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

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

[Out]

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

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Rubi [A]  time = 0.148446, antiderivative size = 29, 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} \[ a \sin (c) \text{CosIntegral}(d x)+a \cos (c) \text{Si}(d x)-\frac{b \cos (c+d x)}{d} \]

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 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{(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*}

Mathematica [A]  time = 0.0388962, size = 40, normalized size = 1.38 \[ a \sin (c) \text{CosIntegral}(d x)+a \cos (c) \text{Si}(d x)+\frac{b \sin (c) \sin (d x)}{d}-\frac{b \cos (c) \cos (d x)}{d} \]

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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Maple [A]  time = 0.009, size = 31, normalized size = 1.1 \begin{align*} -{\frac{b\cos \left ( dx+c \right ) }{d}}+a \left ({\it Si} \left ( dx \right ) \cos \left ( c \right ) +{\it Ci} \left ( dx \right ) \sin \left ( c \right ) \right ) \end{align*}

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 = 1.34586, size = 705, normalized size = 24.31 \begin{align*} \text{result too large to display} \end{align*}

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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Fricas [A]  time = 1.65953, size = 158, normalized size = 5.45 \begin{align*} \frac{2 \, a d \cos \left (c\right ) \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 \left (c\right )}{2 \, d} \end{align*}

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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Sympy [A]  time = 5.16126, size = 37, normalized size = 1.28 \begin{align*} - a \left (- \sin{\left (c \right )} \operatorname{Ci}{\left (d x \right )} - \cos{\left (c \right )} \operatorname{Si}{\left (d x \right )}\right ) - b \left (\begin{cases} - x \sin{\left (c \right )} & \text{for}\: d = 0 \\\frac{\cos{\left (c + d x \right )}}{d} & \text{otherwise} \end{cases}\right ) \end{align*}

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError