\(\int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx\) [7]

Optimal result

Integrand size = 15, antiderivative size = 89 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=-\frac {a d \cos (c+d x)}{2 x}+b d \cos (c) \operatorname {CosIntegral}(d x)-\frac {1}{2} a d^2 \operatorname {CosIntegral}(d x) \sin (c)-\frac {a \sin (c+d x)}{2 x^2}-\frac {b \sin (c+d x)}{x}-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-b d \sin (c) \text {Si}(d x) \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=-\frac {a d x \cos (c+d x)+d x^2 \operatorname {CosIntegral}(d x) (-2 b \cos (c)+a d \sin (c))+a \sin (c+d x)+2 b x \sin (c+d x)+d x^2 (a d \cos (c)+2 b \sin (c)) \text {Si}(d x)}{2 x^2} \] Input:

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

Output:

-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
 

Rubi [A] (verified)

Time = 0.45 (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

\(\displaystyle -\frac {1}{2} a d^2 \sin (c) \operatorname {CosIntegral}(d x)-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{2 x^2}-\frac {a d \cos (c+d x)}{2 x}+b d \cos (c) \operatorname {CosIntegral}(d x)-b d \sin (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{x}\)

Input:

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

Output:

-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]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.84 (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	\(-\frac {\cos \left (c \right ) \operatorname {expIntegral}_{1}\left (i d x \right ) b d}{2}+\frac {i \cos \left (c \right ) \operatorname {expIntegral}_{1}\left (i d x \right ) a \,d^{2}}{4}-\frac {\cos \left (c \right ) \operatorname {expIntegral}_{1}\left (-i d x \right ) b d}{2}-\frac {i \cos \left (c \right ) \operatorname {expIntegral}_{1}\left (-i d x \right ) a \,d^{2}}{4}+\frac {i \sin \left (c \right ) \operatorname {expIntegral}_{1}\left (i d x \right ) b d}{2}+\frac {\sin \left (c \right ) \operatorname {expIntegral}_{1}\left (i d x \right ) a \,d^{2}}{4}-\frac {i \sin \left (c \right ) \operatorname {expIntegral}_{1}\left (-i d x \right ) b d}{2}+\frac {\sin \left (c \right ) \operatorname {expIntegral}_{1}\left (-i d x \right ) a \,d^{2}}{4}-\frac {a d \cos \left (d x +c \right )}{2 x}+\frac {\left (-4 d^{3} x^{3} b -2 a \,d^{3} x^{2}\right ) \sin \left (d x +c \right )}{4 d^{3} x^{4}}\)	\(164\)
meijerg	\(\frac {d^{2} b \sin \left (c \right ) \sqrt {\pi }\, \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{4 \sqrt {d^{2}}}+\frac {d b \cos \left (c \right ) \sqrt {\pi }\, \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, x d}+\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{4}+\frac {a \sin \left (c \right ) \sqrt {\pi }\, d^{2} \left (-\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+\ln \left (d^{2}\right )\right )}{\sqrt {\pi }}+\frac {-6 x^{2} d^{2}+4}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \cos \left (d x \right )}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, x d}-\frac {4 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a \cos \left (c \right ) \sqrt {\pi }\, d^{2} \left (-\frac {4 \cos \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{x^{2} d^{2} \sqrt {\pi }}-\frac {4 \,\operatorname {Si}\left (d x \right )}{\sqrt {\pi }}\right )}{8}\)	\(311\)

Input:

int((b*x+a)*sin(d*x+c)/x^3,x,method=_RETURNVERBOSE)
 

Output:

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)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=-\frac {a d x \cos \left (d x + c\right ) + {\left (a d^{2} x^{2} \operatorname {Si}\left (d x\right ) - 2 \, b d x^{2} \operatorname {Ci}\left (d x\right )\right )} \cos \left (c\right ) + {\left (2 \, b x + a\right )} \sin \left (d x + c\right ) + {\left (a d^{2} x^{2} \operatorname {Ci}\left (d x\right ) + 2 \, b d x^{2} \operatorname {Si}\left (d x\right )\right )} \sin \left (c\right )}{2 \, x^{2}} \] Input:

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

Output:

-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
 

Sympy [F]

\[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=\int \frac {\left (a + b x\right ) \sin {\left (c + d x \right )}}{x^{3}}\, dx \] Input:

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=-\frac {{\left ({\left (a {\left (-i \, \Gamma \left (-2, i \, d x\right ) + i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - a {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3} + 2 \, {\left (b {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - b {\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{2}\right )} x^{2} + 2 \, b \cos \left (d x + c\right )}{2 \, d x^{2}} \] Input:

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

Output:

-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)
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.13 (sec) , antiderivative size = 796, normalized size of antiderivative = 8.94 \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=\text {Too large to display} \] Input:

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

Output:

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*...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=\int \frac {\sin \left (c+d\,x\right )\,\left (a+b\,x\right )}{x^3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b x) \sin (c+d x)}{x^3} \, dx=\frac {-2 \cos \left (d x +c \right ) b -2 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x^{2}+x^{2}}d x \right ) a \,d^{2} x^{2}-8 \left (\int \frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} x^{3}+x^{3}}d x \right ) b \,x^{2}-\sin \left (d x +c \right ) a d -a \,d^{2} x -2 b}{2 d \,x^{2}} \] Input:

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

Output:

( - 2*cos(c + d*x)*b - 2*int(tan((c + d*x)/2)**2/(tan((c + d*x)/2)**2*x**2 
 + x**2),x)*a*d**2*x**2 - 8*int(1/(tan((c + d*x)/2)**2*x**3 + x**3),x)*b*x 
**2 - sin(c + d*x)*a*d - a*d**2*x - 2*b)/(2*d*x**2)