\(\int (a+b \sin (c+\frac {d}{x}))^2 \, dx\) [295]

Optimal result

Integrand size = 14, antiderivative size = 94 \[ \int \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=a^2 x-2 a b d \cos (c) \operatorname {CosIntegral}\left (\frac {d}{x}\right )-b^2 d \operatorname {CosIntegral}\left (\frac {2 d}{x}\right ) \sin (2 c)+2 a b x \sin \left (c+\frac {d}{x}\right )+b^2 x \sin ^2\left (c+\frac {d}{x}\right )+2 a b d \sin (c) \text {Si}\left (\frac {d}{x}\right )-b^2 d \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.12 \[ \int \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=\frac {1}{2} \left (2 a^2 x+b^2 x-b^2 x \cos \left (2 \left (c+\frac {d}{x}\right )\right )-4 a b d \cos (c) \operatorname {CosIntegral}\left (\frac {d}{x}\right )-2 b^2 d \operatorname {CosIntegral}\left (\frac {2 d}{x}\right ) \sin (2 c)+4 a b x \sin \left (c+\frac {d}{x}\right )+4 a b d \sin (c) \text {Si}\left (\frac {d}{x}\right )-2 b^2 d \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right )\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3842, 3042, 3798, 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.

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

\(\Big \downarrow \) 3842

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3798

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

a^2*x - 2*a*b*d*Cos[c]*CosIntegral[d/x] - b^2*d*CosIntegral[(2*d)/x]*Sin[2 
*c] + 2*a*b*x*Sin[c + d/x] + b^2*x*Sin[c + d/x]^2 + 2*a*b*d*Sin[c]*SinInte 
gral[d/x] - b^2*d*Cos[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ 
m, 0] || NeQ[a^2 - b^2, 0])
 

rule 3842

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_S 
ymbol] :> Simp[1/(n*f)   Subst[Int[x^(1/n - 1)*(a + b*Sin[c + d*x])^p, x], 
x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && Intege 
rQ[1/n]
 

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06

method	result	size
parts	\(a^{2} x -b^{2} d \left (-\frac {x}{2 d}+\frac {\cos \left (\frac {2 d}{x}+2 c \right ) x}{2 d}+\operatorname {Si}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )+\operatorname {Ci}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )\right )-2 a b d \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\operatorname {Si}\left (\frac {d}{x}\right ) \sin \left (c \right )+\operatorname {Ci}\left (\frac {d}{x}\right ) \cos \left (c \right )\right )\)	\(100\)
derivativedivides	\(-d \left (-\frac {a^{2} x}{d}+2 a b \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\operatorname {Si}\left (\frac {d}{x}\right ) \sin \left (c \right )+\operatorname {Ci}\left (\frac {d}{x}\right ) \cos \left (c \right )\right )-\frac {b^{2} x}{2 d}-\frac {b^{2} \left (-\frac {2 \cos \left (\frac {2 d}{x}+2 c \right ) x}{d}-4 \,\operatorname {Si}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )-4 \,\operatorname {Ci}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )\right )}{4}\right )\)	\(110\)
default	\(-d \left (-\frac {a^{2} x}{d}+2 a b \left (-\frac {\sin \left (c +\frac {d}{x}\right ) x}{d}-\operatorname {Si}\left (\frac {d}{x}\right ) \sin \left (c \right )+\operatorname {Ci}\left (\frac {d}{x}\right ) \cos \left (c \right )\right )-\frac {b^{2} x}{2 d}-\frac {b^{2} \left (-\frac {2 \cos \left (\frac {2 d}{x}+2 c \right ) x}{d}-4 \,\operatorname {Si}\left (\frac {2 d}{x}\right ) \cos \left (2 c \right )-4 \,\operatorname {Ci}\left (\frac {2 d}{x}\right ) \sin \left (2 c \right )\right )}{4}\right )\)	\(110\)
risch	\(\frac {\pi \,\operatorname {csgn}\left (\frac {d}{x}\right ) {\mathrm e}^{-2 i c} b^{2} d}{2}-\operatorname {Si}\left (\frac {2 d}{x}\right ) {\mathrm e}^{-2 i c} b^{2} d +\frac {i {\mathrm e}^{-2 i c} \operatorname {expIntegral}_{1}\left (-\frac {2 i d}{x}\right ) b^{2} d}{2}-\frac {i d \,b^{2} \operatorname {expIntegral}_{1}\left (-\frac {2 i d}{x}\right ) {\mathrm e}^{2 i c}}{2}+a b d \,\operatorname {expIntegral}_{1}\left (-\frac {i d}{x}\right ) {\mathrm e}^{i c}-i \pi \,\operatorname {csgn}\left (\frac {d}{x}\right ) {\mathrm e}^{-i c} a b d +2 i \operatorname {Si}\left (\frac {d}{x}\right ) {\mathrm e}^{-i c} a b d +\operatorname {expIntegral}_{1}\left (-\frac {i d}{x}\right ) {\mathrm e}^{-i c} a b d +a^{2} x +\frac {b^{2} x}{2}+2 a b x \sin \left (\frac {c x +d}{x}\right )-\frac {b^{2} x \cos \left (\frac {2 c x +2 d}{x}\right )}{2}\)	\(194\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=-b^{2} x \cos \left (\frac {c x + d}{x}\right )^{2} - 2 \, a b d \cos \left (c\right ) \operatorname {Ci}\left (\frac {d}{x}\right ) - b^{2} d \operatorname {Ci}\left (\frac {2 \, d}{x}\right ) \sin \left (2 \, c\right ) - b^{2} d \cos \left (2 \, c\right ) \operatorname {Si}\left (\frac {2 \, d}{x}\right ) + 2 \, a b d \sin \left (c\right ) \operatorname {Si}\left (\frac {d}{x}\right ) + 2 \, a b x \sin \left (\frac {c x + d}{x}\right ) + {\left (a^{2} + b^{2}\right )} x \] Input:

