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

Optimal. Leaf size=169 \[ -\frac{2 a^2+b^2}{8 x^4}-\frac{1}{2} a b d^2 \sin (c) \text{CosIntegral}\left (d x^2\right )-\frac{1}{2} a b d^2 \cos (c) \text{Si}\left (d x^2\right )-\frac{a b \sin \left (c+d x^2\right )}{2 x^4}-\frac{a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac{1}{2} b^2 d^2 \cos (2 c) \text{CosIntegral}\left (2 d x^2\right )-\frac{1}{2} b^2 d^2 \sin (2 c) \text{Si}\left (2 d x^2\right )-\frac{b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}+\frac{b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4} \]

[Out]

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

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Rubi [A]  time = 0.290192, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3403, 6, 3380, 3297, 3303, 3299, 3302, 3379} \[ -\frac{2 a^2+b^2}{8 x^4}-\frac{1}{2} a b d^2 \sin (c) \text{CosIntegral}\left (d x^2\right )-\frac{1}{2} a b d^2 \cos (c) \text{Si}\left (d x^2\right )-\frac{a b \sin \left (c+d x^2\right )}{2 x^4}-\frac{a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac{1}{2} b^2 d^2 \cos (2 c) \text{CosIntegral}\left (2 d x^2\right )-\frac{1}{2} b^2 d^2 \sin (2 c) \text{Si}\left (2 d x^2\right )-\frac{b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}+\frac{b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])^2/x^5,x]

[Out]

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

Rule 3403

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

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]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

