3.3.89 \(\int (e+f x) (a+b \sin (c+\frac {d}{x})) \, dx\) [289]

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

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 = {3512, 14, 3378, 3384, 3380, 3383} \begin {gather*} 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 ) \end {gather*}

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 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 3378

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 3380

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 3383

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 3384

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 3512

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]

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 79, normalized size = 0.67 \begin {gather*} \frac {1}{2} \left (b d f x \cos \left (c+\frac {d}{x}\right )+b d \text {Ci}\left (\frac {d}{x}\right ) (-2 e \cos (c)+d f \sin (c))+x (2 e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )+b d (d f \cos (c)+2 e \sin (c)) \text {Si}\left (\frac {d}{x}\right )\right ) \end {gather*}

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 = 0.06, size = 115, normalized size = 0.97

method	result	size
derivativedivides	\(-d \left (-\frac {a f \,x^{2}}{2 d}-\frac {a e x}{d}+b f d \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x^{2}}{2 d^{2}}-\frac {\cos \left (c +\frac {d}{x}\right ) x}{2 d}-\frac {\sinIntegral \left (\frac {d}{x}\right ) \cos \left (c \right )}{2}-\frac {\cosineIntegral \left (\frac {d}{x}\right ) \sin \left (c \right )}{2}\right )+b e \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\sinIntegral \left (\frac {d}{x}\right ) \sin \left (c \right )+\cosineIntegral \left (\frac {d}{x}\right ) \cos \left (c \right )\right )\right )\)	\(115\)
default	\(-d \left (-\frac {a f \,x^{2}}{2 d}-\frac {a e x}{d}+b f d \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x^{2}}{2 d^{2}}-\frac {\cos \left (c +\frac {d}{x}\right ) x}{2 d}-\frac {\sinIntegral \left (\frac {d}{x}\right ) \cos \left (c \right )}{2}-\frac {\cosineIntegral \left (\frac {d}{x}\right ) \sin \left (c \right )}{2}\right )+b e \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\sinIntegral \left (\frac {d}{x}\right ) \sin \left (c \right )+\cosineIntegral \left (\frac {d}{x}\right ) \cos \left (c \right )\right )\right )\)	\(115\)
risch	\(a e x +\frac {a f \,x^{2}}{2}+\frac {b d e \,{\mathrm e}^{-i c} \expIntegral \left (1, \frac {i d}{x}\right )}{2}-\frac {i b \,d^{2} f \,{\mathrm e}^{-i c} \expIntegral \left (1, \frac {i d}{x}\right )}{4}+\frac {b d e \,{\mathrm e}^{i c} \expIntegral \left (1, -\frac {i d}{x}\right )}{2}+\frac {i b \,d^{2} f \,{\mathrm e}^{i c} \expIntegral \left (1, -\frac {i d}{x}\right )}{4}+\frac {\cos \left (\frac {c x +d}{x}\right ) b d f x}{2}+\frac {b x \left (f x +2 e \right ) \sin \left (\frac {c x +d}{x}\right )}{2}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.35, size = 155, normalized size = 1.31 \begin {gather*} \frac {1}{2} \, a f x^{2} + \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 - \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 + a x e \end {gather*}

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 + 1/4*(((-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 - 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 + a*x*e

