3.81 \(\int (a+b \sec ^2(e+f x))^{3/2} \sin ^3(e+f x) \, dx\)

Optimal. Leaf size=162 \[ \frac{b (3 a-2 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{(3 a-2 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{\sqrt{b} (3 a-2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f} \]

[Out]

((3*a - 2*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) + ((3*a - 2*b)*b*Sec[e
+ f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(2*a*f) - ((3*a - 2*b)*Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2))/(3*a*f) +
(Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(5/2))/(3*a*f)

________________________________________________________________________________________

Rubi [A]  time = 0.142408, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4134, 453, 277, 195, 217, 206} \[ \frac{b (3 a-2 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}-\frac{(3 a-2 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{\sqrt{b} (3 a-2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a - 2*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) + ((3*a - 2*b)*b*Sec[e
+ f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(2*a*f) - ((3*a - 2*b)*Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2))/(3*a*f) +
(Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(5/2))/(3*a*f)

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rule 453

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

Rule 277

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin ^3(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \left (a+b x^2\right )^{3/2}}{x^4} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}+\frac{(3 a-2 b) \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{3/2}}{x^2} \, dx,x,\sec (e+f x)\right )}{3 a f}\\ &=-\frac{(3 a-2 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}+\frac{((3 a-2 b) b) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\sec (e+f x)\right )}{a f}\\ &=\frac{(3 a-2 b) b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}-\frac{(3 a-2 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}+\frac{((3 a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac{(3 a-2 b) b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}-\frac{(3 a-2 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}+\frac{((3 a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}\\ &=\frac{(3 a-2 b) \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac{(3 a-2 b) b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{2 a f}-\frac{(3 a-2 b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 a f}+\frac{\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{5/2}}{3 a f}\\ \end{align*}

Mathematica [A]  time = 0.790841, size = 164, normalized size = 1.01 \[ \frac{\sqrt{2} \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (3 \sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )^{5/2}-(3 a-2 b) \left (\sqrt{-a \sin ^2(e+f x)+a+b} \left (-a \sin ^2(e+f x)+a+4 b\right )-3 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{-a \sin ^2(e+f x)+a+b}}{\sqrt{b}}\right )\right )\right )}{3 b f (a \cos (2 (e+f x))+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2)*(3*Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)^(5/2) - (3*a
- 2*b)*(-3*b^(3/2)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]] + Sqrt[a + b - a*Sin[e + f*x]^2]*(a + 4*b
- a*Sin[e + f*x]^2))))/(3*b*f*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.42, size = 1913, 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]

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

[Out]

-1/12/f/b^(1/2)/(a+b)^(9/2)*(-1+cos(f*x+e))^3*(6*cos(f*x+e)^2*b^(13/2)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(
f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))
^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)+18*cos(f*x+e)^2*b^(11/2)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*
((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/
2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a-18*cos(f*x+e)^2*b^(11/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a
*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a
+b)^(1/2)+b)/sin(f*x+e)^2)*a+18*cos(f*x+e)^2*b^(9/2)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1
/2)+b)/sin(f*x+e)^2)*a^2-18*cos(f*x+e)^2*b^(9/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+
b)/sin(f*x+e)^2)*a^2+6*cos(f*x+e)^2*b^(7/2)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/
(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/si
n(f*x+e)^2)*a^3-6*cos(f*x+e)^2*b^(7/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+co
s(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x
+e)^2)*a^3-6*cos(f*x+e)^2*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)
-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*(a+b)^(7/2)*b^3+8*cos(f*x+e)^3*b^(5/2)*((b+a*cos
(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)+8*cos(f*x+e)^2*b^(5/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1
/2)*(a+b)^(7/2)+3*cos(f*x+e)*b^(5/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)+3*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)*(a+b)^(7/2)*a+2*cos(f*x+e)^5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*
b^(3/2)*(a+b)^(7/2)*a+2*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)*(a+b)^(7/2)*a-6*cos(f
*x+e)^2*b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)*a^2+3*cos(f*x+e)*b^(3/2)*((b+a*cos(f*x
+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)*a+9*cos(f*x+e)^2*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(f*x+e))*(cos(f
*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*(a+b)^(7/2)*a^
2*b+3*cos(f*x+e)^2*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin
(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*(a+b)^(7/2)*a*b^2+2*cos(f*x+e)^5*b^(1/2)*((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)*a^2+2*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(3/
2)*(a+b)^(7/2)*a+2*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*(a+b)^(7/2)*a^2+2*cos(f*x+
e)^3*b^(3/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)*a-6*cos(f*x+e)^3*b^(1/2)*((b+a*cos(f*x+e)
^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(7/2)*a^2-6*cos(f*x+e)^2*b^(13/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x
+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)
^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)+3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(5/2)*(a+b)^(7/2))*cos(f*x
+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)*4^(1/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)/sin(f*x+e)^6

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.58215, size = 710, normalized size = 4.38 \begin{align*} \left [-\frac{3 \,{\left (3 \, a - 2 \, b\right )} \sqrt{b} \cos \left (f x + e\right ) \log \left (\frac{a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \,{\left (2 \, a \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, f \cos \left (f x + e\right )}, -\frac{3 \,{\left (3 \, a - 2 \, b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) \cos \left (f x + e\right ) -{\left (2 \, a \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, f \cos \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(3*a - 2*b)*sqrt(b)*cos(f*x + e)*log((a*cos(f*x + e)^2 - 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f
*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) - 2*(2*a*cos(f*x + e)^4 - 2*(3*a - 4*b)*cos(f*x + e)^2 + 3*b)*s
qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/6*(3*(3*a - 2*b)*sqrt(-b)*arctan(sqrt(-b)*sqr
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)*cos(f*x + e) - (2*a*cos(f*x + e)^4 - 2*(3*a - 4*b)*co
s(f*x + e)^2 + 3*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e))]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.41367, size = 219, normalized size = 1.35 \begin{align*} \frac{{\left (2 \,{\left (a \cos \left (f x + e\right )^{2} + b\right )}^{\frac{3}{2}} a f^{4} - 6 \, \sqrt{a \cos \left (f x + e\right )^{2} + b} a^{2} f^{4} + 6 \, \sqrt{a \cos \left (f x + e\right )^{2} + b} a b f^{4} + \frac{3 \, \sqrt{a \cos \left (f x + e\right )^{2} + b} a b f^{4}}{\cos \left (f x + e\right )^{2}} - \frac{3 \,{\left (3 \, a^{2} b f^{4} - 2 \, a b^{2} f^{4}\right )} \arctan \left (\frac{\sqrt{a \cos \left (f x + e\right )^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{6 \, a f^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(a*cos(f*x + e)^2 + b)^(3/2)*a*f^4 - 6*sqrt(a*cos(f*x + e)^2 + b)*a^2*f^4 + 6*sqrt(a*cos(f*x + e)^2 + b
)*a*b*f^4 + 3*sqrt(a*cos(f*x + e)^2 + b)*a*b*f^4/cos(f*x + e)^2 - 3*(3*a^2*b*f^4 - 2*a*b^2*f^4)*arctan(sqrt(a*
cos(f*x + e)^2 + b)/sqrt(-b))/sqrt(-b))*sgn(cos(f*x + e))/(a*f^5)