3.16 \(\int \frac{(a+b x)^2 \sin (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=175 \[ -\frac{1}{6} a^2 d^3 \cos (c) \text{CosIntegral}(d x)+\frac{1}{6} a^2 d^3 \sin (c) \text{Si}(d x)+\frac{a^2 d^2 \sin (c+d x)}{6 x}-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a^2 d \cos (c+d x)}{6 x^2}-a b d^2 \sin (c) \text{CosIntegral}(d x)-a b d^2 \cos (c) \text{Si}(d x)-\frac{a b \sin (c+d x)}{x^2}-\frac{a b d \cos (c+d x)}{x}+b^2 d \cos (c) \text{CosIntegral}(d x)-b^2 d \sin (c) \text{Si}(d x)-\frac{b^2 \sin (c+d x)}{x} \]

[Out]

-(a^2*d*Cos[c + d*x])/(6*x^2) - (a*b*d*Cos[c + d*x])/x + b^2*d*Cos[c]*CosIntegral[d*x] - (a^2*d^3*Cos[c]*CosIn
tegral[d*x])/6 - a*b*d^2*CosIntegral[d*x]*Sin[c] - (a^2*Sin[c + d*x])/(3*x^3) - (a*b*Sin[c + d*x])/x^2 - (b^2*
Sin[c + d*x])/x + (a^2*d^2*Sin[c + d*x])/(6*x) - a*b*d^2*Cos[c]*SinIntegral[d*x] - b^2*d*Sin[c]*SinIntegral[d*
x] + (a^2*d^3*Sin[c]*SinIntegral[d*x])/6

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Rubi [A]  time = 0.410262, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {6742, 3297, 3303, 3299, 3302} \[ -\frac{1}{6} a^2 d^3 \cos (c) \text{CosIntegral}(d x)+\frac{1}{6} a^2 d^3 \sin (c) \text{Si}(d x)+\frac{a^2 d^2 \sin (c+d x)}{6 x}-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a^2 d \cos (c+d x)}{6 x^2}-a b d^2 \sin (c) \text{CosIntegral}(d x)-a b d^2 \cos (c) \text{Si}(d x)-\frac{a b \sin (c+d x)}{x^2}-\frac{a b d \cos (c+d x)}{x}+b^2 d \cos (c) \text{CosIntegral}(d x)-b^2 d \sin (c) \text{Si}(d x)-\frac{b^2 \sin (c+d x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Sin[c + d*x])/x^4,x]

[Out]

-(a^2*d*Cos[c + d*x])/(6*x^2) - (a*b*d*Cos[c + d*x])/x + b^2*d*Cos[c]*CosIntegral[d*x] - (a^2*d^3*Cos[c]*CosIn
tegral[d*x])/6 - a*b*d^2*CosIntegral[d*x]*Sin[c] - (a^2*Sin[c + d*x])/(3*x^3) - (a*b*Sin[c + d*x])/x^2 - (b^2*
Sin[c + d*x])/x + (a^2*d^2*Sin[c + d*x])/(6*x) - a*b*d^2*Cos[c]*SinIntegral[d*x] - b^2*d*Sin[c]*SinIntegral[d*
x] + (a^2*d^3*Sin[c]*SinIntegral[d*x])/6

