3.1086 \(\int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=180 \[ -\frac{\left (-9 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 d \left (a^2-b^2\right )^{3/2}}+\frac{a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{3 a x}{b^4}-\frac{\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac{3 \cos (c+d x)}{2 b^3 d} \]

[Out]

(3*a*x)/b^4 - ((6*a^4 - 9*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*(a^2 - b^2)^
(3/2)*d) + (3*Cos[c + d*x])/(2*b^3*d) - (Cos[c + d*x]*Sin[c + d*x]^2)/(2*b*d*(a + b*Sin[c + d*x])^2) + (a*(3*a
^2 - 2*b^2)*Cos[c + d*x])/(2*b^3*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.562332, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {2889, 3048, 3032, 3023, 2735, 2660, 618, 204} \[ -\frac{\left (-9 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 d \left (a^2-b^2\right )^{3/2}}+\frac{a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac{3 a x}{b^4}-\frac{\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac{3 \cos (c+d x)}{2 b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

(3*a*x)/b^4 - ((6*a^4 - 9*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*(a^2 - b^2)^
(3/2)*d) + (3*Cos[c + d*x])/(2*b^3*d) - (Cos[c + d*x]*Sin[c + d*x]^2)/(2*b*d*(a + b*Sin[c + d*x])^2) + (a*(3*a
^2 - 2*b^2)*Cos[c + d*x])/(2*b^3*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

Rule 2889

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f,
m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])

Rule 3048

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

Rule 3032

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \frac{\sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^3} \, dx\\ &=-\frac{\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac{\int \frac{\sin (c+d x) \left (-2 \left (a^2-b^2\right )+3 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac{a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\int \frac{b \left (3 a^4-5 a^2 b^2+2 b^4\right )+3 a \left (a^2-b^2\right )^2 \sin (c+d x)-3 b \left (a^2-b^2\right )^2 \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{3 \cos (c+d x)}{2 b^3 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac{a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\int \frac{b^2 \left (3 a^4-5 a^2 b^2+2 b^4\right )+6 a b \left (a^2-b^2\right )^2 \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{3 a x}{b^4}+\frac{3 \cos (c+d x)}{2 b^3 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac{a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\left (6 a^4-9 a^2 b^2+2 b^4\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )}\\ &=\frac{3 a x}{b^4}+\frac{3 \cos (c+d x)}{2 b^3 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac{a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac{\left (6 a^4-9 a^2 b^2+2 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^4 \left (a^2-b^2\right ) d}\\ &=\frac{3 a x}{b^4}+\frac{3 \cos (c+d x)}{2 b^3 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac{a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac{\left (2 \left (6 a^4-9 a^2 b^2+2 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^4 \left (a^2-b^2\right ) d}\\ &=\frac{3 a x}{b^4}-\frac{\left (6 a^4-9 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2} d}+\frac{3 \cos (c+d x)}{2 b^3 d}-\frac{\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac{a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.13289, size = 159, normalized size = 0.88 \[ \frac{-\frac{2 \left (-9 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a b \left (5 a^2-4 b^2\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))}-\frac{a^2 b \cos (c+d x)}{(a+b \sin (c+d x))^2}+6 a (c+d x)+2 b \cos (c+d x)}{2 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

(6*a*(c + d*x) - (2*(6*a^4 - 9*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^
(3/2) + 2*b*Cos[c + d*x] - (a^2*b*Cos[c + d*x])/(a + b*Sin[c + d*x])^2 + (a*b*(5*a^2 - 4*b^2)*Cos[c + d*x])/((
a - b)*(a + b)*(a + b*Sin[c + d*x])))/(2*b^4*d)

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Maple [B]  time = 0.139, size = 711, normalized size = 4. \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)^2*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x)

[Out]

