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

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

[Out]

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

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Rubi [A]  time = 0.279383, 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} \[ -\frac{\left (a^2 \left (-\left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{5/2}}+\frac{(b c-a d)^2 \cos (e+f x)}{2 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}-\frac{\left (3 a c d+b \left (c^2-4 d^2\right )\right ) (b c-a d) \cos (e+f x)}{2 d f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

Mathematica [A]  time = 0.922089, size = 202, normalized size = 1.03 \[ \frac{\frac{2 \left (a^2 \left (2 c^2+d^2\right )-6 a b c d+b^2 \left (c^2+2 d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\left (c^2-d^2\right )^{5/2}}-\frac{\left (-3 a^2 c d^2+2 a b d \left (c^2+2 d^2\right )+b^2 \left (c^3-4 c d^2\right )\right ) \cos (e+f x)}{d (c-d)^2 (c+d)^2 (c+d \sin (e+f x))}+\frac{(b c-a d)^2 \cos (e+f x)}{d (c-d) (c+d) (c+d \sin (e+f x))^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

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Maple [B]  time = 0.088, 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((a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.1914, size = 2167, normalized size = 11.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.34841, 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((a+b*sin(f*x+e))^2/(c+d*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(c) + arctan((c*
tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((c^4 - 2*c^2*d^2 + d^4)*sqrt(c^2 - d^2)) + (b^2*c^5*tan(1/2*f*x +
1/2*e)^3 - 6*a*b*c^4*d*tan(1/2*f*x + 1/2*e)^3 + 5*a^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 2*b^2*c^3*d^2*tan(1/2*
f*x + 1/2*e)^3 - 2*a^2*c*d^4*tan(1/2*f*x + 1/2*e)^3 - 4*a*b*c^5*tan(1/2*f*x + 1/2*e)^2 + 4*a^2*c^4*d*tan(1/2*f
*x + 1/2*e)^2 + 3*b^2*c^4*d*tan(1/2*f*x + 1/2*e)^2 - 10*a*b*c^3*d^2*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*c^2*d^3*tan
(1/2*f*x + 1/2*e)^2 + 6*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e)^2 - 4*a*b*c*d^4*tan(1/2*f*x + 1/2*e)^2 - 2*a^2*d^5*ta
n(1/2*f*x + 1/2*e)^2 - b^2*c^5*tan(1/2*f*x + 1/2*e) - 10*a*b*c^4*d*tan(1/2*f*x + 1/2*e) + 11*a^2*c^3*d^2*tan(1
/2*f*x + 1/2*e) + 10*b^2*c^3*d^2*tan(1/2*f*x + 1/2*e) - 8*a*b*c^2*d^3*tan(1/2*f*x + 1/2*e) - 2*a^2*c*d^4*tan(1
/2*f*x + 1/2*e) - 4*a*b*c^5 + 4*a^2*c^4*d + 3*b^2*c^4*d - 2*a*b*c^3*d^2 - a^2*c^2*d^3)/((c^6 - 2*c^4*d^2 + c^2
*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f