3.17 \(\int \frac{(a+b x)^2 \sin (c+d x)}{x^5} \, dx\)

Optimal. Leaf size=248 \[ \frac{1}{24} a^2 d^4 \sin (c) \text{CosIntegral}(d x)+\frac{1}{24} a^2 d^4 \cos (c) \text{Si}(d x)+\frac{a^2 d^2 \sin (c+d x)}{24 x^2}+\frac{a^2 d^3 \cos (c+d x)}{24 x}-\frac{a^2 \sin (c+d x)}{4 x^4}-\frac{a^2 d \cos (c+d x)}{12 x^3}-\frac{1}{3} a b d^3 \cos (c) \text{CosIntegral}(d x)+\frac{1}{3} a b d^3 \sin (c) \text{Si}(d x)+\frac{a b d^2 \sin (c+d x)}{3 x}-\frac{2 a b \sin (c+d x)}{3 x^3}-\frac{a b d \cos (c+d x)}{3 x^2}-\frac{1}{2} b^2 d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} b^2 d^2 \cos (c) \text{Si}(d x)-\frac{b^2 \sin (c+d x)}{2 x^2}-\frac{b^2 d \cos (c+d x)}{2 x} \]

[Out]

-(a^2*d*Cos[c + d*x])/(12*x^3) - (a*b*d*Cos[c + d*x])/(3*x^2) - (b^2*d*Cos[c + d*x])/(2*x) + (a^2*d^3*Cos[c +
d*x])/(24*x) - (a*b*d^3*Cos[c]*CosIntegral[d*x])/3 - (b^2*d^2*CosIntegral[d*x]*Sin[c])/2 + (a^2*d^4*CosIntegra
l[d*x]*Sin[c])/24 - (a^2*Sin[c + d*x])/(4*x^4) - (2*a*b*Sin[c + d*x])/(3*x^3) - (b^2*Sin[c + d*x])/(2*x^2) + (
a^2*d^2*Sin[c + d*x])/(24*x^2) + (a*b*d^2*Sin[c + d*x])/(3*x) - (b^2*d^2*Cos[c]*SinIntegral[d*x])/2 + (a^2*d^4
*Cos[c]*SinIntegral[d*x])/24 + (a*b*d^3*Sin[c]*SinIntegral[d*x])/3

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Rubi [A]  time = 0.480212, antiderivative size = 248, normalized size of antiderivative = 1., number of steps used = 20, 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} \[ \frac{1}{24} a^2 d^4 \sin (c) \text{CosIntegral}(d x)+\frac{1}{24} a^2 d^4 \cos (c) \text{Si}(d x)+\frac{a^2 d^2 \sin (c+d x)}{24 x^2}+\frac{a^2 d^3 \cos (c+d x)}{24 x}-\frac{a^2 \sin (c+d x)}{4 x^4}-\frac{a^2 d \cos (c+d x)}{12 x^3}-\frac{1}{3} a b d^3 \cos (c) \text{CosIntegral}(d x)+\frac{1}{3} a b d^3 \sin (c) \text{Si}(d x)+\frac{a b d^2 \sin (c+d x)}{3 x}-\frac{2 a b \sin (c+d x)}{3 x^3}-\frac{a b d \cos (c+d x)}{3 x^2}-\frac{1}{2} b^2 d^2 \sin (c) \text{CosIntegral}(d x)-\frac{1}{2} b^2 d^2 \cos (c) \text{Si}(d x)-\frac{b^2 \sin (c+d x)}{2 x^2}-\frac{b^2 d \cos (c+d x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Sin[c + d*x])/x^5,x]

[Out]

