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

Optimal. Leaf size=106 \[ -\frac {1}{6} a d^3 \cos (c) \text {Ci}(d x)+\frac {1}{6} a d^3 \sin (c) \text {Si}(d x)+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {a d \cos (c+d x)}{6 x^2}+b d \cos (c) \text {Ci}(d x)-b d \sin (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3339, 3297, 3303, 3299, 3302} \[ -\frac {1}{6} a d^3 \cos (c) \text {CosIntegral}(d x)+\frac {1}{6} a d^3 \sin (c) \text {Si}(d x)+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {a d \cos (c+d x)}{6 x^2}+b d \cos (c) \text {CosIntegral}(d x)-b d \sin (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*Sin[c + d*x])/x^4,x]

[Out]

-(a*d*Cos[c + d*x])/(6*x^2) + b*d*Cos[c]*CosIntegral[d*x] - (a*d^3*Cos[c]*CosIntegral[d*x])/6 - (a*Sin[c + d*x
])/(3*x^3) - (b*Sin[c + d*x])/x + (a*d^2*Sin[c + d*x])/(6*x) - b*d*Sin[c]*SinIntegral[d*x] + (a*d^3*Sin[c]*Sin
Integral[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 3339

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 95, normalized size = 0.90 \[ \frac {d x^3 \cos (c) \left (6 b-a d^2\right ) \text {Ci}(d x)+d x^3 \sin (c) \left (a d^2-6 b\right ) \text {Si}(d x)+a d^2 x^2 \sin (c+d x)-2 a \sin (c+d x)-a d x \cos (c+d x)-6 b x^2 \sin (c+d x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 105, normalized size = 0.99 \[ \frac {2 \, {\left (a d^{3} - 6 \, b d\right )} x^{3} \sin \relax (c) \operatorname {Si}\left (d x\right ) - 2 \, a d x \cos \left (d x + c\right ) - {\left ({\left (a d^{3} - 6 \, b d\right )} x^{3} \operatorname {Ci}\left (d x\right ) + {\left (a d^{3} - 6 \, b d\right )} x^{3} \operatorname {Ci}\left (-d x\right )\right )} \cos \relax (c) + 2 \, {\left ({\left (a d^{2} - 6 \, b\right )} x^{2} - 2 \, a\right )} \sin \left (d x + c\right )}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.42, size = 834, normalized size = 7.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(a*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a*d^3*x^3*real_part(cos_integral(-d
*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^
3*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a*d^3*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*t
an(1/2*c) - a*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 - a*d^3*x^3*real_part(cos_integral(-d*x))*ta
n(1/2*d*x)^2 + a*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a*d^3*x^3*real_part(cos_integral(-d*x))*t
an(1/2*c)^2 - 6*b*d*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 6*b*d*x^3*real_part(cos_int
egral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^3*x^3*imag_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^3*x^3*i
mag_part(cos_integral(-d*x))*tan(1/2*c) + 4*a*d^3*x^3*sin_integral(d*x)*tan(1/2*c) - 12*b*d*x^3*imag_part(cos_
integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 12*b*d*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)
- 24*b*d*x^3*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) - a*d^3*x^3*real_part(cos_integral(d*x)) - a*d^3*x^3*
real_part(cos_integral(-d*x)) + 6*b*d*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 6*b*d*x^3*real_part(co
s_integral(-d*x))*tan(1/2*d*x)^2 - 4*a*d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) - 6*b*d*x^3*real_part(cos_integral(d*
x))*tan(1/2*c)^2 - 6*b*d*x^3*real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 4*a*d^2*x^2*tan(1/2*d*x)*tan(1/2*c)^
2 - 12*b*d*x^3*imag_part(cos_integral(d*x))*tan(1/2*c) + 12*b*d*x^3*imag_part(cos_integral(-d*x))*tan(1/2*c) -
24*b*d*x^3*sin_integral(d*x)*tan(1/2*c) - 2*a*d*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + 6*b*d*x^3*real_part(cos_integ
ral(d*x)) + 6*b*d*x^3*real_part(cos_integral(-d*x)) + 4*a*d^2*x^2*tan(1/2*d*x) + 4*a*d^2*x^2*tan(1/2*c) + 24*b
*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 24*b*x^2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*a*d*x*tan(1/2*d*x)^2 + 8*a*d*x*tan(1/2
*d*x)*tan(1/2*c) + 2*a*d*x*tan(1/2*c)^2 - 24*b*x^2*tan(1/2*d*x) - 24*b*x^2*tan(1/2*c) + 8*a*tan(1/2*d*x)^2*tan
(1/2*c) + 8*a*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a*d*x - 8*a*tan(1/2*d*x) - 8*a*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 = 0.03, size = 102, normalized size = 0.96 \[ d^{3} \left (\frac {b \left (-\frac {\sin \left (d x +c \right )}{x d}-\Si \left (d x \right ) \sin \relax (c )+\Ci \left (d x \right ) \cos \relax (c )\right )}{d^{2}}+a \left (-\frac {\sin \left (d x +c \right )}{3 x^{3} d^{3}}-\frac {\cos \left (d x +c \right )}{6 x^{2} d^{2}}+\frac {\sin \left (d x +c \right )}{6 x d}+\frac {\Si \left (d x \right ) \sin \relax (c )}{6}-\frac {\Ci \left (d x \right ) \cos \relax (c )}{6}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^3*(1/d^2*b*(-sin(d*x+c)/x/d-Si(d*x)*sin(c)+Ci(d*x)*cos(c))+a*(-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 = 1.13, size = 123, normalized size = 1.16 \[ -\frac {{\left ({\left (a {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) + a {\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{5} - {\left (6 \, b {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) - b {\left (6 i \, \Gamma \left (-3, i \, d x\right ) - 6 i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{3}\right )} x^{3} + 2 \, b d x \cos \left (d x + c\right ) + 4 \, b \sin \left (d x + c\right )}{2 \, d^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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 \[ \int \frac {\left (a + b x^{2}\right ) \sin {\left (c + d x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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