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

Optimal. Leaf size=132 \[ -\frac {1}{6} a d^3 \cos (c) \text {Ci}(d x)+\frac {1}{6} a d^3 \sin (c) \text {Si}(d x)+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {a d \cos (c+d x)}{6 x^2}-\frac {1}{2} b d^2 \sin (c) \text {Ci}(d x)-\frac {1}{2} b d^2 \cos (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{2 x^2}-\frac {b d \cos (c+d x)}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6742, 3297, 3303, 3299, 3302} \[ -\frac {1}{6} a d^3 \cos (c) \text {CosIntegral}(d x)+\frac {1}{6} a d^3 \sin (c) \text {Si}(d x)+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {a d \cos (c+d x)}{6 x^2}-\frac {1}{2} b d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} b d^2 \cos (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{2 x^2}-\frac {b d \cos (c+d x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Sin[c + d*x])/x^4,x]

[Out]

-(a*d*Cos[c + d*x])/(6*x^2) - (b*d*Cos[c + d*x])/(2*x) - (a*d^3*Cos[c]*CosIntegral[d*x])/6 - (b*d^2*CosIntegra
l[d*x]*Sin[c])/2 - (a*Sin[c + d*x])/(3*x^3) - (b*Sin[c + d*x])/(2*x^2) + (a*d^2*Sin[c + d*x])/(6*x) - (b*d^2*C
os[c]*SinIntegral[d*x])/2 + (a*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) \sin (c+d x)}{x^4} \, dx &=\int \left (\frac {a \sin (c+d x)}{x^4}+\frac {b \sin (c+d x)}{x^3}\right ) \, dx\\ &=a \int \frac {\sin (c+d x)}{x^4} \, dx+b \int \frac {\sin (c+d x)}{x^3} \, dx\\ &=-\frac {a \sin (c+d x)}{3 x^3}-\frac {b \sin (c+d x)}{2 x^2}+\frac {1}{3} (a d) \int \frac {\cos (c+d x)}{x^3} \, dx+\frac {1}{2} (b d) \int \frac {\cos (c+d x)}{x^2} \, dx\\ &=-\frac {a d \cos (c+d x)}{6 x^2}-\frac {b d \cos (c+d x)}{2 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {b \sin (c+d x)}{2 x^2}-\frac {1}{6} \left (a d^2\right ) \int \frac {\sin (c+d x)}{x^2} \, dx-\frac {1}{2} \left (b d^2\right ) \int \frac {\sin (c+d x)}{x} \, dx\\ &=-\frac {a d \cos (c+d x)}{6 x^2}-\frac {b d \cos (c+d x)}{2 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {b \sin (c+d x)}{2 x^2}+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {1}{6} \left (a d^3\right ) \int \frac {\cos (c+d x)}{x} \, dx-\frac {1}{2} \left (b d^2 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx-\frac {1}{2} \left (b d^2 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {a d \cos (c+d x)}{6 x^2}-\frac {b d \cos (c+d x)}{2 x}-\frac {1}{2} b d^2 \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{3 x^3}-\frac {b \sin (c+d x)}{2 x^2}+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {1}{2} b d^2 \cos (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}-\frac {b d \cos (c+d x)}{2 x}-\frac {1}{6} a d^3 \cos (c) \text {Ci}(d x)-\frac {1}{2} b d^2 \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{3 x^3}-\frac {b \sin (c+d x)}{2 x^2}+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {1}{2} b d^2 \cos (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 = 0.34, size = 110, normalized size = 0.83 \[ -\frac {d^2 x^3 \text {Ci}(d x) (a d \cos (c)+3 b \sin (c))+d^2 x^3 \text {Si}(d x) (3 b \cos (c)-a d \sin (c))-a d^2 x^2 \sin (c+d x)+2 a \sin (c+d x)+a d x \cos (c+d x)+3 b d x^2 \cos (c+d x)+3 b x \sin (c+d x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(a*d*x*Cos[c + d*x] + 3*b*d*x^2*Cos[c + d*x] + d^2*x^3*CosIntegral[d*x]*(a*d*Cos[c] + 3*b*Sin[c]) + 2*a*S
in[c + d*x] + 3*b*x*Sin[c + d*x] - a*d^2*x^2*Sin[c + d*x] + d^2*x^3*(3*b*Cos[c] - a*d*Sin[c])*SinIntegral[d*x]
)/x^3