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Fricas [A]
time = 0.37, size = 138, normalized size = 1.17 \begin {gather*} \frac {1}{2} \, b d f x \cos \left (\frac {c x + d}{x}\right ) + \frac {1}{2} \, a f x^{2} + a x e + \frac {1}{2} \, {\left (b d^{2} f \operatorname {Si}\left (\frac {d}{x}\right ) - b d \operatorname {Ci}\left (\frac {d}{x}\right ) e - b d \operatorname {Ci}\left (-\frac {d}{x}\right ) e\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 x e\right )} \sin \left (\frac {c x + d}{x}\right ) \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + \frac {d}{x} \right )}\right ) \left (e + f x\right )\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (112) = 224\).
time = 5.75, size = 530, normalized size = 4.49 \begin {gather*} \frac {b c^{2} d^{3} f \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right ) \sin \left (c\right ) - b c^{2} d^{3} f \cos \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right ) - 2 \, b c^{2} d^{2} \cos \left (c\right ) \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right ) e - \frac {2 \, {\left (c x + d\right )} b c d^{3} f \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right ) \sin \left (c\right )}{x} + \frac {2 \, {\left (c x + d\right )} b c d^{3} f \cos \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right )}{x} - 2 \, b c^{2} d^{2} e \sin \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right ) - b c d^{3} f \cos \left (\frac {c x + d}{x}\right ) + \frac {4 \, {\left (c x + d\right )} b c d^{2} \cos \left (c\right ) \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right ) e}{x} + \frac {{\left (c x + d\right )}^{2} b d^{3} f \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right ) \sin \left (c\right )}{x^{2}} - \frac {{\left (c x + d\right )}^{2} b d^{3} f \cos \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right )}{x^{2}} + \frac {4 \, {\left (c x + d\right )} b c d^{2} e \sin \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right )}{x} + \frac {{\left (c x + d\right )} b d^{3} f \cos \left (\frac {c x + d}{x}\right )}{x} - \frac {2 \, {\left (c x + d\right )}^{2} b d^{2} \cos \left (c\right ) \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right ) e}{x^{2}} + b d^{3} f \sin \left (\frac {c x + d}{x}\right ) - 2 \, b c d^{2} e \sin \left (\frac {c x + d}{x}\right ) - \frac {2 \, {\left (c x + d\right )}^{2} b d^{2} e \sin \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right )}{x^{2}} + a d^{3} f - 2 \, a c d^{2} e + \frac {2 \, {\left (c x + d\right )} b d^{2} e \sin \left (\frac {c x + d}{x}\right )}{x} + \frac {2 \, {\left (c x + d\right )} a d^{2} e}{x}}{2 \, {\left (c^{2} - \frac {2 \, {\left (c x + d\right )} c}{x} + \frac {{\left (c x + d\right )}^{2}}{x^{2}}\right )} d} \end {gather*}

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]

1/2*(b*c^2*d^3*f*cos_integral(-c + (c*x + d)/x)*sin(c) - b*c^2*d^3*f*cos(c)*sin_integral(c - (c*x + d)/x) - 2*
b*c^2*d^2*cos(c)*cos_integral(-c + (c*x + d)/x)*e - 2*(c*x + d)*b*c*d^3*f*cos_integral(-c + (c*x + d)/x)*sin(c
)/x + 2*(c*x + d)*b*c*d^3*f*cos(c)*sin_integral(c - (c*x + d)/x)/x - 2*b*c^2*d^2*e*sin(c)*sin_integral(c - (c*
x + d)/x) - b*c*d^3*f*cos((c*x + d)/x) + 4*(c*x + d)*b*c*d^2*cos(c)*cos_integral(-c + (c*x + d)/x)*e/x + (c*x
+ d)^2*b*d^3*f*cos_integral(-c + (c*x + d)/x)*sin(c)/x^2 - (c*x + d)^2*b*d^3*f*cos(c)*sin_integral(c - (c*x +
d)/x)/x^2 + 4*(c*x + d)*b*c*d^2*e*sin(c)*sin_integral(c - (c*x + d)/x)/x + (c*x + d)*b*d^3*f*cos((c*x + d)/x)/
x - 2*(c*x + d)^2*b*d^2*cos(c)*cos_integral(-c + (c*x + d)/x)*e/x^2 + b*d^3*f*sin((c*x + d)/x) - 2*b*c*d^2*e*s
in((c*x + d)/x) - 2*(c*x + d)^2*b*d^2*e*sin(c)*sin_integral(c - (c*x + d)/x)/x^2 + a*d^3*f - 2*a*c*d^2*e + 2*(
c*x + d)*b*d^2*e*sin((c*x + d)/x)/x + 2*(c*x + d)*a*d^2*e/x)/((c^2 - 2*(c*x + d)*c/x + (c*x + d)^2/x^2)*d)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (e+f\,x\right )\,\left (a+b\,\sin \left (c+\frac {d}{x}\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)*(a + b*sin(c + d/x)),x)

[Out]

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

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