3.106 \(\int \frac{\cot ^4(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=135 \[ \frac{9 \cot (e+f x)}{8 f \sqrt{a \sin (e+f x)+a}}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{8 \sqrt{a} f}-\frac{\cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}+\frac{\cot (e+f x) \csc (e+f x)}{12 f \sqrt{a \sin (e+f x)+a}} \]

[Out]

(-7*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/(8*Sqrt[a]*f) + (9*Cot[e + f*x])/(8*f*Sqrt[a + a
*Sin[e + f*x]]) + (Cot[e + f*x]*Csc[e + f*x])/(12*f*Sqrt[a + a*Sin[e + f*x]]) - (Cot[e + f*x]*Csc[e + f*x]^2)/
(3*f*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.621613, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2718, 2649, 206, 3044, 2984, 2985, 2773} \[ \frac{9 \cot (e+f x)}{8 f \sqrt{a \sin (e+f x)+a}}-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{8 \sqrt{a} f}-\frac{\cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}+\frac{\cot (e+f x) \csc (e+f x)}{12 f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

(-7*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/(8*Sqrt[a]*f) + (9*Cot[e + f*x])/(8*f*Sqrt[a + a
*Sin[e + f*x]]) + (Cot[e + f*x]*Csc[e + f*x])/(12*f*Sqrt[a + a*Sin[e + f*x]]) - (Cot[e + f*x]*Csc[e + f*x]^2)/
(3*f*Sqrt[a + a*Sin[e + f*x]])

Rule 2718

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Int[(a + b*Sin[e + f*x
])^m, x] + Int[((a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2))/Sin[e + f*x]^4, x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] &&  !LtQ[m, -1]

Rule 2649

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

Rule 206

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

Rule 3044

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

Rule 2984

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

Rule 2985

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

Rule 2773

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

Rubi steps

\begin{align*} \int \frac{\cot ^4(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx &=\int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx+\int \frac{\csc ^4(e+f x) \left (1-2 \sin ^2(e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{\cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{\int \frac{\csc ^3(e+f x) \left (-\frac{a}{2}-\frac{7}{2} a \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{3 a}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{\cot (e+f x) \csc (e+f x)}{12 f \sqrt{a+a \sin (e+f x)}}-\frac{\cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{\int \frac{\csc ^2(e+f x) \left (-\frac{27 a^2}{4}-\frac{3}{4} a^2 \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{6 a^2}\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{9 \cot (e+f x)}{8 f \sqrt{a+a \sin (e+f x)}}+\frac{\cot (e+f x) \csc (e+f x)}{12 f \sqrt{a+a \sin (e+f x)}}-\frac{\cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{\int \frac{\csc (e+f x) \left (\frac{21 a^3}{8}-\frac{27}{8} a^3 \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{6 a^3}\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{9 \cot (e+f x)}{8 f \sqrt{a+a \sin (e+f x)}}+\frac{\cot (e+f x) \csc (e+f x)}{12 f \sqrt{a+a \sin (e+f x)}}-\frac{\cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{7 \int \csc (e+f x) \sqrt{a+a \sin (e+f x)} \, dx}{16 a}-\int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{9 \cot (e+f x)}{8 f \sqrt{a+a \sin (e+f x)}}+\frac{\cot (e+f x) \csc (e+f x)}{12 f \sqrt{a+a \sin (e+f x)}}-\frac{\cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{8 f}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}\\ &=-\frac{7 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{8 \sqrt{a} f}+\frac{9 \cot (e+f x)}{8 f \sqrt{a+a \sin (e+f x)}}+\frac{\cot (e+f x) \csc (e+f x)}{12 f \sqrt{a+a \sin (e+f x)}}-\frac{\cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 0.579591, size = 292, normalized size = 2.16 \[ \frac{\csc ^9\left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-36 \sin \left (\frac{1}{2} (e+f x)\right )-46 \sin \left (\frac{3}{2} (e+f x)\right )+54 \sin \left (\frac{5}{2} (e+f x)\right )+36 \cos \left (\frac{1}{2} (e+f x)\right )-46 \cos \left (\frac{3}{2} (e+f x)\right )-54 \cos \left (\frac{5}{2} (e+f x)\right )-63 \sin (e+f x) \log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )+63 \sin (e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )+21 \sin (3 (e+f x)) \log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-21 \sin (3 (e+f x)) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )\right )}{24 f \sqrt{a (\sin (e+f x)+1)} \left (\csc ^2\left (\frac{1}{4} (e+f x)\right )-\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

