3.325 \(\int \frac{\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=180 \[ \frac{\left (a^2 A+a b B-2 A b^2\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{2 b \left (3 a^2 A b-2 a^3 B+a b^2 B-2 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x (2 A b-a B)}{a^3} \]

[Out]

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

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Rubi [A]  time = 0.568946, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4030, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (a^2 A+a b B-2 A b^2\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{2 b \left (3 a^2 A b-2 a^3 B+a b^2 B-2 A b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x (2 A b-a B)}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4030

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

Rule 4104

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]

Rule 3919

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 3831

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

Rule 2659

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 208

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

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\frac{b (A b-a B) \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 (-a^2 A+2 A b^2-a b B+a (A b-a B) \sec (c+d x)-b (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A-2 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-\left (a^2-b^2\right ) (2 A b-a B)+a b (A b-a B) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=-\frac{(2 A b-a B) x}{a^3}+\frac{\left (a^2 A-2 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (b \left (3 a^2 A b-2 A b^3-2 a^3 B+a b^2 B\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac{(2 A b-a B) x}{a^3}+\frac{\left (a^2 A-2 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (3 a^2 A b-2 A b^3-2 a^3 B+a b^2 B\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac{(2 A b-a B) x}{a^3}+\frac{\left (a^2 A-2 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 \left (3 a^2 A b-2 A b^3-2 a^3 B+a b^2 B\right )\right ) \operatorname{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^3 \left (a^2-b^2\right ) d}\\ &=-\frac{(2 A b-a B) x}{a^3}+\frac{2 b \left (3 a^2 A b-2 A b^3-2 a^3 B+a b^2 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (a^2 A-2 A b^2+a b B\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.09333, size = 221, normalized size = 1.23 \[ \frac{(a \cos (c+d x)+b) (A+B \sec (c+d x)) \left (\frac{2 b \left (-3 a^2 A b+2 a^3 B-a b^2 B+2 A b^3\right ) \sec (c+d x) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a b^2 (a B-A b) \tan (c+d x)}{(a-b) (a+b)}+(c+d x) (a B-2 A b) \sec (c+d x) (a \cos (c+d x)+b)+a A \tan (c+d x) (a \cos (c+d x)+b)\right )}{a^3 d (a+b \sec (c+d x))^2 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*(A + B*Sec[c + d*x])*((-2*A*b + a*B)*(c + d*x)*(b + a*Cos[c + d*x])*Sec[c + d*x] + (2*b*
(-3*a^2*A*b + 2*A*b^3 + 2*a^3*B - a*b^2*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c +
d*x])*Sec[c + d*x])/(a^2 - b^2)^(3/2) + (a*b^2*(-(A*b) + a*B)*Tan[c + d*x])/((a - b)*(a + b)) + a*A*(b + a*Co
s[c + d*x])*Tan[c + d*x]))/(a^3*d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^2)

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Maple [B]  time = 0.116, size = 453, normalized size = 2.5 \begin{align*} 2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-4\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) b}{d{a}^{3}}}+2\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) A}{d{a}^{2} \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}-2\,{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) B}{ad \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+6\,{\frac{A{b}^{2}}{ad \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-4\,{\frac{A{b}^{4}}{d{a}^{3} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-4\,{\frac{Bb}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{B{b}^{3}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.660387, size = 1715, normalized size = 9.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(B*a^6 - 2*A*a^5*b - 2*B*a^4*b^2 + 4*A*a^3*b^3 + B*a^2*b^4 - 2*A*a*b^5)*d*x*cos(d*x + c) + 2*(B*a^5*b
- 2*A*a^4*b^2 - 2*B*a^3*b^3 + 4*A*a^2*b^4 + B*a*b^5 - 2*A*b^6)*d*x + (2*B*a^3*b^2 - 3*A*a^2*b^3 - B*a*b^4 + 2*
A*b^5 + (2*B*a^4*b - 3*A*a^3*b^2 - B*a^2*b^3 + 2*A*a*b^4)*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*(A*a^5*b + B*a^4*b^2 - 3*A*a^3*b^3 - B*a^2*b^4 + 2*A*a*b^5 + (A*a
^6 - 2*A*a^4*b^2 + A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^8 - 2*a^6*b^2 + a^4*b^4)*d*cos(d*x + c) + (a^7*b
- 2*a^5*b^3 + a^3*b^5)*d), ((B*a^6 - 2*A*a^5*b - 2*B*a^4*b^2 + 4*A*a^3*b^3 + B*a^2*b^4 - 2*A*a*b^5)*d*x*cos(d
*x + c) + (B*a^5*b - 2*A*a^4*b^2 - 2*B*a^3*b^3 + 4*A*a^2*b^4 + B*a*b^5 - 2*A*b^6)*d*x - (2*B*a^3*b^2 - 3*A*a^2
*b^3 - B*a*b^4 + 2*A*b^5 + (2*B*a^4*b - 3*A*a^3*b^2 - B*a^2*b^3 + 2*A*a*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*ar
ctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (A*a^5*b + B*a^4*b^2 - 3*A*a^3*b^3 -
B*a^2*b^4 + 2*A*a*b^5 + (A*a^6 - 2*A*a^4*b^2 + A*a^2*b^4)*cos(d*x + c))*sin(d*x + c))/((a^8 - 2*a^6*b^2 + a^4
*b^4)*d*cos(d*x + c) + (a^7*b - 2*a^5*b^3 + a^3*b^5)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \cos{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.39815, size = 505, normalized size = 2.81 \begin{align*} -\frac{\frac{2 \,{\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2} - B a b^{3} + 2 \, A b^{4}\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^{5} - a^{3} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{2 \,{\left (A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}{\left (a^{4} - a^{2} b^{2}\right )}} - \frac{{\left (B a - 2 \, A b\right )}{\left (d x + c\right )}}{a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(2*B*a^3*b - 3*A*a^2*b^2 - B*a*b^3 + 2*A*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-
(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^5 - a^3*b^2)*sqrt(-a^2 + b^2)) - 2*(A
*a^3*tan(1/2*d*x + 1/2*c)^3 - A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - B*a*b^2*tan(1/
2*d*x + 1/2*c)^3 + 2*A*b^3*tan(1/2*d*x + 1/2*c)^3 - A*a^3*tan(1/2*d*x + 1/2*c) - A*a^2*b*tan(1/2*d*x + 1/2*c)
+ A*a*b^2*tan(1/2*d*x + 1/2*c) - B*a*b^2*tan(1/2*d*x + 1/2*c) + 2*A*b^3*tan(1/2*d*x + 1/2*c))/((a*tan(1/2*d*x
+ 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*b*tan(1/2*d*x + 1/2*c)^2 - a - b)*(a^4 - a^2*b^2)) - (B*a - 2*A*b)*(
d*x + c)/a^3)/d