3.80 \(\int \frac{\sin ^5(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=264 \[ -\frac{\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 f (a-b)^5}-\frac{b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 f (a-b)^5 \left (a+b \sec ^2(e+f x)-b\right )}-\frac{b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac{\sqrt{b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 f (a-b)^{11/2}}+\frac{(10 a-b) \cos ^3(e+f x)}{15 f (a-b)^4}-\frac{\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2} \]

[Out]

-(Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*(a - b)^(11/2)*f) - ((5*a^2
+ 20*a*b + 2*b^2)*Cos[e + f*x])/(5*(a - b)^5*f) + ((10*a - b)*Cos[e + f*x]^3)/(15*(a - b)^4*f) - Cos[e + f*x]
^5/(5*(a - b)*f*(a - b + b*Sec[e + f*x]^2)^2) - (b*(5*a^2 + 4*b^2)*Sec[e + f*x])/(20*(a - b)^4*f*(a - b + b*Se
c[e + f*x]^2)^2) - (b*(35*a^2 + 40*a*b + 24*b^2)*Sec[e + f*x])/(40*(a - b)^5*f*(a - b + b*Sec[e + f*x]^2))

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Rubi [A]  time = 0.410531, antiderivative size = 264, 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, 462, 456, 1259, 1261, 205} \[ -\frac{\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 f (a-b)^5}-\frac{b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 f (a-b)^5 \left (a+b \sec ^2(e+f x)-b\right )}-\frac{b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac{\sqrt{b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 f (a-b)^{11/2}}+\frac{(10 a-b) \cos ^3(e+f x)}{15 f (a-b)^4}-\frac{\cos ^5(e+f x)}{5 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

-(Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*(a - b)^(11/2)*f) - ((5*a^2
+ 20*a*b + 2*b^2)*Cos[e + f*x])/(5*(a - b)^5*f) + ((10*a - b)*Cos[e + f*x]^3)/(15*(a - b)^4*f) - Cos[e + f*x]
^5/(5*(a - b)*f*(a - b + b*Sec[e + f*x]^2)^2) - (b*(5*a^2 + 4*b^2)*Sec[e + f*x])/(20*(a - b)^4*f*(a - b + b*Se
c[e + f*x]^2)^2) - (b*(35*a^2 + 40*a*b + 24*b^2)*Sec[e + f*x])/(40*(a - b)^5*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 462

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

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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{\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2}{x^6 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac{\operatorname{Subst}\left (\int \frac{-10 a+b+5 (a-b) x^2}{x^4 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \operatorname{Subst}\left (\int \frac{\frac{4 (10 a-b)}{(a-b) b}-\frac{4 \left (5 a^2+4 b^2\right ) x^2}{(a-b)^2 b}+\frac{3 \left (5 a^2+4 b^2\right ) x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{20 (a-b) f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{8 (a-b) (10 a-b) b-8 b \left (5 a^2+10 a b+3 b^2\right ) x^2+\frac{b^2 \left (35 a^2+40 a b+24 b^2\right ) x^4}{a-b}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{40 (a-b)^4 b f}\\ &=-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \left (\frac{8 (10 a-b) b}{x^4}-\frac{8 b \left (5 a^2+20 a b+2 b^2\right )}{(a-b) x^2}+\frac{5 b^2 \left (15 a^2+40 a b+8 b^2\right )}{(a-b) \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{40 (a-b)^4 b f}\\ &=-\frac{\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 (a-b)^5 f}+\frac{(10 a-b) \cos ^3(e+f x)}{15 (a-b)^4 f}-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\left (b \left (15 a^2+40 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^5 f}\\ &=-\frac{\sqrt{b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 (a-b)^{11/2} f}-\frac{\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 (a-b)^5 f}+\frac{(10 a-b) \cos ^3(e+f x)}{15 (a-b)^4 f}-\frac{\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 5.47999, size = 278, normalized size = 1.05 \[ \frac{\frac{(a-b) (5 (5 a+7 b) \cos (3 (e+f x))+3 (b-a) \cos (5 (e+f x)))-30 \cos (e+f x) \left (a^2 \left (-\frac{8 b^2}{((a-b) \cos (2 (e+f x))+a+b)^2}+\frac{18 b}{(a-b) \cos (2 (e+f x))+a+b}+5\right )+16 a b \left (\frac{b}{(a-b) \cos (2 (e+f x))+a+b}+2\right )+11 b^2\right )}{(a-b)^5}+\frac{30 \sqrt{b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b}-\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{11/2}}+\frac{30 \sqrt{b} \left (15 a^2+40 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b}+\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{11/2}}}{240 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

