\(\int (e+f x) (a+b \sin (c+\frac {d}{x}))^2 \, dx\) [294]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [C] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 20, antiderivative size = 254 \[ \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=a^2 e x+\frac {1}{2} a^2 f x^2+a b d f x \cos \left (c+\frac {d}{x}\right )-2 a b d e \cos (c) \operatorname {CosIntegral}\left (\frac {d}{x}\right )-b^2 d^2 f \cos (2 c) \operatorname {CosIntegral}\left (\frac {2 d}{x}\right )+a b d^2 f \operatorname {CosIntegral}\left (\frac {d}{x}\right ) \sin (c)-b^2 d e \operatorname {CosIntegral}\left (\frac {2 d}{x}\right ) \sin (2 c)+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+b^2 d f x \cos \left (c+\frac {d}{x}\right ) \sin \left (c+\frac {d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )+a b d^2 f \cos (c) \text {Si}\left (\frac {d}{x}\right )+2 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )-b^2 d e \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )+b^2 d^2 f \sin (2 c) \text {Si}\left (\frac {2 d}{x}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.99 \[ \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=\frac {1}{4} \left (4 a^2 e x+2 b^2 e x+2 a^2 f x^2+b^2 f x^2+4 a b d f x \cos \left (c+\frac {d}{x}\right )-2 b^2 e x \cos \left (2 \left (c+\frac {d}{x}\right )\right )-b^2 f x^2 \cos \left (2 \left (c+\frac {d}{x}\right )\right )+4 a b d \operatorname {CosIntegral}\left (\frac {d}{x}\right ) (-2 e \cos (c)+d f \sin (c))-4 b^2 d \operatorname {CosIntegral}\left (\frac {2 d}{x}\right ) (d f \cos (2 c)+e \sin (2 c))+8 a b e x \sin \left (c+\frac {d}{x}\right )+4 a b f x^2 \sin \left (c+\frac {d}{x}\right )+2 b^2 d f x \sin \left (2 \left (c+\frac {d}{x}\right )\right )+4 a b d^2 f \cos (c) \text {Si}\left (\frac {d}{x}\right )+8 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )-4 b^2 d e \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )+4 b^2 d^2 f \sin (2 c) \text {Si}\left (\frac {2 d}{x}\right )\right ) \] Input:

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

Output:

(4*a^2*e*x + 2*b^2*e*x + 2*a^2*f*x^2 + b^2*f*x^2 + 4*a*b*d*f*x*Cos[c + d/x 
] - 2*b^2*e*x*Cos[2*(c + d/x)] - b^2*f*x^2*Cos[2*(c + d/x)] + 4*a*b*d*CosI 
ntegral[d/x]*(-2*e*Cos[c] + d*f*Sin[c]) - 4*b^2*d*CosIntegral[(2*d)/x]*(d* 
f*Cos[2*c] + e*Sin[2*c]) + 8*a*b*e*x*Sin[c + d/x] + 4*a*b*f*x^2*Sin[c + d/ 
x] + 2*b^2*d*f*x*Sin[2*(c + d/x)] + 4*a*b*d^2*f*Cos[c]*SinIntegral[d/x] + 
8*a*b*d*e*Sin[c]*SinIntegral[d/x] - 4*b^2*d*e*Cos[2*c]*SinIntegral[(2*d)/x 
] + 4*b^2*d^2*f*Sin[2*c]*SinIntegral[(2*d)/x])/4
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3912, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3912

\(\displaystyle -\int \left (f \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 x^3+e \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 x^2\right )d\frac {1}{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle a^2 e x+\frac {1}{2} a^2 f x^2+a b d^2 f \sin (c) \operatorname {CosIntegral}\left (\frac {d}{x}\right )-2 a b d e \cos (c) \operatorname {CosIntegral}\left (\frac {d}{x}\right )+a b d^2 f \cos (c) \text {Si}\left (\frac {d}{x}\right )+2 a b d e \sin (c) \text {Si}\left (\frac {d}{x}\right )+2 a b e x \sin \left (c+\frac {d}{x}\right )+a b f x^2 \sin \left (c+\frac {d}{x}\right )+a b d f x \cos \left (c+\frac {d}{x}\right )-b^2 d^2 f \cos (2 c) \operatorname {CosIntegral}\left (\frac {2 d}{x}\right )-b^2 d e \sin (2 c) \operatorname {CosIntegral}\left (\frac {2 d}{x}\right )+b^2 d^2 f \sin (2 c) \text {Si}\left (\frac {2 d}{x}\right )-b^2 d e \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )+b^2 e x \sin ^2\left (c+\frac {d}{x}\right )+\frac {1}{2} b^2 f x^2 \sin ^2\left (c+\frac {d}{x}\right )+b^2 d f x \sin \left (c+\frac {d}{x}\right ) \cos \left (c+\frac {d}{x}\right )\)

