3.124 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=209 \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}-\frac{128 c^2 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}-\frac{(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac{4 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}+\frac{32 c (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f} \]
[Out]
(-128*(3*A - 13*B)*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) + (32*(3*A - 13*B)*c*Sec[e + f*x]
^3*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*f) - (4*(3*A - 13*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^3*f
) - ((3*A - 13*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9/2))/(15*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin
[e + f*x])^(13/2))/(5*a^3*c^3*f)
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Rubi [A] time = 0.567435, antiderivative size = 209, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used =
{2967, 2855, 2674, 2673} \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}-\frac{128 c^2 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}-\frac{(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac{4 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}+\frac{32 c (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^3,x]
[Out]
(-128*(3*A - 13*B)*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) + (32*(3*A - 13*B)*c*Sec[e + f*x]
^3*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*f) - (4*(3*A - 13*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^3*f
) - ((3*A - 13*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9/2))/(15*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin
[e + f*x])^(13/2))/(5*a^3*c^3*f)
Rule 2967
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Rule 2855
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
Rule 2674
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
Rule 2673
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Rubi steps
\begin{align*} \int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^3} \, dx &=\frac{\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{13/2} \, dx}{a^3 c^3}\\ &=-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}-\frac{(3 A-13 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{10 a^3 c^2}\\ &=-\frac{(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}-\frac{(2 (3 A-13 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{5 a^3 c}\\ &=-\frac{4 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac{(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}-\frac{(16 (3 A-13 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^3}\\ &=\frac{32 (3 A-13 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac{4 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac{(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}+\frac{(64 (3 A-13 B) c) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a^3}\\ &=-\frac{128 (3 A-13 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac{32 (3 A-13 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac{4 (3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac{(3 A-13 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{5 a^3 c^3 f}\\ \end{align*}
Mathematica [A] time = 2.76265, size = 158, normalized size = 0.76 \[ -\frac{c^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) ((2200 B-540 A) \cos (2 (e+f x))+1410 A \sin (e+f x)-30 A \sin (3 (e+f x))+1092 A-6390 B \sin (e+f x)+170 B \sin (3 (e+f x))+5 B \cos (4 (e+f x))-4557 B)}{60 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^3,x]
[Out]
-(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(1092*A - 4557*B + (-540*A + 2200*B)*Cos[
2*(e + f*x)] + 5*B*Cos[4*(e + f*x)] + 1410*A*Sin[e + f*x] - 6390*B*Sin[e + f*x] - 30*A*Sin[3*(e + f*x)] + 170*
B*Sin[3*(e + f*x)]))/(60*a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3)
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Maple [A] time = 1.097, size = 121, normalized size = 0.6 \begin{align*}{\frac{2\,{c}^{4} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \left ( -15\,A+85\,B \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( 180\,A-820\,B \right ) \sin \left ( fx+e \right ) +5\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( -135\,A+545\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+204\,A-844\,B \right ) }{15\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x)
[Out]
2/15*c^4/a^3*(-1+sin(f*x+e))/(1+sin(f*x+e))^2*((-15*A+85*B)*sin(f*x+e)*cos(f*x+e)^2+(180*A-820*B)*sin(f*x+e)+5
*B*cos(f*x+e)^4+(-135*A+545*B)*cos(f*x+e)^2+204*A-844*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
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Maxima [B] time = 1.6508, size = 1153, normalized size = 5.52 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
2/15*(3*(23*c^(7/2) + 110*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 318*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 590*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1065*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 1
220*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 1540*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1220*c^(7
/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 1065*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 590*c^(7/2)*sin(f
*x + e)^9/(cos(f*x + e) + 1)^9 + 318*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 110*c^(7/2)*sin(f*x + e)^