(Csc[(e + f*x)/2]^9*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(36*Cos[(e + f*x)/2] - 46*Cos[(3*(e + f*x))/2] - 54*
Cos[(5*(e + f*x))/2] - 36*Sin[(e + f*x)/2] - 63*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[e + f*x] + 63
*Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x] - 46*Sin[(3*(e + f*x))/2] + 54*Sin[(5*(e + f*x))/2]
+ 21*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[3*(e + f*x)] - 21*Log[1 - Cos[(e + f*x)/2] + Sin[(e + f
*x)/2]]*Sin[3*(e + f*x)]))/(24*f*(Csc[(e + f*x)/4]^2 - Sec[(e + f*x)/4]^2)^3*Sqrt[a*(1 + Sin[e + f*x])])

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Maple [A]  time = 0.638, size = 144, normalized size = 1.1 \begin{align*}{\frac{1+\sin \left ( fx+e \right ) }{24\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}\cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( -21\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }}{\sqrt{a}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{3}{a}^{3}+27\, \left ( -a \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{5/2}\sqrt{a}-56\, \left ( -a \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}{a}^{3/2}+21\,\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{5/2} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/2),x)

[Out]

1/24*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(-21*arctanh((-a*(-1+sin(f*x+e)))^(1/2)/a^(1/2))*sin(f*x+e)^3*a
^3+27*(-a*(-1+sin(f*x+e)))^(5/2)*a^(1/2)-56*(-a*(-1+sin(f*x+e)))^(3/2)*a^(3/2)+21*(-a*(-1+sin(f*x+e)))^(1/2)*a
^(5/2))/sin(f*x+e)^3/a^(7/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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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(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.46294, size = 987, normalized size = 7.31 \begin{align*} \frac{21 \,{\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right ) + 1\right )} \sqrt{a} \log \left (\frac{a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 3\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 3\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a} - 9 \, a \cos \left (f x + e\right ) +{\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right ) - 4 \,{\left (27 \, \cos \left (f x + e\right )^{3} + 25 \, \cos \left (f x + e\right )^{2} -{\left (27 \, \cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) - 17\right )} \sin \left (f x + e\right ) - 19 \, \cos \left (f x + e\right ) - 17\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{96 \,{\left (a f \cos \left (f x + e\right )^{4} - 2 \, a f \cos \left (f x + e\right )^{2} + a f -{\left (a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )^{2} - a f \cos \left (f x + e\right ) - a f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/96*(21*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 - (cos(f*x + e)^3 + cos(f*x + e)^2 - cos(f*x + e) - 1)*sin(f*x + e
) + 1)*sqrt(a)*log((a*cos(f*x + e)^3 - 7*a*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + (cos(f*x + e) + 3)*sin(f*x + e
) - 2*cos(f*x + e) - 3)*sqrt(a*sin(f*x + e) + a)*sqrt(a) - 9*a*cos(f*x + e) + (a*cos(f*x + e)^2 + 8*a*cos(f*x
+ e) - a)*sin(f*x + e) - a)/(cos(f*x + e)^3 + cos(f*x + e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) - cos(f*x + e
) - 1)) - 4*(27*cos(f*x + e)^3 + 25*cos(f*x + e)^2 - (27*cos(f*x + e)^2 + 2*cos(f*x + e) - 17)*sin(f*x + e) -
19*cos(f*x + e) - 17)*sqrt(a*sin(f*x + e) + a))/(a*f*cos(f*x + e)^4 - 2*a*f*cos(f*x + e)^2 + a*f - (a*f*cos(f*
x + e)^3 + a*f*cos(f*x + e)^2 - a*f*cos(f*x + e) - a*f)*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{4}{\left (e + f x \right )}}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**4/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral(cot(e + f*x)**4/sqrt(a*(sin(e + f*x) + 1)), x)

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Giac [B]  time = 2.37322, size = 819, normalized size = 6.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/48*(sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a)*((2*tan(1/2*f*x + 1/2*e)/(a*sgn(tan(1/2*f*x + 1/2*e) + 1)) - 3/(a*sgn
(tan(1/2*f*x + 1/2*e) + 1)))*tan(1/2*f*x + 1/2*e) - 22/(a*sgn(tan(1/2*f*x + 1/2*e) + 1))) - (210*sqrt(2)*sqrt(
a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 105*sqrt(2)*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 294*sq
rt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 147*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) - 128*sqrt(2)
*sqrt(-a) - 186*sqrt(-a))*sgn(tan(1/2*f*x + 1/2*e) + 1)/(5*sqrt(2)*sqrt(-a)*sqrt(a) + 7*sqrt(-a)*sqrt(a)) + 42
*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*f
*x + 1/2*e) + 1)) - 21*log(abs(-sqrt(a)*tan(1/2*f*x + 1/2*e) + sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a)))/(sqrt(a)*s
gn(tan(1/2*f*x + 1/2*e) + 1)) - 2*(3*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^5 + 1
8*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^4*sqrt(a) - 48*(sqrt(a)*tan(1/2*f*x + 1/
2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(3/2) - 3*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x +
1/2*e)^2 + a))*a^2 + 22*a^(5/2))/(((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a)^
3*sgn(tan(1/2*f*x + 1/2*e) + 1)))/f