Input:

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

Output:

a^2*e*x + (a^2*f*x^2)/2 + a*b*d*f*x*Cos[c + d/x] - 2*a*b*d*e*Cos[c]*CosInt 
egral[d/x] - b^2*d^2*f*Cos[2*c]*CosIntegral[(2*d)/x] + a*b*d^2*f*CosIntegr 
al[d/x]*Sin[c] - b^2*d*e*CosIntegral[(2*d)/x]*Sin[2*c] + 2*a*b*e*x*Sin[c + 
 d/x] + a*b*f*x^2*Sin[c + d/x] + b^2*d*f*x*Cos[c + d/x]*Sin[c + d/x] + b^2 
*e*x*Sin[c + d/x]^2 + (b^2*f*x^2*Sin[c + d/x]^2)/2 + a*b*d^2*f*Cos[c]*SinI 
ntegral[d/x] + 2*a*b*d*e*Sin[c]*SinIntegral[d/x] - b^2*d*e*Cos[2*c]*SinInt 
egral[(2*d)/x] + b^2*d^2*f*Sin[2*c]*SinIntegral[(2*d)/x]
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3912
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f 
_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Simp[1/(n*f)   Subst[Int[ExpandIntegra 
nd[(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]
 
Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.98

method result size
parts \(a^{2} \left (\frac {1}{2} f \,x^{2}+e x \right )-b^{2} d \left (-\frac {e x}{2 d}-\frac {e \left (-\frac {2 \cos \left (\frac {2 d}{x}+2 c \right ) x}{d}-4 \,\operatorname {Si}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )-4 \,\operatorname {Ci}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )\right )}{4}-\frac {f \,x^{2}}{4 d}-\frac {d f \left (-\frac {\cos \left (\frac {2 d}{x}+2 c \right ) x^{2}}{d^{2}}+\frac {2 \sin \left (\frac {2 d}{x}+2 c \right ) x}{d}+4 \,\operatorname {Si}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )-4 \,\operatorname {Ci}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )\right )}{4}\right )-2 a b d \left (e \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\operatorname {Si}\left (\frac {d}{x}\right ) \sin \left (c \right )+\operatorname {Ci}\left (\frac {d}{x}\right ) \cos \left (c \right )\right )+d f \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x^{2}}{2 d^{2}}-\frac {\cos \left (c +\frac {d}{x}\right ) x}{2 d}-\frac {\operatorname {Si}\left (\frac {d}{x}\right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (\frac {d}{x}\right ) \sin \left (c \right )}{2}\right )\right )\) \(249\)
derivativedivides \(-d \left (-\frac {a^{2} f \,x^{2}}{2 d}-\frac {a^{2} e x}{d}+2 f b d a \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x^{2}}{2 d^{2}}-\frac {\cos \left (c +\frac {d}{x}\right ) x}{2 d}-\frac {\operatorname {Si}\left (\frac {d}{x}\right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (\frac {d}{x}\right ) \sin \left (c \right )}{2}\right )+2 a b e \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\operatorname {Si}\left (\frac {d}{x}\right ) \sin \left (c \right )+\operatorname {Ci}\left (\frac {d}{x}\right ) \cos \left (c \right )\right )-\frac {b^{2} f \,x^{2}}{4 d}-\frac {b^{2} f d \left (-\frac {\cos \left (\frac {2 d}{x}+2 c \right ) x^{2}}{d^{2}}+\frac {2 \sin \left (\frac {2 d}{x}+2 c \right ) x}{d}+4 \,\operatorname {Si}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )-4 \,\operatorname {Ci}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )\right )}{4}-\frac {b^{2} e x}{2 d}-\frac {b^{2} e \left (-\frac {2 \cos \left (\frac {2 d}{x}+2 c \right ) x}{d}-4 \,\operatorname {Si}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )-4 \,\operatorname {Ci}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )\right )}{4}\right )\) \(265\)
