Integrand size = 15, antiderivative size = 89 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=-\frac {a d \cos (c+d x)}{2 x}+b d \cos (c) \operatorname {CosIntegral}(d x)-\frac {1}{2} a d^2 \operatorname {CosIntegral}(d x) \sin (c)-\frac {a \sin (c+d x)}{2 x^2}-\frac {b \sin (c+d x)}{x}-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-b d \sin (c) \text {Si}(d x) \]
b*d*Ci(d*x)*cos(c)-1/2*a*d*cos(d*x+c)/x-1/2*a*d^2*cos(c)*Si(d*x)-1/2*a*d^2 *Ci(d*x)*sin(c)-b*d*Si(d*x)*sin(c)-1/2*a*sin(d*x+c)/x^2-b*sin(d*x+c)/x
Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=-\frac {a d x \cos (c+d x)+d x^2 \operatorname {CosIntegral}(d x) (-2 b \cos (c)+a d \sin (c))+a \sin (c+d x)+2 b x \sin (c+d x)+d x^2 (a d \cos (c)+2 b \sin (c)) \text {Si}(d x)}{2 x^2} \]
-1/2*(a*d*x*Cos[c + d*x] + d*x^2*CosIntegral[d*x]*(-2*b*Cos[c] + a*d*Sin[c ]) + a*Sin[c + d*x] + 2*b*x*Sin[c + d*x] + d*x^2*(a*d*Cos[c] + 2*b*Sin[c]) *SinIntegral[d*x])/x^2
Time = 0.43 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {a \sin (c+d x)}{x^3}+\frac {b \sin (c+d x)}{x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{2} a d^2 \sin (c) \operatorname {CosIntegral}(d x)-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{2 x^2}-\frac {a d \cos (c+d x)}{2 x}+b d \cos (c) \operatorname {CosIntegral}(d x)-b d \sin (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{x}\) |
-1/2*(a*d*Cos[c + d*x])/x + b*d*Cos[c]*CosIntegral[d*x] - (a*d^2*CosIntegr al[d*x]*Sin[c])/2 - (a*Sin[c + d*x])/(2*x^2) - (b*Sin[c + d*x])/x - (a*d^2 *Cos[c]*SinIntegral[d*x])/2 - b*d*Sin[c]*SinIntegral[d*x]
3.1.7.3.1 Defintions of rubi rules used
Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(d^{2} \left (a \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{2}\right )+\frac {b \left (-\frac {\sin \left (d x +c \right )}{d x}-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )\right )}{d}\right )\) | \(88\) |
| default | \(d^{2} \left (a \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{2}\right )+\frac {b \left (-\frac {\sin \left (d x +c \right )}{d x}-\operatorname {Si}\left (d x \right ) \sin \left (c \right )+\operatorname {Ci}\left (d x \right ) \cos \left (c \right )\right )}{d}\right )\) | \(88\) |
| risch | \(-\frac {\operatorname {Ei}_{1}\left (-i d x \right ) \cos \left (c \right ) b d}{2}-\frac {\cos \left (c \right ) \operatorname {Ei}_{1}\left (i d x \right ) b d}{2}-\frac {i \operatorname {Ei}_{1}\left (-i d x \right ) \cos \left (c \right ) a \,d^{2}}{4}+\frac {i \cos \left (c \right ) \operatorname {Ei}_{1}\left (i d x \right ) a \,d^{2}}{4}-\frac {i \operatorname {Ei}_{1}\left (-i d x \right ) \sin \left (c \right ) b d}{2}+\frac {i \sin \left (c \right ) \operatorname {Ei}_{1}\left (i d x \right ) b d}{2}+\frac {\operatorname {Ei}_{1}\left (-i d x \right ) \sin \left (c \right ) a \,d^{2}}{4}+\frac {\sin \left (c \right ) \operatorname {Ei}_{1}\left (i d x \right ) a \,d^{2}}{4}-\frac {a d \cos \left (d x +c \right )}{2 x}+\frac {\left (-4 d^{3} x^{3} b -2 a \,d^{3} x^{2}\right ) \sin \left (d x +c \right )}{4 d^{3} x^{4}}\) | \(164\) |
| meijerg | \(\frac {d^{2} b \sqrt {\pi }\, \sin \left (c \right ) \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{4 \sqrt {d^{2}}}+\frac {d b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}+\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{4}+\frac {a \sqrt {\pi }\, \sin \left (c \right ) d^{2} \left (-\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+\ln \left (d^{2}\right )\right )}{\sqrt {\pi }}+\frac {-6 d^{2} x^{2}+4}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \cos \left (d x \right )}{\sqrt {\pi }\, d^{2} x^{2}}+\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}-\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a \sqrt {\pi }\, \cos \left (c \right ) d^{2} \left (-\frac {4 \cos \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{d^{2} x^{2} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (d x \right )}{\sqrt {\pi }}\right )}{8}\) | \(311\) |
