\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx\) [125]

Optimal result

Integrand size = 38, antiderivative size = 160 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {32 (A-11 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f} \]

[Out]

-32/15*(A-11*B)*c*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/f+8/5*(A-11*B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/a
^3/f-1/5*(A-11*B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(7/2)/a^3/c/f-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(11/2)/a
^3/c^3/f

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3046, 2934, 2753, 2752} \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}-\frac {(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}+\frac {8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {32 c (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f} \]

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x])^3,x]

[Out]

(-32*(A - 11*B)*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) + (8*(A - 11*B)*Sec[e + f*x]^3*(c - c*
Sin[e + f*x])^(5/2))/(5*a^3*f) - ((A - 11*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^3*c*f) - ((A - B)
*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(11/2))/(5*a^3*c^3*f)

Rule 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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]

Rule 2753

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 2934

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 3046

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]))

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{11/2} \, dx}{a^3 c^3} \\ & = -\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}-\frac {(A-11 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{10 a^3 c^2} \\ & = -\frac {(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}-\frac {(4 (A-11 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^3 c} \\ & = \frac {8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}+\frac {(16 (A-11 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a^3} \\ & = -\frac {32 (A-11 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.57 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (58 A-488 B-30 (A-8 B) \cos (2 (e+f x))+5 (8 A-133 B) \sin (e+f x)+15 B \sin (3 (e+f x)))}{30 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3} \]

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x])^3,x]

[Out]

-1/30*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(58*A - 488*B - 30*(A - 8*B)*Cos[2*(
e + f*x)] + 5*(8*A - 133*B)*Sin[e + f*x] + 15*B*Sin[3*(e + f*x)]))/(a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]
)*(1 + Sin[e + f*x])^3)

Maple [A] (verified)

Time = 118.78 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {2 c^{3} \left (\sin \left (f x +e \right )-1\right ) \left (15 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-15 A +120 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (10 A -170 B \right ) \sin \left (f x +e \right )+22 A -182 B \right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(105\)

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2/15*c^3/a^3*(sin(f*x+e)-1)/(1+sin(f*x+e))^2*(15*B*cos(f*x+e)^2*sin(f*x+e)+(-15*A+120*B)*cos(f*x+e)^2+(10*A-17
0*B)*sin(f*x+e)+22*A-182*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, {\left (A - 8 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (11 \, A - 91 \, B\right )} c^{2} - 5 \, {\left (3 \, B c^{2} \cos \left (f x + e\right )^{2} + 2 \, {\left (A - 17 \, B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/15*(15*(A - 8*B)*c^2*cos(f*x + e)^2 - 2*(11*A - 91*B)*c^2 - 5*(3*B*c^2*cos(f*x + e)^2 + 2*(A - 17*B)*c^2)*s
in(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))

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 761 vs. \(2 (144) = 288\).

Time = 0.31 (sec) , antiderivative size = 761, normalized size of antiderivative = 4.76 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/15*((7*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 95*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 80*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 250*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 120*c^(5
/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 250*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 80*c^(5/2)*sin(f*x
+ e)^7/(cos(f*x + e) + 1)^7 + 95*c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 20*c^(5/2)*sin(f*x + e)^9/(cos
(f*x + e) + 1)^9 + 7*c^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*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)^(5/2)) - 2*(31*c^(5/2) + 155*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 395*c^(5/2)*sin(f*x + e)^2/(cos(f*
x + e) + 1)^2 + 680*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1030*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + 1050*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 1030*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 6
80*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 395*c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 155*c^(5/2)
*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 31*c^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*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/(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)^(5/2)))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (144) = 288\).

Time = 0.51 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.81 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\frac {4 \, \sqrt {2} \sqrt {c} {\left (\frac {15 \, B c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a^{3} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}} + \frac {4 \, A c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 29 \, B c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {20 \, A c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {130 \, B c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {40 \, A c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - \frac {200 \, B c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - \frac {90 \, B c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} - \frac {15 \, B c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}}}{a^{3} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{5}}\right )}}{15 \, f} \]

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

4/15*sqrt(2)*sqrt(c)*(15*B*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a^3*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/
(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)) + (4*A*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 29*B*c^2*sgn(sin(-
1/4*pi + 1/2*f*x + 1/2*e)) + 20*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))
/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 130*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f
*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 40*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1
/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 200*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) -
1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 90*B*c^2*(cos(-1/4*pi + 1/2
*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 - 15*B*c^2*(co
s(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4
)/(a^3*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)^5))/f

Mupad [B] (verification not implemented)

Time = 22.84 (sec) , antiderivative size = 904, normalized size of antiderivative = 5.65 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x))^3,x)

[Out]

((c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((2*B*c^2)/(a^3*f) - (B*c^2*exp(e*1i
+ f*x*1i)*2i)/(a^3*f)))/(exp(e*1i + f*x*1i) - 1i) - (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 -
(exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((c^2*(A*2i - B*7i)*1i)/(3*a^3*f) - (2*c^2*(7*A - 12*B))/(3*a^3*f) + (c^2*(A
*23i - B*28i)*2i)/(3*a^3*f) - (c^2*(42*A - 67*B))/(15*a^3*f) + (2*B*c^2)/(3*a^3*f)))/((exp(e*1i + f*x*1i) - 1i
)*(exp(e*1i + f*x*1i) + 1i)^3) + (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)
*1i)/2))^(1/2)*((c^2*(A*1i - B*4i)*4i)/(a^3*f) + (4*B*c^2)/(a^3*f)))/((exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*
x*1i) + 1i)) - (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((8
*c^2*(A*1i - B*1i))/(a^3*f) + (c^2*(A*1i - B*3i))/(2*a^3*f) + (c^2*(A*11i - B*1i))/(10*a^3*f) + (c^2*(12*A - 1
7*B)*1i)/(4*a^3*f) + (c^2*(52*A - 47*B)*1i)/(4*a^3*f)))/((exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^4
) + (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((c^2*(A*1i -
B*4i))/(a^3*f) + (c^2*(A*5i - B*4i))/(3*a^3*f) + (c^2*(A - 2*B)*8i)/(a^3*f)))/((exp(e*1i + f*x*1i) - 1i)*(exp(
e*1i + f*x*1i) + 1i)^2) + (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)
)^(1/2)*((c^2*(A*2i - B*5i)*1i)/(5*a^3*f) - (c^2*(4*A - 3*B))/(a^3*f) - (c^2*(2*A - 5*B))/(5*a^3*f) + (c^2*(A*
4i - B*3i)*1i)/(a^3*f) - (c^2*(10*A - 11*B))/(5*a^3*f) + (c^2*(A*10i - B*11i)*1i)/(5*a^3*f) + (2*B*c^2)/(5*a^3
*f)))/((exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^5)