default \(-d \left (-\frac {a^{2} f \,x^{2}}{2 d}-\frac {a^{2} e x}{d}+2 f b d a \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x^{2}}{2 d^{2}}-\frac {\cos \left (c +\frac {d}{x}\right ) x}{2 d}-\frac {\operatorname {Si}\left (\frac {d}{x}\right ) \cos \left (c \right )}{2}-\frac {\operatorname {Ci}\left (\frac {d}{x}\right ) \sin \left (c \right )}{2}\right )+2 a b e \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\operatorname {Si}\left (\frac {d}{x}\right ) \sin \left (c \right )+\operatorname {Ci}\left (\frac {d}{x}\right ) \cos \left (c \right )\right )-\frac {b^{2} f \,x^{2}}{4 d}-\frac {b^{2} f d \left (-\frac {\cos \left (\frac {2 d}{x}+2 c \right ) x^{2}}{d^{2}}+\frac {2 \sin \left (\frac {2 d}{x}+2 c \right ) x}{d}+4 \,\operatorname {Si}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )-4 \,\operatorname {Ci}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )\right )}{4}-\frac {b^{2} e x}{2 d}-\frac {b^{2} e \left (-\frac {2 \cos \left (\frac {2 d}{x}+2 c \right ) x}{d}-4 \,\operatorname {Si}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )-4 \,\operatorname {Ci}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )\right )}{4}\right )\) \(265\)
risch \(a^{2} e x +\frac {a^{2} f \,x^{2}}{2}+a b d e \,{\mathrm e}^{-i c} \operatorname {expIntegral}_{1}\left (\frac {i d}{x}\right )+\frac {i {\mathrm e}^{-2 i c} \operatorname {expIntegral}_{1}\left (\frac {2 i d}{x}\right ) b^{2} d e}{2}+\frac {b^{2} e x}{2}+\frac {b^{2} f \,x^{2}}{4}+\frac {{\mathrm e}^{-2 i c} \operatorname {expIntegral}_{1}\left (\frac {2 i d}{x}\right ) b^{2} d^{2} f}{2}-\frac {i a b \,d^{2} f \,{\mathrm e}^{-i c} \operatorname {expIntegral}_{1}\left (\frac {i d}{x}\right )}{2}+\frac {{\mathrm e}^{2 i c} \operatorname {expIntegral}_{1}\left (-\frac {2 i d}{x}\right ) b^{2} d^{2} f}{2}-\frac {i {\mathrm e}^{2 i c} \operatorname {expIntegral}_{1}\left (-\frac {2 i d}{x}\right ) b^{2} d e}{2}+a b d e \,{\mathrm e}^{i c} \operatorname {expIntegral}_{1}\left (-\frac {i d}{x}\right )+\frac {i a b \,d^{2} f \,{\mathrm e}^{i c} \operatorname {expIntegral}_{1}\left (-\frac {i d}{x}\right )}{2}+a b d x f \cos \left (\frac {c x +d}{x}\right )+a b x \left (f x +2 e \right ) \sin \left (\frac {c x +d}{x}\right )-\frac {b^{2} x \left (f x +2 e \right ) \cos \left (\frac {2 c x +2 d}{x}\right )}{4}+\frac {d \,b^{2} x f \sin \left (\frac {2 c x +2 d}{x}\right )}{2}\) \(287\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.94 \[ \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=a b d f x \cos \left (\frac {c x + d}{x}\right ) + \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} f x^{2} + {\left (a^{2} + b^{2}\right )} e x - \frac {1}{2} \, {\left (b^{2} f x^{2} + 2 \, b^{2} e x\right )} \cos \left (\frac {c x + d}{x}\right )^{2} - {\left (b^{2} d^{2} f \operatorname {Ci}\left (\frac {2 \, d}{x}\right ) + b^{2} d e \operatorname {Si}\left (\frac {2 \, d}{x}\right )\right )} \cos \left (2 \, c\right ) + {\left (a b d^{2} f \operatorname {Si}\left (\frac {d}{x}\right ) - 2 \, a b d e \operatorname {Ci}\left (\frac {d}{x}\right )\right )} \cos \left (c\right ) + {\left (b^{2} d^{2} f \operatorname {Si}\left (\frac {2 \, d}{x}\right ) - b^{2} d e \operatorname {Ci}\left (\frac {2 \, d}{x}\right )\right )} \sin \left (2 \, c\right ) + {\left (a b d^{2} f \operatorname {Ci}\left (\frac {d}{x}\right ) + 2 \, a b d e \operatorname {Si}\left (\frac {d}{x}\right )\right )} \sin \left (c\right ) + {\left (b^{2} d f x \cos \left (\frac {c x + d}{x}\right ) + a b f x^{2} + 2 \, a b e x\right )} \sin \left (\frac {c x + d}{x}\right ) \] Input:

