\(\int \frac {\sin (a+b x)}{(c+d x)^2} \, dx\) [6]

Optimal result

Integrand size = 14, antiderivative size = 72 \[ \int \frac {\sin (a+b x)}{(c+d x)^2} \, dx=\frac {b \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right )}{d^2}-\frac {\sin (a+b x)}{d (c+d x)}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {\sin (a+b x)}{(c+d x)^2} \, dx=\frac {b \cos \left (a-\frac {b c}{d}\right ) \operatorname {CosIntegral}\left (b \left (\frac {c}{d}+x\right )\right )-\frac {d \sin (a+b x)}{c+d x}-b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )}{d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3778, 3042, 3784, 3042, 3780, 3783}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

\(\displaystyle \frac {b \int \frac {\cos (a+b x)}{c+d x}dx}{d}-\frac {\sin (a+b x)}{d (c+d x)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

\(\displaystyle \frac {b \left (\cos \left (a-\frac {b c}{d}\right ) \int \frac {\sin \left (\frac {b c}{d}+b x+\frac {\pi }{2}\right )}{c+d x}dx-\frac {\sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d}\right )}{d}-\frac {\sin (a+b x)}{d (c+d x)}\)

\(\Big \downarrow \) 3783

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3778

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] - Simp[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 3780

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte 
gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
 

rule 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3784

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* 
e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[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]
 

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56

method	result	size
derivativedivides	\(b \left (-\frac {\sin \left (b x +a \right )}{\left (-a d +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {\operatorname {Si}\left (-b x -a -\frac {-a d +b c}{d}\right ) \sin \left (\frac {-a d +b c}{d}\right )}{d}+\frac {\operatorname {Ci}\left (b x +a +\frac {-a d +b c}{d}\right ) \cos \left (\frac {-a d +b c}{d}\right )}{d}}{d}\right )\)	\(112\)
default	\(b \left (-\frac {\sin \left (b x +a \right )}{\left (-a d +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {\operatorname {Si}\left (-b x -a -\frac {-a d +b c}{d}\right ) \sin \left (\frac {-a d +b c}{d}\right )}{d}+\frac {\operatorname {Ci}\left (b x +a +\frac {-a d +b c}{d}\right ) \cos \left (\frac {-a d +b c}{d}\right )}{d}}{d}\right )\)	\(112\)
risch	\(-\frac {b \,{\mathrm e}^{\frac {i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (-i b x -i a -\frac {-i a d +i b c}{d}\right )}{2 d^{2}}-\frac {b \,{\mathrm e}^{-\frac {i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (i b x +i a -\frac {i \left (a d -b c \right )}{d}\right )}{2 d^{2}}-\frac {\left (-2 d x b -2 b c \right ) \sin \left (b x +a \right )}{2 d \left (d x +c \right ) \left (-d x b -b c \right )}\)	\(138\)

Input:

int(sin(b*x+a)/(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

b*(-sin(b*x+a)/(-a*d+b*c+d*(b*x+a))/d+(-Si(-b*x-a-(-a*d+b*c)/d)*sin((-a*d+ 
b*c)/d)/d+Ci(b*x+a+(-a*d+b*c)/d)*cos((-a*d+b*c)/d)/d)/d)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.35 \[ \int \frac {\sin (a+b x)}{(c+d x)^2} \, dx=\frac {{\left (b d x + b c\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) - {\left (b d x + b c\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) - d \sin \left (b x + a\right )}{d^{3} x + c d^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.28 \[ \int \frac {\sin (a+b x)}{(c+d x)^2} \, dx=-\frac {b^{2} {\left (i \, E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b^{2} {\left (E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )}{2 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \] Input:

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

Output:

-1/2*(b^2*(I*exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - I*exp_ 
integral_e(2, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b 
^2*(exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e( 
2, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d))/((b*c*d + (b* 
x + a)*d^2 - a*d^2)*b)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 521 vs. \(2 (72) = 144\).

Time = 0.42 (sec) , antiderivative size = 521, normalized size of antiderivative = 7.24 \[ \int \frac {\sin (a+b x)}{(c+d x)^2} \, dx=\frac {{\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) + b^{3} c \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) - a b^{2} d \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) + {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) + b^{3} c \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) - a b^{2} d \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) + b^{2} d \sin \left (-\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )}}{d}\right )\right )} d^{2}}{{\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} d^{4} + b c d^{4} - a d^{5}\right )} b} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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