3.288 \(\int (e+f x)^2 (a+b \sin (c+\frac {d}{x})) \, dx\)

Optimal. Leaf size=224 \[ a e^2 x+a e f x^2+\frac {1}{3} a f^2 x^3+\frac {1}{6} b d^3 f^2 \cos (c) \text {Ci}\left (\frac {d}{x}\right )+b d^2 e f \sin (c) \text {Ci}\left (\frac {d}{x}\right )-b d e^2 \cos (c) \text {Ci}\left (\frac {d}{x}\right )-\frac {1}{6} b d^3 f^2 \sin (c) \text {Si}\left (\frac {d}{x}\right )+b d^2 e f \cos (c) \text {Si}\left (\frac {d}{x}\right )-\frac {1}{6} b d^2 f^2 x \sin \left (c+\frac {d}{x}\right )+b d e^2 \sin (c) \text {Si}\left (\frac {d}{x}\right )+b e^2 x \sin \left (c+\frac {d}{x}\right )+b e f x^2 \sin \left (c+\frac {d}{x}\right )+b d e f x \cos \left (c+\frac {d}{x}\right )+\frac {1}{3} b f^2 x^3 \sin \left (c+\frac {d}{x}\right )+\frac {1}{6} b d f^2 x^2 \cos \left (c+\frac {d}{x}\right ) \]

[Out]

a*e^2*x+a*e*f*x^2+1/3*a*f^2*x^3-b*d*e^2*Ci(d/x)*cos(c)+1/6*b*d^3*f^2*Ci(d/x)*cos(c)+b*d*e*f*x*cos(c+d/x)+1/6*b
*d*f^2*x^2*cos(c+d/x)+b*d^2*e*f*cos(c)*Si(d/x)+b*d^2*e*f*Ci(d/x)*sin(c)+b*d*e^2*Si(d/x)*sin(c)-1/6*b*d^3*f^2*S
i(d/x)*sin(c)+b*e^2*x*sin(c+d/x)-1/6*b*d^2*f^2*x*sin(c+d/x)+b*e*f*x^2*sin(c+d/x)+1/3*b*f^2*x^3*sin(c+d/x)

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Rubi [A]  time = 0.46, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3431, 14, 3297, 3303, 3299, 3302} \[ a e^2 x+a e f x^2+\frac {1}{3} a f^2 x^3+b d^2 e f \sin (c) \text {CosIntegral}\left (\frac {d}{x}\right )+\frac {1}{6} b d^3 f^2 \cos (c) \text {CosIntegral}\left (\frac {d}{x}\right )-b d e^2 \cos (c) \text {CosIntegral}\left (\frac {d}{x}\right )+b d^2 e f \cos (c) \text {Si}\left (\frac {d}{x}\right )-\frac {1}{6} b d^3 f^2 \sin (c) \text {Si}\left (\frac {d}{x}\right )-\frac {1}{6} b d^2 f^2 x \sin \left (c+\frac {d}{x}\right )+b d e^2 \sin (c) \text {Si}\left (\frac {d}{x}\right )+b e^2 x \sin \left (c+\frac {d}{x}\right )+b e f x^2 \sin \left (c+\frac {d}{x}\right )+b d e f x \cos \left (c+\frac {d}{x}\right )+\frac {1}{3} b f^2 x^3 \sin \left (c+\frac {d}{x}\right )+\frac {1}{6} b d f^2 x^2 \cos \left (c+\frac {d}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^2*(a + b*Sin[c + d/x]),x]

[Out]

a*e^2*x + a*e*f*x^2 + (a*f^2*x^3)/3 + b*d*e*f*x*Cos[c + d/x] + (b*d*f^2*x^2*Cos[c + d/x])/6 - b*d*e^2*Cos[c]*C
osIntegral[d/x] + (b*d^3*f^2*Cos[c]*CosIntegral[d/x])/6 + b*d^2*e*f*CosIntegral[d/x]*Sin[c] + b*e^2*x*Sin[c +
d/x] - (b*d^2*f^2*x*Sin[c + d/x])/6 + b*e*f*x^2*Sin[c + d/x] + (b*f^2*x^3*Sin[c + d/x])/3 + b*d^2*e*f*Cos[c]*S
inIntegral[d/x] + b*d*e^2*Sin[c]*SinIntegral[d/x] - (b*d^3*f^2*Sin[c]*SinIntegral[d/x])/6

