\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx\) [850]

Optimal result

Integrand size = 45, antiderivative size = 206 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {-i A-B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[Out]

8/15*A*tan(f*x+e)/a^2/c^2/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)+1/5*(-I*A-B)/f/(a+I*a*tan(f*x+e)
)^(5/2)/(c-I*c*tan(f*x+e))^(5/2)+1/5*I*A/c/f/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(3/2)+4/15*A*tan(f*x+
e)/a/c/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3669, 79, 47, 40, 39} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}-\frac {B+i A}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \]

[In]

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

[Out]

-1/5*(I*A + B)/(f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)) + ((I/5)*A)/(c*f*(a + I*a*Tan[e +
f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (4*A*Tan[e + f*x])/(15*a*c*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*
c*Tan[e + f*x])^(3/2)) + (8*A*Tan[e + f*x])/(15*a^2*c^2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]
])

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-x)*(a + b*x)^(m + 1)*((c + d*x)^(m
+ 1)/(2*a*c*(m + 1))), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /;
FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 3669

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^{7/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {(a A) \text {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {(4 A) \text {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f} \\ & = -\frac {i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {(8 A) \text {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f} \\ & = -\frac {i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.44 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\cos ^4(e+f x) \left (-3 B+15 A \tan (e+f x)+20 A \tan ^3(e+f x)+8 A \tan ^5(e+f x)\right )}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(Cos[e + f*x]^4*(-3*B + 15*A*Tan[e + f*x] + 20*A*Tan[e + f*x]^3 + 8*A*Tan[e + f*x]^5))/(15*a^2*c^2*f*Sqrt[a +
I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.60

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 A \tan \left (f x +e \right )^{7}+28 A \tan \left (f x +e \right )^{5}+35 A \tan \left (f x +e \right )^{3}-3 B \tan \left (f x +e \right )^{2}+15 A \tan \left (f x +e \right )-3 B \right )}{15 f \,a^{3} c^{3} \left (i-\tan \left (f x +e \right )\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(124\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 A \tan \left (f x +e \right )^{7}+28 A \tan \left (f x +e \right )^{5}+35 A \tan \left (f x +e \right )^{3}-3 B \tan \left (f x +e \right )^{2}+15 A \tan \left (f x +e \right )-3 B \right )}{15 f \,a^{3} c^{3} \left (i-\tan \left (f x +e \right )\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(124\)
parts	\(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right ) \left (8 \tan \left (f x +e \right )^{4}+20 \tan \left (f x +e \right )^{2}+15\right )}{15 f \,a^{3} c^{3} \left (i+\tan \left (f x +e \right )\right )^{4} \left (i-\tan \left (f x +e \right )\right )^{4}}-\frac {B \left (1+\tan \left (f x +e \right )^{2}\right ) \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}}{5 f \,c^{3} a^{3} \left (i+\tan \left (f x +e \right )\right )^{4} \left (i-\tan \left (f x +e \right )\right )^{4}}\)	\(184\)

[In]

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

[Out]

1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^3/c^3*(8*A*tan(f*x+e)^7+28*A*tan(f*x+e)^5+35*A
*tan(f*x+e)^3-3*B*tan(f*x+e)^2+15*A*tan(f*x+e)-3*B)/(I-tan(f*x+e))^4/(I+tan(f*x+e))^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.90 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (3 \, {\left (i \, A + B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} + 2 \, {\left (14 i \, A + 9 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + 5 \, {\left (35 i \, A + 9 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 96 \, B e^{\left (7 i \, f x + 7 i \, e\right )} + 60 \, B e^{\left (6 i \, f x + 6 i \, e\right )} - 96 \, B e^{\left (5 i \, f x + 5 i \, e\right )} + 5 \, {\left (-35 i \, A + 9 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-14 i \, A + 9 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{480 \, a^{3} c^{3} f} \]

[In]

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

[Out]

