\(\int \frac {\sin (c+d x)}{x (a+b x)^3} \, dx\) [37]

Optimal result

Integrand size = 17, antiderivative size = 261 \[ \int \frac {\sin (c+d x)}{x (a+b x)^3} \, dx=\frac {d \cos (c+d x)}{2 a b (a+b x)}-\frac {d \cos \left (c-\frac {a d}{b}\right ) \operatorname {CosIntegral}\left (\frac {a d}{b}+d x\right )}{a^2 b}+\frac {\operatorname {CosIntegral}(d x) \sin (c)}{a^3}-\frac {\operatorname {CosIntegral}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{a^3}+\frac {d^2 \operatorname {CosIntegral}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{2 a b^2}+\frac {\sin (c+d x)}{2 a (a+b x)^2}+\frac {\sin (c+d x)}{a^2 (a+b x)}+\frac {\cos (c) \text {Si}(d x)}{a^3}-\frac {\cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{a^3}+\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{2 a b^2}+\frac {d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{a^2 b} \] Output:

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

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 449, normalized size of antiderivative = 1.72 \[ \int \frac {\sin (c+d x)}{x (a+b x)^3} \, dx=\frac {a^3 b d \cos (c+d x)+a^2 b^2 d x \cos (c+d x)+2 b^2 (a+b x)^2 \operatorname {CosIntegral}(d x) \sin (c)+(a+b x)^2 \operatorname {CosIntegral}\left (d \left (\frac {a}{b}+x\right )\right ) \left (-2 a b d \cos \left (c-\frac {a d}{b}\right )+\left (-2 b^2+a^2 d^2\right ) \sin \left (c-\frac {a d}{b}\right )\right )+3 a^2 b^2 \sin (c+d x)+2 a b^3 x \sin (c+d x)+2 a^2 b^2 \cos (c) \text {Si}(d x)+4 a b^3 x \cos (c) \text {Si}(d x)+2 b^4 x^2 \cos (c) \text {Si}(d x)-2 a^2 b^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+a^4 d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )-4 a b^3 x \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+2 a^3 b d^2 x \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )-2 b^4 x^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+a^2 b^2 d^2 x^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+2 a^3 b d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+4 a^2 b^2 d x \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )+2 a b^3 d x^2 \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )}{2 a^3 b^2 (a+b x)^2} \] Input:

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

Output:

(a^3*b*d*Cos[c + d*x] + a^2*b^2*d*x*Cos[c + d*x] + 2*b^2*(a + b*x)^2*CosIn 
tegral[d*x]*Sin[c] + (a + b*x)^2*CosIntegral[d*(a/b + x)]*(-2*a*b*d*Cos[c 
- (a*d)/b] + (-2*b^2 + a^2*d^2)*Sin[c - (a*d)/b]) + 3*a^2*b^2*Sin[c + d*x] 
 + 2*a*b^3*x*Sin[c + d*x] + 2*a^2*b^2*Cos[c]*SinIntegral[d*x] + 4*a*b^3*x* 
Cos[c]*SinIntegral[d*x] + 2*b^4*x^2*Cos[c]*SinIntegral[d*x] - 2*a^2*b^2*Co 
s[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + a^4*d^2*Cos[c - (a*d)/b]*SinInte 
gral[d*(a/b + x)] - 4*a*b^3*x*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + 
2*a^3*b*d^2*x*Cos[c - (a*d)/b]*SinIntegral[d*(a/b + x)] - 2*b^4*x^2*Cos[c 
- (a*d)/b]*SinIntegral[d*(a/b + x)] + a^2*b^2*d^2*x^2*Cos[c - (a*d)/b]*Sin 
Integral[d*(a/b + x)] + 2*a^3*b*d*Sin[c - (a*d)/b]*SinIntegral[d*(a/b + x) 
] + 4*a^2*b^2*d*x*Sin[c - (a*d)/b]*SinIntegral[d*(a/b + x)] + 2*a*b^3*d*x^ 
2*Sin[c - (a*d)/b]*SinIntegral[d*(a/b + x)])/(2*a^3*b^2*(a + b*x)^2)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, 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.