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 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 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]

Rubi steps

\begin {align*} \int (e+f x)^2 \left (a+b \sin \left (c+\frac {d}{x}\right )\right ) \, dx &=-\operatorname {Subst}\left (\int \left (\frac {f^2 (a+b \sin (c+d x))}{x^4}+\frac {2 e f (a+b \sin (c+d x))}{x^3}+\frac {e^2 (a+b \sin (c+d x))}{x^2}\right ) \, dx,x,\frac {1}{x}\right )\\ &=-\left (e^2 \operatorname {Subst}\left (\int \frac {a+b \sin (c+d x)}{x^2} \, dx,x,\frac {1}{x}\right )\right )-(2 e f) \operatorname {Subst}\left (\int \frac {a+b \sin (c+d x)}{x^3} \, dx,x,\frac {1}{x}\right )-f^2 \operatorname {Subst}\left (\int \frac {a+b \sin (c+d x)}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=-\left (e^2 \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 )-(2 e 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 )-f^2 \operatorname {Subst}\left (\int \left (\frac {a}{x^4}+\frac {b \sin (c+d x)}{x^4}\right ) \, dx,x,\frac {1}{x}\right )\\ &=a e^2 x+a e f x^2+\frac {1}{3} a f^2 x^3-\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^2} \, dx,x,\frac {1}{x}\right )-(2 b e f) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^3} \, dx,x,\frac {1}{x}\right )-\left (b f^2\right ) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=a e^2 x+a e f x^2+\frac {1}{3} a f^2 x^3+b e^2 x \sin \left (c+\frac {d}{x}\right )+b e f x^2 \sin \left (c+\frac {d}{x}\right )+\frac {1}{3} b f^2 x^3 \sin \left (c+\frac {d}{x}\right )-\left (b d e^2\right ) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x} \, dx,x,\frac {1}{x}\right )-(b d e f) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x^2} \, dx,x,\frac {1}{x}\right )-\frac {1}{3} \left (b d f^2\right ) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=a e^2 x+a e f x^2+\frac {1}{3} a f^2 x^3+b d e f x \cos \left (c+\frac {d}{x}\right )+\frac {1}{6} b d f^2 x^2 \cos \left (c+\frac {d}{x}\right )+b e^2 x \sin \left (c+\frac {d}{x}\right )+b e f x^2 \sin \left (c+\frac {d}{x}\right )+\frac {1}{3} b f^2 x^3 \sin \left (c+\frac {d}{x}\right )+\left (b d^2 e f\right ) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x} \, dx,x,\frac {1}{x}\right )+\frac {1}{6} \left (b d^2 f^2\right ) \operatorname {Subst}\left (\int \frac {\sin (c+d x)}{x^2} \, dx,x,\frac {1}{x}\right )-\left (b d e^2 \cos (c)\right ) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,\frac {1}{x}\right )+\left (b d e^2 \sin (c)\right ) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=a e^2 x+a e f x^2+\frac {1}{3} a f^2 x^3+b d e f x \cos \left (c+\frac {d}{x}\right )+\frac {1}{6} b d f^2 x^2 \cos \left (c+\frac {d}{x}\right )-b d e^2 \cos (c) \text {Ci}\left (\frac {d}{x}\right )+b e^2 x \sin \left (c+\frac {d}{x}\right )-\frac {1}{6} b d^2 f^2 x \sin \left (c+\frac {d}{x}\right )+b e f x^2 \sin \left (c+\frac {d}{x}\right )+\frac {1}{3} b f^2 x^3 \sin \left (c+\frac {d}{x}\right )+b d e^2 \sin (c) \text {Si}\left (\frac {d}{x}\right )+\frac {1}{6} \left (b d^3 f^2\right ) \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{x} \, dx,x,\frac {1}{x}\right )+\left (b d^2 e f \cos (c)\right ) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,\frac {1}{x}\right )+\left (b d^2 e f \sin (c)\right ) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=a e^2 x+a e f x^2+\frac {1}{3} a f^2 x^3+b d e f x \cos \left (c+\frac {d}{x}\right )+\frac {1}{6} b d f^2 x^2 \cos \left (c+\frac {d}{x}\right )-b d e^2 \cos (c) \text {Ci}\left (\frac {d}{x}\right )+b d^2 e f \text {Ci}\left (\frac {d}{x}\right ) \sin (c)+b e^2 x \sin \left (c+\frac {d}{x}\right )-\frac {1}{6} b d^2 f^2 x \sin \left (c+\frac {d}{x}\right )+b e f x^2 \sin \left (c+\frac {d}{x}\right )+\frac {1}{3} b f^2 x^3 \sin \left (c+\frac {d}{x}\right )+b d^2 e f \cos (c) \text {Si}\left (\frac {d}{x}\right )+b d e^2 \sin (c) \text {Si}\left (\frac {d}{x}\right )+\frac {1}{6} \left (b d^3 f^2 \cos (c)\right ) \operatorname {Subst}\left (\int \frac {\cos (d x)}{x} \, dx,x,\frac {1}{x}\right )-\frac {1}{6} \left (b d^3 f^2 \sin (c)\right ) \operatorname {Subst}\left (\int \frac {\sin (d x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=a e^2 x+a e f x^2+\frac {1}{3} a f^2 x^3+b d e f x \cos \left (c+\frac {d}{x}\right )+\frac {1}{6} b d f^2 x^2 \cos \left (c+\frac {d}{x}\right )-b d e^2 \cos (c) \text {Ci}\left (\frac {d}{x}\right )+\frac {1}{6} b d^3 f^2 \cos (c) \text {Ci}\left (\frac {d}{x}\right )+b d^2 e f \text {Ci}\left (\frac {d}{x}\right ) \sin (c)+b e^2 x \sin \left (c+\frac {d}{x}\right )-\frac {1}{6} b d^2 f^2 x \sin \left (c+\frac {d}{x}\right )+b e f x^2 \sin \left (c+\frac {d}{x}\right )+\frac {1}{3} b f^2 x^3 \sin \left (c+\frac {d}{x}\right )+b d^2 e f \cos (c) \text {Si}\left (\frac {d}{x}\right )+b d e^2 \sin (c) \text {Si}\left (\frac {d}{x}\right )-\frac {1}{6} b d^3 f^2 \sin (c) \text {Si}\left (\frac {d}{x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 150, normalized size = 0.67 \[ \frac {1}{6} \left (x \left (2 a \left (3 e^2+3 e f x+f^2 x^2\right )+b \sin \left (c+\frac {d}{x}\right ) \left (-f^2 \left (d^2-2 x^2\right )+6 e^2+6 e f x\right )+b d f (6 e+f x) \cos \left (c+\frac {d}{x}\right )\right )+b d \text {Ci}\left (\frac {d}{x}\right ) \left (\cos (c) \left (d^2 f^2-6 e^2\right )+6 d e f \sin (c)\right )-b d \text {Si}\left (\frac {d}{x}\right ) \left (\sin (c) \left (d^2 f^2-6 e^2\right )-6 d e f \cos (c)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^2*(a + b*Sin[c + d/x]),x]

[Out]

(b*d*CosIntegral[d/x]*((-6*e^2 + d^2*f^2)*Cos[c] + 6*d*e*f*Sin[c]) + x*(2*a*(3*e^2 + 3*e*f*x + f^2*x^2) + b*d*
f*(6*e + f*x)*Cos[c + d/x] + b*(6*e^2 + 6*e*f*x - f^2*(d^2 - 2*x^2))*Sin[c + d/x]) - b*d*(-6*d*e*f*Cos[c] + (-
6*e^2 + d^2*f^2)*Sin[c])*SinIntegral[d/x])/6

