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

Optimal. Leaf size=149 \[ \frac {1}{24} a d^4 \sin (c) \text {Ci}(d x)+\frac {1}{24} a d^4 \cos (c) \text {Si}(d x)+\frac {a d^3 \cos (c+d x)}{24 x}+\frac {a d^2 \sin (c+d x)}{24 x^2}-\frac {a \sin (c+d x)}{4 x^4}-\frac {a d \cos (c+d x)}{12 x^3}-\frac {1}{2} b d^2 \sin (c) \text {Ci}(d x)-\frac {1}{2} b d^2 \cos (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{2 x^2}-\frac {b d \cos (c+d x)}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ \frac {1}{24} a d^4 \sin (c) \text {CosIntegral}(d x)+\frac {1}{24} a d^4 \cos (c) \text {Si}(d x)+\frac {a d^2 \sin (c+d x)}{24 x^2}+\frac {a d^3 \cos (c+d x)}{24 x}-\frac {a \sin (c+d x)}{4 x^4}-\frac {a d \cos (c+d x)}{12 x^3}-\frac {1}{2} b d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} b d^2 \cos (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{2 x^2}-\frac {b d \cos (c+d x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*Sin[c + d*x])/x^5,x]

[Out]

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

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

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Mathematica [A]  time = 0.25, size = 125, normalized size = 0.84 \[ \frac {d^2 x^4 \sin (c) \left (a d^2-12 b\right ) \text {Ci}(d x)+d^2 x^4 \cos (c) \left (a d^2-12 b\right ) \text {Si}(d x)+a d^3 x^3 \cos (c+d x)+a d^2 x^2 \sin (c+d x)-6 a \sin (c+d x)-2 a d x \cos (c+d x)-12 b d x^3 \cos (c+d x)-12 b x^2 \sin (c+d x)}{24 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*d*x*Cos[c + d*x] - 12*b*d*x^3*Cos[c + d*x] + a*d^3*x^3*Cos[c + d*x] + d^2*(-12*b + a*d^2)*x^4*CosIntegra
l[d*x]*Sin[c] - 6*a*Sin[c + d*x] - 12*b*x^2*Sin[c + d*x] + a*d^2*x^2*Sin[c + d*x] + d^2*(-12*b + a*d^2)*x^4*Co
s[c]*SinIntegral[d*x])/(24*x^4)