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3303

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{(a+b x)^2 \sin (c+d x)}{x^4} \, dx &=\int \left (\frac{a^2 \sin (c+d x)}{x^4}+\frac{2 a b \sin (c+d x)}{x^3}+\frac{b^2 \sin (c+d x)}{x^2}\right ) \, dx\\ &=a^2 \int \frac{\sin (c+d x)}{x^4} \, dx+(2 a b) \int \frac{\sin (c+d x)}{x^3} \, dx+b^2 \int \frac{\sin (c+d x)}{x^2} \, dx\\ &=-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a b \sin (c+d x)}{x^2}-\frac{b^2 \sin (c+d x)}{x}+\frac{1}{3} \left (a^2 d\right ) \int \frac{\cos (c+d x)}{x^3} \, dx+(a b d) \int \frac{\cos (c+d x)}{x^2} \, dx+\left (b^2 d\right ) \int \frac{\cos (c+d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{6 x^2}-\frac{a b d \cos (c+d x)}{x}-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a b \sin (c+d x)}{x^2}-\frac{b^2 \sin (c+d x)}{x}-\frac{1}{6} \left (a^2 d^2\right ) \int \frac{\sin (c+d x)}{x^2} \, dx-\left (a b d^2\right ) \int \frac{\sin (c+d x)}{x} \, dx+\left (b^2 d \cos (c)\right ) \int \frac{\cos (d x)}{x} \, dx-\left (b^2 d \sin (c)\right ) \int \frac{\sin (d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{6 x^2}-\frac{a b d \cos (c+d x)}{x}+b^2 d \cos (c) \text{Ci}(d x)-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a b \sin (c+d x)}{x^2}-\frac{b^2 \sin (c+d x)}{x}+\frac{a^2 d^2 \sin (c+d x)}{6 x}-b^2 d \sin (c) \text{Si}(d x)-\frac{1}{6} \left (a^2 d^3\right ) \int \frac{\cos (c+d x)}{x} \, dx-\left (a b d^2 \cos (c)\right ) \int \frac{\sin (d x)}{x} \, dx-\left (a b d^2 \sin (c)\right ) \int \frac{\cos (d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{6 x^2}-\frac{a b d \cos (c+d x)}{x}+b^2 d \cos (c) \text{Ci}(d x)-a b d^2 \text{Ci}(d x) \sin (c)-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a b \sin (c+d x)}{x^2}-\frac{b^2 \sin (c+d x)}{x}+\frac{a^2 d^2 \sin (c+d x)}{6 x}-a b d^2 \cos (c) \text{Si}(d x)-b^2 d \sin (c) \text{Si}(d x)-\frac{1}{6} \left (a^2 d^3 \cos (c)\right ) \int \frac{\cos (d x)}{x} \, dx+\frac{1}{6} \left (a^2 d^3 \sin (c)\right ) \int \frac{\sin (d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{6 x^2}-\frac{a b d \cos (c+d x)}{x}+b^2 d \cos (c) \text{Ci}(d x)-\frac{1}{6} a^2 d^3 \cos (c) \text{Ci}(d x)-a b d^2 \text{Ci}(d x) \sin (c)-\frac{a^2 \sin (c+d x)}{3 x^3}-\frac{a b \sin (c+d x)}{x^2}-\frac{b^2 \sin (c+d x)}{x}+\frac{a^2 d^2 \sin (c+d x)}{6 x}-a b d^2 \cos (c) \text{Si}(d x)-b^2 d \sin (c) \text{Si}(d x)+\frac{1}{6} a^2 d^3 \sin (c) \text{Si}(d x)\\ \end{align*}

Mathematica [A]  time = 0.53178, size = 154, normalized size = 0.88 \[ -\frac{d x^3 \text{CosIntegral}(d x) \left (\cos (c) \left (a^2 d^2-6 b^2\right )+6 a b d \sin (c)\right )+d x^3 \text{Si}(d x) \left (-a^2 d^2 \sin (c)+6 a b d \cos (c)+6 b^2 \sin (c)\right )-a^2 d^2 x^2 \sin (c+d x)+2 a^2 \sin (c+d x)+a^2 d x \cos (c+d x)+6 a b d x^2 \cos (c+d x)+6 a b x \sin (c+d x)+6 b^2 x^2 \sin (c+d x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*Sin[c + d*x])/x^4,x]

[Out]

-(a^2*d*x*Cos[c + d*x] + 6*a*b*d*x^2*Cos[c + d*x] + d*x^3*CosIntegral[d*x]*((-6*b^2 + a^2*d^2)*Cos[c] + 6*a*b*
d*Sin[c]) + 2*a^2*Sin[c + d*x] + 6*a*b*x*Sin[c + d*x] + 6*b^2*x^2*Sin[c + d*x] - a^2*d^2*x^2*Sin[c + d*x] + d*
x^3*(6*a*b*d*Cos[c] + 6*b^2*Sin[c] - a^2*d^2*Sin[c])*SinIntegral[d*x])/(6*x^3)

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Maple [A]  time = 0.015, size = 158, normalized size = 0.9 \begin{align*}{d}^{3} \left ({\frac{{b}^{2}}{{d}^{2}} \left ( -{\frac{\sin \left ( dx+c \right ) }{dx}}-{\it Si} \left ( dx \right ) \sin \left ( c \right ) +{\it Ci} \left ( dx \right ) \cos \left ( c \right ) \right ) }+2\,{\frac{ab}{d} \left ( -1/2\,{\frac{\sin \left ( dx+c \right ) }{{d}^{2}{x}^{2}}}-1/2\,{\frac{\cos \left ( dx+c \right ) }{dx}}-1/2\,{\it Si} \left ( dx \right ) \cos \left ( c \right ) -1/2\,{\it Ci} \left ( dx \right ) \sin \left ( c \right ) \right ) }+{a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) }{3\,{d}^{3}{x}^{3}}}-{\frac{\cos \left ( dx+c \right ) }{6\,{d}^{2}{x}^{2}}}+{\frac{\sin \left ( dx+c \right ) }{6\,dx}}+{\frac{{\it Si} \left ( dx \right ) \sin \left ( c \right ) }{6}}-{\frac{{\it Ci} \left ( dx \right ) \cos \left ( c \right ) }{6}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*sin(d*x+c)/x^4,x)

