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

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

[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

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Rubi [A]  time = 0.457095, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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 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)^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*}

Mathematica [A]  time = 0.597292, size = 150, normalized size = 0.67 \[ \frac{1}{6} \left (x \left (2 a \left (3 e^2+3 e f x+f^2 x^2\right )+b \sin \left (c+\frac{d}{x}\right ) \left (-f^2 \left (d^2-2 x^2\right )+6 e^2+6 e f x\right )+b d f (6 e+f x) \cos \left (c+\frac{d}{x}\right )\right )+b d \text{CosIntegral}\left (\frac{d}{x}\right ) \left (\cos (c) \left (d^2 f^2-6 e^2\right )+6 d e f \sin (c)\right )-b d \text{Si}\left (\frac{d}{x}\right ) \left (\sin (c) \left (d^2 f^2-6 e^2\right )-6 d e f \cos (c)\right )\right ) \]

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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Maple [A]  time = 0.037, size = 209, normalized size = 0.9 \begin{align*} -d \left ( -{\frac{a{e}^{2}x}{d}}-{\frac{aef{x}^{2}}{d}}-{\frac{a{f}^{2}{x}^{3}}{3\,d}}+b{e}^{2} \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 ) +2\,befd \left ( -1/2\,{\frac{{x}^{2}}{{d}^{2}}\sin \left ( c+{\frac{d}{x}} \right ) }-1/2\,{\frac{x}{d}\cos \left ( c+{\frac{d}{x}} \right ) }-1/2\,{\it Si} \left ({\frac{d}{x}} \right ) \cos \left ( c \right ) -1/2\,{\it Ci} \left ({\frac{d}{x}} \right ) \sin \left ( c \right ) \right ) +b{f}^{2}{d}^{2} \left ( -{\frac{{x}^{3}}{3\,{d}^{3}}\sin \left ( c+{\frac{d}{x}} \right ) }-{\frac{{x}^{2}}{6\,{d}^{2}}\cos \left ( c+{\frac{d}{x}} \right ) }+{\frac{x}{6\,d}\sin \left ( c+{\frac{d}{x}} \right ) }+{\frac{\sin \left ( c \right ) }{6}{\it Si} \left ({\frac{d}{x}} \right ) }-{\frac{\cos \left ( c \right ) }{6}{\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)^2*(a+b*sin(c+d/x)),x)

[Out]

-d*(-a*e^2*x/d-a/d*e*f*x^2-1/3*a*f^2/d*x^3+b*e^2*(-sin(c+d/x)*x/d-Si(d/x)*sin(c)+Ci(d/x)*cos(c))+2*b*e*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))+b*f^2*d^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 = 1.34175, size = 348, normalized size = 1.55 \begin{align*} \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 \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^{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 \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 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 \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^{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 \end{align*}

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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Fricas [A]  time = 1.40155, size = 571, normalized size = 2.55 \begin{align*} \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 \left (c\right ) + \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 \left (c\right ) + \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 ) \end{align*}

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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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 )^{2}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}^{2}{\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)^2*(a+b*sin(c+d/x)),x, algorithm="giac")

[Out]

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