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {b \sin (c+d x)}{a^3 (a+b x)}+\frac {\sin (c+d x)}{a^3 x}-\frac {b \sin (c+d x)}{a^2 (a+b x)^2}-\frac {b \sin (c+d x)}{a (a+b x)^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sin \left (c-\frac {a d}{b}\right ) \operatorname {CosIntegral}\left (x d+\frac {a d}{b}\right )}{a^3}-\frac {\cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^3}+\frac {\sin (c) \operatorname {CosIntegral}(d x)}{a^3}+\frac {\cos (c) \text {Si}(d x)}{a^3}-\frac {d \cos \left (c-\frac {a d}{b}\right ) \operatorname {CosIntegral}\left (x d+\frac {a d}{b}\right )}{a^2 b}+\frac {d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{a^2 b}+\frac {\sin (c+d x)}{a^2 (a+b x)}+\frac {d^2 \sin \left (c-\frac {a d}{b}\right ) \operatorname {CosIntegral}\left (x d+\frac {a d}{b}\right )}{2 a b^2}+\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{2 a b^2}+\frac {\sin (c+d x)}{2 a (a+b x)^2}+\frac {d \cos (c+d x)}{2 a b (a+b x)}\)

Input:

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

Output:

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

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 = 1.26 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )+\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{a^{3}}-\frac {d^{2} b \left (-\frac {\sin \left (d x +c \right )}{2 \left (a d -b c +b \left (d x +c \right )\right )^{2} b}+\frac {-\frac {\cos \left (d x +c \right )}{\left (a d -b c +b \left (d x +c \right )\right ) b}-\frac {\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}}{b}}{2 b}\right )}{a}-\frac {b \left (\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}\right )}{a^{3}}-\frac {d b \left (-\frac {\sin \left (d x +c \right )}{\left (a d -b c +b \left (d x +c \right )\right ) b}+\frac {\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}+\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}}{b}\right )}{a^{2}}\)	\(359\)
default	\(\frac {\operatorname {Si}\left (d x \right ) \cos \left (c \right )+\operatorname {Ci}\left (d x \right ) \sin \left (c \right )}{a^{3}}-\frac {d^{2} b \left (-\frac {\sin \left (d x +c \right )}{2 \left (a d -b c +b \left (d x +c \right )\right )^{2} b}+\frac {-\frac {\cos \left (d x +c \right )}{\left (a d -b c +b \left (d x +c \right )\right ) b}-\frac {\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}}{b}}{2 b}\right )}{a}-\frac {b \left (\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}-\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}\right )}{a^{3}}-\frac {d b \left (-\frac {\sin \left (d x +c \right )}{\left (a d -b c +b \left (d x +c \right )\right ) b}+\frac {\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}+\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}}{b}\right )}{a^{2}}\)	\(359\)
risch	\(\frac {i {\mathrm e}^{\frac {i \left (a d -b c \right )}{b}} \operatorname {expIntegral}_{1}\left (i d x +i c +\frac {i \left (a d -b c \right )}{b}\right )}{2 a^{3}}-\frac {i {\mathrm e}^{-\frac {i \left (a d -b c \right )}{b}} \operatorname {expIntegral}_{1}\left (-i d x -i c -\frac {i a d -i b c}{b}\right )}{2 a^{3}}+\frac {i d^{2} {\mathrm e}^{-\frac {i \left (a d -b c \right )}{b}} \operatorname {expIntegral}_{1}\left (-i d x -i c -\frac {i a d -i b c}{b}\right )}{4 b^{2} a}+\frac {d \,{\mathrm e}^{-\frac {i \left (a d -b c \right )}{b}} \operatorname {expIntegral}_{1}\left (-i d x -i c -\frac {i a d -i b c}{b}\right )}{2 b \,a^{2}}-\frac {{\mathrm e}^{-i c} \pi \,\operatorname {csgn}\left (d x \right )}{2 a^{3}}+\frac {{\mathrm e}^{-i c} \operatorname {Si}\left (d x \right )}{a^{3}}-\frac {i {\mathrm e}^{-i c} \operatorname {expIntegral}_{1}\left (-i d x \right )}{2 a^{3}}+\frac {{\mathrm e}^{\frac {i \left (a d -b c \right )}{b}} \operatorname {expIntegral}_{1}\left (i d x +i c +\frac {i \left (a d -b c \right )}{b}\right ) d}{2 a^{2} b}+\frac {i {\mathrm e}^{i c} \operatorname {expIntegral}_{1}\left (-i d x \right )}{2 a^{3}}-\frac {i {\mathrm e}^{\frac {i \left (a d -b c \right )}{b}} \operatorname {expIntegral}_{1}\left (i d x +i c +\frac {i \left (a d -b c \right )}{b}\right ) d^{2}}{4 a \,b^{2}}-\frac {b^{2} \cos \left (d x +c \right ) d^{3} x^{3}}{2 a \left (b x +a \right )^{2} \left (-x^{2} d^{2} b^{2}-2 a b \,d^{2} x -a^{2} d^{2}\right )}-\frac {3 b \cos \left (d x +c \right ) d^{3} x^{2}}{2 \left (b x +a \right )^{2} \left (-x^{2} d^{2} b^{2}-2 a b \,d^{2} x -a^{2} d^{2}\right )}-\frac {3 a \cos \left (d x +c \right ) d^{3} x}{2 \left (b x +a \right )^{2} \left (-x^{2} d^{2} b^{2}-2 a b \,d^{2} x -a^{2} d^{2}\right )}-\frac {a^{2} \cos \left (d x +c \right ) d^{3}}{2 b \left (b x +a \right )^{2} \left (-x^{2} d^{2} b^{2}-2 a b \,d^{2} x -a^{2} d^{2}\right )}-\frac {\sin \left (d x +c \right ) d^{2} x^{3} b^{3}}{a^{2} \left (b x +a \right )^{2} \left (-x^{2} d^{2} b^{2}-2 a b \,d^{2} x -a^{2} d^{2}\right )}-\frac {7 \sin \left (d x +c \right ) b^{2} d^{2} x^{2}}{2 a \left (b x +a \right )^{2} \left (-x^{2} d^{2} b^{2}-2 a b \,d^{2} x -a^{2} d^{2}\right )}-\frac {4 \sin \left (d x +c \right ) b \,d^{2} x}{\left (b x +a \right )^{2} \left (-x^{2} d^{2} b^{2}-2 a b \,d^{2} x -a^{2} d^{2}\right )}-\frac {3 a \sin \left (d x +c \right ) d^{2}}{2 \left (b x +a \right )^{2} \left (-x^{2} d^{2} b^{2}-2 a b \,d^{2} x -a^{2} d^{2}\right )}\)	\(798\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 391, normalized size of antiderivative = 1.50 \[ \int \frac {\sin (c+d x)}{x (a+b x)^3} \, dx=\frac {2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )} \operatorname {Ci}\left (d x\right ) \sin \left (c\right ) + 2 \, {\left (b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2}\right )} \cos \left (c\right ) \operatorname {Si}\left (d x\right ) + {\left (a^{2} b^{2} d x + a^{3} b d\right )} \cos \left (d x + c\right ) - {\left (2 \, {\left (a b^{3} d x^{2} + 2 \, a^{2} b^{2} d x + a^{3} b d\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) - {\left (a^{4} d^{2} - 2 \, a^{2} b^{2} + {\left (a^{2} b^{2} d^{2} - 2 \, b^{4}\right )} x^{2} + 2 \, {\left (a^{3} b d^{2} - 2 \, a b^{3}\right )} x\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) + {\left (2 \, a b^{3} x + 3 \, a^{2} b^{2}\right )} \sin \left (d x + c\right ) - {\left ({\left (a^{4} d^{2} - 2 \, a^{2} b^{2} + {\left (a^{2} b^{2} d^{2} - 2 \, b^{4}\right )} x^{2} + 2 \, {\left (a^{3} b d^{2} - 2 \, a b^{3}\right )} x\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + 2 \, {\left (a b^{3} d x^{2} + 2 \, a^{2} b^{2} d x + a^{3} b d\right )} \operatorname {Si}\left (\frac {b d x + a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{2 \, {\left (a^{3} b^{4} x^{2} + 2 \, a^{4} b^{3} x + a^{5} b^{2}\right )}} \] Input:

