3.717 \(\int \frac{(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=196 \[ -\frac{\left (a^2 \left (-\left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac{(b c-a d)^2 \cos (e+f x)}{2 b f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2}+\frac{\left (a^2 d+3 a b c-4 b^2 d\right ) (b c-a d) \cos (e+f x)}{2 b f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))} \]

[Out]

-(((6*a*b*c*d - a^2*(2*c^2 + d^2) - b^2*(c^2 + 2*d^2))*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2
- b^2)^(5/2)*f)) + ((b*c - a*d)^2*Cos[e + f*x])/(2*b*(a^2 - b^2)*f*(a + b*Sin[e + f*x])^2) + ((b*c - a*d)*(3*
a*b*c + a^2*d - 4*b^2*d)*Cos[e + f*x])/(2*b*(a^2 - b^2)^2*f*(a + b*Sin[e + f*x]))

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Rubi [A]  time = 0.285129, antiderivative size = 196, 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 = {2790, 2754, 12, 2660, 618, 204} \[ -\frac{\left (a^2 \left (-\left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac{(b c-a d)^2 \cos (e+f x)}{2 b f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2}+\frac{\left (a^2 d+3 a b c-4 b^2 d\right ) (b c-a d) \cos (e+f x)}{2 b f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((6*a*b*c*d - a^2*(2*c^2 + d^2) - b^2*(c^2 + 2*d^2))*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2
- b^2)^(5/2)*f)) + ((b*c - a*d)^2*Cos[e + f*x])/(2*b*(a^2 - b^2)*f*(a + b*Sin[e + f*x])^2) + ((b*c - a*d)*(3*
a*b*c + a^2*d - 4*b^2*d)*Cos[e + f*x])/(2*b*(a^2 - b^2)^2*f*(a + b*Sin[e + f*x]))

Rule 2790

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

Rule 2754

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 + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.922206, size = 204, normalized size = 1.04 \[ \frac{\frac{2 \left (a^2 \left (2 c^2+d^2\right )-6 a b c d+b^2 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac{\left (2 a^2 b c d+a^3 d^2-a b^2 \left (3 c^2+4 d^2\right )+4 b^3 c d\right ) \cos (e+f x)}{b (a-b)^2 (a+b)^2 (a+b \sin (e+f x))}+\frac{(b c-a d)^2 \cos (e+f x)}{b (a-b) (a+b) (a+b \sin (e+f x))^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(-6*a*b*c*d + a^2*(2*c^2 + d^2) + b^2*(c^2 + 2*d^2))*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(a^
2 - b^2)^(5/2) + ((b*c - a*d)^2*Cos[e + f*x])/((a - b)*b*(a + b)*(a + b*Sin[e + f*x])^2) - ((2*a^2*b*c*d + 4*b
^3*c*d + a^3*d^2 - a*b^2*(3*c^2 + 4*d^2))*Cos[e + f*x])/((a - b)^2*b*(a + b)^2*(a + b*Sin[e + f*x])))/(2*f)