-(a^2*d*Cos[c + d*x])/(12*x^3) - (a*b*d*Cos[c + d*x])/(3*x^2) - (b^2*d*Cos[c + d*x])/(2*x) + (a^2*d^3*Cos[c +
d*x])/(24*x) - (a*b*d^3*Cos[c]*CosIntegral[d*x])/3 - (b^2*d^2*CosIntegral[d*x]*Sin[c])/2 + (a^2*d^4*CosIntegra
l[d*x]*Sin[c])/24 - (a^2*Sin[c + d*x])/(4*x^4) - (2*a*b*Sin[c + d*x])/(3*x^3) - (b^2*Sin[c + d*x])/(2*x^2) + (
a^2*d^2*Sin[c + d*x])/(24*x^2) + (a*b*d^2*Sin[c + d*x])/(3*x) - (b^2*d^2*Cos[c]*SinIntegral[d*x])/2 + (a^2*d^4
*Cos[c]*SinIntegral[d*x])/24 + (a*b*d^3*Sin[c]*SinIntegral[d*x])/3

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^5} \, dx &=\int \left (\frac{a^2 \sin (c+d x)}{x^5}+\frac{2 a b \sin (c+d x)}{x^4}+\frac{b^2 \sin (c+d x)}{x^3}\right ) \, dx\\ &=a^2 \int \frac{\sin (c+d x)}{x^5} \, dx+(2 a b) \int \frac{\sin (c+d x)}{x^4} \, dx+b^2 \int \frac{\sin (c+d x)}{x^3} \, dx\\ &=-\frac{a^2 \sin (c+d x)}{4 x^4}-\frac{2 a b \sin (c+d x)}{3 x^3}-\frac{b^2 \sin (c+d x)}{2 x^2}+\frac{1}{4} \left (a^2 d\right ) \int \frac{\cos (c+d x)}{x^4} \, dx+\frac{1}{3} (2 a b d) \int \frac{\cos (c+d x)}{x^3} \, dx+\frac{1}{2} \left (b^2 d\right ) \int \frac{\cos (c+d x)}{x^2} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{12 x^3}-\frac{a b d \cos (c+d x)}{3 x^2}-\frac{b^2 d \cos (c+d x)}{2 x}-\frac{a^2 \sin (c+d x)}{4 x^4}-\frac{2 a b \sin (c+d x)}{3 x^3}-\frac{b^2 \sin (c+d x)}{2 x^2}-\frac{1}{12} \left (a^2 d^2\right ) \int \frac{\sin (c+d x)}{x^3} \, dx-\frac{1}{3} \left (a b d^2\right ) \int \frac{\sin (c+d x)}{x^2} \, dx-\frac{1}{2} \left (b^2 d^2\right ) \int \frac{\sin (c+d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{12 x^3}-\frac{a b d \cos (c+d x)}{3 x^2}-\frac{b^2 d \cos (c+d x)}{2 x}-\frac{a^2 \sin (c+d x)}{4 x^4}-\frac{2 a b \sin (c+d x)}{3 x^3}-\frac{b^2 \sin (c+d x)}{2 x^2}+\frac{a^2 d^2 \sin (c+d x)}{24 x^2}+\frac{a b d^2 \sin (c+d x)}{3 x}-\frac{1}{24} \left (a^2 d^3\right ) \int \frac{\cos (c+d x)}{x^2} \, dx-\frac{1}{3} \left (a b d^3\right ) \int \frac{\cos (c+d x)}{x} \, dx-\frac{1}{2} \left (b^2 d^2 \cos (c)\right ) \int \frac{\sin (d x)}{x} \, dx-\frac{1}{2} \left (b^2 d^2 \sin (c)\right ) \int \frac{\cos (d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{12 x^3}-\frac{a b d \cos (c+d x)}{3 x^2}-\frac{b^2 d \cos (c+d x)}{2 x}+\frac{a^2 d^3 \cos (c+d x)}{24 x}-\frac{1}{2} b^2 d^2 \text{Ci}(d x) \sin (c)-\frac{a^2 \sin (c+d x)}{4 x^4}-\frac{2 a b \sin (c+d x)}{3 x^3}-\frac{b^2 \sin (c+d x)}{2 x^2}+\frac{a^2 d^2 \sin (c+d x)}{24 x^2}+\frac{a b d^2 \sin (c+d x)}{3 x}-\frac{1}{2} b^2 d^2 \cos (c) \text{Si}(d x)+\frac{1}{24} \left (a^2 d^4\right ) \int \frac{\sin (c+d x)}{x} \, dx-\frac{1}{3} \left (a b d^3 \cos (c)\right ) \int \frac{\cos (d x)}{x} \, dx+\frac{1}{3} \left (a b d^3 \sin (c)\right ) \int \frac{\sin (d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{12 x^3}-\frac{a b d \cos (c+d x)}{3 x^2}-\frac{b^2 d \cos (c+d x)}{2 x}+\frac{a^2 d^3 \cos (c+d x)}{24 x}-\frac{1}{3} a b d^3 \cos (c) \text{Ci}(d x)-\frac{1}{2} b^2 d^2 \text{Ci}(d x) \sin (c)-\frac{a^2 \sin (c+d x)}{4 x^4}-\frac{2 a b \sin (c+d x)}{3 x^3}-\frac{b^2 \sin (c+d x)}{2 x^2}+\frac{a^2 d^2 \sin (c+d x)}{24 x^2}+\frac{a b d^2 \sin (c+d x)}{3 x}-\frac{1}{2} b^2 d^2 \cos (c) \text{Si}(d x)+\frac{1}{3} a b d^3 \sin (c) \text{Si}(d x)+\frac{1}{24} \left (a^2 d^4 \cos (c)\right ) \int \frac{\sin (d x)}{x} \, dx+\frac{1}{24} \left (a^2 d^4 \sin (c)\right ) \int \frac{\cos (d x)}{x} \, dx\\ &=-\frac{a^2 d \cos (c+d x)}{12 x^3}-\frac{a b d \cos (c+d x)}{3 x^2}-\frac{b^2 d \cos (c+d x)}{2 x}+\frac{a^2 d^3 \cos (c+d x)}{24 x}-\frac{1}{3} a b d^3 \cos (c) \text{Ci}(d x)-\frac{1}{2} b^2 d^2 \text{Ci}(d x) \sin (c)+\frac{1}{24} a^2 d^4 \text{Ci}(d x) \sin (c)-\frac{a^2 \sin (c+d x)}{4 x^4}-\frac{2 a b \sin (c+d x)}{3 x^3}-\frac{b^2 \sin (c+d x)}{2 x^2}+\frac{a^2 d^2 \sin (c+d x)}{24 x^2}+\frac{a b d^2 \sin (c+d x)}{3 x}-\frac{1}{2} b^2 d^2 \cos (c) \text{Si}(d x)+\frac{1}{24} a^2 d^4 \cos (c) \text{Si}(d x)+\frac{1}{3} a b d^3 \sin (c) \text{Si}(d x)\\ \end{align*}