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fricas [A]  time = 0.73, size = 224, normalized size = 1.00 \[ \frac {1}{3} \, a f^{2} x^{3} + a e f x^{2} + a e^{2} x + \frac {1}{12} \, {\left (12 \, b d^{2} e f \operatorname {Si}\left (\frac {d}{x}\right ) + {\left (b d^{3} f^{2} - 6 \, b d e^{2}\right )} \operatorname {Ci}\left (\frac {d}{x}\right ) + {\left (b d^{3} f^{2} - 6 \, b d e^{2}\right )} \operatorname {Ci}\left (-\frac {d}{x}\right )\right )} \cos \relax (c) + \frac {1}{6} \, {\left (b d f^{2} x^{2} + 6 \, b d e f x\right )} \cos \left (\frac {c x + d}{x}\right ) + \frac {1}{6} \, {\left (3 \, b d^{2} e f \operatorname {Ci}\left (\frac {d}{x}\right ) + 3 \, b d^{2} e f \operatorname {Ci}\left (-\frac {d}{x}\right ) - {\left (b d^{3} f^{2} - 6 \, b d e^{2}\right )} \operatorname {Si}\left (\frac {d}{x}\right )\right )} \sin \relax (c) + \frac {1}{6} \, {\left (2 \, b f^{2} x^{3} + 6 \, b e f x^{2} - {\left (b d^{2} f^{2} - 6 \, b e^{2}\right )} x\right )} \sin \left (\frac {c x + d}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*(a+b*sin(c+d/x)),x, algorithm="fricas")

[Out]

1/3*a*f^2*x^3 + a*e*f*x^2 + a*e^2*x + 1/12*(12*b*d^2*e*f*sin_integral(d/x) + (b*d^3*f^2 - 6*b*d*e^2)*cos_integ
ral(d/x) + (b*d^3*f^2 - 6*b*d*e^2)*cos_integral(-d/x))*cos(c) + 1/6*(b*d*f^2*x^2 + 6*b*d*e*f*x)*cos((c*x + d)/
x) + 1/6*(3*b*d^2*e*f*cos_integral(d/x) + 3*b*d^2*e*f*cos_integral(-d/x) - (b*d^3*f^2 - 6*b*d*e^2)*sin_integra
l(d/x))*sin(c) + 1/6*(2*b*f^2*x^3 + 6*b*e*f*x^2 - (b*d^2*f^2 - 6*b*e^2)*x)*sin((c*x + d)/x)

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giac [B]  time = 1.55, size = 1264, normalized size = 5.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*(a+b*sin(c+d/x)),x, algorithm="giac")

[Out]

