3.1.36 \(\int \sin (\frac {a+b x}{c+d x}) \, dx\) [36]

3.1.36.1 Optimal result

Integrand size = 14, antiderivative size = 100 \[ \int \sin \left (\frac {a+b x}{c+d x}\right ) \, dx=\frac {(b c-a d) \cos \left (\frac {b}{d}\right ) \operatorname {CosIntegral}\left (\frac {b c-a d}{d (c+d x)}\right )}{d^2}+\frac {(c+d x) \sin \left (\frac {a+b x}{c+d x}\right )}{d}+\frac {(b c-a d) \sin \left (\frac {b}{d}\right ) \text {Si}\left (\frac {b c-a d}{d (c+d x)}\right )}{d^2} \]

output

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

3.1.36.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.80 (sec) , antiderivative size = 324, normalized size of antiderivative = 3.24 \[ \int \sin \left (\frac {a+b x}{c+d x}\right ) \, dx=\frac {i c d e^{-\frac {i (a+b x)}{c+d x}}-i c d e^{\frac {i (a+b x)}{c+d x}}+2 d^2 x \cos \left (\frac {-b c+a d}{d (c+d x)}\right ) \sin \left (\frac {b}{d}\right )+2 d^2 x \cos \left (\frac {b}{d}\right ) \sin \left (\frac {-b c+a d}{d (c+d x)}\right )+(b c-a d) \left (\operatorname {CosIntegral}\left (\frac {-b c+a d}{d (c+d x)}\right ) \left (\cos \left (\frac {b}{d}\right )-i \sin \left (\frac {b}{d}\right )\right )+\operatorname {CosIntegral}\left (\frac {b c-a d}{c d+d^2 x}\right ) \left (\cos \left (\frac {b}{d}\right )+i \sin \left (\frac {b}{d}\right )\right )+i \cos \left (\frac {b}{d}\right ) \text {Si}\left (\frac {-b c+a d}{d (c+d x)}\right )-\sin \left (\frac {b}{d}\right ) \text {Si}\left (\frac {-b c+a d}{d (c+d x)}\right )+i \cos \left (\frac {b}{d}\right ) \text {Si}\left (\frac {b c-a d}{c d+d^2 x}\right )+\sin \left (\frac {b}{d}\right ) \text {Si}\left (\frac {b c-a d}{c d+d^2 x}\right )\right )}{2 d^2} \]

input

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

output

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

3.1.36.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {5074, 3042, 3778, 3042, 3784, 25, 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.

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

\(\Big \downarrow \) 5074

\(\displaystyle -\frac {\int (c+d x)^2 \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int (c+d x)^2 \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )d\frac {1}{c+d x}}{d}\)

\(\Big \downarrow \) 3778

\(\displaystyle -\frac {-\frac {(b c-a d) \int (c+d x) \cos \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )d\frac {1}{c+d x}}{d}-\left ((c+d x) \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {(b c-a d) \int (c+d x) \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}+\frac {\pi }{2}\right )d\frac {1}{c+d x}}{d}-\left ((c+d x) \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )\right )}{d}\)

\(\Big \downarrow \) 3784

\(\displaystyle -\frac {-\frac {(b c-a d) \left (\cos \left (\frac {b}{d}\right ) \int (c+d x) \cos \left (\frac {b c-a d}{d (c+d x)}\right )d\frac {1}{c+d x}-\sin \left (\frac {b}{d}\right ) \int -\left ((c+d x) \sin \left (\frac {b c-a d}{d (c+d x)}\right )\right )d\frac {1}{c+d x}\right )}{d}-\left ((c+d x) \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-\frac {(b c-a d) \left (\sin \left (\frac {b}{d}\right ) \int (c+d x) \sin \left (\frac {b c-a d}{d (c+d x)}\right )d\frac {1}{c+d x}+\cos \left (\frac {b}{d}\right ) \int (c+d x) \cos \left (\frac {b c-a d}{d (c+d x)}\right )d\frac {1}{c+d x}\right )}{d}-\left ((c+d x) \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {(b c-a d) \left (\sin \left (\frac {b}{d}\right ) \int (c+d x) \sin \left (\frac {b c-a d}{d (c+d x)}\right )d\frac {1}{c+d x}+\cos \left (\frac {b}{d}\right ) \int (c+d x) \sin \left (\frac {b c-a d}{d (c+d x)}+\frac {\pi }{2}\right )d\frac {1}{c+d x}\right )}{d}-\left ((c+d x) \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )\right )}{d}\)

\(\Big \downarrow \) 3780

\(\displaystyle -\frac {-\frac {(b c-a d) \left (\cos \left (\frac {b}{d}\right ) \int (c+d x) \sin \left (\frac {b c-a d}{d (c+d x)}+\frac {\pi }{2}\right )d\frac {1}{c+d x}+\sin \left (\frac {b}{d}\right ) \text {Si}\left (\frac {b c-a d}{d (c+d x)}\right )\right )}{d}-\left ((c+d x) \sin \left (\frac {b}{d}-\frac {b c-a d}{d (c+d x)}\right )\right )}{d}\)

\(\Big \downarrow \) 3783

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

input

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

output

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

3.1.36.3.1 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]
 

rule 5074

Int[Sin[((e_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_))]^(n_.), x_Symbol] 
 :> Simp[-d^(-1)   Subst[Int[Sin[b*(e/d) - e*(b*c - a*d)*(x/d)]^n/x^2, x], 
x, 1/(c + d*x)], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && NeQ[b*c - a* 
d, 0]
 

