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

Optimal. Leaf size=29 \[ a \sin (c) \text {Ci}(d x)+a \cos (c) \text {Si}(d x)-\frac {b \cos (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 29, normalized size of antiderivative = 1.00, 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} \[ a \sin (c) \text {CosIntegral}(d x)+a \cos (c) \text {Si}(d x)-\frac {b \cos (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Sin[c + d*x])/x,x]

[Out]

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

Rule 2638

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

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]

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 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 40, normalized size = 1.38 \[ a \sin (c) \text {Ci}(d x)+a \cos (c) \text {Si}(d x)+\frac {b \sin (c) \sin (d x)}{d}-\frac {b \cos (c) \cos (d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 44, normalized size = 1.52 \[ \frac {2 \, a d \cos \relax (c) \operatorname {Si}\left (d x\right ) - 2 \, b \cos \left (d x + c\right ) + {\left (a d \operatorname {Ci}\left (d x\right ) + a d \operatorname {Ci}\left (-d x\right )\right )} \sin \relax (c)}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.48, size = 339, normalized size = 11.69 \[ -\frac {a d \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d \Re \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d \Re \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - a d \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 2 \, a d \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d \Im \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - a d \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d \Re \left (\operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d \Re \left (\operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - a d \Im \left (\operatorname {Ci}\left (d x\right ) \right ) + a d \Im \left (\operatorname {Ci}\left (-d x\right ) \right ) - 2 \, a d \operatorname {Si}\left (d x\right ) - 2 \, b \tan \left (\frac {1}{2} \, d x\right )^{2} - 8 \, b \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, b \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b}{2 \, {\left (d \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d \tan \left (\frac {1}{2} \, d x\right )^{2} + d \tan \left (\frac {1}{2} \, c\right )^{2} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*d*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d*imag_part(cos_integral(-d*x))*tan(1/2
*d*x)^2*tan(1/2*c)^2 + 2*a*d*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d*real_part(cos_integral(d*x)
)*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - a*d*imag_part(co
s_integral(d*x))*tan(1/2*d*x)^2 + a*d*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a*d*sin_integral(d*x)*t
an(1/2*d*x)^2 + a*d*imag_part(cos_integral(d*x))*tan(1/2*c)^2 - a*d*imag_part(cos_integral(-d*x))*tan(1/2*c)^2
+ 2*a*d*sin_integral(d*x)*tan(1/2*c)^2 + 2*b*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d*real_part(cos_integral(d*x))
*tan(1/2*c) - 2*a*d*real_part(cos_integral(-d*x))*tan(1/2*c) - a*d*imag_part(cos_integral(d*x)) + a*d*imag_par
t(cos_integral(-d*x)) - 2*a*d*sin_integral(d*x) - 2*b*tan(1/2*d*x)^2 - 8*b*tan(1/2*d*x)*tan(1/2*c) - 2*b*tan(1
/2*c)^2 + 2*b)/(d*tan(1/2*d*x)^2*tan(1/2*c)^2 + d*tan(1/2*d*x)^2 + d*tan(1/2*c)^2 + d)