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fricas [A]  time = 0.49, size = 137, normalized size = 1.04 \[ -\frac {2 \, {\left (3 \, b d x^{2} + a d x\right )} \cos \left (d x + c\right ) + {\left (a d^{3} x^{3} \operatorname {Ci}\left (d x\right ) + a d^{3} x^{3} \operatorname {Ci}\left (-d x\right ) + 6 \, b d^{2} x^{3} \operatorname {Si}\left (d x\right )\right )} \cos \relax (c) - 2 \, {\left (a d^{2} x^{2} - 3 \, b x - 2 \, a\right )} \sin \left (d x + c\right ) - {\left (2 \, a d^{3} x^{3} \operatorname {Si}\left (d x\right ) - 3 \, b d^{2} x^{3} \operatorname {Ci}\left (d x\right ) - 3 \, b d^{2} x^{3} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.54, size = 961, normalized size = 7.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+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) + 3*b*d^2*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 3*b*d^2*x^3*imag_part(cos_i
ntegral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 6*b*d^2*x^3*sin_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 - a*d^3*x^3*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 6*
b*d^2*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 6*b*d^2*x^3*real_part(cos_integral(-d*x))*t
an(1/2*d*x)^2*tan(1/2*c) + a*d^3*x^3*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a*d^3*x^3*real_part(cos_integ
ral(-d*x))*tan(1/2*c)^2 - 3*b*d^2*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 3*b*d^2*x^3*imag_part(cos_
integral(-d*x))*tan(1/2*d*x)^2 - 6*b*d^2*x^3*sin_integral(d*x)*tan(1/2*d*x)^2 + 2*a*d^3*x^3*imag_part(cos_inte
gral(d*x))*tan(1/2*c) - 2*a*d^3*x^3*imag_part(cos_integral(-d*x))*tan(1/2*c) + 4*a*d^3*x^3*sin_integral(d*x)*t
an(1/2*c) + 3*b*d^2*x^3*imag_part(cos_integral(d*x))*tan(1/2*c)^2 - 3*b*d^2*x^3*imag_part(cos_integral(-d*x))*
tan(1/2*c)^2 + 6*b*d^2*x^3*sin_integral(d*x)*tan(1/2*c)^2 - 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^2*x^3*real_part(cos_integral(d*x))*tan(1/2*c) - 6*b*d^2*x^3*real_part(c
os_integral(-d*x))*tan(1/2*c) - 4*a*d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) - 4*a*d^2*x^2*tan(1/2*d*x)*tan(1/2*c)^2
- 6*b*d*x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 3*b*d^2*x^3*imag_part(cos_integral(d*x)) + 3*b*d^2*x^3*imag_part(cos
_integral(-d*x)) - 6*b*d^2*x^3*sin_integral(d*x) - 2*a*d*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + 4*a*d^2*x^2*tan(1/2*d
*x) + 6*b*d*x^2*tan(1/2*d*x)^2 + 4*a*d^2*x^2*tan(1/2*c) + 24*b*d*x^2*tan(1/2*d*x)*tan(1/2*c) + 6*b*d*x^2*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) + 12*b*x*tan(1/2*d*x)^2*tan(1/2*c) + 2*a*d*
x*tan(1/2*c)^2 + 12*b*x*tan(1/2*d*x)*tan(1/2*c)^2 - 6*b*d*x^2 + 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 - 12*b*x*tan(1/2*d*x) - 12*b*x*tan(1/2*c) - 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 = 0.04, size = 117, normalized size = 0.89 \[ d^{3} \left (\frac {b \left (-\frac {\sin \left (d x +c \right )}{2 x^{2} d^{2}}-\frac {\cos \left (d x +c \right )}{2 x d}-\frac {\Si \left (d x \right ) \cos \relax (c )}{2}-\frac {\Ci \left (d x \right ) \sin \relax (c )}{2}\right )}{d}+a \left (-\frac {\sin \left (d x +c \right )}{3 x^{3} d^{3}}-\frac {\cos \left (d x +c \right )}{6 x^{2} d^{2}}+\frac {\sin \left (d x +c \right )}{6 x d}+\frac {\Si \left (d x \right ) \sin \relax (c )}{6}-\frac {\Ci \left (d x \right ) \cos \relax (c )}{6}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^3*(b/d*(-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*(-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 = 2.14, size = 110, normalized size = 0.83 \[ -\frac {{\left ({\left (a {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) + a {\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{4} + {\left (b {\left (3 i \, \Gamma \left (-3, i \, d x\right ) - 3 i \, \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) + 3 \, b {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{3}\right )} x^{3} + 2 \, b \cos \left (d x + c\right )}{2 \, d x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+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^4 + (b*(3*I*gamma(-3, I*d*x) - 3*I*gamma(-3, -I*d*x))*cos(c) + 3*b*(gamma(-3, I*d*x) + gamma(-3, -I*d*x))*
sin(c))*d^3)*x^3 + 2*b*cos(d*x + c))/(d*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x))/x^4,x)

[Out]

int((sin(c + d*x)*(a + b*x))/x^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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