\begin{align*} \int \frac{\left (a+b \sin \left (c+d x^2\right )\right )^2}{x^5} \, dx &=\int \left (\frac{a^2}{x^5}+\frac{b^2}{2 x^5}-\frac{b^2 \cos \left (2 c+2 d x^2\right )}{2 x^5}+\frac{2 a b \sin \left (c+d x^2\right )}{x^5}\right ) \, dx\\ &=\int \left (\frac{a^2+\frac{b^2}{2}}{x^5}-\frac{b^2 \cos \left (2 c+2 d x^2\right )}{2 x^5}+\frac{2 a b \sin \left (c+d x^2\right )}{x^5}\right ) \, dx\\ &=-\frac{2 a^2+b^2}{8 x^4}+(2 a b) \int \frac{\sin \left (c+d x^2\right )}{x^5} \, dx-\frac{1}{2} b^2 \int \frac{\cos \left (2 c+2 d x^2\right )}{x^5} \, dx\\ &=-\frac{2 a^2+b^2}{8 x^4}+(a b) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{x^3} \, dx,x,x^2\right )-\frac{1}{4} b^2 \operatorname{Subst}\left (\int \frac{\cos (2 c+2 d x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac{2 a^2+b^2}{8 x^4}+\frac{b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4}-\frac{a b \sin \left (c+d x^2\right )}{2 x^4}+\frac{1}{2} (a b d) \operatorname{Subst}\left (\int \frac{\cos (c+d x)}{x^2} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 d\right ) \operatorname{Subst}\left (\int \frac{\sin (2 c+2 d x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac{2 a^2+b^2}{8 x^4}-\frac{a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac{b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4}-\frac{a b \sin \left (c+d x^2\right )}{2 x^4}-\frac{b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}-\frac{1}{2} \left (a b d^2\right ) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{x} \, dx,x,x^2\right )+\frac{1}{2} \left (b^2 d^2\right ) \operatorname{Subst}\left (\int \frac{\cos (2 c+2 d x)}{x} \, dx,x,x^2\right )\\ &=-\frac{2 a^2+b^2}{8 x^4}-\frac{a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac{b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4}-\frac{a b \sin \left (c+d x^2\right )}{2 x^4}-\frac{b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}-\frac{1}{2} \left (a b d^2 \cos (c)\right ) \operatorname{Subst}\left (\int \frac{\sin (d x)}{x} \, dx,x,x^2\right )+\frac{1}{2} \left (b^2 d^2 \cos (2 c)\right ) \operatorname{Subst}\left (\int \frac{\cos (2 d x)}{x} \, dx,x,x^2\right )-\frac{1}{2} \left (a b d^2 \sin (c)\right ) \operatorname{Subst}\left (\int \frac{\cos (d x)}{x} \, dx,x,x^2\right )-\frac{1}{2} \left (b^2 d^2 \sin (2 c)\right ) \operatorname{Subst}\left (\int \frac{\sin (2 d x)}{x} \, dx,x,x^2\right )\\ &=-\frac{2 a^2+b^2}{8 x^4}-\frac{a b d \cos \left (c+d x^2\right )}{2 x^2}+\frac{b^2 \cos \left (2 \left (c+d x^2\right )\right )}{8 x^4}+\frac{1}{2} b^2 d^2 \cos (2 c) \text{Ci}\left (2 d x^2\right )-\frac{1}{2} a b d^2 \text{Ci}\left (d x^2\right ) \sin (c)-\frac{a b \sin \left (c+d x^2\right )}{2 x^4}-\frac{b^2 d \sin \left (2 \left (c+d x^2\right )\right )}{4 x^2}-\frac{1}{2} a b d^2 \cos (c) \text{Si}\left (d x^2\right )-\frac{1}{2} b^2 d^2 \sin (2 c) \text{Si}\left (2 d x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.46481, size = 158, normalized size = 0.93 \[ -\frac{2 a^2+4 a b d^2 x^4 \sin (c) \text{CosIntegral}\left (d x^2\right )+4 a b d^2 x^4 \cos (c) \text{Si}\left (d x^2\right )+4 a b \sin \left (c+d x^2\right )+4 a b d x^2 \cos \left (c+d x^2\right )-4 b^2 d^2 x^4 \cos (2 c) \text{CosIntegral}\left (2 d x^2\right )+4 b^2 d^2 x^4 \sin (2 c) \text{Si}\left (2 d x^2\right )+2 b^2 d x^2 \sin \left (2 \left (c+d x^2\right )\right )-b^2 \cos \left (2 \left (c+d x^2\right )\right )+b^2}{8 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.227, size = 262, normalized size = 1.6 \begin{align*}{\frac{-{\frac{i}{4}}ab{{\rm e}^{-i \left ( d{x}^{2}+c \right ) }}}{{x}^{4}}}-{\frac{bad{{\rm e}^{-i \left ( d{x}^{2}+c \right ) }}}{4\,{x}^{2}}}+{\frac{i}{4}}ab{{\rm e}^{-ic}}{d}^{2}{\it Ei} \left ( 1,id{x}^{2} \right ) -{\frac{{a}^{2}}{4\,{x}^{4}}}-{\frac{{b}^{2}}{8\,{x}^{4}}}+{\frac{{b}^{2}{{\rm e}^{-2\,i \left ( d{x}^{2}+c \right ) }}}{16\,{x}^{4}}}-{\frac{{\frac{i}{8}}{b}^{2}d{{\rm e}^{-2\,i \left ( d{x}^{2}+c \right ) }}}{{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{-2\,ic}}{d}^{2}{\it Ei} \left ( 1,2\,id{x}^{2} \right ) }{4}}+{\frac{{b}^{2}{{\rm e}^{2\,i \left ( d{x}^{2}+c \right ) }}}{16\,{x}^{4}}}+{\frac{{\frac{i}{8}}{b}^{2}d{{\rm e}^{2\,i \left ( d{x}^{2}+c \right ) }}}{{x}^{2}}}-{\frac{{b}^{2}{{\rm e}^{2\,ic}}{d}^{2}{\it Ei} \left ( 1,-2\,id{x}^{2} \right ) }{4}}+{\frac{{\frac{i}{4}}ab{{\rm e}^{i \left ( d{x}^{2}+c \right ) }}}{{x}^{4}}}-{\frac{bad{{\rm e}^{i \left ( d{x}^{2}+c \right ) }}}{4\,{x}^{2}}}-{\frac{i}{4}}ab{{\rm e}^{ic}}{d}^{2}{\it Ei} \left ( 1,-id{x}^{2} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I*a*b/x^4*exp(-I*(d*x^2+c))-1/4*a*b*d/x^2*exp(-I*(d*x^2+c))+1/4*I*a*b*exp(-I*c)*d^2*Ei(1,I*d*x^2)-1/4/x^4
*a^2-1/8*b^2/x^4+1/16*b^2/x^4*exp(-2*I*(d*x^2+c))-1/8*I*b^2*d/x^2*exp(-2*I*(d*x^2+c))-1/4*b^2*exp(-2*I*c)*d^2*
Ei(1,2*I*d*x^2)+1/16*b^2/x^4*exp(2*I*(d*x^2+c))+1/8*I*b^2*d/x^2*exp(2*I*(d*x^2+c))-1/4*b^2*exp(2*I*c)*d^2*Ei(1
,-2*I*d*x^2)+1/4*I*a*b/x^4*exp(I*(d*x^2+c))-1/4*a*b*d/x^2*exp(I*(d*x^2+c))-1/4*I*a*b*exp(I*c)*d^2*Ei(1,-I*d*x^
2)

