3.73 \(\int \frac{\csc ^5(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=210 \[ -\frac{3 \left (a^2-8 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^4 f}-\frac{3 b (3 a-4 b) \sec (e+f x)}{8 a^3 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac{3 \sqrt{b} (a-2 b) \sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 a^4 f}-\frac{(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a+b \sec ^2(e+f x)-b\right )} \]

[Out]

(-3*(a - 2*b)*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*a^4*f) - (3*(a^2 - 8*a*b + 8*
b^2)*ArcTanh[Cos[e + f*x]])/(8*a^4*f) - ((5*a - 6*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f*(a - b + b*Sec[e + f*
x]^2)) - (Cot[e + f*x]^3*Csc[e + f*x])/(4*a*f*(a - b + b*Sec[e + f*x]^2)) - (3*(3*a - 4*b)*b*Sec[e + f*x])/(8*
a^3*f*(a - b + b*Sec[e + f*x]^2))

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Rubi [A]  time = 0.275453, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3664, 470, 527, 522, 207, 205} \[ -\frac{3 \left (a^2-8 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^4 f}-\frac{3 b (3 a-4 b) \sec (e+f x)}{8 a^3 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac{3 \sqrt{b} (a-2 b) \sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 a^4 f}-\frac{(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a+b \sec ^2(e+f x)-b\right )}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a+b \sec ^2(e+f x)-b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(-3*(a - 2*b)*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*a^4*f) - (3*(a^2 - 8*a*b + 8*
b^2)*ArcTanh[Cos[e + f*x]])/(8*a^4*f) - ((5*a - 6*b)*Cot[e + f*x]*Csc[e + f*x])/(8*a^2*f*(a - b + b*Sec[e + f*
x]^2)) - (Cot[e + f*x]^3*Csc[e + f*x])/(4*a*f*(a - b + b*Sec[e + f*x]^2)) - (3*(3*a - 4*b)*b*Sec[e + f*x])/(8*
a^3*f*(a - b + b*Sec[e + f*x]^2))

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 207

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\csc ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^3 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-a+b+(-4 a+5 b) x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{4 a f}\\ &=-\frac{(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 (a-2 b) (a-b)+3 (5 a-6 b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{8 a^2 f}\\ &=-\frac{(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{3 (3 a-4 b) b \sec (e+f x)}{8 a^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-6 (a-4 b) (a-b)^2+6 (3 a-4 b) (a-b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{16 a^3 (a-b) f}\\ &=-\frac{(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{3 (3 a-4 b) b \sec (e+f x)}{8 a^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{(3 (a-2 b) (a-b) b) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{2 a^4 f}+\frac{\left (3 \left (a^2-8 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{8 a^4 f}\\ &=-\frac{3 (a-2 b) \sqrt{a-b} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 a^4 f}-\frac{3 \left (a^2-8 a b+8 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 a^4 f}-\frac{(5 a-6 b) \cot (e+f x) \csc (e+f x)}{8 a^2 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\cot ^3(e+f x) \csc (e+f x)}{4 a f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{3 (3 a-4 b) b \sec (e+f x)}{8 a^3 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 6.33467, size = 392, normalized size = 1.87 \[ \frac{b^2 \cos (e+f x)-a b \cos (e+f x)}{a^3 f (a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b)}+\frac{3 \left (a^2-8 a b+8 b^2\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{8 a^4 f}-\frac{3 \left (a^2-8 a b+8 b^2\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{8 a^4 f}+\frac{(8 b-3 a) \csc ^2\left (\frac{1}{2} (e+f x)\right )}{32 a^3 f}+\frac{(3 a-8 b) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{32 a^3 f}+\frac{3 \sqrt{b} (a-2 b) \sqrt{a-b} \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (e+f x)\right ) \left (\sqrt{a-b} \cos \left (\frac{1}{2} (e+f x)\right )-\sqrt{a} \sin \left (\frac{1}{2} (e+f x)\right )\right )}{\sqrt{b}}\right )}{2 a^4 f}+\frac{3 \sqrt{b} (a-2 b) \sqrt{a-b} \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (e+f x)\right ) \left (\sqrt{a-b} \cos \left (\frac{1}{2} (e+f x)\right )+\sqrt{a} \sin \left (\frac{1}{2} (e+f x)\right )\right )}{\sqrt{b}}\right )}{2 a^4 f}-\frac{\csc ^4\left (\frac{1}{2} (e+f x)\right )}{64 a^2 f}+\frac{\sec ^4\left (\frac{1}{2} (e+f x)\right )}{64 a^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^5/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(3*(a - 2*b)*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] - Sqrt[a]*Sin[(e + f*x
)/2]))/Sqrt[b]])/(2*a^4*f) + (3*(a - 2*b)*Sqrt[a - b]*Sqrt[b]*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f
*x)/2] + Sqrt[a]*Sin[(e + f*x)/2]))/Sqrt[b]])/(2*a^4*f) + (-(a*b*Cos[e + f*x]) + b^2*Cos[e + f*x])/(a^3*f*(a +
b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])) + ((-3*a + 8*b)*Csc[(e + f*x)/2]^2)/(32*a^3*f) - Csc[(e + f*x)/
2]^4/(64*a^2*f) - (3*(a^2 - 8*a*b + 8*b^2)*Log[Cos[(e + f*x)/2]])/(8*a^4*f) + (3*(a^2 - 8*a*b + 8*b^2)*Log[Sin
[(e + f*x)/2]])/(8*a^4*f) + ((3*a - 8*b)*Sec[(e + f*x)/2]^2)/(32*a^3*f) + Sec[(e + f*x)/2]^4/(64*a^2*f)

