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

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

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 175, normalized size of antiderivative = 1.00, 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 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 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 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 6742

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

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*}

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Mathematica [A]  time = 0.57, size = 154, normalized size = 0.88 \[ -\frac {d x^3 \text {Ci}(d x) \left (\cos (c) \left (a^2 d^2-6 b^2\right )+6 a b d \sin (c)\right )+d x^3 \text {Si}(d x) \left (-a^2 d^2 \sin (c)+6 a b d \cos (c)+6 b^2 \sin (c)\right )-a^2 d^2 x^2 \sin (c+d x)+2 a^2 \sin (c+d x)+a^2 d x \cos (c+d x)+6 a b d x^2 \cos (c+d x)+6 a b x \sin (c+d x)+6 b^2 x^2 \sin (c+d x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(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])/x^3

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fricas [A]  time = 0.58, size = 186, normalized size = 1.06 \[ -\frac {2 \, {\left (6 \, a b d x^{2} + a^{2} d x\right )} \cos \left (d x + c\right ) + {\left (12 \, a b d^{2} x^{3} \operatorname {Si}\left (d x\right ) + {\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x^{3} \operatorname {Ci}\left (d x\right ) + {\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x^{3} \operatorname {Ci}\left (-d x\right )\right )} \cos \relax (c) + 2 \, {\left (6 \, a b x - {\left (a^{2} d^{2} - 6 \, b^{2}\right )} x^{2} + 2 \, a^{2}\right )} \sin \left (d x + c\right ) + 2 \, {\left (3 \, a b d^{2} x^{3} \operatorname {Ci}\left (d x\right ) + 3 \, a b d^{2} x^{3} \operatorname {Ci}\left (-d x\right ) - {\left (a^{2} d^{3} - 6 \, b^{2} d\right )} x^{3} \operatorname {Si}\left (d x\right )\right )} \sin \relax (c)}{12 \, x^{3}} \]

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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giac [C]  time = 0.79, size = 1400, normalized size = 8.00 \[ \text {result too large to display} \]

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)

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maple [A]  time = 0.04, size = 158, normalized size = 0.90 \[ d^{3} \left (\frac {b^{2} \left (-\frac {\sin \left (d x +c \right )}{x d}-\Si \left (d x \right ) \sin \relax (c )+\Ci \left (d x \right ) \cos \relax (c )\right )}{d^{2}}+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )}{2 x^{2} d^{2}}-\frac {\cos \left (d x +c \right )}{2 x d}-\frac {\Si \left (d x \right ) \cos \relax (c )}{2}-\frac {\Ci \left (d x \right ) \sin \relax (c )}{2}\right )}{d}+a^{2} \left (-\frac {\sin \left (d x +c \right )}{3 x^{3} d^{3}}-\frac {\cos \left (d x +c \right )}{6 x^{2} d^{2}}+\frac {\sin \left (d x +c \right )}{6 x d}+\frac {\Si \left (d x \right ) \sin \relax (c )}{6}-\frac {\Ci \left (d x \right ) \cos \relax (c )}{6}\right )\right ) \]

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 = 5.67, size = 188, normalized size = 1.07 \[ -\frac {{\left ({\left (a^{2} {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) + a^{2} {\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{5} + {\left (a b {\left (6 i \, \Gamma \left (-3, i \, d x\right ) - 6 i \, \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) + 6 \, a b {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{4} - {\left (6 \, b^{2} {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) - b^{2} {\left (6 i \, \Gamma \left (-3, i \, d x\right ) - 6 i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{3}\right )} x^{3} + 4 \, b^{2} \sin \left (d x + c\right ) + 2 \, {\left (b^{2} d x + 2 \, a b d\right )} \cos \left (d x + c\right )}{2 \, d^{2} x^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,{\left (a+b\,x\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x)^2)/x^4,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{2} \sin {\left (c + d x \right )}}{x^{4}}\, dx \]

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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