3.219 \(\int \sin (\coth (a+b x)) \, dx\)

Optimal. Leaf size=77 \[ -\frac{\sin (1) \text{CosIntegral}(1-\coth (a+b x))}{2 b}-\frac{\sin (1) \text{CosIntegral}(\coth (a+b x)+1)}{2 b}+\frac{\cos (1) \text{Si}(1-\coth (a+b x))}{2 b}+\frac{\cos (1) \text{Si}(\coth (a+b x)+1)}{2 b} \]

[Out]

-(CosIntegral[1 - Coth[a + b*x]]*Sin[1])/(2*b) - (CosIntegral[1 + Coth[a + b*x]]*Sin[1])/(2*b) + (Cos[1]*SinIn
tegral[1 - Coth[a + b*x]])/(2*b) + (Cos[1]*SinIntegral[1 + Coth[a + b*x]])/(2*b)

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Rubi [A]  time = 0.144187, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {3333, 3303, 3299, 3302} \[ -\frac{\sin (1) \text{CosIntegral}(1-\coth (a+b x))}{2 b}-\frac{\sin (1) \text{CosIntegral}(\coth (a+b x)+1)}{2 b}+\frac{\cos (1) \text{Si}(1-\coth (a+b x))}{2 b}+\frac{\cos (1) \text{Si}(\coth (a+b x)+1)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[Coth[a + b*x]],x]

[Out]

-(CosIntegral[1 - Coth[a + b*x]]*Sin[1])/(2*b) - (CosIntegral[1 + Coth[a + b*x]]*Sin[1])/(2*b) + (Cos[1]*SinIn
tegral[1 - Coth[a + b*x]])/(2*b) + (Cos[1]*SinIntegral[1 + Coth[a + b*x]])/(2*b)

Rule 3333

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*Sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[ExpandIntegrand[Sin[c + d*x], (a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IGtQ[n, 0] && (EqQ[n, 2] || EqQ[p, -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 \sin (\coth (a+b x)) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin (x)}{1-x^2} \, dx,x,\coth (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\sin (x)}{2 (1-x)}+\frac{\sin (x)}{2 (1+x)}\right ) \, dx,x,\coth (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sin (x)}{1-x} \, dx,x,\coth (a+b x)\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\sin (x)}{1+x} \, dx,x,\coth (a+b x)\right )}{2 b}\\ &=-\frac{\cos (1) \operatorname{Subst}\left (\int \frac{\sin (1-x)}{1-x} \, dx,x,\coth (a+b x)\right )}{2 b}+\frac{\cos (1) \operatorname{Subst}\left (\int \frac{\sin (1+x)}{1+x} \, dx,x,\coth (a+b x)\right )}{2 b}+\frac{\sin (1) \operatorname{Subst}\left (\int \frac{\cos (1-x)}{1-x} \, dx,x,\coth (a+b x)\right )}{2 b}-\frac{\sin (1) \operatorname{Subst}\left (\int \frac{\cos (1+x)}{1+x} \, dx,x,\coth (a+b x)\right )}{2 b}\\ &=-\frac{\text{Ci}(1-\coth (a+b x)) \sin (1)}{2 b}-\frac{\text{Ci}(1+\coth (a+b x)) \sin (1)}{2 b}+\frac{\cos (1) \text{Si}(1-\coth (a+b x))}{2 b}+\frac{\cos (1) \text{Si}(1+\coth (a+b x))}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.128473, size = 59, normalized size = 0.77 \[ -\frac{\sin (1) \text{CosIntegral}(1-\coth (a+b x))+\sin (1) \text{CosIntegral}(\coth (a+b x)+1)-\cos (1) (\text{Si}(1-\coth (a+b x))+\text{Si}(\coth (a+b x)+1))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[Coth[a + b*x]],x]

[Out]

-(CosIntegral[1 - Coth[a + b*x]]*Sin[1] + CosIntegral[1 + Coth[a + b*x]]*Sin[1] - Cos[1]*(SinIntegral[1 - Coth
[a + b*x]] + SinIntegral[1 + Coth[a + b*x]]))/(2*b)

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Maple [A]  time = 0.016, size = 58, normalized size = 0.8 \begin{align*}{\frac{1}{b} \left ({\frac{{\it Si} \left ({\rm coth} \left (bx+a\right )+1 \right ) \cos \left ( 1 \right ) }{2}}-{\frac{{\it Ci} \left ({\rm coth} \left (bx+a\right )+1 \right ) \sin \left ( 1 \right ) }{2}}-{\frac{{\it Si} \left ({\rm coth} \left (bx+a\right )-1 \right ) \cos \left ( 1 \right ) }{2}}-{\frac{{\it Ci} \left ({\rm coth} \left (bx+a\right )-1 \right ) \sin \left ( 1 \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(coth(b*x+a)),x)

[Out]

1/b*(1/2*Si(coth(b*x+a)+1)*cos(1)-1/2*Ci(coth(b*x+a)+1)*sin(1)-1/2*Si(coth(b*x+a)-1)*cos(1)-1/2*Ci(coth(b*x+a)
-1)*sin(1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (\coth \left (b x + a\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(coth(b*x+a)),x, algorithm="maxima")

[Out]

integrate(sin(coth(b*x + a)), x)

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Fricas [B]  time = 2.43796, size = 464, normalized size = 6.03 \begin{align*} -\frac{\operatorname{Ci}\left (\frac{2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (1\right ) + \operatorname{Ci}\left (-\frac{2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (1\right ) + \operatorname{Ci}\left (\frac{2}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (1\right ) + \operatorname{Ci}\left (-\frac{2}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (1\right ) - 2 \, \cos \left (1\right ) \operatorname{Si}\left (\frac{2 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) + 2 \, \cos \left (1\right ) \operatorname{Si}\left (\frac{2}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(coth(b*x+a)),x, algorithm="fricas")

[Out]

-1/4*(cos_integral(2*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) - 1))*sin(1) + cos_integral(-2*e^(2*b*x + 2*a)/(e^(2*b*x
+ 2*a) - 1))*sin(1) + cos_integral(2/(e^(2*b*x + 2*a) - 1))*sin(1) + cos_integral(-2/(e^(2*b*x + 2*a) - 1))*s
in(1) - 2*cos(1)*sin_integral(2*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) - 1)) + 2*cos(1)*sin_integral(2/(e^(2*b*x + 2
*a) - 1)))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (\coth{\left (a + b x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(coth(b*x+a)),x)

[Out]

Integral(sin(coth(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (\coth \left (b x + a\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(coth(b*x+a)),x, algorithm="giac")

[Out]

integrate(sin(coth(b*x + a)), x)