1/6*(b*c^3*d^4*f^2*cos(c)*cos_integral(-c + (c*x + d)/x) + b*c^3*d^4*f^2*sin(c)*sin_integral(c - (c*x + d)/x)
- 3*(c*x + d)*b*c^2*d^4*f^2*cos(c)*cos_integral(-c + (c*x + d)/x)/x + 6*b*c^3*d^3*f*cos_integral(-c + (c*x + d
)/x)*e*sin(c) - 6*b*c^3*d^3*f*cos(c)*e*sin_integral(c - (c*x + d)/x) - 3*(c*x + d)*b*c^2*d^4*f^2*sin(c)*sin_in
tegral(c - (c*x + d)/x)/x + 3*(c*x + d)^2*b*c*d^4*f^2*cos(c)*cos_integral(-c + (c*x + d)/x)/x^2 - 18*(c*x + d)
*b*c^2*d^3*f*cos_integral(-c + (c*x + d)/x)*e*sin(c)/x + b*c^2*d^4*f^2*sin((c*x + d)/x) + 18*(c*x + d)*b*c^2*d
^3*f*cos(c)*e*sin_integral(c - (c*x + d)/x)/x + 3*(c*x + d)^2*b*c*d^4*f^2*sin(c)*sin_integral(c - (c*x + d)/x)
/x^2 + b*c*d^4*f^2*cos((c*x + d)/x) - (c*x + d)^3*b*d^4*f^2*cos(c)*cos_integral(-c + (c*x + d)/x)/x^3 - 6*b*c^
3*d^2*cos(c)*cos_integral(-c + (c*x + d)/x)*e^2 - 6*b*c^2*d^3*f*cos((c*x + d)/x)*e + 18*(c*x + d)^2*b*c*d^3*f*
cos_integral(-c + (c*x + d)/x)*e*sin(c)/x^2 - 2*(c*x + d)*b*c*d^4*f^2*sin((c*x + d)/x)/x - 18*(c*x + d)^2*b*c*
d^3*f*cos(c)*e*sin_integral(c - (c*x + d)/x)/x^2 - (c*x + d)^3*b*d^4*f^2*sin(c)*sin_integral(c - (c*x + d)/x)/
x^3 - 6*b*c^3*d^2*e^2*sin(c)*sin_integral(c - (c*x + d)/x) - (c*x + d)*b*d^4*f^2*cos((c*x + d)/x)/x + 18*(c*x
+ d)*b*c^2*d^2*cos(c)*cos_integral(-c + (c*x + d)/x)*e^2/x + 12*(c*x + d)*b*c*d^3*f*cos((c*x + d)/x)*e/x - 6*(
c*x + d)^3*b*d^3*f*cos_integral(-c + (c*x + d)/x)*e*sin(c)/x^3 - 2*b*d^4*f^2*sin((c*x + d)/x) + (c*x + d)^2*b*
d^4*f^2*sin((c*x + d)/x)/x^2 + 6*b*c*d^3*f*e*sin((c*x + d)/x) + 6*(c*x + d)^3*b*d^3*f*cos(c)*e*sin_integral(c
- (c*x + d)/x)/x^3 + 18*(c*x + d)*b*c^2*d^2*e^2*sin(c)*sin_integral(c - (c*x + d)/x)/x - 2*a*d^4*f^2 - 18*(c*x
+ d)^2*b*c*d^2*cos(c)*cos_integral(-c + (c*x + d)/x)*e^2/x^2 + 6*a*c*d^3*f*e - 6*(c*x + d)^2*b*d^3*f*cos((c*x
+ d)/x)*e/x^2 - 6*b*c^2*d^2*e^2*sin((c*x + d)/x) - 6*(c*x + d)*b*d^3*f*e*sin((c*x + d)/x)/x - 18*(c*x + d)^2*
b*c*d^2*e^2*sin(c)*sin_integral(c - (c*x + d)/x)/x^2 - 6*a*c^2*d^2*e^2 + 6*(c*x + d)^3*b*d^2*cos(c)*cos_integr
al(-c + (c*x + d)/x)*e^2/x^3 - 6*(c*x + d)*a*d^3*f*e/x + 12*(c*x + d)*b*c*d^2*e^2*sin((c*x + d)/x)/x + 6*(c*x
+ d)^3*b*d^2*e^2*sin(c)*sin_integral(c - (c*x + d)/x)/x^3 + 12*(c*x + d)*a*c*d^2*e^2/x - 6*(c*x + d)^2*b*d^2*e
^2*sin((c*x + d)/x)/x^2 - 6*(c*x + d)^2*a*d^2*e^2/x^2)/((c^3 - 3*(c*x + d)*c^2/x + 3*(c*x + d)^2*c/x^2 - (c*x
+ d)^3/x^3)*d)