Mathematica [A]  time = 0.453075, size = 204, normalized size = 0.82 \[ \frac{d^2 x^4 \text{CosIntegral}(d x) \left (\sin (c) \left (a^2 d^2-12 b^2\right )-8 a b d \cos (c)\right )+d^2 x^4 \text{Si}(d x) \left (a^2 d^2 \cos (c)+8 a b d \sin (c)-12 b^2 \cos (c)\right )+a^2 d^2 x^2 \sin (c+d x)+a^2 d^3 x^3 \cos (c+d x)-6 a^2 \sin (c+d x)-2 a^2 d x \cos (c+d x)+8 a b d^2 x^3 \sin (c+d x)-8 a b d x^2 \cos (c+d x)-16 a b x \sin (c+d x)-12 b^2 x^2 \sin (c+d x)-12 b^2 d x^3 \cos (c+d x)}{24 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*Sin[c + d*x])/x^5,x]

[Out]

(-2*a^2*d*x*Cos[c + d*x] - 8*a*b*d*x^2*Cos[c + d*x] - 12*b^2*d*x^3*Cos[c + d*x] + a^2*d^3*x^3*Cos[c + d*x] + d
^2*x^4*CosIntegral[d*x]*(-8*a*b*d*Cos[c] + (-12*b^2 + a^2*d^2)*Sin[c]) - 6*a^2*Sin[c + d*x] - 16*a*b*x*Sin[c +
d*x] - 12*b^2*x^2*Sin[c + d*x] + a^2*d^2*x^2*Sin[c + d*x] + 8*a*b*d^2*x^3*Sin[c + d*x] + d^2*x^4*(-12*b^2*Cos
[c] + a^2*d^2*Cos[c] + 8*a*b*d*Sin[c])*SinIntegral[d*x])/(24*x^4)