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

Output:

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

Sympy [F]

\[ \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=\int \left (a + b \sin {\left (c + \frac {d}{x} \right )}\right )^{2} \left (e + f x\right )\, dx \] Input:

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.26 \[ \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=\frac {1}{2} \, a^{2} f x^{2} - {\left ({\left ({\left ({\rm Ei}\left (\frac {i \, d}{x}\right ) + {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \cos \left (c\right ) - {\left (-i \, {\rm Ei}\left (\frac {i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \sin \left (c\right )\right )} d - 2 \, x \sin \left (\frac {c x + d}{x}\right )\right )} a b e - \frac {1}{2} \, {\left ({\left ({\left (-i \, {\rm Ei}\left (\frac {2 i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\rm Ei}\left (\frac {2 i \, d}{x}\right ) + {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \sin \left (2 \, c\right )\right )} d + x \cos \left (\frac {2 \, {\left (c x + d\right )}}{x}\right ) - x\right )} b^{2} e + \frac {1}{2} \, {\left ({\left ({\left (-i \, {\rm Ei}\left (\frac {i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \cos \left (c\right ) + {\left ({\rm Ei}\left (\frac {i \, d}{x}\right ) + {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \sin \left (c\right )\right )} d^{2} + 2 \, d x \cos \left (\frac {c x + d}{x}\right ) + 2 \, x^{2} \sin \left (\frac {c x + d}{x}\right )\right )} a b f - \frac {1}{4} \, {\left (2 \, {\left ({\left ({\rm Ei}\left (\frac {2 i \, d}{x}\right ) + {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \cos \left (2 \, c\right ) + {\left (i \, {\rm Ei}\left (\frac {2 i \, d}{x}\right ) - i \, {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \sin \left (2 \, c\right )\right )} d^{2} + x^{2} \cos \left (\frac {2 \, {\left (c x + d\right )}}{x}\right ) - 2 \, d x \sin \left (\frac {2 \, {\left (c x + d\right )}}{x}\right ) - x^{2}\right )} b^{2} f + a^{2} e x \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1125 vs. \(2 (250) = 500\).

Time = 0.14 (sec) , antiderivative size = 1125, normalized size of antiderivative = 4.43 \[ \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*(4*b^2*c^2*d^3*f*cos(2*c)*cos_integral(-2*c + 2*(c*x + d)/x) - 4*a*b* 
c^2*d^3*f*cos_integral(-c + (c*x + d)/x)*sin(c) + 4*b^2*c^2*d^3*f*sin(2*c) 
*sin_integral(2*c - 2*(c*x + d)/x) + 4*a*b*c^2*d^3*f*cos(c)*sin_integral(c 
 - (c*x + d)/x) + 8*a*b*c^2*d^2*e*cos(c)*cos_integral(-c + (c*x + d)/x) - 
8*(c*x + d)*b^2*c*d^3*f*cos(2*c)*cos_integral(-2*c + 2*(c*x + d)/x)/x + 4* 
b^2*c^2*d^2*e*cos_integral(-2*c + 2*(c*x + d)/x)*sin(2*c) + 8*(c*x + d)*a* 
b*c*d^3*f*cos_integral(-c + (c*x + d)/x)*sin(c)/x - 4*b^2*c^2*d^2*e*cos(2* 
c)*sin_integral(2*c - 2*(c*x + d)/x) - 8*(c*x + d)*b^2*c*d^3*f*sin(2*c)*si 
n_integral(2*c - 2*(c*x + d)/x)/x - 8*(c*x + d)*a*b*c*d^3*f*cos(c)*sin_int 
egral(c - (c*x + d)/x)/x + 8*a*b*c^2*d^2*e*sin(c)*sin_integral(c - (c*x + 
d)/x) + 4*a*b*c*d^3*f*cos((c*x + d)/x) - 16*(c*x + d)*a*b*c*d^2*e*cos(c)*c 
os_integral(-c + (c*x + d)/x)/x + 4*(c*x + d)^2*b^2*d^3*f*cos(2*c)*cos_int 
egral(-2*c + 2*(c*x + d)/x)/x^2 - 8*(c*x + d)*b^2*c*d^2*e*cos_integral(-2* 
c + 2*(c*x + d)/x)*sin(2*c)/x - 4*(c*x + d)^2*a*b*d^3*f*cos_integral(-c + 
(c*x + d)/x)*sin(c)/x^2 + 2*b^2*c*d^3*f*sin(2*(c*x + d)/x) + 8*(c*x + d)*b 
^2*c*d^2*e*cos(2*c)*sin_integral(2*c - 2*(c*x + d)/x)/x + 4*(c*x + d)^2*b^ 
2*d^3*f*sin(2*c)*sin_integral(2*c - 2*(c*x + d)/x)/x^2 + 4*(c*x + d)^2*a*b 
*d^3*f*cos(c)*sin_integral(c - (c*x + d)/x)/x^2 - 16*(c*x + d)*a*b*c*d^2*e 
*sin(c)*sin_integral(c - (c*x + d)/x)/x - 2*b^2*c*d^2*e*cos(2*(c*x + d)/x) 
 + b^2*d^3*f*cos(2*(c*x + d)/x) - 4*(c*x + d)*a*b*d^3*f*cos((c*x + d)/x...
 