((30*Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(11/2
) + (30*Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(1
1/2) + (-30*Cos[e + f*x]*(11*b^2 + 16*a*b*(2 + b/(a + b + (a - b)*Cos[2*(e + f*x)])) + a^2*(5 - (8*b^2)/(a + b
+ (a - b)*Cos[2*(e + f*x)])^2 + (18*b)/(a + b + (a - b)*Cos[2*(e + f*x)]))) + (a - b)*(5*(5*a + 7*b)*Cos[3*(e
+ f*x)] + 3*(-a + b)*Cos[5*(e + f*x)]))/(a - b)^5)/(240*f)

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Maple [B]  time = 0.089, size = 844, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x)

[Out]

-1/5/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b^2)*cos(f*x+e)^5*a^2+2/5/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b
^2)*cos(f*x+e)^5*a*b-1/5/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b^2)*cos(f*x+e)^5*b^2+2/3/f/(a^3-3*a^2*b+3*a*b
^2-b^3)/(a^2-2*a*b+b^2)*a^2*cos(f*x+e)^3-1/3/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b^2)*cos(f*x+e)^3*a*b-1/3/
f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b^2)*b^2*cos(f*x+e)^3-1/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b^2)*a^2
*cos(f*x+e)-4/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b^2)*a*cos(f*x+e)*b-1/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*
a*b+b^2)*cos(f*x+e)*b^2-9/8/f*b/(a-b)^5/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*cos(f*x+e)^3*a^3+1/8/f*b^2/(a-b)^5
/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*cos(f*x+e)^3*a^2+1/f*b^3/(a-b)^5/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*a*co
s(f*x+e)^3-7/8/f*b^2/(a-b)^5/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*cos(f*x+e)*a^2-1/f*b^3/(a-b)^5/(a*cos(f*x+e)^
2-cos(f*x+e)^2*b+b)^2*cos(f*x+e)*a+15/8/f*b/(a-b)^5/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))*a
^2+5/f*b^2/(a-b)^5/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))*a+1/f*b^3/(a-b)^5/(b*(a-b))^(1/2)*
arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))

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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(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.15777, size = 2313, normalized size = 8.76 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[-1/240*(48*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 16*(10*a^4 - 31*a^3*b + 33*a^2*b^2 -
13*a*b^3 + b^4)*cos(f*x + e)^7 + 16*(15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^5 + 50*(1
5*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^3 + 15*((15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b
^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b^3 + 8*b^4 + 2*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)
^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(
f*x + e)^2 + b)) + 30*(15*a^2*b^2 + 40*a*b^3 + 8*b^4)*cos(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3
+ 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 +
15*a^2*b^5 - 6*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2 - 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7
)*f), -1/120*(24*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 8*(10*a^4 - 31*a^3*b + 33*a^2*b^
2 - 13*a*b^3 + b^4)*cos(f*x + e)^7 + 8*(15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^5 + 25
*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^3 + 15*((15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 +
8*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b^3 + 8*b^4 + 2*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x +
e)^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 15*(15*a^2*b^2 + 40*a*b^3 + 8*b^4)*co
s(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e
)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 + 15*a^2*b^5 - 6*a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2
- 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2*b^5 + 5*a*b^6 - b^7)*f)]

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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(sin(f*x+e)**5/(a+b*tan(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.67518, size = 1195, normalized size = 4.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/120*(15*(15*a^2*b + 40*a*b^2 + 8*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)))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*sqrt(a*b - b^2)) + 30*(9
*a^3*b + 6*a^2*b^2 + 27*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 32*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) - 40*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 27*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2
- 54*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 24*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 +
48*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 9*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 16*a^2
*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 8*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a^5 - 5*
a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*(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)^2) - 16*(8*a^2 + 59*a*b + 23*b^2
- 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 250*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 70*b^2*(cos(f
*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 320*a*b*(cos(f*x + e) - 1
)^2/(cos(f*x + e) + 1)^2 + 140*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 270*a*b*(cos(f*x + e) - 1)^3/(c
os(f*x + e) + 1)^3 - 90*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 45*a*b*(cos(f*x + e) - 1)^4/(cos(f*x +
e) + 1)^4 + 45*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a
*b^4 - b^5)*((cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 1)^5))/f