3.58 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=102 \[ \frac{2 (3 A-2 B) \tan (e+f x)}{15 a c^3 f}+\frac{(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac{(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.257064, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2967, 2859, 2672, 3767, 8} \[ \frac{2 (3 A-2 B) \tan (e+f x)}{15 a c^3 f}+\frac{(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac{(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + B)*Sec[e + f*x])/(5*a*c*f*(c - c*Sin[e + f*x])^2) + ((3*A - 2*B)*Sec[e + f*x])/(15*a*f*(c^3 - c^3*Sin[e
+ f*x])) + (2*(3*A - 2*B)*Tan[e + f*x])/(15*a*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 2859

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*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 2672

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)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), 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] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^3} \, dx &=\frac{\int \frac{\sec ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx}{a c}\\ &=\frac{(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac{(3 A-2 B) \int \frac{\sec ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{5 a c^2}\\ &=\frac{(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac{(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac{(2 (3 A-2 B)) \int \sec ^2(e+f x) \, dx}{15 a c^3}\\ &=\frac{(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac{(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}-\frac{(2 (3 A-2 B)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{15 a c^3 f}\\ &=\frac{(A+B) \sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2}+\frac{(3 A-2 B) \sec (e+f x)}{15 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac{2 (3 A-2 B) \tan (e+f x)}{15 a c^3 f}\\ \end{align*}

Mathematica [A]  time = 0.842252, size = 157, normalized size = 1.54 \[ -\frac{\cos (e+f x) (5 (B-9 A) \cos (e+f x)+32 (3 A-2 B) \cos (2 (e+f x))+120 A \sin (e+f x)+36 A \sin (2 (e+f x))-24 A \sin (3 (e+f x))+9 A \cos (3 (e+f x))-80 B \sin (e+f x)-4 B \sin (2 (e+f x))+16 B \sin (3 (e+f x))+B (-\cos (3 (e+f x)))+80 B)}{240 a c^3 f (\sin (e+f x)-1)^3 (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[e + f*x]*(80*B + 5*(-9*A + B)*Cos[e + f*x] + 32*(3*A - 2*B)*Cos[2*(e + f*x)] + 9*A*Cos[3*(e + f*x)] - B*
Cos[3*(e + f*x)] + 120*A*Sin[e + f*x] - 80*B*Sin[e + f*x] + 36*A*Sin[2*(e + f*x)] - 4*B*Sin[2*(e + f*x)] - 24*
A*Sin[3*(e + f*x)] + 16*B*Sin[3*(e + f*x)]))/(240*a*c^3*f*(-1 + Sin[e + f*x])^3*(1 + Sin[e + f*x]))

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Maple [A]  time = 0.085, size = 145, normalized size = 1.4 \begin{align*} 2\,{\frac{1}{af{c}^{3}} \left ( -1/5\,{\frac{2\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-1/4\,{\frac{4\,A+4\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-1/2\,{\frac{5/2\,A+3/2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{1}{\tan \left ( 1/2\,fx+e/2 \right ) -1} \left ({\frac{7\,A}{8}}+B/8 \right ) }-1/3\,{\frac{9/2\,A+7/2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-{\frac{A/8-B/8}{\tan \left ( 1/2\,fx+e/2 \right ) +1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a/c^3*(-1/5*(2*A+2*B)/(tan(1/2*f*x+1/2*e)-1)^5-1/4*(4*A+4*B)/(tan(1/2*f*x+1/2*e)-1)^4-1/2*(5/2*A+3/2*B)/(t
an(1/2*f*x+1/2*e)-1)^2-(7/8*A+1/8*B)/(tan(1/2*f*x+1/2*e)-1)-1/3*(9/2*A+7/2*B)/(tan(1/2*f*x+1/2*e)-1)^3-(1/8*A-
1/8*B)/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.02646, size = 571, normalized size = 5.6 \begin{align*} -\frac{2 \,{\left (\frac{B{\left (\frac{4 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{20 \, \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}} - 1\right )}}{a c^{3} - \frac{4 \, a c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{5 \, a c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{5 \, a c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{4 \, a c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac{a c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}} + \frac{3 \, A{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - 2\right )}}{a c^{3} - \frac{4 \, a c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{5 \, a c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{5 \, a c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{4 \, a c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac{a c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}\right )}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(B*(4*sin(f*x + e)/(cos(f*x + e) + 1) - 20*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 20*sin(f*x + e)^3/(cos(
f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)/(a*c^3 - 4*a*c^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 5*a*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*a*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4*a*c^3*sin(f*x
+ e)^5/(cos(f*x + e) + 1)^5 - a*c^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6) + 3*A*(3*sin(f*x + e)/(cos(f*x + e)
+ 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 10*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos
(f*x + e) + 1)^5 - 2)/(a*c^3 - 4*a*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a*c^3*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 - 5*a*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4*a*c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - a*c^3*sin
(f*x + e)^6/(cos(f*x + e) + 1)^6))/f

