3.558 \(\int \frac{(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac{\left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac{3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

-((3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqrt[2]*a^
(5/2)*f) - (3*(c - d)*(c + 3*d)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((c - d)*Cos[e + f*x]*(c +
d*Sin[e + f*x]))/(4*f*(a + a*Sin[e + f*x])^(5/2))

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Rubi [A]  time = 0.229836, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2760, 2750, 2649, 206} \[ -\frac{\left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac{3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

-((3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqrt[2]*a^
(5/2)*f) - (3*(c - d)*(c + 3*d)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((c - d)*Cos[e + f*x]*(c +
d*Sin[e + f*x]))/(4*f*(a + a*Sin[e + f*x])^(5/2))

Rule 2760

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

Rule 2750

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)]

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])

Rubi steps

\begin{align*} \int \frac{(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{\int \frac{-\frac{1}{2} a \left (3 c^2+7 c d-2 d^2\right )-\frac{1}{2} a d (c+7 d) \sin (e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}}+\frac{\left (3 c^2+10 c d+19 d^2\right ) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{32 a^2}\\ &=-\frac{3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}}-\frac{\left (3 c^2+10 c d+19 d^2\right ) \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 )}{16 a^2 f}\\ &=-\frac{\left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac{(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.553697, size = 252, normalized size = 1.71 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (2 \left (3 c^2+10 c d-13 d^2\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(1+i) (-1)^{3/4} \left (3 c^2+10 c d+19 d^2\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )+8 (c-d)^2 \sin \left (\frac{1}{2} (e+f x)\right )-(c-d) (3 c+13 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-4 (c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{16 f (a (\sin (e+f x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(c - d)^2*Sin[(e + f*x)/2] - 4*(c - d)^2*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2]) + 2*(3*c^2 + 10*c*d - 13*d^2)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (c - d)*(3
*c + 13*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 + I)*(-1)^(3/4)*(3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(1/2
+ I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4))/(16*f*(a*(1 + Sin[e + f*
x]))^(5/2))

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Maple [B]  time = 0.929, size = 379, normalized size = 2.6 \begin{align*} -{\frac{1}{ \left ( 32+32\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 2\,\sin \left ( fx+e \right ) \sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( 3\,{c}^{2}+10\,cd+19\,{d}^{2} \right ) -\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{2} \left ( 3\,{c}^{2}+10\,cd+19\,{d}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+6\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}{c}^{2}+20\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}cd+38\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2}{d}^{2}+20\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2}{c}^{2}+24\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2}cd-44\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2}{d}^{2}-6\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a}{c}^{2}-20\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a}cd+26\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}\sqrt{a}{d}^{2} \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }{a}^{-{\frac{9}{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((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x)

[Out]

-1/32*(2*sin(f*x+e)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*(3*c^2+10*c*d+19*d^2)-2^(1
/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*(3*c^2+10*c*d+19*d^2)*cos(f*x+e)^2+6*2^(1/2)*arcta
nh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2+20*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a
^(1/2))*a^2*c*d+38*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2+20*(a-a*sin(f*x+e))^(1/
2)*a^(3/2)*c^2+24*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d-44*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^2-6*(a-a*sin(f*x+e))^
(3/2)*a^(1/2)*c^2-20*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d+26*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^2)*(-a*(-1+sin(f*x
+e)))^(1/2)/a^(9/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^2/(a*sin(f*x + e) + a)^(5/2), x)

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Fricas [B]  time = 1.87514, size = 1234, normalized size = 8.39 \begin{align*} \frac{\sqrt{2}{\left ({\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} - 40 \, c d - 76 \, d^{2} - 2 \,{\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right ) +{\left ({\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} - 40 \, c d - 76 \, d^{2} - 2 \,{\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \,{\left ({\left (3 \, c^{2} + 10 \, c d - 13 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, c^{2} - 8 \, c d + 4 \, d^{2} +{\left (7 \, c^{2} + 2 \, c d - 9 \, d^{2}\right )} \cos \left (f x + e\right ) -{\left (4 \, c^{2} - 8 \, c d + 4 \, d^{2} -{\left (3 \, c^{2} + 10 \, c d - 13 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{64 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/64*(sqrt(2)*((3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e)^3 + 3*(3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*c^2
- 40*c*d - 76*d^2 - 2*(3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e) + ((3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*
c^2 - 40*c*d - 76*d^2 - 2*(3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2
- 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*
x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(
(3*c^2 + 10*c*d - 13*d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 + 2*c*d - 9*d^2)*cos(f*x + e) - (4*c
^2 - 8*c*d + 4*d^2 - (3*c^2 + 10*c*d - 13*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^3*f*co
s(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))

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Sympy [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((c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 4.75015, size = 1540, normalized size = 10.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/16*(sqrt(2)*(3*c^2 + 10*c*d + 19*d^2)*arctan(-1/2*sqrt(2)*(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))/(sqrt(-a)*a^2*sgn(tan(1/2*f*x + 1/2*e) + 1)) + 2*(29*(sqrt(a)*tan(1/2*f*
x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^7*c^2 - 10*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x
+ 1/2*e)^2 + a))^7*c*d - 19*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^7*d^2 + 75*(s
qrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^6*sqrt(a)*c^2 + 58*(sqrt(a)*tan(1/2*f*x + 1/
2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^6*sqrt(a)*c*d - 133*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*
f*x + 1/2*e)^2 + a))^6*sqrt(a)*d^2 + 55*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^5*
a*c^2 + 34*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^5*a*c*d - 89*(sqrt(a)*tan(1/2*f
*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^5*a*d^2 - 91*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*
f*x + 1/2*e)^2 + a))^4*a^(3/2)*c^2 - 26*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^4*
a^(3/2)*c*d + 117*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^4*a^(3/2)*d^2 - (sqrt(a)
*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*a^2*c^2 + 18*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqr
t(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*a^2*c*d - 17*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2
+ a))^3*a^2*d^2 + 65*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(5/2)*c^2 - 18*(s
qrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(5/2)*c*d - 47*(sqrt(a)*tan(1/2*f*x + 1/
2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(5/2)*d^2 - 27*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f
*x + 1/2*e)^2 + a))*a^3*c^2 - 26*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*a^3*c*d +
53*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*a^3*d^2 + 7*a^(7/2)*c^2 + 2*a^(7/2)*c*
d - 9*a^(7/2)*d^2)/(((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 2*(sqrt(a)*tan(1/
2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a)^4*a^2*sgn(tan(1/2*f*x + 1/2*e) + 1)))/f