Mupad [F(-1)]

Timed out. \[ \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=\int \left (e+f\,x\right )\,{\left (a+b\,\sin \left (c+\frac {d}{x}\right )\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (e+f x) \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*cos((c*x + d)/x)*sin((c*x + d)/x)*tan((c*x + d)/(2*x))**4*b**2*d*f*x 
 - 4*cos((c*x + d)/x)*sin((c*x + d)/x)*tan((c*x + d)/(2*x))**2*b**2*d*f*x 
- 2*cos((c*x + d)/x)*sin((c*x + d)/x)*b**2*d*f*x + 6*cos((c*x + d)/x)*tan( 
(c*x + d)/(2*x))**4*a*b*d*f*x + 8*cos((c*x + d)/x)*tan((c*x + d)/(2*x))**4 
*b**2*f*x**2 + 12*cos((c*x + d)/x)*tan((c*x + d)/(2*x))**2*a*b*d*f*x + 16* 
cos((c*x + d)/x)*tan((c*x + d)/(2*x))**2*b**2*f*x**2 + 6*cos((c*x + d)/x)* 
a*b*d*f*x + 8*cos((c*x + d)/x)*b**2*f*x**2 + 6*int(sin((c*x + d)/x)**2,x)* 
tan((c*x + d)/(2*x))**4*b**2*e + 12*int(sin((c*x + d)/x)**2,x)*tan((c*x + 
d)/(2*x))**2*b**2*e + 6*int(sin((c*x + d)/x)**2,x)*b**2*e - 6*int(sin((c*x 
 + d)/x)/x,x)*tan((c*x + d)/(2*x))**4*a*b*d**2*f - 12*int(sin((c*x + d)/x) 
/x,x)*tan((c*x + d)/(2*x))**2*a*b*d**2*f - 6*int(sin((c*x + d)/x)/x,x)*a*b 
*d**2*f + 32*int(tan((c*x + d)/(2*x))/(tan((c*x + d)/(2*x))**4 + 2*tan((c* 
x + d)/(2*x))**2 + 1),x)*tan((c*x + d)/(2*x))**4*b**2*d*f + 64*int(tan((c* 
x + d)/(2*x))/(tan((c*x + d)/(2*x))**4 + 2*tan((c*x + d)/(2*x))**2 + 1),x) 
*tan((c*x + d)/(2*x))**2*b**2*d*f + 32*int(tan((c*x + d)/(2*x))/(tan((c*x 
+ d)/(2*x))**4 + 2*tan((c*x + d)/(2*x))**2 + 1),x)*b**2*d*f - 16*int(1/(ta 
n((c*x + d)/(2*x))**4*x + 2*tan((c*x + d)/(2*x))**2*x + x),x)*tan((c*x + d 
)/(2*x))**4*b**2*d**2*f - 32*int(1/(tan((c*x + d)/(2*x))**4*x + 2*tan((c*x 
 + d)/(2*x))**2*x + x),x)*tan((c*x + d)/(2*x))**2*b**2*d**2*f - 16*int(1/( 
tan((c*x + d)/(2*x))**4*x + 2*tan((c*x + d)/(2*x))**2*x + x),x)*b**2*d*...