d^2*(a*(-1/2*sin(d*x+c)/d^2/x^2-1/2*cos(d*x+c)/d/x-1/2*Si(d*x)*cos(c)-1/2* Ci(d*x)*sin(c))+1/d*b*(-sin(d*x+c)/d/x-Si(d*x)*sin(c)+Ci(d*x)*cos(c)))
Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=-\frac {a d x \cos \left (d x + c\right ) + {\left (a d^{2} x^{2} \operatorname {Si}\left (d x\right ) - 2 \, b d x^{2} \operatorname {Ci}\left (d x\right )\right )} \cos \left (c\right ) + {\left (2 \, b x + a\right )} \sin \left (d x + c\right ) + {\left (a d^{2} x^{2} \operatorname {Ci}\left (d x\right ) + 2 \, b d x^{2} \operatorname {Si}\left (d x\right )\right )} \sin \left (c\right )}{2 \, x^{2}} \]
-1/2*(a*d*x*cos(d*x + c) + (a*d^2*x^2*sin_integral(d*x) - 2*b*d*x^2*cos_in tegral(d*x))*cos(c) + (2*b*x + a)*sin(d*x + c) + (a*d^2*x^2*cos_integral(d *x) + 2*b*d*x^2*sin_integral(d*x))*sin(c))/x^2
\[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=\int \frac {\left (a + b x\right ) \sin {\left (c + d x \right )}}{x^{3}}\, dx \]
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=-\frac {{\left ({\left (a {\left (-i \, \Gamma \left (-2, i \, d x\right ) + i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - a {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3} + 2 \, {\left (b {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - b {\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{2}\right )} x^{2} + 2 \, b \cos \left (d x + c\right )}{2 \, d x^{2}} \]
-1/2*(((a*(-I*gamma(-2, I*d*x) + I*gamma(-2, -I*d*x))*cos(c) - a*(gamma(-2 , I*d*x) + gamma(-2, -I*d*x))*sin(c))*d^3 + 2*(b*(gamma(-2, I*d*x) + gamma (-2, -I*d*x))*cos(c) - b*(I*gamma(-2, I*d*x) - I*gamma(-2, -I*d*x))*sin(c) )*d^2)*x^2 + 2*b*cos(d*x + c))/(d*x^2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.30 (sec) , antiderivative size = 796, normalized size of antiderivative = 8.94 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=\text {Too large to display} \]
1/4*(a*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a* d^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^2*x^2*real_p art(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^2*x^2*real_part(c os_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*b*d*x^2*real_part(cos_int egral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*b*d*x^2*real_part(cos_integral (-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^2*x^2*imag_part(cos_integral(d*x ))*tan(1/2*d*x)^2 + a*d^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a*d^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2 - 4*b*d*x^2*imag_part(cos_ integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*b*d*x^2*imag_part(cos_integra l(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 8*b*d*x^2*sin_integral(d*x)*tan(1/2*d *x)^2*tan(1/2*c) + a*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*c)^2 - a *d^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a*d^2*x^2*sin_inte gral(d*x)*tan(1/2*c)^2 + 2*b*d*x^2*real_part(cos_integral(d*x))*tan(1/2*d* x)^2 + 2*b*d*x^2*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a*d^2*x^ 2*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^2*x^2*real_part(cos_inte gral(-d*x))*tan(1/2*c) - 2*b*d*x^2*real_part(cos_integral(d*x))*tan(1/2*c) ^2 - 2*b*d*x^2*real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 2*a*d*x*tan(1/ 2*d*x)^2*tan(1/2*c)^2 - a*d^2*x^2*imag_part(cos_integral(d*x)) + a*d^2*x^2 *imag_part(cos_integral(-d*x)) - 2*a*d^2*x^2*sin_integral(d*x) - 4*b*d*...
Timed out. \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=\int \frac {\sin \left (c+d\,x\right )\,\left (a+b\,x\right )}{x^3} \,d x \]