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Maxima [C]  time = 1.22135, size = 174, normalized size = 1.03 \begin{align*} \frac{1}{2} \,{\left ({\left (i \, \Gamma \left (-2, i \, d x^{2}\right ) - i \, \Gamma \left (-2, -i \, d x^{2}\right )\right )} \cos \left (c\right ) +{\left (\Gamma \left (-2, i \, d x^{2}\right ) + \Gamma \left (-2, -i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} a b d^{2} - \frac{{\left ({\left (4 \,{\left (\Gamma \left (-2, 2 i \, d x^{2}\right ) + \Gamma \left (-2, -2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) -{\left (4 i \, \Gamma \left (-2, 2 i \, d x^{2}\right ) - 4 i \, \Gamma \left (-2, -2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right )\right )} d^{2} x^{4} + 1\right )} b^{2}}{8 \, x^{4}} - \frac{a^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.16567, size = 510, normalized size = 3.02 \begin{align*} -\frac{2 \, b^{2} d^{2} x^{4} \sin \left (2 \, c\right ) \operatorname{Si}\left (2 \, d x^{2}\right ) + 2 \, a b d^{2} x^{4} \cos \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) + 2 \, a b d x^{2} \cos \left (d x^{2} + c\right ) - b^{2} \cos \left (d x^{2} + c\right )^{2} + a^{2} + b^{2} -{\left (b^{2} d^{2} x^{4} \operatorname{Ci}\left (2 \, d x^{2}\right ) + b^{2} d^{2} x^{4} \operatorname{Ci}\left (-2 \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) + 2 \,{\left (b^{2} d x^{2} \cos \left (d x^{2} + c\right ) + a b\right )} \sin \left (d x^{2} + c\right ) +{\left (a b d^{2} x^{4} \operatorname{Ci}\left (d x^{2}\right ) + a b d^{2} x^{4} \operatorname{Ci}\left (-d x^{2}\right )\right )} \sin \left (c\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*b^2*d^2*x^4*sin(2*c)*sin_integral(2*d*x^2) + 2*a*b*d^2*x^4*cos(c)*sin_integral(d*x^2) + 2*a*b*d*x^2*co
s(d*x^2 + c) - b^2*cos(d*x^2 + c)^2 + a^2 + b^2 - (b^2*d^2*x^4*cos_integral(2*d*x^2) + b^2*d^2*x^4*cos_integra
l(-2*d*x^2))*cos(2*c) + 2*(b^2*d*x^2*cos(d*x^2 + c) + a*b)*sin(d*x^2 + c) + (a*b*d^2*x^4*cos_integral(d*x^2) +
a*b*d^2*x^4*cos_integral(-d*x^2))*sin(c))/x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sin{\left (c + d x^{2} \right )}\right )^{2}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.4964, size = 605, normalized size = 3.58 \begin{align*} \frac{4 \,{\left (d x^{2} + c\right )}^{2} b^{2} d^{3} \cos \left (2 \, c\right ) \operatorname{Ci}\left (2 \, d x^{2}\right ) - 8 \,{\left (d x^{2} + c\right )} b^{2} c d^{3} \cos \left (2 \, c\right ) \operatorname{Ci}\left (2 \, d x^{2}\right ) + 4 \, b^{2} c^{2} d^{3} \cos \left (2 \, c\right ) \operatorname{Ci}\left (2 \, d x^{2}\right ) - 4 \,{\left (d x^{2} + c\right )}^{2} a b d^{3} \operatorname{Ci}\left (d x^{2}\right ) \sin \left (c\right ) + 8 \,{\left (d x^{2} + c\right )} a b c d^{3} \operatorname{Ci}\left (d x^{2}\right ) \sin \left (c\right ) - 4 \, a b c^{2} d^{3} \operatorname{Ci}\left (d x^{2}\right ) \sin \left (c\right ) - 4 \,{\left (d x^{2} + c\right )}^{2} a b d^{3} \cos \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) + 8 \,{\left (d x^{2} + c\right )} a b c d^{3} \cos \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) - 4 \, a b c^{2} d^{3} \cos \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) + 4 \,{\left (d x^{2} + c\right )}^{2} b^{2} d^{3} \sin \left (2 \, c\right ) \operatorname{Si}\left (-2 \, d x^{2}\right ) - 8 \,{\left (d x^{2} + c\right )} b^{2} c d^{3} \sin \left (2 \, c\right ) \operatorname{Si}\left (-2 \, d x^{2}\right ) + 4 \, b^{2} c^{2} d^{3} \sin \left (2 \, c\right ) \operatorname{Si}\left (-2 \, d x^{2}\right ) - 4 \,{\left (d x^{2} + c\right )} a b d^{3} \cos \left (d x^{2} + c\right ) + 4 \, a b c d^{3} \cos \left (d x^{2} + c\right ) - 2 \,{\left (d x^{2} + c\right )} b^{2} d^{3} \sin \left (2 \, d x^{2} + 2 \, c\right ) + 2 \, b^{2} c d^{3} \sin \left (2 \, d x^{2} + 2 \, c\right ) + b^{2} d^{3} \cos \left (2 \, d x^{2} + 2 \, c\right ) - 4 \, a b d^{3} \sin \left (d x^{2} + c\right ) - 2 \, a^{2} d^{3} - b^{2} d^{3}}{8 \,{\left ({\left (d x^{2} + c\right )}^{2} - 2 \,{\left (d x^{2} + c\right )} c + c^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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