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maple [A]  time = 0.07, size = 209, normalized size = 0.93 \[ -d \left (-\frac {a \,e^{2} x}{d}-\frac {a e f \,x^{2}}{d}-\frac {a \,f^{2} x^{3}}{3 d}+b \,e^{2} \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\Si \left (\frac {d}{x}\right ) \sin \relax (c )+\Ci \left (\frac {d}{x}\right ) \cos \relax (c )\right )+2 b d e f \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x^{2}}{2 d^{2}}-\frac {\cos \left (c +\frac {d}{x}\right ) x}{2 d}-\frac {\Si \left (\frac {d}{x}\right ) \cos \relax (c )}{2}-\frac {\Ci \left (\frac {d}{x}\right ) \sin \relax (c )}{2}\right )+b \,d^{2} f^{2} \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x^{3}}{3 d^{3}}-\frac {\cos \left (c +\frac {d}{x}\right ) x^{2}}{6 d^{2}}+\frac {\sin \left (c +\frac {d}{x}\right ) x}{6 d}+\frac {\Si \left (\frac {d}{x}\right ) \sin \relax (c )}{6}-\frac {\Ci \left (\frac {d}{x}\right ) \cos \relax (c )}{6}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*(a+b*sin(c+d/x)),x)

[Out]

-d*(-a*e^2*x/d-a/d*e*f*x^2-1/3*a/d*f^2*x^3+b*e^2*(-sin(c+d/x)*x/d-Si(d/x)*sin(c)+Ci(d/x)*cos(c))+2*b*d*e*f*(-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))+b*d^2*f^2*(-1/3*sin(c+d/x)*x^3
/d^3-1/6*cos(c+d/x)*x^2/d^2+1/6*sin(c+d/x)*x/d+1/6*Si(d/x)*sin(c)-1/6*Ci(d/x)*cos(c)))

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maxima [C]  time = 0.55, size = 258, normalized size = 1.15 \[ \frac {1}{3} \, a f^{2} x^{3} + a e f x^{2} - \frac {1}{2} \, {\left ({\left ({\left ({\rm Ei}\left (\frac {i \, d}{x}\right ) + {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \cos \relax (c) - {\left (-i \, {\rm Ei}\left (\frac {i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \sin \relax (c)\right )} d - 2 \, x \sin \left (\frac {c x + d}{x}\right )\right )} b e^{2} + \frac {1}{2} \, {\left ({\left ({\left (-i \, {\rm Ei}\left (\frac {i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \cos \relax (c) + {\left ({\rm Ei}\left (\frac {i \, d}{x}\right ) + {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \sin \relax (c)\right )} d^{2} + 2 \, d x \cos \left (\frac {c x + d}{x}\right ) + 2 \, x^{2} \sin \left (\frac {c x + d}{x}\right )\right )} b e f + \frac {1}{12} \, {\left ({\left ({\left ({\rm Ei}\left (\frac {i \, d}{x}\right ) + {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \cos \relax (c) + {\left (i \, {\rm Ei}\left (\frac {i \, d}{x}\right ) - i \, {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \sin \relax (c)\right )} d^{3} + 2 \, d x^{2} \cos \left (\frac {c x + d}{x}\right ) - 2 \, {\left (d^{2} x - 2 \, x^{3}\right )} \sin \left (\frac {c x + d}{x}\right )\right )} b f^{2} + a e^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*(a+b*sin(c+d/x)),x, algorithm="maxima")

[Out]

1/3*a*f^2*x^3 + a*e*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^2 + 1/2*(((-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*e*f + 1/12*(((Ei(I*d/x) + Ei(-I*d/x))*cos(c) + (I*Ei(I*
d/x) - I*Ei(-I*d/x))*sin(c))*d^3 + 2*d*x^2*cos((c*x + d)/x) - 2*(d^2*x - 2*x^3)*sin((c*x + d)/x))*b*f^2 + a*e^
2*x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e+f\,x\right )}^2\,\left (a+b\,\sin \left (c+\frac {d}{x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^2*(a + b*sin(c + d/x)),x)

[Out]

int((e + f*x)^2*(a + b*sin(c + d/x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + \frac {d}{x} \right )}\right ) \left (e + f x\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*(a+b*sin(c+d/x)),x)

[Out]

Integral((a + b*sin(c + d/x))*(e + f*x)**2, x)

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