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Maple [B]  time = 0.109, size = 428, normalized size = 2. \begin{align*}{\frac{1}{16\,f{a}^{2} \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}+{\frac{3}{16\,f{a}^{2} \left ( \cos \left ( fx+e \right ) +1 \right ) }}-{\frac{b}{2\,f{a}^{3} \left ( \cos \left ( fx+e \right ) +1 \right ) }}-{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) +1 \right ) }{16\,f{a}^{2}}}+{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) +1 \right ) b}{2\,f{a}^{3}}}-{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) +1 \right ){b}^{2}}{2\,f{a}^{4}}}-{\frac{b\cos \left ( fx+e \right ) }{2\,f{a}^{2} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }}+{\frac{{b}^{2}\cos \left ( fx+e \right ) }{2\,f{a}^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }}+{\frac{3\,b}{2\,f{a}^{2}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}}-{\frac{9\,{b}^{2}}{2\,f{a}^{3}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}}+3\,{\frac{{b}^{3}}{f{a}^{4}\sqrt{b \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \cos \left ( fx+e \right ) }{\sqrt{b \left ( a-b \right ) }}} \right ) }-{\frac{1}{16\,f{a}^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2}}}+{\frac{3}{16\,f{a}^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) }}-{\frac{b}{2\,f{a}^{3} \left ( \cos \left ( fx+e \right ) -1 \right ) }}+{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) -1 \right ) }{16\,f{a}^{2}}}-{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) -1 \right ) b}{2\,f{a}^{3}}}+{\frac{3\,\ln \left ( \cos \left ( fx+e \right ) -1 \right ){b}^{2}}{2\,f{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x)

[Out]

