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

Optimal result

Integrand size = 14, antiderivative size = 104 \[ \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx=-\frac {b \cos (a+b x)}{2 d^2 (c+d x)}-\frac {b^2 \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{2 d^3}-\frac {\sin (a+b x)}{2 d (c+d x)^2}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{2 d^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx=-\frac {b^2 \operatorname {CosIntegral}\left (b \left (\frac {c}{d}+x\right )\right ) \sin \left (a-\frac {b c}{d}\right )+\frac {d (b (c+d x) \cos (a+b x)+d \sin (a+b x))}{(c+d x)^2}+b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )}{2 d^3} \] Input:

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

Output:

-1/2*(b^2*CosIntegral[b*(c/d + x)]*Sin[a - (b*c)/d] + (d*(b*(c + d*x)*Cos[ 
a + b*x] + d*Sin[a + b*x]))/(c + d*x)^2 + b^2*Cos[a - (b*c)/d]*SinIntegral 
[b*(c/d + x)])/d^3
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3778, 3042, 3778, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

Input:

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

Output:

-1/2*Sin[a + b*x]/(d*(c + d*x)^2) + (b*(-(Cos[a + b*x]/(d*(c + d*x))) - (b 
*((CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/d + (Cos[a - (b*c)/d]*SinI 
ntegral[(b*c)/d + b*x])/d))/d))/(2*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 = 0.97 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.44

method	result	size
derivativedivides	\(b^{2} \left (-\frac {\sin \left (b x +a \right )}{2 \left (-a d +b c +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (-a d +b c +d \left (b x +a \right )\right ) d}-\frac {-\frac {\operatorname {Si}\left (-b x -a -\frac {-a d +b c}{d}\right ) \cos \left (\frac {-a d +b c}{d}\right )}{d}-\frac {\operatorname {Ci}\left (b x +a +\frac {-a d +b c}{d}\right ) \sin \left (\frac {-a d +b c}{d}\right )}{d}}{d}}{2 d}\right )\)	\(150\)
default	\(b^{2} \left (-\frac {\sin \left (b x +a \right )}{2 \left (-a d +b c +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (-a d +b c +d \left (b x +a \right )\right ) d}-\frac {-\frac {\operatorname {Si}\left (-b x -a -\frac {-a d +b c}{d}\right ) \cos \left (\frac {-a d +b c}{d}\right )}{d}-\frac {\operatorname {Ci}\left (b x +a +\frac {-a d +b c}{d}\right ) \sin \left (\frac {-a d +b c}{d}\right )}{d}}{d}}{2 d}\right )\)	\(150\)
risch	\(-\frac {i b^{2} {\mathrm e}^{\frac {i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (-i b x -i a -\frac {-i a d +i b c}{d}\right )}{4 d^{3}}+\frac {i b^{2} {\mathrm e}^{-\frac {i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (i b x +i a -\frac {i \left (a d -b c \right )}{d}\right )}{4 d^{3}}+\frac {i \left (2 i b^{3} d^{3} x^{3}+6 i b^{3} c \,d^{2} x^{2}+6 i b^{3} c^{2} d x +2 i b^{3} c^{3}\right ) \cos \left (b x +a \right )}{4 d^{2} \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {\left (2 x^{2} d^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}\right ) \sin \left (b x +a \right )}{4 d \left (d x +c \right )^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}\)	\(271\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.57 \[ \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx=-\frac {d^{2} \sin \left (b x + a\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.91 \[ \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx=-\frac {b^{3} {\left (i \, E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b^{3} {\left (E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )}{2 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \] Input:

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

Output:

-1/2*(b^3*(I*exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - I*exp_ 
integral_e(3, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b 
^3*(exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e( 
3, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d))/((b^2*c^2*d - 
 2*a*b*c*d^2 + (b*x + a)^2*d^3 + a^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))* 
b)
 

Giac [C] (verification not implemented)

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

Time = 0.52 (sec) , antiderivative size = 5727, normalized size of antiderivative = 55.07 \[ \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan( 
1/2*a)^2*tan(1/2*b*c/d)^2 - b^2*d^2*x^2*imag_part(cos_integral(-b*x - b*c/ 
d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^2*d^2*x^2*sin_integ 
ral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^2* 
d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*t 
an(1/2*b*c/d) + 2*b^2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/ 
2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d) - 2*b^2*d^2*x^2*real_part(cos_integra 
l(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d)^2 - 2*b^2*d^2*x^2 
*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b 
*c/d)^2 + 2*b^2*c*d*x*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2* 
tan(1/2*a)^2*tan(1/2*b*c/d)^2 - 2*b^2*c*d*x*imag_part(cos_integral(-b*x - 
b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 4*b^2*c*d*x*sin_int 
egral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - b^2* 
d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2 + 
 b^2*d^2*x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2* 
a)^2 - 2*b^2*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2* 
a)^2 + 4*b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*t 
an(1/2*a)*tan(1/2*b*c/d) - 4*b^2*d^2*x^2*imag_part(cos_integral(-b*x - b*c 
/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 8*b^2*d^2*x^2*sin_integral 
((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 4*b^2*c*d*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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