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Maple [A]  time = 0.019, size = 201, normalized size = 0.8 \begin{align*}{d}^{4} \left ({\frac{{b}^{2}}{{d}^{2}} \left ( -{\frac{\sin \left ( dx+c \right ) }{2\,{d}^{2}{x}^{2}}}-{\frac{\cos \left ( dx+c \right ) }{2\,dx}}-{\frac{{\it Si} \left ( dx \right ) \cos \left ( c \right ) }{2}}-{\frac{{\it Ci} \left ( dx \right ) \sin \left ( c \right ) }{2}} \right ) }+2\,{\frac{ab}{d} \left ( -1/3\,{\frac{\sin \left ( dx+c \right ) }{{d}^{3}{x}^{3}}}-1/6\,{\frac{\cos \left ( dx+c \right ) }{{d}^{2}{x}^{2}}}+1/6\,{\frac{\sin \left ( dx+c \right ) }{dx}}+1/6\,{\it Si} \left ( dx \right ) \sin \left ( c \right ) -1/6\,{\it Ci} \left ( dx \right ) \cos \left ( c \right ) \right ) }+{a}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) }{4\,{x}^{4}{d}^{4}}}-{\frac{\cos \left ( dx+c \right ) }{12\,{d}^{3}{x}^{3}}}+{\frac{\sin \left ( dx+c \right ) }{24\,{d}^{2}{x}^{2}}}+{\frac{\cos \left ( dx+c \right ) }{24\,dx}}+{\frac{{\it Si} \left ( dx \right ) \cos \left ( c \right ) }{24}}+{\frac{{\it Ci} \left ( dx \right ) \sin \left ( c \right ) }{24}} \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^5,x)

[Out]

d^4*(1/d^2*b^2*(-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))+2/d*a*b*(-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))+a^2*(-1/
4*sin(d*x+c)/x^4/d^4-1/12*cos(d*x+c)/x^3/d^3+1/24*sin(d*x+c)/x^2/d^2+1/24*cos(d*x+c)/x/d+1/24*Si(d*x)*cos(c)+1
/24*Ci(d*x)*sin(c)))

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Maxima [C]  time = 7.13207, size = 254, normalized size = 1.02 \begin{align*} -\frac{{\left ({\left (a^{2}{\left (i \, \Gamma \left (-4, i \, d x\right ) - i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2}{\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} -{\left (8 \, a b{\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \cos \left (c\right ) - a b{\left (8 i \, \Gamma \left (-4, i \, d x\right ) - 8 i \, \Gamma \left (-4, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{5} +{\left (b^{2}{\left (-12 i \, \Gamma \left (-4, i \, d x\right ) + 12 i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \left (c\right ) - 12 \, b^{2}{\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{4}\right )} x^{4} + 6 \, 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^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^5,x, algorithm="maxima")

[Out]

-1/2*(((a^2*(I*gamma(-4, I*d*x) - I*gamma(-4, -I*d*x))*cos(c) + a^2*(gamma(-4, I*d*x) + gamma(-4, -I*d*x))*sin
(c))*d^6 - (8*a*b*(gamma(-4, I*d*x) + gamma(-4, -I*d*x))*cos(c) - a*b*(8*I*gamma(-4, I*d*x) - 8*I*gamma(-4, -I
*d*x))*sin(c))*d^5 + (b^2*(-12*I*gamma(-4, I*d*x) + 12*I*gamma(-4, -I*d*x))*cos(c) - 12*b^2*(gamma(-4, I*d*x)
+ gamma(-4, -I*d*x))*sin(c))*d^4)*x^4 + 6*b^2*sin(d*x + c) + 2*(b^2*d*x + 2*a*b*d)*cos(d*x + c))/(d^2*x^4)

