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

Optimal. Leaf size=69 \[ -\frac{a \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^2}-\frac{a \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^2}-\frac{\cos (c+d x)}{b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.165794, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6742, 2638, 3303, 3299, 3302} \[ -\frac{a \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (\frac{a d}{b}+d x\right )}{b^2}-\frac{a \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (x d+\frac{a d}{b}\right )}{b^2}-\frac{\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sin[c + d*x])/(a + b*x),x]

[Out]

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[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 3299

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

Rule 3302

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

Rubi steps

\begin{align*} \int \frac{x \sin (c+d x)}{a+b x} \, dx &=\int \left (\frac{\sin (c+d x)}{b}-\frac{a \sin (c+d x)}{b (a+b x)}\right ) \, dx\\ &=\frac{\int \sin (c+d x) \, dx}{b}-\frac{a \int \frac{\sin (c+d x)}{a+b x} \, dx}{b}\\ &=-\frac{\cos (c+d x)}{b d}-\frac{\left (a \cos \left (c-\frac{a d}{b}\right )\right ) \int \frac{\sin \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{b}-\frac{\left (a \sin \left (c-\frac{a d}{b}\right )\right ) \int \frac{\cos \left (\frac{a d}{b}+d x\right )}{a+b x} \, dx}{b}\\ &=-\frac{\cos (c+d x)}{b d}-\frac{a \text{Ci}\left (\frac{a d}{b}+d x\right ) \sin \left (c-\frac{a d}{b}\right )}{b^2}-\frac{a \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (\frac{a d}{b}+d x\right )}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.190586, size = 63, normalized size = 0.91 \[ -\frac{a d \sin \left (c-\frac{a d}{b}\right ) \text{CosIntegral}\left (d \left (\frac{a}{b}+x\right )\right )+a d \cos \left (c-\frac{a d}{b}\right ) \text{Si}\left (d \left (\frac{a}{b}+x\right )\right )+b \cos (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.009, size = 180, normalized size = 2.6 \begin{align*}{\frac{1}{{d}^{2}} \left ( -{\frac{d\cos \left ( dx+c \right ) }{b}}-{\frac{ \left ( da-cb \right ) d}{b} \left ({\frac{1}{b}{\it Si} \left ( dx+c+{\frac{da-cb}{b}} \right ) \cos \left ({\frac{da-cb}{b}} \right ) }-{\frac{1}{b}{\it Ci} \left ( dx+c+{\frac{da-cb}{b}} \right ) \sin \left ({\frac{da-cb}{b}} \right ) } \right ) }-dc \left ({\frac{1}{b}{\it Si} \left ( dx+c+{\frac{da-cb}{b}} \right ) \cos \left ({\frac{da-cb}{b}} \right ) }-{\frac{1}{b}{\it Ci} \left ( dx+c+{\frac{da-cb}{b}} \right ) \sin \left ({\frac{da-cb}{b}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [C]  time = 2.48401, size = 1048, normalized size = 15.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.7011, size = 251, normalized size = 3.64 \begin{align*} -\frac{2 \, a d \cos \left (-\frac{b c - a d}{b}\right ) \operatorname{Si}\left (\frac{b d x + a d}{b}\right ) + 2 \, b \cos \left (d x + c\right ) -{\left (a d \operatorname{Ci}\left (\frac{b d x + a d}{b}\right ) + a d \operatorname{Ci}\left (-\frac{b d x + a d}{b}\right )\right )} \sin \left (-\frac{b c - a d}{b}\right )}{2 \, b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sin{\left (c + d x \right )}}{a + b x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sin(c + d*x)/(a + b*x), x)

________________________________________________________________________________________

Giac [C]  time = 1.15187, size = 849, normalized size = 12.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b)^2 - a*imag_part(cos_integral(-d*x - a
*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a*sin_integral((b*d*x + a*d)/b)*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + 2*a*r
eal_part(cos_integral(d*x + a*d/b))*tan(1/2*c)^2*tan(1/2*a*d/b) + 2*a*real_part(cos_integral(-d*x - a*d/b))*ta
n(1/2*c)^2*tan(1/2*a*d/b) - 2*a*real_part(cos_integral(d*x + a*d/b))*tan(1/2*c)*tan(1/2*a*d/b)^2 - 2*a*real_pa
rt(cos_integral(-d*x - a*d/b))*tan(1/2*c)*tan(1/2*a*d/b)^2 - a*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*c)
^2 + a*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*c)^2 - 2*a*sin_integral((b*d*x + a*d)/b)*tan(1/2*c)^2 + 4
*a*imag_part(cos_integral(d*x + a*d/b))*tan(1/2*c)*tan(1/2*a*d/b) - 4*a*imag_part(cos_integral(-d*x - a*d/b))*
tan(1/2*c)*tan(1/2*a*d/b) + 8*a*sin_integral((b*d*x + a*d)/b)*tan(1/2*c)*tan(1/2*a*d/b) - a*imag_part(cos_inte
gral(d*x + a*d/b))*tan(1/2*a*d/b)^2 + a*imag_part(cos_integral(-d*x - a*d/b))*tan(1/2*a*d/b)^2 - 2*a*sin_integ
ral((b*d*x + a*d)/b)*tan(1/2*a*d/b)^2 + 2*a*real_part(cos_integral(d*x + a*d/b))*tan(1/2*c) + 2*a*real_part(co
s_integral(-d*x - a*d/b))*tan(1/2*c) - 2*a*real_part(cos_integral(d*x + a*d/b))*tan(1/2*a*d/b) - 2*a*real_part
(cos_integral(-d*x - a*d/b))*tan(1/2*a*d/b) + a*imag_part(cos_integral(d*x + a*d/b)) - a*imag_part(cos_integra
l(-d*x - a*d/b)) + 2*a*sin_integral((b*d*x + a*d)/b))/(b^2*tan(1/2*c)^2*tan(1/2*a*d/b)^2 + b^2*tan(1/2*c)^2 +
b^2*tan(1/2*a*d/b)^2 + b^2)