b - 8*C^2*a^9*b - 120*A^2*a^2*b^
8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 11*A^2*a^8*b^2 + 8*C
^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^7*b^3 + 12*C^2*a^8*b^2 + 4*A*C*a^10 - 8*A*C*a^9*b + 48*
A*C*a^2*b^8 - 48*A*C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^5*b^5 + 36*A*C*a^6*b^4 - 32*A*C*a^7*b^3 + 20*A*C*a^8
*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (((8*(2*A*a^15 + 4*C*a^15 - 12*A*a^8*b^7 + 6*A*a^9*b^6 + 28*A*a^10*
b^5 - 14*A*a^11*b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a^10*b^5 + 12*C*a^12*b^3 - 4*C*a^13*b^2 - 8*C*a^14*b)
)/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (8*tan(c/2 + (d*x)/2)*(A*b^2*3i + a^2*((A*1i)/2 + C*1i))*(8*a^13*b -
8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2)))*(A*b
^2*3i + a^2*((A*1i)/2 + C*1i)))/a^4))/a^4))*(A*b^2*3i + a^2*((A*1i)/2 + C*1i))*2i)/(a^4*d) - ((tan(c/2 + (d*x)
/2)*(A*a^4 + 6*A*b^4 - 5*A*a^2*b^2 + 2*C*a^2*b^2 + 3*A*a*b^3 - 3*A*a^3*b))/((a^3*b - a^4)*(a + b)) + (tan(c/2
+ (d*x)/2)^5*(A*a^4 + 6*A*b^4 - 5*A*a^2*b^2 + 2*C*a^2*b^2 - 3*A*a*b^3 + 3*A*a^3*b))/((a^3*b - a^4)*(a + b)) -
(2*tan(c/2 + (d*x)/2)^3*(A*a^4 - 6*A*b^4 + 3*A*a^2*b^2 - 2*C*a^2*b^2))/(a*(a^2*b - a^3)*(a + b)))/(d*(a + b +
tan(c/2 + (d*x)/2)^2*(a + 3*b) - tan(c/2 + (d*x)/2)^4*(a - 3*b) - tan(c/2 + (d*x)/2)^6*(a - b))) + (b*atan(((b
*((8*tan(c/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^2*a*b^9 - 2*A^2*a^9*b - 8*C^2*a^9*b - 120*
A^2*a^2*b^8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 11*A^2*a^8
*b^2 + 8*C^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^7*b^3 + 12*C^2*a^8*b^2 + 4*A*C*a^10 - 8*A*C*a
^9*b + 48*A*C*a^2*b^8 - 48*A*C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^5*b^5 + 36*A*C*a^6*b^4 - 32*A*C*a^7*b^3 +
20*A*C*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (b*((a + b)^3*(a - b)^3)^(1/2)*((8*(2*A*a^15 + 4*C*a^15 -
12*A*a^8*b^7 + 6*A*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11*b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a^10*b^5 + 12
*C*a^12*b^3 - 4*C*a^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (8*b*tan(c/2 + (d*x)/2)*((a +
b)^3*(a - b)^3)^(1/2)*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^
10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2
)))*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*((a + b)^3*(a - b
)^3)^(1/2)*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2)*1i)/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2) + (b*((8
*tan(c/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^2*a*b^9 - 2*A^2*a^9*b - 8*C^2*a^9*b - 120*A^2*
a^2*b^8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 11*A^2*a^8*b^2
+ 8*C^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^7*b^3 + 12*C^2*a^8*b^2 + 4*A*C*a^10 - 8*A*C*a^9*b
+ 48*A*C*a^2*b^8 - 48*A*C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^5*b^5 + 36*A*C*a^6*b^4 - 32*A*C*a^7*b^3 + 20*A
*C*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (b*((a + b)^3*(a - b)^3)^(1/2)*((8*(2*A*a^15 + 4*C*a^15 - 12*
A*a^8*b^7 + 6*A*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11*b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a^10*b^5 + 12*C*a
^12*b^3 - 4*C*a^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (8*b*tan(c/2 + (d*x)/2)*((a + b)^
