3.86 \(\int \frac {(a+b x^3) \sin (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=91 \[ -\frac {1}{6} a d^3 \cos (c) \text {Ci}(d x)+\frac {1}{6} a d^3 \sin (c) \text {Si}(d x)+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {a d \cos (c+d x)}{6 x^2}+b \sin (c) \text {Ci}(d x)+b \cos (c) \text {Si}(d x) \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 11, 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} \[ -\frac {1}{6} a d^3 \cos (c) \text {CosIntegral}(d x)+\frac {1}{6} a d^3 \sin (c) \text {Si}(d x)+\frac {a d^2 \sin (c+d x)}{6 x}-\frac {a \sin (c+d x)}{3 x^3}-\frac {a d \cos (c+d x)}{6 x^2}+b \sin (c) \text {CosIntegral}(d x)+b \cos (c) \text {Si}(d x) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*Sin[c + d*x])/x^4,x]

[Out]

-(a*d*Cos[c + d*x])/(6*x^2) - (a*d^3*Cos[c]*CosIntegral[d*x])/6 + b*CosIntegral[d*x]*Sin[c] - (a*Sin[c + d*x])
/(3*x^3) + (a*d^2*Sin[c + d*x])/(6*x) + b*Cos[c]*SinIntegral[d*x] + (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 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^3\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}\right ) \, dx\\ &=a \int \frac {\sin (c+d x)}{x^4} \, dx+b \int \frac {\sin (c+d x)}{x} \, dx\\ &=-\frac {a \sin (c+d x)}{3 x^3}+\frac {1}{3} (a d) \int \frac {\cos (c+d x)}{x^3} \, dx+(b \cos (c)) \int \frac {\sin (d x)}{x} \, dx+(b \sin (c)) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {a d \cos (c+d x)}{6 x^2}+b \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{3 x^3}+b \cos (c) \text {Si}(d x)-\frac {1}{6} \left (a d^2\right ) \int \frac {\sin (c+d x)}{x^2} \, dx\\ &=-\frac {a d \cos (c+d x)}{6 x^2}+b \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{3 x^3}+\frac {a d^2 \sin (c+d x)}{6 x}+b \cos (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 \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{3 x^3}+\frac {a d^2 \sin (c+d x)}{6 x}+b \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 {1}{6} a d^3 \cos (c) \text {Ci}(d x)+b \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{3 x^3}+\frac {a d^2 \sin (c+d x)}{6 x}+b \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 = 0.23, size = 104, normalized size = 1.14 \[ -\frac {1}{6} a d^3 (\cos (c) \text {Ci}(d x)-\sin (c) \text {Si}(d x))+\frac {a \cos (d x) \left (d^2 x^2 \sin (c)-d x \cos (c)-2 \sin (c)\right )}{6 x^3}+\frac {a \sin (d x) \left (d^2 x^2 \cos (c)+d x \sin (c)-2 \cos (c)\right )}{6 x^3}+b \sin (c) \text {Ci}(d x)+b \cos (c) \text {Si}(d x) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)*Sin[c + d*x])/x^4,x]

[Out]

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

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fricas [A]  time = 0.59, size = 114, normalized size = 1.25 \[ -\frac {2 \, a d x \cos \left (d x + c\right ) + {\left (a d^{3} x^{3} \operatorname {Ci}\left (d x\right ) + a d^{3} x^{3} \operatorname {Ci}\left (-d x\right ) - 12 \, b x^{3} \operatorname {Si}\left (d x\right )\right )} \cos \relax (c) - 2 \, {\left (a d^{2} x^{2} - 2 \, a\right )} \sin \left (d x + c\right ) - 2 \, {\left (a d^{3} x^{3} \operatorname {Si}\left (d x\right ) + 3 \, b x^{3} \operatorname {Ci}\left (d x\right ) + 3 \, b x^{3} \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*sin(d*x+c)/x^4,x, algorithm="fricas")

[Out]

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

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giac [C]  time = 0.50, size = 796, normalized size = 8.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+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 + 2*a*d^3*x^3*imag_part(cos_integral(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)*tan(1/2*c) - 6*b*x^3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*ta
n(1/2*c)^2 + 6*b*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 12*b*x^3*sin_integral(d*x)*ta
n(1/2*d*x)^2*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)) -
4*a*d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 12*b*x^3*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 12*
b*x^3*real_part(cos_integral(-d*x))*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*x^
3*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 6*b*x^3*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 + 12*b*x^
3*sin_integral(d*x)*tan(1/2*d*x)^2 - 6*b*x^3*imag_part(cos_integral(d*x))*tan(1/2*c)^2 + 6*b*x^3*imag_part(cos
_integral(-d*x))*tan(1/2*c)^2 - 12*b*x^3*sin_integral(d*x)*tan(1/2*c)^2 - 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) + 4*a*d^2*x^2*tan(1/2*c) + 12*b*x^3*real_part(cos_integral(d*x))*tan(1/2*c) + 12*b*
x^3*real_part(cos_integral(-d*x))*tan(1/2*c) + 6*b*x^3*imag_part(cos_integral(d*x)) - 6*b*x^3*imag_part(cos_in
tegral(-d*x)) + 12*b*x^3*sin_integral(d*x) + 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 + 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 = 0.04, size = 87, normalized size = 0.96 \[ d^{3} \left (\frac {b \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right )}{d^{3}}+a \left (-\frac {\sin \left (d x +c \right )}{3 x^{3} d^{3}}-\frac {\cos \left (d x +c \right )}{6 x^{2} d^{2}}+\frac {\sin \left (d x +c \right )}{6 x d}+\frac {\Si \left (d x \right ) \sin \relax (c )}{6}-\frac {\Ci \left (d x \right ) \cos \relax (c )}{6}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)*sin(d*x+c)/x^4,x)

[Out]

d^3*(1/d^3*b*(Si(d*x)*cos(c)+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 = 2.55, size = 132, normalized size = 1.45 \[ -\frac {{\left ({\left (a {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) + a {\left (-i \, \Gamma \left (-3, i \, d x\right ) + i \, \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{6} + {\left (b {\left (-6 i \, \Gamma \left (-3, i \, d x\right ) + 6 i \, \Gamma \left (-3, -i \, d x\right )\right )} \cos \relax (c) - 6 \, b {\left (\Gamma \left (-3, i \, d x\right ) + \Gamma \left (-3, -i \, d x\right )\right )} \sin \relax (c)\right )} d^{3}\right )} x^{3} + 2 \, b d x \sin \left (d x + c\right ) + 2 \, {\left (b d^{2} x^{2} - 2 \, b\right )} \cos \left (d x + c\right )}{2 \, d^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+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^6 + (b*(-6*I*gamma(-3, I*d*x) + 6*I*gamma(-3, -I*d*x))*cos(c) - 6*b*(gamma(-3, I*d*x) + gamma(-3, -I*d*x))
*sin(c))*d^3)*x^3 + 2*b*d*x*sin(d*x + c) + 2*(b*d^2*x^2 - 2*b)*cos(d*x + c))/(d^3*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (c+d\,x\right )\,\left (b\,x^3+a\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x^3))/x^4,x)

[Out]

int((sin(c + d*x)*(a + b*x^3))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{3}\right ) \sin {\left (c + d x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)*sin(d*x+c)/x**4,x)

[Out]

Integral((a + b*x**3)*sin(c + d*x)/x**4, x)

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