3.699 \(\int \frac{(c+d \sin (e+f x))^3}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=156 \[ -\frac{d x \left (-2 a^2 d^2+6 a b c d+b^2 \left (-\left (6 c^2+d^2\right )\right )\right )}{2 b^3}+\frac{2 (b c-a d)^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 f \sqrt{a^2-b^2}}-\frac{d^2 (5 b c-2 a d) \cos (e+f x)}{2 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f} \]

[Out]

-(d*(6*a*b*c*d - 2*a^2*d^2 - b^2*(6*c^2 + d^2))*x)/(2*b^3) + (2*(b*c - a*d)^3*ArcTan[(b + a*Tan[(e + f*x)/2])/
Sqrt[a^2 - b^2]])/(b^3*Sqrt[a^2 - b^2]*f) - (d^2*(5*b*c - 2*a*d)*Cos[e + f*x])/(2*b^2*f) - (d^2*Cos[e + f*x]*(
c + d*Sin[e + f*x]))/(2*b*f)

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Rubi [A]  time = 0.363093, antiderivative size = 156, 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 = {2793, 3023, 2735, 2660, 618, 204} \[ -\frac{d x \left (-2 a^2 d^2+6 a b c d+b^2 \left (-\left (6 c^2+d^2\right )\right )\right )}{2 b^3}+\frac{2 (b c-a d)^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 f \sqrt{a^2-b^2}}-\frac{d^2 (5 b c-2 a d) \cos (e+f x)}{2 b^2 f}-\frac{d^2 \cos (e+f x) (c+d \sin (e+f x))}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(6*a*b*c*d - 2*a^2*d^2 - b^2*(6*c^2 + d^2))*x)/(2*b^3) + (2*(b*c - a*d)^3*ArcTan[(b + a*Tan[(e + f*x)/2])/
Sqrt[a^2 - b^2]])/(b^3*Sqrt[a^2 - b^2]*f) - (d^2*(5*b*c - 2*a*d)*Cos[e + f*x])/(2*b^2*f) - (d^2*Cos[e + f*x]*(
c + d*Sin[e + f*x]))/(2*b*f)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

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

Mathematica [A]  time = 0.33562, size = 138, normalized size = 0.88 \[ \frac{2 d (e+f x) \left (2 a^2 d^2-6 a b c d+b^2 \left (6 c^2+d^2\right )\right )+\frac{8 (b c-a d)^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-4 b d^2 (3 b c-a d) \cos (e+f x)-b^2 d^3 \sin (2 (e+f x))}{4 b^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(-6*a*b*c*d + 2*a^2*d^2 + b^2*(6*c^2 + d^2))*(e + f*x) + (8*(b*c - a*d)^3*ArcTan[(b + a*Tan[(e + f*x)/2])
/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 4*b*d^2*(3*b*c - a*d)*Cos[e + f*x] - b^2*d^3*Sin[2*(e + f*x)])/(4*b^3*f)

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Maple [B]  time = 0.076, size = 506, normalized size = 3.2 \begin{align*}{\frac{{d}^{3}}{bf} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{{d}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a}{{b}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-6\,{\frac{{d}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c}{bf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{d}^{3}}{bf}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{a{d}^{3}}{{b}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-6\,{\frac{c{d}^{2}}{bf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{{d}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ){a}^{2}}{{b}^{3}f}}-6\,{\frac{{d}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ac}{{b}^{2}f}}+6\,{\frac{d\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ){c}^{2}}{bf}}+{\frac{{d}^{3}}{bf}\arctan \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) }-2\,{\frac{{a}^{3}{d}^{3}}{{b}^{3}f\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+6\,{\frac{{a}^{2}c{d}^{2}}{{b}^{2}f\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-6\,{\frac{a{c}^{2}d}{bf\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{{c}^{3}}{f\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,fx+e/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sin(f*x+e))^3/(a+b*sin(f*x+e)),x)

[Out]

