\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx\) [282]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {(A-B) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A+3 B) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(2 A+3 B) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )} \]

[Out]

-1/5*(A-B)*cos(f*x+e)/f/(a+a*sin(f*x+e))^3-1/15*(2*A+3*B)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^2-1/15*(2*A+3*B)*cos
(f*x+e)/f/(a^3+a^3*sin(f*x+e))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2829, 2729, 2727} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {(2 A+3 B) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(2 A+3 B) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac {(A-B) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]

[In]

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

[Out]

-1/5*((A - B)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^3) - ((2*A + 3*B)*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x
])^2) - ((2*A + 3*B)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x]))

Rule 2727

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

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2829

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}+\frac {(2 A+3 B) \int \frac {1}{(a+a \sin (e+f x))^2} \, dx}{5 a} \\ & = -\frac {(A-B) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A+3 B) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}+\frac {(2 A+3 B) \int \frac {1}{a+a \sin (e+f x)} \, dx}{15 a^2} \\ & = -\frac {(A-B) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {(2 A+3 B) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(2 A+3 B) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {\cos (e+f x) \left (7 A+3 B+(6 A+9 B) \sin (e+f x)+(2 A+3 B) \sin ^2(e+f x)\right )}{15 a^3 f (1+\sin (e+f x))^3} \]

[In]

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

[Out]

-1/15*(Cos[e + f*x]*(7*A + 3*B + (6*A + 9*B)*Sin[e + f*x] + (2*A + 3*B)*Sin[e + f*x]^2))/(a^3*f*(1 + Sin[e + f
*x])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {2 i \left (20 i A \,{\mathrm e}^{2 i \left (f x +e \right )}+15 i B \,{\mathrm e}^{2 i \left (f x +e \right )}+15 B \,{\mathrm e}^{3 i \left (f x +e \right )}-2 i A -10 A \,{\mathrm e}^{i \left (f x +e \right )}-3 i B -15 B \,{\mathrm e}^{i \left (f x +e \right )}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\)	\(95\)
parallelrisch	\(\frac {-30 A \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-60 A -30 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-80 A -30 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-40 A -30 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-14 A -6 B}{15 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\)	\(98\)
derivativedivides	\(\frac {-\frac {2 A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (8 A -6 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-8 A +8 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A -4 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A +2 B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{3} f}\)	\(114\)
default	\(\frac {-\frac {2 A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (8 A -6 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-8 A +8 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A -4 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A +2 B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{3} f}\)	\(114\)
norman	\(\frac {-\frac {14 A +6 B}{15 f a}-\frac {2 A \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {\left (94 A +36 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 f a}-\frac {\left (20 A +12 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {\left (22 A +6 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}-\frac {\left (8 A +6 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f a}-\frac {\left (4 A +2 B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\)	\(197\)

[In]

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

[Out]

-2/15*I*(20*I*A*exp(2*I*(f*x+e))+15*I*B*exp(2*I*(f*x+e))+15*B*exp(3*I*(f*x+e))-2*I*A-10*A*exp(I*(f*x+e))-3*I*B
-15*B*exp(I*(f*x+e)))/f/a^3/(exp(I*(f*x+e))+I)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.86 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {{\left (2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (3 \, A + 2 \, B\right )} \cos \left (f x + e\right ) - {\left ({\left (2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, A + 3 \, B\right )} \cos \left (f x + e\right ) - 3 \, A + 3 \, B\right )} \sin \left (f x + e\right ) - 3 \, A + 3 \, B}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

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

[Out]

-1/15*((2*A + 3*B)*cos(f*x + e)^3 - 2*(2*A + 3*B)*cos(f*x + e)^2 - 3*(3*A + 2*B)*cos(f*x + e) - ((2*A + 3*B)*c
os(f*x + e)^2 + 3*(2*A + 3*B)*cos(f*x + e) - 3*A + 3*B)*sin(f*x + e) - 3*A + 3*B)/(a^3*f*cos(f*x + e)^3 + 3*a^
3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*
sin(f*x + e))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1015 vs. \(2 (87) = 174\).

Time = 2.37 (sec) , antiderivative size = 1015, normalized size of antiderivative = 9.95 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3,x)

[Out]

Piecewise((-30*A*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3
*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*tan(e
/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3
+ 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*A*tan(e/2 + f*x/2)**2/(15*a**
3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2
+ f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 40*A*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 +
75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*t
an(e/2 + f*x/2) + 15*a**3*f) - 14*A/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*
f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*tan(e/
2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3
+ 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*tan(e/2 + f*x/2)**2/(15*a**3
*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 +
f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 +
75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*ta
n(e/2 + f*x/2) + 15*a**3*f) - 6*B/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*
tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f), Ne(f, 0)), (x*
(A + B*sin(e))/(a*sin(e) + a)**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (96) = 192\).

Time = 0.30 (sec) , antiderivative size = 387, normalized size of antiderivative = 3.79 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (\frac {A {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {3 \, B {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \]

[In]

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

[Out]

-2/15*(A*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos
(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(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) + 3*B*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(co
s(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))/f

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, A + 3 \, B\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \]

[In]

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

[Out]

-2/15*(15*A*tan(1/2*f*x + 1/2*e)^4 + 30*A*tan(1/2*f*x + 1/2*e)^3 + 15*B*tan(1/2*f*x + 1/2*e)^3 + 40*A*tan(1/2*
f*x + 1/2*e)^2 + 15*B*tan(1/2*f*x + 1/2*e)^2 + 20*A*tan(1/2*f*x + 1/2*e) + 15*B*tan(1/2*f*x + 1/2*e) + 7*A + 3
*B)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)

Mupad [B] (verification not implemented)

Time = 13.02 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.47 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {53\,A}{4}+3\,B-4\,A\,\cos \left (e+f\,x\right )+\frac {3\,B\,\cos \left (e+f\,x\right )}{2}+\frac {25\,A\,\sin \left (e+f\,x\right )}{2}+\frac {15\,B\,\sin \left (e+f\,x\right )}{2}-\frac {9\,A\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {3\,B\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {5\,A\,\sin \left (2\,e+2\,f\,x\right )}{4}\right )}{15\,a^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}-\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \]

[In]

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

[Out]

(2*cos(e/2 + (f*x)/2)*((53*A)/4 + 3*B - 4*A*cos(e + f*x) + (3*B*cos(e + f*x))/2 + (25*A*sin(e + f*x))/2 + (15*
B*sin(e + f*x))/2 - (9*A*cos(2*e + 2*f*x))/4 - (3*B*cos(2*e + 2*f*x))/2 - (5*A*sin(2*e + 2*f*x))/4))/(15*a^3*f
*((5*2^(1/2)*cos((3*e)/2 + pi/4 + (3*f*x)/2))/4 - (5*2^(1/2)*cos(e/2 - pi/4 + (f*x)/2))/2 + (2^(1/2)*cos((5*e)
/2 - pi/4 + (5*f*x)/2))/4))