3.85 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac{15 a^3 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{2 \sqrt{2} c^{5/2} f}+\frac{15 a^3 \tan (e+f x)}{4 c^2 f \sqrt{c-c \sec (e+f x)}}+\frac{5 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{4 c f (c-c \sec (e+f x))^{3/2}}-\frac{a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}} \]

[Out]

(-15*a^3*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(2*Sqrt[2]*c^(5/2)*f) - (a*(a + a*
Sec[e + f*x])^2*Tan[e + f*x])/(2*f*(c - c*Sec[e + f*x])^(5/2)) + (5*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(4*
c*f*(c - c*Sec[e + f*x])^(3/2)) + (15*a^3*Tan[e + f*x])/(4*c^2*f*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.351981, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3957, 3956, 3795, 203} \[ -\frac{15 a^3 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{2 \sqrt{2} c^{5/2} f}+\frac{15 a^3 \tan (e+f x)}{4 c^2 f \sqrt{c-c \sec (e+f x)}}+\frac{5 \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{4 c f (c-c \sec (e+f x))^{3/2}}-\frac{a \tan (e+f x) (a \sec (e+f x)+a)^2}{2 f (c-c \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a^3*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(2*Sqrt[2]*c^(5/2)*f) - (a*(a + a*
Sec[e + f*x])^2*Tan[e + f*x])/(2*f*(c - c*Sec[e + f*x])^(5/2)) + (5*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(4*
c*f*(c - c*Sec[e + f*x])^(3/2)) + (15*a^3*Tan[e + f*x])/(4*c^2*f*Sqrt[c - c*Sec[e + f*x]])

Rule 3957

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

Rule 3956

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.)
+ (a_)], x_Symbol] :> Simp[(-2*d*Cot[e + f*x]*(c + d*Csc[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*
x]]), x] + Dist[(2*c*(2*n - 1))/(2*n - 1), Int[(Csc[e + f*x]*(c + d*Csc[e + f*x])^(n - 1))/Sqrt[a + b*Csc[e +
f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^{5/2}} \, dx &=-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}-\frac{(5 a) \int \frac{\sec (e+f x) (a+a \sec (e+f x))^2}{(c-c \sec (e+f x))^{3/2}} \, dx}{4 c}\\ &=-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac{\left (15 a^2\right ) \int \frac{\sec (e+f x) (a+a \sec (e+f x))}{\sqrt{c-c \sec (e+f x)}} \, dx}{8 c^2}\\ &=-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac{15 a^3 \tan (e+f x)}{4 c^2 f \sqrt{c-c \sec (e+f x)}}+\frac{\left (15 a^3\right ) \int \frac{\sec (e+f x)}{\sqrt{c-c \sec (e+f x)}} \, dx}{4 c^2}\\ &=-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac{15 a^3 \tan (e+f x)}{4 c^2 f \sqrt{c-c \sec (e+f x)}}-\frac{\left (15 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 c+x^2} \, dx,x,\frac{c \tan (e+f x)}{\sqrt{c-c \sec (e+f x)}}\right )}{2 c^2 f}\\ &=-\frac{15 a^3 \tan ^{-1}\left (\frac{\sqrt{c} \tan (e+f x)}{\sqrt{2} \sqrt{c-c \sec (e+f x)}}\right )}{2 \sqrt{2} c^{5/2} f}-\frac{a (a+a \sec (e+f x))^2 \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac{5 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{4 c f (c-c \sec (e+f x))^{3/2}}+\frac{15 a^3 \tan (e+f x)}{4 c^2 f \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 4.5866, size = 263, normalized size = 1.51 \[ -\frac{a^3 e^{-\frac{1}{2} i (2 e+f x)} \tan ^5\left (\frac{1}{2} (e+f x)\right ) \sec \left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+1)^3 \left (120 e^{\frac{i e}{2}} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac{1+e^{i (e+f x)}}{\sqrt{2} \sqrt{1+e^{2 i (e+f x)}}}\right )+\left (25 \cos \left (\frac{3}{2} (e+f x)\right )-9 \cos \left (\frac{5}{2} (e+f x)\right )\right ) \csc ^4\left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec (e+f x)} \left (\cos \left (e+\frac{f x}{2}\right )+i \sin \left (e+\frac{f x}{2}\right )\right )\right )}{32 c^2 f (\sec (e+f x)-1)^2 \sqrt{\sec (e+f x)} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sec[(e + f*x)/2]*(1 + Sec[e + f*x])^3*(120*E^((I/2)*e)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*S
qrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[(1 + E^(I*(e + f*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])] + (25*Cos[
(3*(e + f*x))/2] - 9*Cos[(5*(e + f*x))/2])*Csc[(e + f*x)/2]^4*Sqrt[Sec[e + f*x]]*(Cos[e + (f*x)/2] + I*Sin[e +
(f*x)/2]))*Tan[(e + f*x)/2]^5)/(32*c^2*E^((I/2)*(2*e + f*x))*f*(-1 + Sec[e + f*x])^2*Sqrt[Sec[e + f*x]]*Sqrt[
c - c*Sec[e + f*x]])