2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)+6/d/b^4*a*arctan(tan(1/2*d*x+1/2*c))+3/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/
2*d*x+1/2*c)*b+a)^2*a^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3-2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2
*a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3+4/d/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/
2*d*x+1/2*c)^2*a^4+5/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2*a^
2-6/d*b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2+13/d/b^2/(tan(1/2*d
*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)-10/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(
1/2*d*x+1/2*c)*b+a)^2*a/(a^2-b^2)*tan(1/2*d*x+1/2*c)+4/d/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)
^2*a^4/(a^2-b^2)-3/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^2/(a^2-b^2)-6/d/b^4/(a^2-b^2)^(3/
2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*a^4+9/d/b^2/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/
2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*a^2-2/d/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(
1/2))

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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)^2*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.90576, size = 1975, normalized size = 10.97 \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)^2*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(12*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*x*cos(d*x + c)^2 + 4*(a^4*b^3 - 2*a^2*b^5 + b^7)*cos(d*x + c)^3 - 12*
(a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*x - (6*a^6 - 3*a^4*b^2 - 7*a^2*b^4 + 2*b^6 - (6*a^4*b^2 - 9*a^2*b^4 + 2*b^
6)*cos(d*x + c)^2 + 2*(6*a^5*b - 9*a^3*b^3 + 2*a*b^5)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*
x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))
/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*(6*a^6*b - 9*a^4*b^3 + a^2*b^5 + 2*b^7)*cos(d*x +
c) - 2*(12*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*x + (9*a^5*b^2 - 17*a^3*b^4 + 8*a*b^6)*cos(d*x + c))*sin(d*x + c))/
((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c)^2 - 2*(a^5*b^5 - 2*a^3*b^7 + a*b^9)*d*sin(d*x + c) - (a^6*b^4 - a
^4*b^6 - a^2*b^8 + b^10)*d), 1/2*(6*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*x*cos(d*x + c)^2 + 2*(a^4*b^3 - 2*a^2*b^5
+ b^7)*cos(d*x + c)^3 - 6*(a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*x - (6*a^6 - 3*a^4*b^2 - 7*a^2*b^4 + 2*b^6 - (6*
a^4*b^2 - 9*a^2*b^4 + 2*b^6)*cos(d*x + c)^2 + 2*(6*a^5*b - 9*a^3*b^3 + 2*a*b^5)*sin(d*x + c))*sqrt(a^2 - b^2)*
arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (6*a^6*b - 9*a^4*b^3 + a^2*b^5 + 2*b^7)*cos(d*x
+ c) - (12*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*x + (9*a^5*b^2 - 17*a^3*b^4 + 8*a*b^6)*cos(d*x + c))*sin(d*x + c))
/((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c)^2 - 2*(a^5*b^5 - 2*a^3*b^7 + a*b^9)*d*sin(d*x + c) - (a^6*b^4 -
a^4*b^6 - a^2*b^8 + b^10)*d)]

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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(cos(d*x+c)**2*sin(d*x+c)**2/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.38279, size = 408, normalized size = 2.27 \begin{align*} -\frac{\frac{{\left (6 \, a^{4} - 9 \, a^{2} b^{2} + 2 \, b^{4}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt{a^{2} - b^{2}}} - \frac{3 \, a^{3} b \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} + 4 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 5 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 13 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 10 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a^{4} - 3 \, a^{2} b^{2}}{{\left (a^{2} b^{3} - b^{5}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}^{2}} - \frac{3 \,{\left (d x + c\right )} a}{b^{4}} - \frac{2}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} b^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((6*a^4 - 9*a^2*b^2 + 2*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/s
qrt(a^2 - b^2)))/((a^2*b^4 - b^6)*sqrt(a^2 - b^2)) - (3*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^3*tan(1/2*d*x + 1
/2*c)^3 + 4*a^4*tan(1/2*d*x + 1/2*c)^2 + 5*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*b^4*tan(1/2*d*x + 1/2*c)^2 + 13*
a^3*b*tan(1/2*d*x + 1/2*c) - 10*a*b^3*tan(1/2*d*x + 1/2*c) + 4*a^4 - 3*a^2*b^2)/((a^2*b^3 - b^5)*(a*tan(1/2*d*
x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) - 3*(d*x + c)*a/b^4 - 2/((tan(1/2*d*x + 1/2*c)^2 + 1)*b^3))/d