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

Optimal. Leaf size=94 \[ a^2 x-2 a b d \cos (c) \text {Ci}\left (\frac {d}{x}\right )-b^2 d \text {Ci}\left (\frac {2 d}{x}\right ) \sin (2 c)+2 a b x \sin \left (c+\frac {d}{x}\right )+b^2 x \sin ^2\left (c+\frac {d}{x}\right )+2 a b d \sin (c) \text {Si}\left (\frac {d}{x}\right )-b^2 d \cos (2 c) \text {Si}\left (\frac {2 d}{x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3442, 3398, 3378, 3384, 3380, 3383, 3394, 12} \begin {gather*} a^2 x-2 a b d \cos (c) \text {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) \text {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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d/x])^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3394

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]^
n/(d*(m + 1))), x] - Dist[f*(n/(d*(m + 1))), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 3398

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 3442

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Dist[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] && In
tegerQ[1/n]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 105, normalized size = 1.12 \begin {gather*} \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) \text {Ci}\left (\frac {d}{x}\right )-2 b^2 d \text {Ci}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d/x])^2,x]

[Out]

(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

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Maple [A]
time = 0.07, size = 110, normalized size = 1.17

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.34, size = 137, normalized size = 1.46 \begin {gather*} -{\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((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))*cos(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

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Fricas [A]
time = 0.38, size = 130, normalized size = 1.38 \begin {gather*} -b^{2} x \cos \left (\frac {c x + d}{x}\right )^{2} - 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 - {\left (a b d \operatorname {Ci}\left (\frac {d}{x}\right ) + a b d \operatorname {Ci}\left (-\frac {d}{x}\right )\right )} \cos \left (c\right ) - \frac {1}{2} \, {\left (b^{2} d \operatorname {Ci}\left (\frac {2 \, d}{x}\right ) + b^{2} d \operatorname {Ci}\left (-\frac {2 \, d}{x}\right )\right )} \sin \left (2 \, c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*x*cos((c*x + d)/x)^2 - 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*si
n((c*x + d)/x) + (a^2 + b^2)*x - (a*b*d*cos_integral(d/x) + a*b*d*cos_integral(-d/x))*cos(c) - 1/2*(b^2*d*cos_
integral(2*d/x) + b^2*d*cos_integral(-2*d/x))*sin(2*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*sin(c + d/x))**2, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (94) = 188\).
time = 4.84, size = 305, normalized size = 3.24 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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_integral(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+b\,\sin \left (c+\frac {d}{x}\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d/x))^2,x)

[Out]

int((a + b*sin(c + d/x))^2, x)

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