\(\int \frac {\sin (c+d x)}{(a+b x)^2} \, dx\) [30]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {\sin (c+d x)}{(a+b x)^2} \, dx=\frac {d \cos \left (c-\frac {a d}{b}\right ) \operatorname {CosIntegral}\left (d \left (\frac {a}{b}+x\right )\right )-\frac {b \sin (c+d x)}{a+b x}-d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )}{b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

Input:

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

Output:

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

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 = 0.88 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.49

method	result	size
derivativedivides	\(d \left (-\frac {\sin \left (d x +c \right )}{\left (a d -b c +b \left (d x +c \right )\right ) b}+\frac {\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}+\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}}{b}\right )\)	\(107\)
default	\(d \left (-\frac {\sin \left (d x +c \right )}{\left (a d -b c +b \left (d x +c \right )\right ) b}+\frac {\frac {\operatorname {Si}\left (d x +c +\frac {a d -b c}{b}\right ) \sin \left (\frac {a d -b c}{b}\right )}{b}+\frac {\operatorname {Ci}\left (d x +c +\frac {a d -b c}{b}\right ) \cos \left (\frac {a d -b c}{b}\right )}{b}}{b}\right )\)	\(107\)
risch	\(-\frac {d \,{\mathrm e}^{-\frac {i \left (a d -b c \right )}{b}} \operatorname {expIntegral}_{1}\left (-i d x -i c -\frac {i a d -i b c}{b}\right )}{2 b^{2}}-\frac {d \,{\mathrm e}^{\frac {i \left (a d -b c \right )}{b}} \operatorname {expIntegral}_{1}\left (i d x +i c +\frac {i \left (a d -b c \right )}{b}\right )}{2 b^{2}}-\frac {\left (-2 d x b -2 a d \right ) \sin \left (d x +c \right )}{2 b \left (b x +a \right ) \left (-d x b -a d \right )}\)	\(138\)

Input:

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

Output:

d*(-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 = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.33 \[ \int \frac {\sin (c+d x)}{(a+b x)^2} \, dx=\frac {{\left (b d x + a d\right )} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left (b d x + a d\right )} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {b d x + a d}{b}\right ) - b \sin \left (d x + c\right )}{b^{3} x + a b^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.28 \[ \int \frac {\sin (c+d x)}{(a+b x)^2} \, dx=\frac {d^{2} {\left (-i \, E_{2}\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right ) + i \, E_{2}\left (-\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) + d^{2} {\left (E_{2}\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right ) + E_{2}\left (-\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{2 \, {\left ({\left (d x + c\right )} b^{2} - b^{2} c + a b d\right )} d} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (73) = 146\).

Time = 0.15 (sec) , antiderivative size = 518, normalized size of antiderivative = 7.19 \[ \int \frac {\sin (c+d x)}{(a+b x)^2} \, dx=\frac {{\left ({\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d^{2} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) - b c d^{2} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + a d^{3} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + {\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d^{2} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) - b c d^{2} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + a d^{3} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + b d^{2} \sin \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right )\right )} b^{2}}{{\left ({\left (b x + a\right )} b^{4} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b^{5} c + a b^{4} d\right )} d} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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