3.318 \(\int \frac{\cos ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=178 \[ \frac{\left (2 a^2 A-3 a b B+3 A b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac{2 b^3 (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{x \left (a^2+2 b^2\right ) (A b-a B)}{2 a^4}-\frac{(A b-a B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.641879, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4034, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (2 a^2 A-3 a b B+3 A b^2\right ) \sin (c+d x)}{3 a^3 d}+\frac{2 b^3 (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{x \left (a^2+2 b^2\right ) (A b-a B)}{2 a^4}-\frac{(A b-a B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{A \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]

[Out]

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

Rule 4034

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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*(n +
1)*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, m}, x] && NeQ[A*b
- a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

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 ^3(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx &=\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac{\int \frac{\cos ^2(c+d x) \left (3 (A b-a B)-2 a A \sec (c+d x)-2 A b \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a}\\ &=-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\int \frac{\cos (c+d x) \left (2 \left (2 a^2 A+3 A b^2-3 a b B\right )+a (A b+3 a B) \sec (c+d x)-3 b (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^2}\\ &=\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac{\int \frac{3 \left (a^2+2 b^2\right ) (A b-a B)+3 a b (A b-a B) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^3}\\ &=-\frac{\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (b^3 (A b-a B)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4}\\ &=-\frac{\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (b^2 (A b-a B)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^4}\\ &=-\frac{\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (2 b^2 (A b-a B)\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^4 d}\\ &=-\frac{\left (a^2+2 b^2\right ) (A b-a B) x}{2 a^4}+\frac{2 b^3 (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 \sqrt{a-b} \sqrt{a+b} d}+\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \sin (c+d x)}{3 a^3 d}-\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{A \cos ^2(c+d x) \sin (c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.487204, size = 152, normalized size = 0.85 \[ \frac{6 \left (a^2+2 b^2\right ) (c+d x) (a B-A b)+3 a \left (3 a^2 A-4 a b B+4 A b^2\right ) \sin (c+d x)-\frac{24 b^3 (A b-a B) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+3 a^2 (a B-A b) \sin (2 (c+d x))+a^3 A \sin (3 (c+d x))}{12 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x]),x]

[Out]

(6*(a^2 + 2*b^2)*(-(A*b) + a*B)*(c + d*x) - (24*b^3*(A*b - a*B)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 -
b^2]])/Sqrt[a^2 - b^2] + 3*a*(3*a^2*A + 4*A*b^2 - 4*a*b*B)*Sin[c + d*x] + 3*a^2*(-(A*b) + a*B)*Sin[2*(c + d*x
)] + a^3*A*Sin[3*(c + d*x)])/(12*a^4*d)

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Maple [B]  time = 0.102, size = 641, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)

[Out]

2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^
5*A*b+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A*b^2-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d
*x+1/2*c)^5*B-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*B*b+4/3/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*t
an(1/2*d*x+1/2*c)^3*A+4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A*b^2-4/d/a^2/(1+tan(1/2*d*x+1/2
*c)^2)^3*tan(1/2*d*x+1/2*c)^3*B*b+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A+2/d/a^3/(1+tan(1/2*d*x
+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b^2-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*B*b-1/d/a^2/(1+tan
(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*A*b+1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)*B-1/d/a^2*A*ar
ctan(tan(1/2*d*x+1/2*c))*b-2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*A*b^3+1/d/a*B*arctan(tan(1/2*d*x+1/2*c))+2/d/a^3
*arctan(tan(1/2*d*x+1/2*c))*B*b^2+2/d*b^4/a^4/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b
))^(1/2))*A-2/d*b^3/a^3/((a+b)*(a-b))^(1/2)*arctanh((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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.583899, size = 1177, normalized size = 6.61 \begin{align*} \left [\frac{3 \,{\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x - 3 \,{\left (B a b^{3} - A b^{4}\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 (4 \, A a^{5} - 6 \, B a^{4} b + 2 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \,{\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - a^{4} b^{2}\right )} d}, \frac{3 \,{\left (B a^{5} - A a^{4} b + B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} d x - 6 \,{\left (B a b^{3} - A b^{4}\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 (4 \, A a^{5} - 6 \, B a^{4} b + 2 \, A a^{3} b^{2} + 6 \, B a^{2} b^{3} - 6 \, A a b^{4} + 2 \,{\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

[1/6*(3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*d*x - 3*(B*a*b^3 - A*b^4)*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)) + (4*A*a^5 - 6*B*a^4*b + 2*A*a^3*b^2 + 6*B*
a^2*b^3 - 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c)^2 + 3*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d
*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d), 1/6*(3*(B*a^5 - A*a^4*b + B*a^3*b^2 - A*a^2*b^3 - 2*B*a*b^4 + 2*A*
b^5)*d*x - 6*(B*a*b^3 - A*b^4)*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))) + (4*A*a^5 - 6*B*a^4*b + 2*A*a^3*b^2 + 6*B*a^2*b^3 - 6*A*a*b^4 + 2*(A*a^5 - A*a^3*b^2)*cos(d*x + c
)^2 + 3*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*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 ^{3}{\left (c + d x \right )}}{a + b \sec{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*cos(c + d*x)**3/(a + b*sec(c + d*x)), x)

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Giac [B]  time = 1.22299, size = 486, normalized size = 2.73 \begin{align*} \frac{\frac{3 \,{\left (B a^{3} - A a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )}{\left (d x + c\right )}}{a^{4}} - \frac{12 \,{\left (B a b^{3} - 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 )}}{\sqrt{-a^{2} + b^{2}} a^{4}} + \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A b^{2} \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 )}^{3} a^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(3*(B*a^3 - A*a^2*b + 2*B*a*b^2 - 2*A*b^3)*(d*x + c)/a^4 - 12*(B*a*b^3 - 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)))/(sqrt(-
a^2 + b^2)*a^4) + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b*tan(1/2*d*x + 1
/2*c)^5 - 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 1
2*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) + 3*B*a^2*tan(
1/2*d*x + 1/2*c) - 3*A*a*b*tan(1/2*d*x + 1/2*c) - 6*B*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c))
/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3))/d