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Fricas [A]  time = 1.70371, size = 572, normalized size = 2.31 \begin{align*} -\frac{2 \,{\left (8 \, a b d x^{2} + 2 \, a^{2} d x -{\left (a^{2} d^{3} - 12 \, b^{2} d\right )} x^{3}\right )} \cos \left (d x + c\right ) + 2 \,{\left (4 \, a b d^{3} x^{4} \operatorname{Ci}\left (d x\right ) + 4 \, a b d^{3} x^{4} \operatorname{Ci}\left (-d x\right ) -{\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname{Si}\left (d x\right )\right )} \cos \left (c\right ) - 2 \,{\left (8 \, a b d^{2} x^{3} - 16 \, a b x +{\left (a^{2} d^{2} - 12 \, b^{2}\right )} x^{2} - 6 \, a^{2}\right )} \sin \left (d x + c\right ) -{\left (16 \, a b d^{3} x^{4} \operatorname{Si}\left (d x\right ) +{\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname{Ci}\left (d x\right ) +{\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname{Ci}\left (-d x\right )\right )} \sin \left (c\right )}{48 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^5,x, algorithm="fricas")

[Out]

-1/48*(2*(8*a*b*d*x^2 + 2*a^2*d*x - (a^2*d^3 - 12*b^2*d)*x^3)*cos(d*x + c) + 2*(4*a*b*d^3*x^4*cos_integral(d*x
) + 4*a*b*d^3*x^4*cos_integral(-d*x) - (a^2*d^4 - 12*b^2*d^2)*x^4*sin_integral(d*x))*cos(c) - 2*(8*a*b*d^2*x^3
- 16*a*b*x + (a^2*d^2 - 12*b^2)*x^2 - 6*a^2)*sin(d*x + c) - (16*a*b*d^3*x^4*sin_integral(d*x) + (a^2*d^4 - 12
*b^2*d^2)*x^4*cos_integral(d*x) + (a^2*d^4 - 12*b^2*d^2)*x^4*cos_integral(-d*x))*sin(c))/x^4

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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^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*sin(d*x+c)/x**5,x)

[Out]

Integral((a + b*x)**2*sin(c + d*x)/x**5, x)

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Giac [C]  time = 1.16252, size = 2311, normalized size = 9.32 \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^5,x, algorithm="giac")

[Out]