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

Output:

1/2*(2*(b^4*x^2 + 2*a*b^3*x + a^2*b^2)*cos_integral(d*x)*sin(c) + 2*(b^4*x 
^2 + 2*a*b^3*x + a^2*b^2)*cos(c)*sin_integral(d*x) + (a^2*b^2*d*x + a^3*b* 
d)*cos(d*x + c) - (2*(a*b^3*d*x^2 + 2*a^2*b^2*d*x + a^3*b*d)*cos_integral( 
(b*d*x + a*d)/b) - (a^4*d^2 - 2*a^2*b^2 + (a^2*b^2*d^2 - 2*b^4)*x^2 + 2*(a 
^3*b*d^2 - 2*a*b^3)*x)*sin_integral((b*d*x + a*d)/b))*cos(-(b*c - a*d)/b) 
+ (2*a*b^3*x + 3*a^2*b^2)*sin(d*x + c) - ((a^4*d^2 - 2*a^2*b^2 + (a^2*b^2* 
d^2 - 2*b^4)*x^2 + 2*(a^3*b*d^2 - 2*a*b^3)*x)*cos_integral((b*d*x + a*d)/b 
) + 2*(a*b^3*d*x^2 + 2*a^2*b^2*d*x + a^3*b*d)*sin_integral((b*d*x + a*d)/b 
))*sin(-(b*c - a*d)/b))/(a^3*b^4*x^2 + 2*a^4*b^3*x + a^5*b^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [C] (verification not implemented)

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

Time = 0.46 (sec) , antiderivative size = 17806, normalized size of antiderivative = 68.22 \[ \int \frac {\sin (c+d x)}{x (a+b x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(a^2*b^2*d^2*x^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2*t 
an(1/2*c)^2*tan(1/2*a*d/b)^2 - a^2*b^2*d^2*x^2*imag_part(cos_integral(-d*x 
 - a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a^2*b^2*d^2*x^ 
2*sin_integral((b*d*x + a*d)/b)*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b) 
^2 + 2*a^2*b^2*d^2*x^2*real_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2 
*tan(1/2*c)^2*tan(1/2*a*d/b) + 2*a^2*b^2*d^2*x^2*real_part(cos_integral(-d 
*x - a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b) - 2*a^2*b^2*d^2*x^ 
2*real_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)*tan(1/2*a 
*d/b)^2 - 2*a^2*b^2*d^2*x^2*real_part(cos_integral(-d*x - a*d/b))*tan(1/2* 
d*x)^2*tan(1/2*c)*tan(1/2*a*d/b)^2 + 2*a^3*b*d^2*x*imag_part(cos_integral( 
d*x + a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 - 2*a^3*b*d^2*x 
*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2 
*a*d/b)^2 - 2*a*b^3*d*x^2*real_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x 
)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 - 2*a*b^3*d*x^2*real_part(cos_integral(- 
d*x - a*d/b))*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 4*a^3*b*d^2*x 
*sin_integral((b*d*x + a*d)/b)*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(1/2*a*d/b)^ 
2 - a^2*b^2*d^2*x^2*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*d*x)^2*ta 
n(1/2*c)^2 + a^2*b^2*d^2*x^2*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2 
*d*x)^2*tan(1/2*c)^2 - 2*a^2*b^2*d^2*x^2*sin_integral((b*d*x + a*d)/b)*tan 
(1/2*d*x)^2*tan(1/2*c)^2 + 4*a^2*b^2*d^2*x^2*imag_part(cos_integral(d*x...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sin (c+d x)}{x (a+b x)^3} \, dx=\int \frac {\sin \left (d x +c \right )}{b^{3} x^{4}+3 a \,b^{2} x^{3}+3 a^{2} b \,x^{2}+a^{3} x}d x \] Input:

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

Output:

int(sin(c + d*x)/(a**3*x + 3*a**2*b*x**2 + 3*a*b**2*x**3 + b**3*x**4),x)