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

Optimal. Leaf size=118 \[ a e x+\frac{1}{2} a f x^2+\frac{1}{2} b d^2 f \sin (c) \text{CosIntegral}\left (\frac{d}{x}\right )-b d e \cos (c) \text{CosIntegral}\left (\frac{d}{x}\right )+\frac{1}{2} b d^2 f \cos (c) \text{Si}\left (\frac{d}{x}\right )+b d e \sin (c) \text{Si}\left (\frac{d}{x}\right )+b e x \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b f x^2 \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b d f x \cos \left (c+\frac{d}{x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.239125, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3431, 14, 3297, 3303, 3299, 3302} \[ a e x+\frac{1}{2} a f x^2+\frac{1}{2} b d^2 f \sin (c) \text{CosIntegral}\left (\frac{d}{x}\right )-b d e \cos (c) \text{CosIntegral}\left (\frac{d}{x}\right )+\frac{1}{2} b d^2 f \cos (c) \text{Si}\left (\frac{d}{x}\right )+b d e \sin (c) \text{Si}\left (\frac{d}{x}\right )+b e x \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b f x^2 \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b d f x \cos \left (c+\frac{d}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)*(a + b*Sin[c + d/x]),x]

[Out]

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

Rule 3431

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x])^p, x^(1/n - 1)*(g - (e*h)/f + (h*x^(1/n))/f)^m,
x], x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && IntegerQ[1/n]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 3297

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] - Dist[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 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 (e+f x) \left (a+b \sin \left (c+\frac{d}{x}\right )\right ) \, dx &=-\operatorname{Subst}\left (\int \left (\frac{f (a+b \sin (c+d x))}{x^3}+\frac{e (a+b \sin (c+d x))}{x^2}\right ) \, dx,x,\frac{1}{x}\right )\\ &=-\left (e \operatorname{Subst}\left (\int \frac{a+b \sin (c+d x)}{x^2} \, dx,x,\frac{1}{x}\right )\right )-f \operatorname{Subst}\left (\int \frac{a+b \sin (c+d x)}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=-\left (e \operatorname{Subst}\left (\int \left (\frac{a}{x^2}+\frac{b \sin (c+d x)}{x^2}\right ) \, dx,x,\frac{1}{x}\right )\right )-f \operatorname{Subst}\left (\int \left (\frac{a}{x^3}+\frac{b \sin (c+d x)}{x^3}\right ) \, dx,x,\frac{1}{x}\right )\\ &=a e x+\frac{1}{2} a f x^2-(b e) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{x^2} \, dx,x,\frac{1}{x}\right )-(b f) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=a e x+\frac{1}{2} a f x^2+b e x \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b f x^2 \sin \left (c+\frac{d}{x}\right )-(b d e) \operatorname{Subst}\left (\int \frac{\cos (c+d x)}{x} \, dx,x,\frac{1}{x}\right )-\frac{1}{2} (b d f) \operatorname{Subst}\left (\int \frac{\cos (c+d x)}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=a e x+\frac{1}{2} a f x^2+\frac{1}{2} b d f x \cos \left (c+\frac{d}{x}\right )+b e x \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b f x^2 \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} \left (b d^2 f\right ) \operatorname{Subst}\left (\int \frac{\sin (c+d x)}{x} \, dx,x,\frac{1}{x}\right )-(b d e \cos (c)) \operatorname{Subst}\left (\int \frac{\cos (d x)}{x} \, dx,x,\frac{1}{x}\right )+(b d e \sin (c)) \operatorname{Subst}\left (\int \frac{\sin (d x)}{x} \, dx,x,\frac{1}{x}\right )\\ &=a e x+\frac{1}{2} a f x^2+\frac{1}{2} b d f x \cos \left (c+\frac{d}{x}\right )-b d e \cos (c) \text{Ci}\left (\frac{d}{x}\right )+b e x \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b f x^2 \sin \left (c+\frac{d}{x}\right )+b d e \sin (c) \text{Si}\left (\frac{d}{x}\right )+\frac{1}{2} \left (b d^2 f \cos (c)\right ) \operatorname{Subst}\left (\int \frac{\sin (d x)}{x} \, dx,x,\frac{1}{x}\right )+\frac{1}{2} \left (b d^2 f \sin (c)\right ) \operatorname{Subst}\left (\int \frac{\cos (d x)}{x} \, dx,x,\frac{1}{x}\right )\\ &=a e x+\frac{1}{2} a f x^2+\frac{1}{2} b d f x \cos \left (c+\frac{d}{x}\right )-b d e \cos (c) \text{Ci}\left (\frac{d}{x}\right )+\frac{1}{2} b d^2 f \text{Ci}\left (\frac{d}{x}\right ) \sin (c)+b e x \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b f x^2 \sin \left (c+\frac{d}{x}\right )+\frac{1}{2} b d^2 f \cos (c) \text{Si}\left (\frac{d}{x}\right )+b d e \sin (c) \text{Si}\left (\frac{d}{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.213352, size = 79, normalized size = 0.67 \[ \frac{1}{2} \left (x (2 e+f x) \left (a+b \sin \left (c+\frac{d}{x}\right )\right )+b d \text{CosIntegral}\left (\frac{d}{x}\right ) (d f \sin (c)-2 e \cos (c))+b d \text{Si}\left (\frac{d}{x}\right ) (d f \cos (c)+2 e \sin (c))+b d f x \cos \left (c+\frac{d}{x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 115, normalized size = 1. \begin{align*} -d \left ( -{\frac{aex}{d}}-{\frac{af{x}^{2}}{2\,d}}+be \left ( -{\frac{x}{d}\sin \left ( c+{\frac{d}{x}} \right ) }-{\it Si} \left ({\frac{d}{x}} \right ) \sin \left ( c \right ) +{\it Ci} \left ({\frac{d}{x}} \right ) \cos \left ( c \right ) \right ) +bfd \left ( -{\frac{{x}^{2}}{2\,{d}^{2}}\sin \left ( c+{\frac{d}{x}} \right ) }-{\frac{x}{2\,d}\cos \left ( c+{\frac{d}{x}} \right ) }-{\frac{\cos \left ( c \right ) }{2}{\it Si} \left ({\frac{d}{x}} \right ) }-{\frac{\sin \left ( c \right ) }{2}{\it Ci} \left ({\frac{d}{x}} \right ) } \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.22079, size = 207, normalized size = 1.75 \begin{align*} \frac{1}{2} \, a f x^{2} - \frac{1}{2} \,{\left ({\left ({\left ({\rm Ei}\left (\frac{i \, d}{x}\right ) +{\rm Ei}\left (-\frac{i \, d}{x}\right )\right )} \cos \left (c\right ) -{\left (-i \,{\rm Ei}\left (\frac{i \, d}{x}\right ) + i \,{\rm Ei}\left (-\frac{i \, d}{x}\right )\right )} \sin \left (c\right )\right )} d - 2 \, x \sin \left (\frac{c x + d}{x}\right )\right )} b e + \frac{1}{4} \,{\left ({\left ({\left (-i \,{\rm Ei}\left (\frac{i \, d}{x}\right ) + i \,{\rm Ei}\left (-\frac{i \, d}{x}\right )\right )} \cos \left (c\right ) +{\left ({\rm Ei}\left (\frac{i \, d}{x}\right ) +{\rm Ei}\left (-\frac{i \, d}{x}\right )\right )} \sin \left (c\right )\right )} d^{2} + 2 \, d x \cos \left (\frac{c x + d}{x}\right ) + 2 \, x^{2} \sin \left (\frac{c x + d}{x}\right )\right )} b f + a e x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*f*x^2 - 1/2*(((Ei(I*d/x) + Ei(-I*d/x))*cos(c) - (-I*Ei(I*d/x) + I*Ei(-I*d/x))*sin(c))*d - 2*x*sin((c*x +
d)/x))*b*e + 1/4*(((-I*Ei(I*d/x) + I*Ei(-I*d/x))*cos(c) + (Ei(I*d/x) + Ei(-I*d/x))*sin(c))*d^2 + 2*d*x*cos((c
*x + d)/x) + 2*x^2*sin((c*x + d)/x))*b*f + a*e*x