-1/48*(a^2*d^4*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^4*x^4*imag_part(cos_integr
al(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^4*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a^2*d^
4*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^4*x^4*real_part(cos_integral(-d*x))*tan
(1/2*d*x)^2*tan(1/2*c) - 8*a*b*d^3*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 8*a*b*d^3*x^
4*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^4*x^4*imag_part(cos_integral(d*x))*tan(1/2
*d*x)^2 + a^2*d^4*x^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a^2*d^4*x^4*sin_integral(d*x)*tan(1/2*d
*x)^2 - 16*a*b*d^3*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 16*a*b*d^3*x^4*imag_part(cos_i
ntegral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 32*a*b*d^3*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) + a^2*d^
4*x^4*imag_part(cos_integral(d*x))*tan(1/2*c)^2 - a^2*d^4*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a
^2*d^4*x^4*sin_integral(d*x)*tan(1/2*c)^2 - 12*b^2*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2
*c)^2 + 12*b^2*d^2*x^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 24*b^2*d^2*x^4*sin_integral
(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 + 8*a*b*d^3*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 8*a*b*d^3*x^4*
real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a^2*d^4*x^4*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d
^4*x^4*real_part(cos_integral(-d*x))*tan(1/2*c) + 24*b^2*d^2*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*t
an(1/2*c) + 24*b^2*d^2*x^4*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 8*a*b*d^3*x^4*real_part(c
os_integral(d*x))*tan(1/2*c)^2 - 8*a*b*d^3*x^4*real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 2*a^2*d^3*x^3*tan(
1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^4*x^4*imag_part(cos_integral(d*x)) + a^2*d^4*x^4*imag_part(cos_integral(-d*x))
- 2*a^2*d^4*x^4*sin_integral(d*x) + 12*b^2*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 12*b^2*d^2*x
^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 + 24*b^2*d^2*x^4*sin_integral(d*x)*tan(1/2*d*x)^2 - 16*a*b*d^3
*x^4*imag_part(cos_integral(d*x))*tan(1/2*c) + 16*a*b*d^3*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c) - 32*a*
b*d^3*x^4*sin_integral(d*x)*tan(1/2*c) - 12*b^2*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*c)^2 + 12*b^2*d^2
*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 - 24*b^2*d^2*x^4*sin_integral(d*x)*tan(1/2*c)^2 + 8*a*b*d^3*x^
4*real_part(cos_integral(d*x)) + 8*a*b*d^3*x^4*real_part(cos_integral(-d*x)) + 2*a^2*d^3*x^3*tan(1/2*d*x)^2 +
24*b^2*d^2*x^4*real_part(cos_integral(d*x))*tan(1/2*c) + 24*b^2*d^2*x^4*real_part(cos_integral(-d*x))*tan(1/2*
c) + 8*a^2*d^3*x^3*tan(1/2*d*x)*tan(1/2*c) + 32*a*b*d^2*x^3*tan(1/2*d*x)^2*tan(1/2*c) + 2*a^2*d^3*x^3*tan(1/2*
c)^2 + 32*a*b*d^2*x^3*tan(1/2*d*x)*tan(1/2*c)^2 + 24*b^2*d*x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*b^2*d^2*x^4*im
ag_part(cos_integral(d*x)) - 12*b^2*d^2*x^4*imag_part(cos_integral(-d*x)) + 24*b^2*d^2*x^4*sin_integral(d*x) +
4*a^2*d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^2*x^2*tan(1/2*d*x)*tan(1/2*c)^2 + 16*a*b*d*x^2*tan(1/2*d*x)
^2*tan(1/2*c)^2 - 2*a^2*d^3*x^3 - 32*a*b*d^2*x^3*tan(1/2*d*x) - 24*b^2*d*x^3*tan(1/2*d*x)^2 - 32*a*b*d^2*x^3*t
an(1/2*c) - 96*b^2*d*x^3*tan(1/2*d*x)*tan(1/2*c) - 24*b^2*d*x^3*tan(1/2*c)^2 + 4*a^2*d*x*tan(1/2*d*x)^2*tan(1/
2*c)^2 - 4*a^2*d^2*x^2*tan(1/2*d*x) - 16*a*b*d*x^2*tan(1/2*d*x)^2 - 4*a^2*d^2*x^2*tan(1/2*c) - 64*a*b*d*x^2*ta
n(1/2*d*x)*tan(1/2*c) - 48*b^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) - 16*a*b*d*x^2*tan(1/2*c)^2 - 48*b^2*x^2*tan(1/2*
d*x)*tan(1/2*c)^2 + 24*b^2*d*x^3 - 4*a^2*d*x*tan(1/2*d*x)^2 - 16*a^2*d*x*tan(1/2*d*x)*tan(1/2*c) - 64*a*b*x*ta
n(1/2*d*x)^2*tan(1/2*c) - 4*a^2*d*x*tan(1/2*c)^2 - 64*a*b*x*tan(1/2*d*x)*tan(1/2*c)^2 + 16*a*b*d*x^2 + 48*b^2*
x^2*tan(1/2*d*x) + 48*b^2*x^2*tan(1/2*c) - 24*a^2*tan(1/2*d*x)^2*tan(1/2*c) - 24*a^2*tan(1/2*d*x)*tan(1/2*c)^2
+ 4*a^2*d*x + 64*a*b*x*tan(1/2*d*x) + 64*a*b*x*tan(1/2*c) + 24*a^2*tan(1/2*d*x) + 24*a^2*tan(1/2*c))/(x^4*tan
(1/2*d*x)^2*tan(1/2*c)^2 + x^4*tan(1/2*d*x)^2 + x^4*tan(1/2*c)^2 + x^4)