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Maple [A]  time = 0.233, size = 206, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}\sin \left ( fx+e \right ) }{4\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ( 15\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}-30\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}\cos \left ( fx+e \right ) -18\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+15\,\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}+34\,\cos \left ( fx+e \right ) -8 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^3/f*(15*cos(f*x+e)^2*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2
)-30*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)-18*cos(f*x
+e)^2+15*arctan(1/(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)+34*cos(f*x+e)-8)*
sin(f*x+e)/cos(f*x+e)^3/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)

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Maxima [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(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.668452, size = 1107, normalized size = 6.36 \begin{align*} \left [-\frac{15 \, \sqrt{2}{\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \sqrt{-c} \log \left (\frac{2 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{-c} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} +{\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \,{\left (9 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{8 \,{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, \frac{15 \, \sqrt{2}{\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) + a^{3}\right )} \sqrt{c} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \,{\left (9 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{4 \,{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(15*sqrt(2)*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^3)*sqrt(-c)*log((2*sqrt(2)*(cos(f*x + e)^2 + co
s(f*x + e))*sqrt(-c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) + (3*c*cos(f*x + e) + c)*sin(f*x + e))/((cos(f*x
+ e) - 1)*sin(f*x + e)))*sin(f*x + e) + 4*(9*a^3*cos(f*x + e)^3 - 8*a^3*cos(f*x + e)^2 - 13*a^3*cos(f*x + e) +
4*a^3)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*
x + e)), 1/4*(15*sqrt(2)*(a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) + a^3)*sqrt(c)*arctan(sqrt(2)*sqrt((c*cos(f*
x + e) - c)/cos(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f*x + e)))*sin(f*x + e) - 2*(9*a^3*cos(f*x + e)^3 - 8*a^3*
cos(f*x + e)^2 - 13*a^3*cos(f*x + e) + 4*a^3)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2
- 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \frac{\sec{\left (e + f x \right )}}{c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec{\left (e + f x \right )} + c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c}}\, dx + \int \frac{3 \sec ^{2}{\left (e + f x \right )}}{c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec{\left (e + f x \right )} + c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c}}\, dx + \int \frac{3 \sec ^{3}{\left (e + f x \right )}}{c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec{\left (e + f x \right )} + c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c}}\, dx + \int \frac{\sec ^{4}{\left (e + f x \right )}}{c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c} \sec{\left (e + f x \right )} + c^{2} \sqrt{- c \sec{\left (e + f x \right )} + c}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(Integral(sec(e + f*x)/(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**2 - 2*c**2*sqrt(-c*sec(e + f*x) + c)
*sec(e + f*x) + c**2*sqrt(-c*sec(e + f*x) + c)), x) + Integral(3*sec(e + f*x)**2/(c**2*sqrt(-c*sec(e + f*x) +
c)*sec(e + f*x)**2 - 2*c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c**2*sqrt(-c*sec(e + f*x) + c)), x) + Int
egral(3*sec(e + f*x)**3/(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**2 - 2*c**2*sqrt(-c*sec(e + f*x) + c)*sec
(e + f*x) + c**2*sqrt(-c*sec(e + f*x) + c)), x) + Integral(sec(e + f*x)**4/(c**2*sqrt(-c*sec(e + f*x) + c)*sec
(e + f*x)**2 - 2*c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c**2*sqrt(-c*sec(e + f*x) + c)), x))

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Giac [A]  time = 2.47044, size = 305, normalized size = 1.75 \begin{align*} -\frac{a^{3} c{\left (\frac{15 \, \sqrt{2} \arctan \left (\frac{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}{\sqrt{c}}\right )}{c^{\frac{7}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} + \frac{8 \, \sqrt{2}}{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} + \frac{\sqrt{2}{\left (7 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} + 9 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c\right )}}{c^{5} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4}}\right )}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^3*c*(15*sqrt(2)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c))/(c^(7/2)*sgn(tan(1/2*f*x + 1/2*e)^2
- 1)*sgn(tan(1/2*f*x + 1/2*e))) + 8*sqrt(2)/(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^3*sgn(tan(1/2*f*x + 1/2*e)^2
- 1)*sgn(tan(1/2*f*x + 1/2*e))) + sqrt(2)*(7*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2) + 9*sqrt(c*tan(1/2*f*x + 1/
2*e)^2 - c)*c)/(c^5*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)*sgn(tan(1/2*f*x + 1/2*e))*tan(1/2*f*x + 1/2*e)^4))/f