1/f*d^3/b/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3+2/f*d^3/b^2/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1
/2*e)^2*a-6/f*d^2/b/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^2*c-1/f*d^3/b/(1+tan(1/2*f*x+1/2*e)^2)^2*tan
(1/2*f*x+1/2*e)+2/f*d^3/b^2/(1+tan(1/2*f*x+1/2*e)^2)^2*a-6/f*d^2/b/(1+tan(1/2*f*x+1/2*e)^2)^2*c+2/f*d^3/b^3*ar
ctan(tan(1/2*f*x+1/2*e))*a^2-6/f*d^2/b^2*arctan(tan(1/2*f*x+1/2*e))*a*c+6/f*d/b*arctan(tan(1/2*f*x+1/2*e))*c^2
+1/f*d^3/b*arctan(tan(1/2*f*x+1/2*e))-2/f/b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2
)^(1/2))*a^3*d^3+6/f/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^2*c*d^2-6/
f/b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a*c^2*d+2/f/(a^2-b^2)^(1/2)*arcta
n(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*c^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59816, size = 1153, normalized size = 7.39 \begin{align*} \left [-\frac{{\left (a^{2} b^{2} - b^{4}\right )} d^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (6 \,{\left (a^{2} b^{2} - b^{4}\right )} c^{2} d - 6 \,{\left (a^{3} b - a b^{3}\right )} c d^{2} +{\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} d^{3}\right )} f x -{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (3 \,{\left (a^{2} b^{2} - b^{4}\right )} c d^{2} -{\left (a^{3} b - a b^{3}\right )} d^{3}\right )} \cos \left (f x + e\right )}{2 \,{\left (a^{2} b^{3} - b^{5}\right )} f}, -\frac{{\left (a^{2} b^{2} - b^{4}\right )} d^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left (6 \,{\left (a^{2} b^{2} - b^{4}\right )} c^{2} d - 6 \,{\left (a^{3} b - a b^{3}\right )} c d^{2} +{\left (2 \, a^{4} - a^{2} b^{2} - b^{4}\right )} d^{3}\right )} f x + 2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (f x + e\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) + 2 \,{\left (3 \,{\left (a^{2} b^{2} - b^{4}\right )} c d^{2} -{\left (a^{3} b - a b^{3}\right )} d^{3}\right )} \cos \left (f x + e\right )}{2 \,{\left (a^{2} b^{3} - b^{5}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((a^2*b^2 - b^4)*d^3*cos(f*x + e)*sin(f*x + e) - (6*(a^2*b^2 - b^4)*c^2*d - 6*(a^3*b - a*b^3)*c*d^2 + (2
*a^4 - a^2*b^2 - b^4)*d^3)*f*x - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-a^2 + b^2)*log(-((2
*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2 - 2*(a*cos(f*x + e)*sin(f*x + e) + b*cos(f*x + e))
*sqrt(-a^2 + b^2))/(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)) + 2*(3*(a^2*b^2 - b^4)*c*d^2 - (a^3*
b - a*b^3)*d^3)*cos(f*x + e))/((a^2*b^3 - b^5)*f), -1/2*((a^2*b^2 - b^4)*d^3*cos(f*x + e)*sin(f*x + e) - (6*(a
^2*b^2 - b^4)*c^2*d - 6*(a^3*b - a*b^3)*c*d^2 + (2*a^4 - a^2*b^2 - b^4)*d^3)*f*x + 2*(b^3*c^3 - 3*a*b^2*c^2*d
+ 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) + 2*(3
*(a^2*b^2 - b^4)*c*d^2 - (a^3*b - a*b^3)*d^3)*cos(f*x + e))/((a^2*b^3 - b^5)*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((c+d*sin(f*x+e))**3/(a+b*sin(f*x+e)),x)

[Out]

Timed out

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Giac [A]  time = 1.37567, size = 340, normalized size = 2.18 \begin{align*} \frac{\frac{{\left (6 \, b^{2} c^{2} d - 6 \, a b c d^{2} + 2 \, a^{2} d^{3} + b^{2} d^{3}\right )}{\left (f x + e\right )}}{b^{3}} + \frac{4 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} b^{3}} + \frac{2 \,{\left (b d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 \, b c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 2 \, a d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - b d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 \, b c d^{2} + 2 \, a d^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((6*b^2*c^2*d - 6*a*b*c*d^2 + 2*a^2*d^3 + b^2*d^3)*(f*x + e)/b^3 + 4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*
d^2 - a^3*d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2))
)/(sqrt(a^2 - b^2)*b^3) + 2*(b*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*b*c*d^2*tan(1/2*f*x + 1/2*e)^2 + 2*a*d^3*tan(1/2
*f*x + 1/2*e)^2 - b*d^3*tan(1/2*f*x + 1/2*e) - 6*b*c*d^2 + 2*a*d^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*b^2))/f