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

Output:

-b^2*x*cos((c*x + d)/x)^2 - 2*a*b*d*cos(c)*cos_integral(d/x) - b^2*d*cos_i 
ntegral(2*d/x)*sin(2*c) - b^2*d*cos(2*c)*sin_integral(2*d/x) + 2*a*b*d*sin 
(c)*sin_integral(d/x) + 2*a*b*x*sin((c*x + d)/x) + (a^2 + b^2)*x
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.46 \[ \int \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=-{\left ({\left ({\left ({\rm Ei}\left (\frac {i \, d}{x}\right ) + {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \cos \left (c\right ) - {\left (-i \, {\rm Ei}\left (\frac {i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {i \, d}{x}\right )\right )} \sin \left (c\right )\right )} d - 2 \, x \sin \left (\frac {c x + d}{x}\right )\right )} a b - \frac {1}{2} \, {\left ({\left ({\left (-i \, {\rm Ei}\left (\frac {2 i \, d}{x}\right ) + i \, {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\rm Ei}\left (\frac {2 i \, d}{x}\right ) + {\rm Ei}\left (-\frac {2 i \, d}{x}\right )\right )} \sin \left (2 \, c\right )\right )} d + x \cos \left (\frac {2 \, {\left (c x + d\right )}}{x}\right ) - x\right )} b^{2} + a^{2} x \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (94) = 188\).

Time = 0.17 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.24 \[ \int \left (a+b \sin \left (c+\frac {d}{x}\right )\right )^2 \, dx=-\frac {4 \, a b c d^{2} \cos \left (c\right ) \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right ) + 2 \, b^{2} c d^{2} \operatorname {Ci}\left (-2 \, c + \frac {2 \, {\left (c x + d\right )}}{x}\right ) \sin \left (2 \, c\right ) - 2 \, b^{2} c d^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (2 \, c - \frac {2 \, {\left (c x + d\right )}}{x}\right ) + 4 \, a b c d^{2} \sin \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right ) - \frac {4 \, {\left (c x + d\right )} a b d^{2} \cos \left (c\right ) \operatorname {Ci}\left (-c + \frac {c x + d}{x}\right )}{x} - \frac {2 \, {\left (c x + d\right )} b^{2} d^{2} \operatorname {Ci}\left (-2 \, c + \frac {2 \, {\left (c x + d\right )}}{x}\right ) \sin \left (2 \, c\right )}{x} + \frac {2 \, {\left (c x + d\right )} b^{2} d^{2} \cos \left (2 \, c\right ) \operatorname {Si}\left (2 \, c - \frac {2 \, {\left (c x + d\right )}}{x}\right )}{x} - \frac {4 \, {\left (c x + d\right )} a b d^{2} \sin \left (c\right ) \operatorname {Si}\left (c - \frac {c x + d}{x}\right )}{x} - b^{2} d^{2} \cos \left (\frac {2 \, {\left (c x + d\right )}}{x}\right ) + 4 \, a b d^{2} \sin \left (\frac {c x + d}{x}\right ) + 2 \, a^{2} d^{2} + b^{2} d^{2}}{2 \, {\left (c - \frac {c x + d}{x}\right )} d} \] Input:

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

Output:

-1/2*(4*a*b*c*d^2*cos(c)*cos_integral(-c + (c*x + d)/x) + 2*b^2*c*d^2*cos_ 
integral(-2*c + 2*(c*x + d)/x)*sin(2*c) - 2*b^2*c*d^2*cos(2*c)*sin_integra 
l(2*c - 2*(c*x + d)/x) + 4*a*b*c*d^2*sin(c)*sin_integral(c - (c*x + d)/x) 
- 4*(c*x + d)*a*b*d^2*cos(c)*cos_integral(-c + (c*x + d)/x)/x - 2*(c*x + d 
)*b^2*d^2*cos_integral(-2*c + 2*(c*x + d)/x)*sin(2*c)/x + 2*(c*x + d)*b^2* 
d^2*cos(2*c)*sin_integral(2*c - 2*(c*x + d)/x)/x - 4*(c*x + d)*a*b*d^2*sin 
(c)*sin_integral(c - (c*x + d)/x)/x - b^2*d^2*cos(2*(c*x + d)/x) + 4*a*b*d 
^2*sin((c*x + d)/x) + 2*a^2*d^2 + b^2*d^2)/((c - (c*x + d)/x)*d)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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