3.998 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=238 \[ \frac{\left (5 a^2 A b^3-3 a^4 b (2 A+C)+a^3 b^2 B+2 a^5 B-2 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x) \left (-a^2 b^2 (5 A+2 C)+3 a^3 b B+a^4 (-C)+2 A b^4\right )}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

[Out]

((5*a^2*A*b^3 - 2*A*b^5 + 2*a^5*B + a^3*b^2*B - 3*a^4*b*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[
a + b]])/(a^3*(a - b)^(5/2)*(a + b)^(5/2)*d) + (A*ArcTanh[Sin[c + d*x]])/(a^3*d) + ((A*b^2 - a*(b*B - a*C))*Si
n[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((2*A*b^4 + 3*a^3*b*B - a^4*C - a^2*b^2*(5*A + 2*C))*
Sin[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.978207, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3055, 3001, 3770, 2659, 205} \[ \frac{\left (5 a^2 A b^3-3 a^4 b (2 A+C)+a^3 b^2 B+2 a^5 B-2 A b^5\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x) \left (-a^2 b^2 (5 A+2 C)+3 a^3 b B+a^4 (-C)+2 A b^4\right )}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^3,x]

[Out]

((5*a^2*A*b^3 - 2*A*b^5 + 2*a^5*B + a^3*b^2*B - 3*a^4*b*(2*A + C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[
a + b]])/(a^3*(a - b)^(5/2)*(a + b)^(5/2)*d) + (A*ArcTanh[Sin[c + d*x]])/(a^3*d) + ((A*b^2 - a*(b*B - a*C))*Si
n[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((2*A*b^4 + 3*a^3*b*B - a^4*C - a^2*b^2*(5*A + 2*C))*
Sin[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 205

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

Rubi steps

\begin{align*} \int \frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (2 A \left (a^2-b^2\right )-2 a (A b-a B+b C) \cos (c+d x)+\left (A b^2-a (b B-a C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (2 A b^4+3 a^3 b B-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (2 A \left (a^2-b^2\right )^2+a \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (2 A b^4+3 a^3 b B-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{A \int \sec (c+d x) \, dx}{a^3}+\frac{\left (5 a^2 A b^3-2 A b^5+2 a^5 B+a^3 b^2 B-3 a^4 b (2 A+C)\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (2 A b^4+3 a^3 b B-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (5 a^2 A b^3-2 A b^5+2 a^5 B+a^3 b^2 B-3 a^4 b (2 A+C)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 \left (a^2-b^2\right )^2 d}\\ &=-\frac{\left (6 a^4 A b-5 a^2 A b^3+2 A b^5-2 a^5 B-a^3 b^2 B+3 a^4 b C\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac{\left (2 A b^4+3 a^3 b B-a^4 C-a^2 b^2 (5 A+2 C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [C]  time = 4.31554, size = 473, normalized size = 1.99 \[ \frac{\cos (c+d x) (A \sec (c+d x)+B+C \cos (c+d x)) \left (-\frac{4 i (\cos (c)-i \sin (c)) \left (5 a^2 A b^3-3 a^4 b (2 A+C)+a^3 b^2 B+2 a^5 B-2 A b^5\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}\right )}{\left (a^2-b^2\right )^2 \sqrt{-\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2}}+\frac{a \left (b \sec (c) \left (b \left (a \sin (2 c+d x) \left (-a^2 b (4 A+3 C)+2 a^3 B+a b^2 B+A b^3\right )+\sin (c+2 d x) \left (a^2 b^2 (5 A+2 C)-3 a^3 b B+a^4 C-2 A b^4\right )\right )+a \sin (d x) \left (a^2 b^2 (16 A+5 C)-10 a^3 b B+4 a^4 C+a b^3 B-7 A b^4\right )\right )-\left (2 a^2+b^2\right ) \tan (c) \left (a^2 b^2 (5 A+2 C)-3 a^3 b B+a^4 C-2 A b^4\right )\right )}{b \left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-4 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{2 a^3 d (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + b*Cos[c + d*x])^3,x]

[Out]

(Cos[c + d*x]*(B + C*Cos[c + d*x] + A*Sec[c + d*x])*(-4*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*A*Log[C
os[(c + d*x)/2] + Sin[(c + d*x)/2]] - ((4*I)*(5*a^2*A*b^3 - 2*A*b^5 + 2*a^5*B + a^3*b^2*B - 3*a^4*b*(2*A + C))
*ArcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a + b*Cos[c])*Tan[(d*x)/2]))/Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])
^2)]]*(Cos[c] - I*Sin[c]))/((a^2 - b^2)^2*Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]) + (a*(b*Sec[c]*(a*(-7*A*
b^4 - 10*a^3*b*B + a*b^3*B + 4*a^4*C + a^2*b^2*(16*A + 5*C))*Sin[d*x] + b*(a*(A*b^3 + 2*a^3*B + a*b^2*B - a^2*
b*(4*A + 3*C))*Sin[2*c + d*x] + (-2*A*b^4 - 3*a^3*b*B + a^4*C + a^2*b^2*(5*A + 2*C))*Sin[c + 2*d*x])) - (2*a^2
+ b^2)*(-2*A*b^4 - 3*a^3*b*B + a^4*C + a^2*b^2*(5*A + 2*C))*Tan[c]))/(b*(a^2 - b^2)^2*(a + b*Cos[c + d*x])^2)
))/(2*a^3*d*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)]))

________________________________________________________________________________________

Maple [B]  time = 0.087, size = 1507, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x)

[Out]

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

________________________________________________________________________________________

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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.30691, size = 842, normalized size = 3.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

((2*B*a^5 - 6*A*a^4*b - 3*C*a^4*b + B*a^3*b^2 + 5*A*a^2*b^3 - 2*A*b^5)*(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^7 - 2*a^5*b^2 + a^3
*b^4)*sqrt(a^2 - b^2)) + A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 +
(2*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 4*B*a^4*b*tan(1/2*d*x + 1/2*c)^3 - C*a^4*b*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3
*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*A*a^2*
b^3*tan(1/2*d*x + 1/2*c)^3 + B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*A*a*b^4
*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^5*tan(1/2*d*x + 1/2*c)^3 + 2*C*a^5*tan(1/2*d*x + 1/2*c) - 4*B*a^4*b*tan(1/2*d*
x + 1/2*c) + C*a^4*b*tan(1/2*d*x + 1/2*c) + 6*A*a^3*b^2*tan(1/2*d*x + 1/2*c) - 3*B*a^3*b^2*tan(1/2*d*x + 1/2*c
) + C*a^3*b^2*tan(1/2*d*x + 1/2*c) + 5*A*a^2*b^3*tan(1/2*d*x + 1/2*c) + B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 2*C*a
^2*b^3*tan(1/2*d*x + 1/2*c) - 3*A*a*b^4*tan(1/2*d*x + 1/2*c) - 2*A*b^5*tan(1/2*d*x + 1/2*c))/((a^6 - 2*a^4*b^2
+ a^2*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 + a + b)^2))/d