3*(a - b)^3)^(1/2)*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b
^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*
(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*((a + b)^3*(a - b)^3)
^(1/2)*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2)*1i)/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))/((16*(108*A
^3*b^11 - 54*A^3*a*b^10 + 8*C^3*a^10*b - 216*A^3*a^2*b^9 + 81*A^3*a^3*b^8 + 63*A^3*a^4*b^7 - 9*A^3*a^5*b^6 + 4
1*A^3*a^6*b^5 - 4*A^3*a^7*b^4 + 4*A^3*a^8*b^3 + 4*C^3*a^6*b^5 - 4*C^3*a^7*b^4 - 12*C^3*a^8*b^3 + 8*C^3*a^9*b^2
+ 8*A*C^2*a^10*b + 2*A^2*C*a^10*b + 36*A*C^2*a^4*b^7 - 30*A*C^2*a^5*b^6 - 96*A*C^2*a^6*b^5 + 52*A*C^2*a^7*b^4
+ 52*A*C^2*a^8*b^3 + 108*A^2*C*a^2*b^9 - 72*A^2*C*a^3*b^8 - 252*A^2*C*a^4*b^7 + 111*A^2*C*a^5*b^6 + 105*A^2*C
*a^6*b^5 - 5*A^2*C*a^7*b^4 + 37*A^2*C*a^8*b^3 - 2*A^2*C*a^9*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (b*((
8*tan(c/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^2*a*b^9 - 2*A^2*a^9*b - 8*C^2*a^9*b - 120*A^2
*a^2*b^8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 11*A^2*a^8*b^
2 + 8*C^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^7*b^3 + 12*C^2*a^8*b^2 + 4*A*C*a^10 - 8*A*C*a^9*
b + 48*A*C*a^2*b^8 - 48*A*C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^5*b^5 + 36*A*C*a^6*b^4 - 32*A*C*a^7*b^3 + 20*
A*C*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (b*((a + b)^3*(a - b)^3)^(1/2)*((8*(2*A*a^15 + 4*C*a^15 - 12
*A*a^8*b^7 + 6*A*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11*b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a^10*b^5 + 12*C*
a^12*b^3 - 4*C*a^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (8*b*tan(c/2 + (d*x)/2)*((a + b)
^3*(a - b)^3)^(1/2)*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*
b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))
*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*((a + b)^3*(a - b)^3
)^(1/2)*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2) + (b*((8*tan(c
/2 + (d*x)/2)*(A^2*a^10 + 72*A^2*b^10 + 4*C^2*a^10 - 72*A^2*a*b^9 - 2*A^2*a^9*b - 8*C^2*a^9*b - 120*A^2*a^2*b^
8 + 120*A^2*a^3*b^7 + 17*A^2*a^4*b^6 - 26*A^2*a^5*b^5 + 23*A^2*a^6*b^4 - 20*A^2*a^7*b^3 + 11*A^2*a^8*b^2 + 8*C
^2*a^4*b^6 - 8*C^2*a^5*b^5 - 20*C^2*a^6*b^4 + 16*C^2*a^7*b^3 + 12*C^2*a^8*b^2 + 4*A*C*a^10 - 8*A*C*a^9*b + 48*
A*C*a^2*b^8 - 48*A*C*a^3*b^7 - 100*A*C*a^4*b^6 + 88*A*C*a^5*b^5 + 36*A*C*a^6*b^4 - 32*A*C*a^7*b^3 + 20*A*C*a^8
*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (b*((a + b)^3*(a - b)^3)^(1/2)*((8*(2*A*a^15 + 4*C*a^15 - 12*A*a^8*
b^7 + 6*A*a^9*b^6 + 28*A*a^10*b^5 - 14*A*a^11*b^4 - 16*A*a^12*b^3 + 6*A*a^13*b^2 - 4*C*a^10*b^5 + 12*C*a^12*b^
3 - 4*C*a^13*b^2 - 8*C*a^14*b))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (8*b*tan(c/2 + (d*x)/2)*((a + b)^3*(a -
b)^3)^(1/2)*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 1
6*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(3*A*b
^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*((a + b)^3*(a - b)^3)^(1/2)
*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*((a + b)^3*(a - b)^
3)^(1/2)*(3*A*b^4 - 2*C*a^4 - 4*A*a^2*b^2 + C*a^2*b^2)*2i)/(d*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))