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Fricas [A]  time = 1.31978, size = 265, normalized size = 2.6 \begin{align*} -\frac{4 \,{\left (3 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} -{\left (2 \,{\left (3 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - 9 \, A + 6 \, B\right )} \sin \left (f x + e\right ) - 6 \, A + 9 \, B}{15 \,{\left (a c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{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))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(4*(3*A - 2*B)*cos(f*x + e)^2 - (2*(3*A - 2*B)*cos(f*x + e)^2 - 9*A + 6*B)*sin(f*x + e) - 6*A + 9*B)/(a*
c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f*x + e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x + e))

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Sympy [A]  time = 31.9582, size = 1732, normalized size = 16.98 \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))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**3,x)

[Out]

Piecewise((-2*A*tan(e/2 + f*x/2)**6/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*
c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) - 2
2*A*tan(e/2 + f*x/2)**5/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e
/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) + 50*A*tan(e/2
+ f*x/2)**4/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)*
*4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) - 60*A*tan(e/2 + f*x/2)**3/
(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c*
*3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) + 10*A*tan(e/2 + f*x/2)**2/(15*a*c**3*f
*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2
+ f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) + 10*A*tan(e/2 + f*x/2)/(15*a*c**3*f*tan(e/2 + f*x/
2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 +
60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) - 10*A/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*
x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 1
5*a*c**3*f) - 7*B*tan(e/2 + f*x/2)**6/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*
a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) +
28*B*tan(e/2 + f*x/2)**5/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan
(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) - 65*B*tan(e/
2 + f*x/2)**4/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2
)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) + 40*B*tan(e/2 + f*x/2)**
3/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*
c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) - 5*B*tan(e/2 + f*x/2)**2/(15*a*c**3*
f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/
2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) - 20*B*tan(e/2 + f*x/2)/(15*a*c**3*f*tan(e/2 + f*x
/2)**6 - 60*a*c**3*f*tan(e/2 + f*x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 +
60*a*c**3*f*tan(e/2 + f*x/2) - 15*a*c**3*f) + 5*B/(15*a*c**3*f*tan(e/2 + f*x/2)**6 - 60*a*c**3*f*tan(e/2 + f*
x/2)**5 + 75*a*c**3*f*tan(e/2 + f*x/2)**4 - 75*a*c**3*f*tan(e/2 + f*x/2)**2 + 60*a*c**3*f*tan(e/2 + f*x/2) - 1
5*a*c**3*f), Ne(f, 0)), (x*(A + B*sin(e))/((a*sin(e) + a)*(-c*sin(e) + c)**3), True))

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Giac [A]  time = 1.22914, size = 239, normalized size = 2.34 \begin{align*} -\frac{\frac{15 \,{\left (A - B\right )}}{a c^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}} + \frac{105 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 15 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 270 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 30 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 360 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 40 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 210 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 50 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 63 \, A - 7 \, B}{a c^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{5}}}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*(A - B)/(a*c^3*(tan(1/2*f*x + 1/2*e) + 1)) + (105*A*tan(1/2*f*x + 1/2*e)^4 + 15*B*tan(1/2*f*x + 1/2*
e)^4 - 270*A*tan(1/2*f*x + 1/2*e)^3 + 30*B*tan(1/2*f*x + 1/2*e)^3 + 360*A*tan(1/2*f*x + 1/2*e)^2 - 40*B*tan(1/
2*f*x + 1/2*e)^2 - 210*A*tan(1/2*f*x + 1/2*e) + 50*B*tan(1/2*f*x + 1/2*e) + 63*A - 7*B)/(a*c^3*(tan(1/2*f*x +
1/2*e) - 1)^5))/f