3.256 \(\int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=198 \[ -\frac{2 a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^3 f (c+d) \sqrt{c^2-d^2}}+\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 f (c+d)}-\frac{a^2 x (-A d+2 B c-2 B d)}{d^3}+\frac{(B c-A d) \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{d f (c+d) (c+d \sin (e+f x))} \]

[Out]

-((a^2*(2*B*c - A*d - 2*B*d)*x)/d^3) - (2*a^2*(c - d)*(A*d*(c + 2*d) - B*(2*c^2 + 2*c*d - d^2))*ArcTan[(d + c*
Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d^3*(c + d)*Sqrt[c^2 - d^2]*f) + (a^2*(A*d - B*(2*c + d))*Cos[e + f*x])/(
d^2*(c + d)*f) + ((B*c - A*d)*Cos[e + f*x]*(a^2 + a^2*Sin[e + f*x]))/(d*(c + d)*f*(c + d*Sin[e + f*x]))

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Rubi [A]  time = 0.580826, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2975, 2968, 3023, 2735, 2660, 618, 204} \[ -\frac{2 a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{d^3 f (c+d) \sqrt{c^2-d^2}}+\frac{a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 f (c+d)}-\frac{a^2 x (-A d+2 B c-2 B d)}{d^3}+\frac{(B c-A d) \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right )}{d f (c+d) (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^2,x]

[Out]

-((a^2*(2*B*c - A*d - 2*B*d)*x)/d^3) - (2*a^2*(c - d)*(A*d*(c + 2*d) - B*(2*c^2 + 2*c*d - d^2))*ArcTan[(d + c*
Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(d^3*(c + d)*Sqrt[c^2 - d^2]*f) + (a^2*(A*d - B*(2*c + d))*Cos[e + f*x])/(
d^2*(c + d)*f) + ((B*c - A*d)*Cos[e + f*x]*(a^2 + a^2*Sin[e + f*x]))/(d*(c + d)*f*(c + d*Sin[e + f*x]))

Rule 2975

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

Rule 2968

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

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

Mathematica [A]  time = 1.00535, size = 192, normalized size = 0.97 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (\frac{2 (c-d) \left (B \left (2 c^2+2 c d-d^2\right )-A d (c+2 d)\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d) \sqrt{c^2-d^2}}+(e+f x) (A d-2 B c+2 B d)-\frac{d (d-c) (A d-B c) \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}-B d \cos (e+f x)\right )}{d^3 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c + d*Sin[e + f*x])^2,x]

[Out]

(a^2*(1 + Sin[e + f*x])^2*((-2*B*c + A*d + 2*B*d)*(e + f*x) + (2*(c - d)*(-(A*d*(c + 2*d)) + B*(2*c^2 + 2*c*d
- d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)*Sqrt[c^2 - d^2]) - B*d*Cos[e + f*x] - (d*(-
c + d)*(-(B*c) + A*d)*Cos[e + f*x])/((c + d)*(c + d*Sin[e + f*x]))))/(d^3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2])^4)

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Maple [B]  time = 0.161, size = 848, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x)

[Out]

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

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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((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.56487, size = 1589, normalized size = 8.03 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*(2*B*a^2*c^3 - A*a^2*c^2*d - (A + 2*B)*a^2*c*d^2)*f*x + (2*B*a^2*c^3 - (A - 2*B)*a^2*c^2*d - (2*A + B
)*a^2*c*d^2 + (2*B*a^2*c^2*d - (A - 2*B)*a^2*c*d^2 - (2*A + B)*a^2*d^3)*sin(f*x + e))*sqrt(-(c - d)/(c + d))*l
og(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*((c^2 + c*d)*cos(f*x + e)*sin(f*x + e) +
(c*d + d^2)*cos(f*x + e))*sqrt(-(c - d)/(c + d)))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 2*
(2*B*a^2*c^2*d - A*a^2*c*d^2 + A*a^2*d^3)*cos(f*x + e) + 2*((2*B*a^2*c^2*d - A*a^2*c*d^2 - (A + 2*B)*a^2*d^3)*
f*x + (B*a^2*c*d^2 + B*a^2*d^3)*cos(f*x + e))*sin(f*x + e))/((c*d^4 + d^5)*f*sin(f*x + e) + (c^2*d^3 + c*d^4)*
f), -((2*B*a^2*c^3 - A*a^2*c^2*d - (A + 2*B)*a^2*c*d^2)*f*x + (2*B*a^2*c^3 - (A - 2*B)*a^2*c^2*d - (2*A + B)*a
^2*c*d^2 + (2*B*a^2*c^2*d - (A - 2*B)*a^2*c*d^2 - (2*A + B)*a^2*d^3)*sin(f*x + e))*sqrt((c - d)/(c + d))*arcta
n(-(c*sin(f*x + e) + d)*sqrt((c - d)/(c + d))/((c - d)*cos(f*x + e))) + (2*B*a^2*c^2*d - A*a^2*c*d^2 + A*a^2*d
^3)*cos(f*x + e) + ((2*B*a^2*c^2*d - A*a^2*c*d^2 - (A + 2*B)*a^2*d^3)*f*x + (B*a^2*c*d^2 + B*a^2*d^3)*cos(f*x
+ e))*sin(f*x + e))/((c*d^4 + d^5)*f*sin(f*x + e) + (c^2*d^3 + c*d^4)*f)]

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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((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.35175, size = 672, normalized size = 3.39 \begin{align*} \frac{\frac{2 \,{\left (2 \, B a^{2} c^{3} - A a^{2} c^{2} d - A a^{2} c d^{2} - 3 \, B a^{2} c d^{2} + 2 \, A a^{2} d^{3} + B a^{2} d^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d}{\sqrt{c^{2} - d^{2}}}\right )\right )}}{{\left (c d^{3} + d^{4}\right )} \sqrt{c^{2} - d^{2}}} - \frac{2 \,{\left (B a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - A a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - B a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + A a^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, B a^{2} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - A a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + A a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B a^{2} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - A a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + B a^{2} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + A a^{2} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, B a^{2} c^{3} - A a^{2} c^{2} d + A a^{2} c d^{2}\right )}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + c\right )}{\left (c^{2} d^{2} + c d^{3}\right )}} - \frac{{\left (2 \, B a^{2} c - A a^{2} d - 2 \, B a^{2} d\right )}{\left (f x + e\right )}}{d^{3}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

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