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Fricas [A]  time = 1.32544, size = 387, normalized size = 3.28 \begin{align*} \frac{1}{2} \, b d f x \cos \left (\frac{c x + d}{x}\right ) + \frac{1}{2} \, a f x^{2} + a e x + \frac{1}{2} \,{\left (b d^{2} f \operatorname{Si}\left (\frac{d}{x}\right ) - b d e \operatorname{Ci}\left (\frac{d}{x}\right ) - b d e \operatorname{Ci}\left (-\frac{d}{x}\right )\right )} \cos \left (c\right ) + \frac{1}{4} \,{\left (b d^{2} f \operatorname{Ci}\left (\frac{d}{x}\right ) + b d^{2} f \operatorname{Ci}\left (-\frac{d}{x}\right ) + 4 \, b d e \operatorname{Si}\left (\frac{d}{x}\right )\right )} \sin \left (c\right ) + \frac{1}{2} \,{\left (b f x^{2} + 2 \, b e x\right )} \sin \left (\frac{c x + d}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*d*f*x*cos((c*x + d)/x) + 1/2*a*f*x^2 + a*e*x + 1/2*(b*d^2*f*sin_integral(d/x) - b*d*e*cos_integral(d/x)
- b*d*e*cos_integral(-d/x))*cos(c) + 1/4*(b*d^2*f*cos_integral(d/x) + b*d^2*f*cos_integral(-d/x) + 4*b*d*e*sin
_integral(d/x))*sin(c) + 1/2*(b*f*x^2 + 2*b*e*x)*sin((c*x + d)/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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