[Out]

d^3*(1/d^2*b^2*(-sin(d*x+c)/x/d-Si(d*x)*sin(c)+Ci(d*x)*cos(c))+2/d*a*b*(-1/2*sin(d*x+c)/x^2/d^2-1/2*cos(d*x+c)
/x/d-1/2*Si(d*x)*cos(c)-1/2*Ci(d*x)*sin(c))+a^2*(-1/3*sin(d*x+c)/x^3/d^3-1/6*cos(d*x+c)/x^2/d^2+1/6*sin(d*x+c)
/x/d+1/6*Si(d*x)*sin(c)-1/6*Ci(d*x)*cos(c)))

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Maxima [C]  time = 6.14697, size = 254, normalized size = 1.45 \begin{align*} -\frac{{\left ({\left (a^{2}{\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2}{\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{5} +{\left (a b{\left (6 i \, \Gamma \left (-3, i \, d x\right ) - 6 i \, \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) + 6 \, a b{\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{4} -{\left (6 \, b^{2}{\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \left (c\right ) - b^{2}{\left (6 i \, \Gamma \left (-3, i \, d x\right ) - 6 i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3}\right )} x^{3} + 4 \, b^{2} \sin \left (d x + c\right ) + 2 \,{\left (b^{2} d x + 2 \, a b d\right )} \cos \left (d x + c\right )}{2 \, d^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^4,x, algorithm="maxima")

[Out]

-1/2*(((a^2*(gamma(-3, I*d*x) + gamma(-3, -I*d*x))*cos(c) + a^2*(-I*gamma(-3, I*d*x) + I*gamma(-3, -I*d*x))*si
n(c))*d^5 + (a*b*(6*I*gamma(-3, I*d*x) - 6*I*gamma(-3, -I*d*x))*cos(c) + 6*a*b*(gamma(-3, I*d*x) + gamma(-3, -
I*d*x))*sin(c))*d^4 - (6*b^2*(gamma(-3, I*d*x) + gamma(-3, -I*d*x))*cos(c) - b^2*(6*I*gamma(-3, I*d*x) - 6*I*g
amma(-3, -I*d*x))*sin(c))*d^3)*x^3 + 4*b^2*sin(d*x + c) + 2*(b^2*d*x + 2*a*b*d)*cos(d*x + c))/(d^2*x^3)

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Fricas [A]  time = 1.77188, size = 497, normalized size = 2.84 \begin{align*} -\frac{2 \,{\left (6 \, a b d x^{2} + a^{2} d x\right )} \cos \left (d x + c\right ) +{\left (12 \, a b d^{2} x^{3} \operatorname{Si}\left (d x\right ) +{\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x^{3} \operatorname{Ci}\left (d x\right ) +{\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x^{3} \operatorname{Ci}\left (-d x\right )\right )} \cos \left (c\right ) + 2 \,{\left (6 \, a b x -{\left (a^{2} d^{2} - 6 \, b^{2}\right )} x^{2} + 2 \, a^{2}\right )} \sin \left (d x + c\right ) + 2 \,{\left (3 \, a b d^{2} x^{3} \operatorname{Ci}\left (d x\right ) + 3 \, a b d^{2} x^{3} \operatorname{Ci}\left (-d x\right ) -{\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x^{3} \operatorname{Si}\left (d x\right )\right )} \sin \left (c\right )}{12 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^4,x, algorithm="fricas")

[Out]

-1/12*(2*(6*a*b*d*x^2 + a^2*d*x)*cos(d*x + c) + (12*a*b*d^2*x^3*sin_integral(d*x) + (a^2*d^3 - 6*b^2*d)*x^3*co
s_integral(d*x) + (a^2*d^3 - 6*b^2*d)*x^3*cos_integral(-d*x))*cos(c) + 2*(6*a*b*x - (a^2*d^2 - 6*b^2)*x^2 + 2*
a^2)*sin(d*x + c) + 2*(3*a*b*d^2*x^3*cos_integral(d*x) + 3*a*b*d^2*x^3*cos_integral(-d*x) - (a^2*d^3 - 6*b^2*d
)*x^3*sin_integral(d*x))*sin(c))/x^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{2} \sin{\left (c + d x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*sin(d*x+c)/x**4,x)

[Out]

Integral((a + b*x)**2*sin(c + d*x)/x**4, x)

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Giac [C]  time = 1.16185, size = 1890, normalized size = 10.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^4,x, algorithm="giac")

[Out]

1/12*(a^2*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a^2*d^3*x^3*real_part(cos_integra
l(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) -
2*a^2*d^3*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^3*x^3*sin_integral(d*x)*tan(1/
2*d*x)^2*tan(1/2*c) + 6*a*b*d^2*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 6*a*b*d^2*x^3*i
mag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*a*b*d^2*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*tan
(1/2*c)^2 - a^2*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 - a^2*d^3*x^3*real_part(cos_integral(-d*x)
)*tan(1/2*d*x)^2 - 12*a*b*d^2*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 12*a*b*d^2*x^3*real
_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + a^2*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 +
a^2*d^3*x^3*real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 6*b^2*d*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)
^2*tan(1/2*c)^2 - 6*b^2*d*x^3*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 6*a*b*d^2*x^3*imag_p
art(cos_integral(d*x))*tan(1/2*d*x)^2 + 6*a*b*d^2*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 12*a*b*d^
2*x^3*sin_integral(d*x)*tan(1/2*d*x)^2 + 2*a^2*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d^3*x^3
*imag_part(cos_integral(-d*x))*tan(1/2*c) + 4*a^2*d^3*x^3*sin_integral(d*x)*tan(1/2*c) - 12*b^2*d*x^3*imag_par
t(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 12*b^2*d*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan
(1/2*c) - 24*b^2*d*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) + 6*a*b*d^2*x^3*imag_part(cos_integral(d*x)
)*tan(1/2*c)^2 - 6*a*b*d^2*x^3*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 12*a*b*d^2*x^3*sin_integral(d*x)*t
an(1/2*c)^2 - a^2*d^3*x^3*real_part(cos_integral(d*x)) - a^2*d^3*x^3*real_part(cos_integral(-d*x)) + 6*b^2*d*x
^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 6*b^2*d*x^3*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 12
*a*b*d^2*x^3*real_part(cos_integral(d*x))*tan(1/2*c) - 12*a*b*d^2*x^3*real_part(cos_integral(-d*x))*tan(1/2*c)
- 4*a^2*d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) - 6*b^2*d*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 - 6*b^2*d*x
^3*real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 4*a^2*d^2*x^2*tan(1/2*d*x)*tan(1/2*c)^2 - 12*a*b*d*x^2*tan(1/2
*d*x)^2*tan(1/2*c)^2 - 6*a*b*d^2*x^3*imag_part(cos_integral(d*x)) + 6*a*b*d^2*x^3*imag_part(cos_integral(-d*x)
) - 12*a*b*d^2*x^3*sin_integral(d*x) - 12*b^2*d*x^3*imag_part(cos_integral(d*x))*tan(1/2*c) + 12*b^2*d*x^3*ima
g_part(cos_integral(-d*x))*tan(1/2*c) - 24*b^2*d*x^3*sin_integral(d*x)*tan(1/2*c) - 2*a^2*d*x*tan(1/2*d*x)^2*t
an(1/2*c)^2 + 6*b^2*d*x^3*real_part(cos_integral(d*x)) + 6*b^2*d*x^3*real_part(cos_integral(-d*x)) + 4*a^2*d^2
*x^2*tan(1/2*d*x) + 12*a*b*d*x^2*tan(1/2*d*x)^2 + 4*a^2*d^2*x^2*tan(1/2*c) + 48*a*b*d*x^2*tan(1/2*d*x)*tan(1/2
*c) + 24*b^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 12*a*b*d*x^2*tan(1/2*c)^2 + 24*b^2*x^2*tan(1/2*d*x)*tan(1/2*c)^2
+ 2*a^2*d*x*tan(1/2*d*x)^2 + 8*a^2*d*x*tan(1/2*d*x)*tan(1/2*c) + 24*a*b*x*tan(1/2*d*x)^2*tan(1/2*c) + 2*a^2*d*
x*tan(1/2*c)^2 + 24*a*b*x*tan(1/2*d*x)*tan(1/2*c)^2 - 12*a*b*d*x^2 - 24*b^2*x^2*tan(1/2*d*x) - 24*b^2*x^2*tan(
1/2*c) + 8*a^2*tan(1/2*d*x)^2*tan(1/2*c) + 8*a^2*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a^2*d*x - 24*a*b*x*tan(1/2*d*x)
- 24*a*b*x*tan(1/2*c) - 8*a^2*tan(1/2*d*x) - 8*a^2*tan(1/2*c))/(x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 + x^3*tan(1/2
*d*x)^2 + x^3*tan(1/2*c)^2 + x^3)