-1/480*(3*(I*A + B)*e^(12*I*f*x + 12*I*e) + 2*(14*I*A + 9*B)*e^(10*I*f*x + 10*I*e) + 5*(35*I*A + 9*B)*e^(8*I*f
*x + 8*I*e) - 96*B*e^(7*I*f*x + 7*I*e) + 60*B*e^(6*I*f*x + 6*I*e) - 96*B*e^(5*I*f*x + 5*I*e) + 5*(-35*I*A + 9*
B)*e^(4*I*f*x + 4*I*e) + 2*(-14*I*A + 9*B)*e^(2*I*f*x + 2*I*e) - 3*I*A + 3*B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)
)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-5*I*f*x - 5*I*e)/(a^3*c^3*f)

Sympy [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral((A + B*tan(e + f*x))/((I*a*(tan(e + f*x) - I))**(5/2)*(-I*c*(tan(e + f*x) + I))**(5/2)), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 334 vs. \(2 (160) = 320\).

Time = 0.59 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (30 \, {\left (5 i \, A - B\right )} \cos \left (4 \, f x + 4 \, e\right ) + 5 \, {\left (5 i \, A - 3 \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) - 30 \, {\left (5 \, A + i \, B\right )} \sin \left (4 \, f x + 4 \, e\right ) - 5 \, {\left (5 \, A + 3 i \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) - 6 \, B\right )} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 5 \, {\left (-5 i \, A - 3 \, B\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 30 \, {\left (-5 i \, A - B\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (30 \, {\left (5 \, A + i \, B\right )} \cos \left (4 \, f x + 4 \, e\right ) + 5 \, {\left (5 \, A + 3 i \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) + 30 \, {\left (5 i \, A - B\right )} \sin \left (4 \, f x + 4 \, e\right ) + 5 \, {\left (5 i \, A - 3 \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, A\right )} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 5 \, {\left (5 \, A - 3 i \, B\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 30 \, {\left (5 \, A - i \, B\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )}{480 \, a^{\frac {5}{2}} c^{\frac {5}{2}} f} \]

[In]

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

[Out]

1/480*((30*(5*I*A - B)*cos(4*f*x + 4*e) + 5*(5*I*A - 3*B)*cos(2*f*x + 2*e) - 30*(5*A + I*B)*sin(4*f*x + 4*e) -
5*(5*A + 3*I*B)*sin(2*f*x + 2*e) - 6*B)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 5*(-5*I*A - 3*
B)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 30*(-5*I*A - B)*cos(1/2*arctan2(sin(2*f*x + 2*e), co
s(2*f*x + 2*e))) + (30*(5*A + I*B)*cos(4*f*x + 4*e) + 5*(5*A + 3*I*B)*cos(2*f*x + 2*e) + 30*(5*I*A - B)*sin(4*
f*x + 4*e) + 5*(5*I*A - 3*B)*sin(2*f*x + 2*e) + 6*A)*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 5*
(5*A - 3*I*B)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 30*(5*A - I*B)*sin(1/2*arctan2(sin(2*f*x
+ 2*e), cos(2*f*x + 2*e))))/(a^(5/2)*c^(5/2)*f)

Giac [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate((B*tan(f*x + e) + A)/((I*a*tan(f*x + e) + a)^(5/2)*(-I*c*tan(f*x + e) + c)^(5/2)), x)

Mupad [B] (verification not implemented)

Time = 10.48 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.21 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (175\,A\,\sin \left (2\,e+2\,f\,x\right )-30\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,125{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,22{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-45\,B\,\cos \left (2\,e+2\,f\,x\right )-18\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )-A\,150{}\mathrm {i}+28\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,12{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{480\,a^3\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^(5/2)*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*cos(2*e + 2*f*x)*125i - 30
*B - A*150i + A*cos(4*e + 4*f*x)*22i + A*cos(6*e + 6*f*x)*3i - 45*B*cos(2*e + 2*f*x) - 18*B*cos(4*e + 4*f*x) -
3*B*cos(6*e + 6*f*x) + 175*A*sin(2*e + 2*f*x) + 28*A*sin(4*e + 4*f*x) + 3*A*sin(6*e + 6*f*x) + B*sin(2*e + 2*
f*x)*15i + B*sin(4*e + 4*f*x)*12i + B*sin(6*e + 6*f*x)*3i))/(480*a^3*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2
*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))