1/16/f/a^2/(cos(f*x+e)+1)^2+3/16/f/a^2/(cos(f*x+e)+1)-1/2/f/a^3/(cos(f*x+e)+1)*b-3/16/f/a^2*ln(cos(f*x+e)+1)+3
/2/f/a^3*ln(cos(f*x+e)+1)*b-3/2/f/a^4*ln(cos(f*x+e)+1)*b^2-1/2/f*b/a^2*cos(f*x+e)/(a*cos(f*x+e)^2-cos(f*x+e)^2
*b+b)+1/2/f*b^2/a^3*cos(f*x+e)/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)+3/2/f*b/a^2/(b*(a-b))^(1/2)*arctan((a-b)*cos(
f*x+e)/(b*(a-b))^(1/2))-9/2/f*b^2/a^3/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))+3/f*b^3/a^4/(b*
(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))-1/16/f/a^2/(cos(f*x+e)-1)^2+3/16/f/a^2/(cos(f*x+e)-1)-1/
2/f/a^3/(cos(f*x+e)-1)*b+3/16/f/a^2*ln(cos(f*x+e)-1)-3/2/f/a^3*ln(cos(f*x+e)-1)*b+3/2/f/a^4*ln(cos(f*x+e)-1)*b
^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.77455, size = 2477, normalized size = 11.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[1/16*(6*(a^3 - 5*a^2*b + 4*a*b^2)*cos(f*x + e)^5 - 2*(5*a^3 - 24*a^2*b + 24*a*b^2)*cos(f*x + e)^3 - 12*((a^2
- 3*a*b + 2*b^2)*cos(f*x + e)^6 - (2*a^2 - 7*a*b + 6*b^2)*cos(f*x + e)^4 + (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^
2 + a*b - 2*b^2)*sqrt(-a*b + b^2)*log(((a - b)*cos(f*x + e)^2 - 2*sqrt(-a*b + b^2)*cos(f*x + e) - b)/((a - b)*
cos(f*x + e)^2 + b)) - 6*(3*a^2*b - 4*a*b^2)*cos(f*x + e) - 3*((a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*x + e)
^6 - (2*a^3 - 19*a^2*b + 40*a*b^2 - 24*b^3)*cos(f*x + e)^4 + a^2*b - 8*a*b^2 + 8*b^3 + (a^3 - 11*a^2*b + 32*a*
b^2 - 24*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*x + e)
^6 - (2*a^3 - 19*a^2*b + 40*a*b^2 - 24*b^3)*cos(f*x + e)^4 + a^2*b - 8*a*b^2 + 8*b^3 + (a^3 - 11*a^2*b + 32*a*
b^2 - 24*b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5
- 3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2), 1/16*(6*(a^3 - 5*a^2*b + 4*a*b^2)*cos(f*x +
e)^5 - 2*(5*a^3 - 24*a^2*b + 24*a*b^2)*cos(f*x + e)^3 + 24*((a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^6 - (2*a^2 - 7*
a*b + 6*b^2)*cos(f*x + e)^4 + (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^2 + a*b - 2*b^2)*sqrt(a*b - b^2)*arctan(sqrt(
a*b - b^2)*cos(f*x + e)/b) - 6*(3*a^2*b - 4*a*b^2)*cos(f*x + e) - 3*((a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*
x + e)^6 - (2*a^3 - 19*a^2*b + 40*a*b^2 - 24*b^3)*cos(f*x + e)^4 + a^2*b - 8*a*b^2 + 8*b^3 + (a^3 - 11*a^2*b +
32*a*b^2 - 24*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^3 - 9*a^2*b + 16*a*b^2 - 8*b^3)*cos(f*
x + e)^6 - (2*a^3 - 19*a^2*b + 40*a*b^2 - 24*b^3)*cos(f*x + e)^4 + a^2*b - 8*a*b^2 + 8*b^3 + (a^3 - 11*a^2*b +
32*a*b^2 - 24*b^3)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f -
(2*a^5 - 3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2)]

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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(csc(f*x+e)**5/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.43885, size = 740, normalized size = 3.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/64*(12*(a^2 - 8*a*b + 8*b^2)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/a^4 - 96*(a^2*b - 3*a*b^2 + 2*b^3)*
arctan(-(a*cos(f*x + e) - b*cos(f*x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + sqrt(a*b - b^2)))/(sqrt(a*b - b^
2)*a^4) - (8*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 16*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - a^2*(c
os(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/a^4 - (a^2 - 8*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 16*a*b*(co
s(f*x + e) - 1)/(cos(f*x + e) + 1) + 18*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 144*a*b*(cos(f*x + e)
- 1)^2/(cos(f*x + e) + 1)^2 + 144*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/(a^4*(co
s(f*x + e) - 1)^2) - 64*(a^2*b - a*b^2 + a^2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 3*a*b^2*(cos(f*x + e) -
1)/(cos(f*x + e) + 1) + 2*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))/((a + 2*a*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*a^4))/f