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fricas [A]  time = 0.87, size = 127, normalized size = 0.85 \[ \frac {2 \, {\left (a d^{4} - 12 \, b d^{2}\right )} x^{4} \cos \relax (c) \operatorname {Si}\left (d x\right ) + 2 \, {\left ({\left (a d^{3} - 12 \, b d\right )} x^{3} - 2 \, a d x\right )} \cos \left (d x + c\right ) + 2 \, {\left ({\left (a d^{2} - 12 \, b\right )} x^{2} - 6 \, a\right )} \sin \left (d x + c\right ) + {\left ({\left (a d^{4} - 12 \, b d^{2}\right )} x^{4} \operatorname {Ci}\left (d x\right ) + {\left (a d^{4} - 12 \, b d^{2}\right )} x^{4} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{48 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.48, size = 1086, normalized size = 7.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(a*d^4*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^4*x^4*imag_part(cos_integral(-
d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^4*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^4*x^4*re
al_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^4*x^4*real_part(cos_integral(-d*x))*tan(1/2*d*x)^
2*tan(1/2*c) - a*d^4*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a*d^4*x^4*imag_part(cos_integral(-d*x))
*tan(1/2*d*x)^2 - 2*a*d^4*x^4*sin_integral(d*x)*tan(1/2*d*x)^2 + a*d^4*x^4*imag_part(cos_integral(d*x))*tan(1/
2*c)^2 - a*d^4*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a*d^4*x^4*sin_integral(d*x)*tan(1/2*c)^2 - 1
2*b*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*b*d^2*x^4*imag_part(cos_integral(-d*
x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 24*b*d^2*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^4*x^4*rea
l_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^4*x^4*real_part(cos_integral(-d*x))*tan(1/2*c) + 24*b*d^2*x^4*rea
l_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 24*b*d^2*x^4*real_part(cos_integral(-d*x))*tan(1/2*d*x)^
2*tan(1/2*c) - 2*a*d^3*x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^4*x^4*imag_part(cos_integral(d*x)) + a*d^4*x^4*im
ag_part(cos_integral(-d*x)) - 2*a*d^4*x^4*sin_integral(d*x) + 12*b*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/
2*d*x)^2 - 12*b*d^2*x^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 + 24*b*d^2*x^4*sin_integral(d*x)*tan(1/2*
d*x)^2 - 12*b*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*c)^2 + 12*b*d^2*x^4*imag_part(cos_integral(-d*x))*t
an(1/2*c)^2 - 24*b*d^2*x^4*sin_integral(d*x)*tan(1/2*c)^2 + 2*a*d^3*x^3*tan(1/2*d*x)^2 + 24*b*d^2*x^4*real_par
t(cos_integral(d*x))*tan(1/2*c) + 24*b*d^2*x^4*real_part(cos_integral(-d*x))*tan(1/2*c) + 8*a*d^3*x^3*tan(1/2*
d*x)*tan(1/2*c) + 2*a*d^3*x^3*tan(1/2*c)^2 + 24*b*d*x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*b*d^2*x^4*imag_part(c
os_integral(d*x)) - 12*b*d^2*x^4*imag_part(cos_integral(-d*x)) + 24*b*d^2*x^4*sin_integral(d*x) + 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 - 2*a*d^3*x^3 - 24*b*d*x^3*tan(1/2*d*x)^2 -
96*b*d*x^3*tan(1/2*d*x)*tan(1/2*c) - 24*b*d*x^3*tan(1/2*c)^2 + 4*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) - 4*a*d^2*x^2*tan(1/2*c) - 48*b*x^2*tan(1/2*d*x)^2*tan(1/2*c) - 48*b*x^2*tan(1/2*d*x)*tan(1/2*
c)^2 + 24*b*d*x^3 - 4*a*d*x*tan(1/2*d*x)^2 - 16*a*d*x*tan(1/2*d*x)*tan(1/2*c) - 4*a*d*x*tan(1/2*c)^2 + 48*b*x^
2*tan(1/2*d*x) + 48*b*x^2*tan(1/2*c) - 24*a*tan(1/2*d*x)^2*tan(1/2*c) - 24*a*tan(1/2*d*x)*tan(1/2*c)^2 + 4*a*d
*x + 24*a*tan(1/2*d*x) + 24*a*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)

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maple [A]  time = 0.03, size = 131, normalized size = 0.88 \[ d^{4} \left (\frac {b \left (-\frac {\sin \left (d x +c \right )}{2 x^{2} d^{2}}-\frac {\cos \left (d x +c \right )}{2 x d}-\frac {\Si \left (d x \right ) \cos \relax (c )}{2}-\frac {\Ci \left (d x \right ) \sin \relax (c )}{2}\right )}{d^{2}}+a \left (-\frac {\sin \left (d x +c \right )}{4 x^{4} d^{4}}-\frac {\cos \left (d x +c \right )}{12 x^{3} d^{3}}+\frac {\sin \left (d x +c \right )}{24 x^{2} d^{2}}+\frac {\cos \left (d x +c \right )}{24 x d}+\frac {\Si \left (d x \right ) \cos \relax (c )}{24}+\frac {\Ci \left (d x \right ) \sin \relax (c )}{24}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^4*(1/d^2*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*(-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 = 1.23, size = 121, normalized size = 0.81 \[ -\frac {{\left ({\left (a {\left (i \, \Gamma \left (-4, i \, d x\right ) - i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) + a {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{6} + {\left (b {\left (-12 i \, \Gamma \left (-4, i \, d x\right ) + 12 i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \relax (c) - 12 \, b {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{4}\right )} x^{4} + 2 \, b d x \cos \left (d x + c\right ) + 6 \, b \sin \left (d x + c\right )}{2 \, d^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,\left (b\,x^2+a\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^2))/x^5,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right ) \sin {\left (c + d x \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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