3.1.36.4 Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(-\left (a d -c b \right ) \left (-\frac {\sin \left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right )}{\left (\left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right ) d -b \right ) d}+\frac {-\frac {\operatorname {Si}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) \sin \left (\frac {b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) \cos \left (\frac {b}{d}\right )}{d}}{d}\right )\)	\(142\)
default	\(-\left (a d -c b \right ) \left (-\frac {\sin \left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right )}{\left (\left (\frac {b}{d}+\frac {a d -c b}{d \left (d x +c \right )}\right ) d -b \right ) d}+\frac {-\frac {\operatorname {Si}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) \sin \left (\frac {b}{d}\right )}{d}+\frac {\operatorname {Ci}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) \cos \left (\frac {b}{d}\right )}{d}}{d}\right )\)	\(142\)
risch	\(\frac {\operatorname {Ei}_{1}\left (-\frac {i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) {\mathrm e}^{\frac {i b}{d}} a}{2 d}-\frac {\operatorname {Ei}_{1}\left (-\frac {i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) {\mathrm e}^{\frac {i b}{d}} c b}{2 d^{2}}-\frac {i {\mathrm e}^{-\frac {i b}{d}} \pi \,\operatorname {csgn}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) a}{2 d}+\frac {i {\mathrm e}^{-\frac {i b}{d}} \pi \,\operatorname {csgn}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) b c}{2 d^{2}}+\frac {i {\mathrm e}^{-\frac {i b}{d}} \operatorname {Si}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) a}{d}-\frac {i {\mathrm e}^{-\frac {i b}{d}} \operatorname {Si}\left (\frac {a d -c b}{d \left (d x +c \right )}\right ) b c}{d^{2}}+\frac {{\mathrm e}^{-\frac {i b}{d}} \operatorname {Ei}_{1}\left (-\frac {i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) a}{2 d}-\frac {{\mathrm e}^{-\frac {i b}{d}} \operatorname {Ei}_{1}\left (-\frac {i \left (a d -c b \right )}{d \left (d x +c \right )}\right ) b c}{2 d^{2}}+\sin \left (\frac {x b +a}{d x +c}\right ) x +\frac {\sin \left (\frac {x b +a}{d x +c}\right ) c}{d}\)	\(331\)

input

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

output

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

3.1.36.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04 \[ \int \sin \left (\frac {a+b x}{c+d x}\right ) \, dx=\frac {{\left (b c - a d\right )} \cos \left (\frac {b}{d}\right ) \operatorname {Ci}\left (-\frac {b c - a d}{d^{2} x + c d}\right ) - {\left (b c - a d\right )} \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (-\frac {b c - a d}{d^{2} x + c d}\right ) + {\left (d^{2} x + c d\right )} \sin \left (\frac {b x + a}{d x + c}\right )}{d^{2}} \]

input

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

output

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

3.1.36.6 Sympy [F]

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

input

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

output

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

3.1.36.7 Maxima [F]

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

input

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

output

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

3.1.36.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 630 vs. \(2 (100) = 200\).

Time = 3.57 (sec) , antiderivative size = 630, normalized size of antiderivative = 6.30 \[ \int \sin \left (\frac {a+b x}{c+d x}\right ) \, dx=\frac {{\left (b^{3} c^{2} \cos \left (\frac {b}{d}\right ) \operatorname {Ci}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) - 2 \, a b^{2} c d \cos \left (\frac {b}{d}\right ) \operatorname {Ci}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) - \frac {{\left (b x + a\right )} b^{2} c^{2} d \cos \left (\frac {b}{d}\right ) \operatorname {Ci}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right )}{d x + c} + a^{2} b d^{2} \cos \left (\frac {b}{d}\right ) \operatorname {Ci}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) + \frac {2 \, {\left (b x + a\right )} a b c d^{2} \cos \left (\frac {b}{d}\right ) \operatorname {Ci}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right )}{d x + c} - \frac {{\left (b x + a\right )} a^{2} d^{3} \cos \left (\frac {b}{d}\right ) \operatorname {Ci}\left (-\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right )}{d x + c} + b^{3} c^{2} \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) - 2 \, a b^{2} c d \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) - \frac {{\left (b x + a\right )} b^{2} c^{2} d \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right )}{d x + c} + a^{2} b d^{2} \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right ) + \frac {2 \, {\left (b x + a\right )} a b c d^{2} \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right )}{d x + c} - \frac {{\left (b x + a\right )} a^{2} d^{3} \sin \left (\frac {b}{d}\right ) \operatorname {Si}\left (\frac {b - \frac {{\left (b x + a\right )} d}{d x + c}}{d}\right )}{d x + c} + b^{2} c^{2} d \sin \left (\frac {b x + a}{d x + c}\right ) - 2 \, a b c d^{2} \sin \left (\frac {b x + a}{d x + c}\right ) + a^{2} d^{3} \sin \left (\frac {b x + a}{d x + c}\right )\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )}}{b d^{2} - \frac {{\left (b x + a\right )} d^{3}}{d x + c}} \]

input

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

output

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

3.1.36.9 Mupad [F(-1)]

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

input

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

output

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