11/(cos(f*x + e) + 1)^11 + 23*c^(7/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12)*A/((a^3 + 5*a^3*sin(f*x + e)/(cos
(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a
^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*(sin(f*x + e)^2/(cos(f*x + e
) + 1)^2 + 1)^(7/2)) - 2*(147*c^(7/2) + 735*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 1992*c^(7/2)*sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 4015*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 6605*c^(7/2)*sin(f*x + e)^4/(cos
(f*x + e) + 1)^4 + 8370*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 9520*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e
) + 1)^6 + 8370*c^(7/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 6605*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8
+ 4015*c^(7/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 1992*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 735
*c^(7/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 147*c^(7/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12)*B/((a^3 +
5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*(sin(
f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(7/2)))/f
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Fricas [A] time = 1.81164, size = 367, normalized size = 1.76 \begin{align*} \frac{2 \,{\left (5 \, B c^{3} \cos \left (f x + e\right )^{4} - 5 \,{\left (27 \, A - 109 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} + 4 \,{\left (51 \, A - 211 \, B\right )} c^{3} - 5 \,{\left ({\left (3 \, A - 17 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 4 \,{\left (9 \, A - 41 \, B\right )} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
2/15*(5*B*c^3*cos(f*x + e)^4 - 5*(27*A - 109*B)*c^3*cos(f*x + e)^2 + 4*(51*A - 211*B)*c^3 - 5*((3*A - 17*B)*c^
3*cos(f*x + e)^2 - 4*(9*A - 41*B)*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a^3*f*cos(f*x + e)^3 - 2*a^3*f
*cos(f*x + e)*sin(f*x + e) - 2*a^3*f*cos(f*x + e))
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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+B*sin(f*x+e))*(c-c*sin(f*x+e))**(7/2)/(a+a*sin(f*x+e))**3,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 2.61968, size = 2021, normalized size = 9.67 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")
[Out]
1/15*((615*sqrt(2)*A*a^12*sqrt(c) - 3895*sqrt(2)*B*a^12*sqrt(c) - 870*A*a^12*sqrt(c) + 5510*B*a^12*sqrt(c) - 4
26*sqrt(2)*A*c^(13/2) + 1986*sqrt(2)*B*c^(13/2) + 600*A*c^(13/2) - 2800*B*c^(13/2))*sgn(tan(1/2*f*x + 1/2*e) -
1)/(29*sqrt(2)*a^3*c^3 - 41*a^3*c^3) + 5*((((3*A*a^9*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 20*B*a^9*c^5*sgn(tan
(1/2*f*x + 1/2*e) - 1))*tan(1/2*f*x + 1/2*e)/c^6 + 3*(A*a^9*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 6*B*a^9*c^5*sg
n(tan(1/2*f*x + 1/2*e) - 1))/c^6)*tan(1/2*f*x + 1/2*e) + 3*(A*a^9*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 6*B*a^9*
c^5*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^6)*tan(1/2*f*x + 1/2*e) + (3*A*a^9*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 20
*B*a^9*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^6)/(c*tan(1/2*f*x + 1/2*e)^2 + c)^(3/2) + 8*(15*(sqrt(c)*tan(1/2*f
*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^9*A*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 45*(sqrt(c)*tan(1/2*
f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^9*B*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) + 45*(sqrt(c)*tan(1/2
*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*A*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 375*(sqrt(c)*t
an(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*B*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 300*(sqr
t(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*A*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 1060*(
sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*B*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 180
*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*A*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1
) + 860*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*B*c^(11/2)*sgn(tan(1/2*f*x + 1/2
*e) - 1) - 918*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*A*c^6*sgn(tan(1/2*f*x + 1
/2*e) - 1) + 3298*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*B*c^6*sgn(tan(1/2*f*x
+ 1/2*e) - 1) + 630*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*A*c^(13/2)*sgn(tan(1
/2*f*x + 1/2*e) - 1) - 2050*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*B*c^(13/2)*s
gn(tan(1/2*f*x + 1/2*e) - 1) + 780*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*A*c^7
*sgn(tan(1/2*f*x + 1/2*e) - 1) - 2820*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*B*
c^7*sgn(tan(1/2*f*x + 1/2*e) - 1) - 900*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*
A*c^(15/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 3020*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2
+ c))^2*B*c^(15/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 255*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/
2*e)^2 + c))*A*c^8*sgn(tan(1/2*f*x + 1/2*e) - 1) - 925*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/
2*e)^2 + c))*B*c^8*sgn(tan(1/2*f*x + 1/2*e) - 1) - 27*A*c^(17/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 97*B*c^(17/2)
*sgn(tan(1/2*f*x + 1/2*e) - 1))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 + 2*(s
qrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)^5*a^3))/f