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maple [A]  time = 0.03, size = 31, normalized size = 1.07 \[ -\frac {b \cos \left (d x +c \right )}{d}+a \left (\Si \left (d x \right ) \cos \relax (c )+\Ci \left (d x \right ) \sin \relax (c )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.64, size = 522, normalized size = 18.00 \[ -\frac {1}{2} \, {\left ({\left (i \, E_{1}\left (i \, d x\right ) - i \, E_{1}\left (-i \, d x\right )\right )} \cos \relax (c) + {\left (E_{1}\left (i \, d x\right ) + E_{1}\left (-i \, d x\right )\right )} \sin \relax (c)\right )} a + \frac {{\left ({\left (i \, E_{1}\left (i \, d x\right ) - i \, E_{1}\left (-i \, d x\right )\right )} \cos \relax (c) + {\left (E_{1}\left (i \, d x\right ) + E_{1}\left (-i \, d x\right )\right )} \sin \relax (c)\right )} b c}{2 \, d} - \frac {{\left (2 \, {\left (d x + c\right )} {\left (\cos \relax (c)^{2} + \sin \relax (c)^{2}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (d x + c\right )} {\left (\cos \relax (c)^{2} + \sin \relax (c)^{2}\right )} \cos \left (d x + c\right ) - {\left (c {\left (E_{2}\left (i \, d x\right ) + E_{2}\left (-i \, d x\right )\right )} \cos \relax (c)^{3} + c {\left (E_{2}\left (i \, d x\right ) + E_{2}\left (-i \, d x\right )\right )} \cos \relax (c) \sin \relax (c)^{2} - c {\left (i \, E_{2}\left (i \, d x\right ) - i \, E_{2}\left (-i \, d x\right )\right )} \sin \relax (c)^{3} + c {\left (E_{2}\left (i \, d x\right ) + E_{2}\left (-i \, d x\right )\right )} \cos \relax (c) - {\left (c {\left (i \, E_{2}\left (i \, d x\right ) - i \, E_{2}\left (-i \, d x\right )\right )} \cos \relax (c)^{2} + c {\left (i \, E_{2}\left (i \, d x\right ) - i \, E_{2}\left (-i \, d x\right )\right )}\right )} \sin \relax (c)\right )} \cos \left (d x + c\right )^{2} - {\left (c {\left (E_{2}\left (i \, d x\right ) + E_{2}\left (-i \, d x\right )\right )} \cos \relax (c)^{3} + c {\left (E_{2}\left (i \, d x\right ) + E_{2}\left (-i \, d x\right )\right )} \cos \relax (c) \sin \relax (c)^{2} - c {\left (i \, E_{2}\left (i \, d x\right ) - i \, E_{2}\left (-i \, d x\right )\right )} \sin \relax (c)^{3} - 2 \, {\left (d x + c\right )} {\left (\cos \relax (c)^{2} + \sin \relax (c)^{2}\right )} \cos \left (d x + c\right ) + c {\left (E_{2}\left (i \, d x\right ) + E_{2}\left (-i \, d x\right )\right )} \cos \relax (c) - {\left (c {\left (i \, E_{2}\left (i \, d x\right ) - i \, E_{2}\left (-i \, d x\right )\right )} \cos \relax (c)^{2} + c {\left (i \, E_{2}\left (i \, d x\right ) - i \, E_{2}\left (-i \, d x\right )\right )}\right )} \sin \relax (c)\right )} \sin \left (d x + c\right )^{2}\right )} b}{4 \, {\left ({\left ({\left (d x + c\right )} {\left (\cos \relax (c)^{2} + \sin \relax (c)^{2}\right )} d - {\left (c \cos \relax (c)^{2} + c \sin \relax (c)^{2}\right )} d\right )} \cos \left (d x + c\right )^{2} + {\left ({\left (d x + c\right )} {\left (\cos \relax (c)^{2} + \sin \relax (c)^{2}\right )} d - {\left (c \cos \relax (c)^{2} + c \sin \relax (c)^{2}\right )} d\right )} \sin \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((I*exp_integral_e(1, I*d*x) - I*exp_integral_e(1, -I*d*x))*cos(c) + (exp_integral_e(1, I*d*x) + exp_inte
gral_e(1, -I*d*x))*sin(c))*a + 1/2*((I*exp_integral_e(1, I*d*x) - I*exp_integral_e(1, -I*d*x))*cos(c) + (exp_i
ntegral_e(1, I*d*x) + exp_integral_e(1, -I*d*x))*sin(c))*b*c/d - 1/4*(2*(d*x + c)*(cos(c)^2 + sin(c)^2)*cos(d*
x + c)^3 + 2*(d*x + c)*(cos(c)^2 + sin(c)^2)*cos(d*x + c) - (c*(exp_integral_e(2, I*d*x) + exp_integral_e(2, -
I*d*x))*cos(c)^3 + c*(exp_integral_e(2, I*d*x) + exp_integral_e(2, -I*d*x))*cos(c)*sin(c)^2 - c*(I*exp_integra
l_e(2, I*d*x) - I*exp_integral_e(2, -I*d*x))*sin(c)^3 + c*(exp_integral_e(2, I*d*x) + exp_integral_e(2, -I*d*x
))*cos(c) - (c*(I*exp_integral_e(2, I*d*x) - I*exp_integral_e(2, -I*d*x))*cos(c)^2 + c*(I*exp_integral_e(2, I*
d*x) - I*exp_integral_e(2, -I*d*x)))*sin(c))*cos(d*x + c)^2 - (c*(exp_integral_e(2, I*d*x) + exp_integral_e(2,
-I*d*x))*cos(c)^3 + c*(exp_integral_e(2, I*d*x) + exp_integral_e(2, -I*d*x))*cos(c)*sin(c)^2 - c*(I*exp_integ
ral_e(2, I*d*x) - I*exp_integral_e(2, -I*d*x))*sin(c)^3 - 2*(d*x + c)*(cos(c)^2 + sin(c)^2)*cos(d*x + c) + c*(
exp_integral_e(2, I*d*x) + exp_integral_e(2, -I*d*x))*cos(c) - (c*(I*exp_integral_e(2, I*d*x) - I*exp_integral
_e(2, -I*d*x))*cos(c)^2 + c*(I*exp_integral_e(2, I*d*x) - I*exp_integral_e(2, -I*d*x)))*sin(c))*sin(d*x + c)^2
)*b/(((d*x + c)*(cos(c)^2 + sin(c)^2)*d - (c*cos(c)^2 + c*sin(c)^2)*d)*cos(d*x + c)^2 + ((d*x + c)*(cos(c)^2 +
sin(c)^2)*d - (c*cos(c)^2 + c*sin(c)^2)*d)*sin(d*x + c)^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ a\,\mathrm {cosint}\left (d\,x\right )\,\sin \relax (c)+a\,\mathrm {sinint}\left (d\,x\right )\,\cos \relax (c)-\frac {b\,\cos \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x))/x,x)

[Out]

a*cosint(d*x)*sin(c) + a*sinint(d*x)*cos(c) - (b*cos(c + d*x))/d

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sympy [A]  time = 6.40, size = 37, normalized size = 1.28 \[ - a \left (- \sin {\relax (c )} \operatorname {Ci}{\left (d x \right )} - \cos {\relax (c )} \operatorname {Si}{\left (d x \right )}\right ) - b \left (\begin {cases} - x \sin {\relax (c )} & \text {for}\: d = 0 \\\frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(-sin(c)*Ci(d*x) - cos(c)*Si(d*x)) - b*Piecewise((-x*sin(c), Eq(d, 0)), (cos(c + d*x)/d, True))

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