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Maple [B]  time = 0.096, size = 1923, normalized size = 9.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*b^3*c^2-4/f/(tan(1/2*f*x+1/2*e)^2
*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*f*x+1/2*e)^2*b^4*c*d-10/f/(tan(1/2*f*x+1/2*e)^2*a
+2*tan(1/2*f*x+1/2*e)*b+a)^2*a^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*b*c*d-6/f/(a^4-2*a^2*b^2+b^4)/(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*b*c*d-6/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2
*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^2*tan(1/2*f*x+1/2*e)^3*b*c*d-10/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f
*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*tan(1/2*f*x+1/2*e)^2*b^2*c*d-1/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+
1/2*e)*b+a)^2*a^3/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*d^2-4/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*
b+a)^2/(a^4-2*a^2*b^2+b^4)*a^3*c*d+4/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)
*a^2*b*c^2+3/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^2*b*d^2-4/f/(tan(1/2*
f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^3*tan(1/2*f*x+1/2*e)^2*c*d+4/f/(tan(1/2*f*x+1
/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^2*tan(1/2*f*x+1/2*e)^2*b*c^2+3/f/(tan(1/2*f*x+1/2*
e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^2*tan(1/2*f*x+1/2*e)^2*b*d^2-2/f/(tan(1/2*f*x+1/2*e)^
2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)/a^2*tan(1/2*f*x+1/2*e)^2*b^5*c^2+2/f/(a^4-2*a^2*b^2+b^4)/(
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))*c^2*a^2+1/f/(a^4-2*a^2*b^2+b^4)/(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*d^2+1/f/(a^4-2*a^2*b^2+b^4)/(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))*c^2*b^2+2/f/(a^4-2*a^2*b^2+b^4)/(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))*b^2*d^2+1/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2
*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^3*tan(1/2*f*x+1/2*e)^3*d^2+7/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b
+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^2*b^3*c^2+6/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2
/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^2*b^3*d^2+11/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*a/(
a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*b^2*c^2+10/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*a/(a^4-
2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*b^2*d^2-8/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b
^2+b^4)*tan(1/2*f*x+1/2*e)*b^3*c*d-2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/a/(a^4-2*a^2*b^2+b^
4)*tan(1/2*f*x+1/2*e)*b^4*c^2-2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*b^
2*c*d+5/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*tan(1/2*f*x+1/2*e)^3*b^2*c
^2+2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*tan(1/2*f*x+1/2*e)^3*b^2*d^2-
2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*f*x+1/2*e)^3*b^4*c^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((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.75024, size = 2137, normalized size = 10.9 \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+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/4*(2*(3*(a^3*b^2 - a*b^4)*c^2 - 2*(a^4*b + a^2*b^3 - 2*b^5)*c*d - (a^5 - 5*a^3*b^2 + 4*a*b^4)*d^2)*cos(f*x
+ e)*sin(f*x + e) - ((2*a^4 + 3*a^2*b^2 + b^4)*c^2 - 6*(a^3*b + a*b^3)*c*d + (a^4 + 3*a^2*b^2 + 2*b^4)*d^2 +
(6*a*b^3*c*d - (2*a^2*b^2 + b^4)*c^2 - (a^2*b^2 + 2*b^4)*d^2)*cos(f*x + e)^2 - 2*(6*a^2*b^2*c*d - (2*a^3*b + a
*b^3)*c^2 - (a^3*b + 2*a*b^3)*d^2)*sin(f*x + e))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*si
n(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*((4*a^4*b - 5*a^2*b^3 + b^5)*c^2 - 2*(2*a^5 - a^3*b^2 - a*b^4)*c*d +
3*(a^4*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*f*cos(f*x + e)^2 - 2*(a^7*b -
3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*f*sin(f*x + e) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*f), -1/2*((3*(a^3*b^2 - a*
b^4)*c^2 - 2*(a^4*b + a^2*b^3 - 2*b^5)*c*d - (a^5 - 5*a^3*b^2 + 4*a*b^4)*d^2)*cos(f*x + e)*sin(f*x + e) - ((2*
a^4 + 3*a^2*b^2 + b^4)*c^2 - 6*(a^3*b + a*b^3)*c*d + (a^4 + 3*a^2*b^2 + 2*b^4)*d^2 + (6*a*b^3*c*d - (2*a^2*b^2
+ b^4)*c^2 - (a^2*b^2 + 2*b^4)*d^2)*cos(f*x + e)^2 - 2*(6*a^2*b^2*c*d - (2*a^3*b + a*b^3)*c^2 - (a^3*b + 2*a*
b^3)*d^2)*sin(f*x + e))*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) + ((4*a^4
*b - 5*a^2*b^3 + b^5)*c^2 - 2*(2*a^5 - a^3*b^2 - a*b^4)*c*d + 3*(a^4*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^6*b^2
- 3*a^4*b^4 + 3*a^2*b^6 - b^8)*f*cos(f*x + e)^2 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*f*sin(f*x + e) -
(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*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))**2/(a+b*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.39004, size = 822, normalized size = 4.19 \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+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

((2*a^2*c^2 + b^2*c^2 - 6*a*b*c*d + a^2*d^2 + 2*b^2*d^2)*(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)))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (5*a^3*b^2*c^2*tan(1/2
*f*x + 1/2*e)^3 - 2*a*b^4*c^2*tan(1/2*f*x + 1/2*e)^3 - 6*a^4*b*c*d*tan(1/2*f*x + 1/2*e)^3 + a^5*d^2*tan(1/2*f*
x + 1/2*e)^3 + 2*a^3*b^2*d^2*tan(1/2*f*x + 1/2*e)^3 + 4*a^4*b*c^2*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*b^3*c^2*tan(1
/2*f*x + 1/2*e)^2 - 2*b^5*c^2*tan(1/2*f*x + 1/2*e)^2 - 4*a^5*c*d*tan(1/2*f*x + 1/2*e)^2 - 10*a^3*b^2*c*d*tan(1
/2*f*x + 1/2*e)^2 - 4*a*b^4*c*d*tan(1/2*f*x + 1/2*e)^2 + 3*a^4*b*d^2*tan(1/2*f*x + 1/2*e)^2 + 6*a^2*b^3*d^2*ta
n(1/2*f*x + 1/2*e)^2 + 11*a^3*b^2*c^2*tan(1/2*f*x + 1/2*e) - 2*a*b^4*c^2*tan(1/2*f*x + 1/2*e) - 10*a^4*b*c*d*t
an(1/2*f*x + 1/2*e) - 8*a^2*b^3*c*d*tan(1/2*f*x + 1/2*e) - a^5*d^2*tan(1/2*f*x + 1/2*e) + 10*a^3*b^2*d^2*tan(1
/2*f*x + 1/2*e) + 4*a^4*b*c^2 - a^2*b^3*c^2 - 4*a^5*c*d - 2*a^3*b^2*c*d + 3*a^4*b*d^2)/((a^6 - 2*a^4*b^2 + a^2
*b^4)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)^2))/f