3.218 \(\int \sin ^2(\coth (a+b x)) \, dx\)

Optimal. Leaf size=115 \[ \frac{\cos (2) \text{CosIntegral}(2-2 \coth (a+b x))}{4 b}-\frac{\cos (2) \text{CosIntegral}(2 \coth (a+b x)+2)}{4 b}+\frac{\sin (2) \text{Si}(2-2 \coth (a+b x))}{4 b}-\frac{\sin (2) \text{Si}(2 \coth (a+b x)+2)}{4 b}-\frac{\log (1-\coth (a+b x))}{4 b}+\frac{\log (\coth (a+b x)+1)}{4 b} \]

[Out]

(Cos[2]*CosIntegral[2 - 2*Coth[a + b*x]])/(4*b) - (Cos[2]*CosIntegral[2 + 2*Coth[a + b*x]])/(4*b) - Log[1 - Co
th[a + b*x]]/(4*b) + Log[1 + Coth[a + b*x]]/(4*b) + (Sin[2]*SinIntegral[2 - 2*Coth[a + b*x]])/(4*b) - (Sin[2]*
SinIntegral[2 + 2*Coth[a + b*x]])/(4*b)

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Rubi [A]  time = 0.262045, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 5, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {6725, 3312, 3303, 3299, 3302} \[ \frac{\cos (2) \text{CosIntegral}(2-2 \coth (a+b x))}{4 b}-\frac{\cos (2) \text{CosIntegral}(2 \coth (a+b x)+2)}{4 b}+\frac{\sin (2) \text{Si}(2-2 \coth (a+b x))}{4 b}-\frac{\sin (2) \text{Si}(2 \coth (a+b x)+2)}{4 b}-\frac{\log (1-\coth (a+b x))}{4 b}+\frac{\log (\coth (a+b x)+1)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[Coth[a + b*x]]^2,x]

[Out]

(Cos[2]*CosIntegral[2 - 2*Coth[a + b*x]])/(4*b) - (Cos[2]*CosIntegral[2 + 2*Coth[a + b*x]])/(4*b) - Log[1 - Co
th[a + b*x]]/(4*b) + Log[1 + Coth[a + b*x]]/(4*b) + (Sin[2]*SinIntegral[2 - 2*Coth[a + b*x]])/(4*b) - (Sin[2]*
SinIntegral[2 + 2*Coth[a + b*x]])/(4*b)

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && 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 \sin ^2(\coth (a+b x)) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^2(x)}{1-x^2} \, dx,x,\coth (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{\sin ^2(x)}{2 (-1+x)}+\frac{\sin ^2(x)}{2 (1+x)}\right ) \, dx,x,\coth (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sin ^2(x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\sin ^2(x)}{1+x} \, dx,x,\coth (a+b x)\right )}{2 b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{2 (-1+x)}-\frac{\cos (2 x)}{2 (-1+x)}\right ) \, dx,x,\coth (a+b x)\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \left (\frac{1}{2 (1+x)}-\frac{\cos (2 x)}{2 (1+x)}\right ) \, dx,x,\coth (a+b x)\right )}{2 b}\\ &=-\frac{\log (1-\coth (a+b x))}{4 b}+\frac{\log (1+\coth (a+b x))}{4 b}+\frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{4 b}-\frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{1+x} \, dx,x,\coth (a+b x)\right )}{4 b}\\ &=-\frac{\log (1-\coth (a+b x))}{4 b}+\frac{\log (1+\coth (a+b x))}{4 b}+\frac{\cos (2) \operatorname{Subst}\left (\int \frac{\cos (2-2 x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{4 b}-\frac{\cos (2) \operatorname{Subst}\left (\int \frac{\cos (2+2 x)}{1+x} \, dx,x,\coth (a+b x)\right )}{4 b}+\frac{\sin (2) \operatorname{Subst}\left (\int \frac{\sin (2-2 x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{4 b}-\frac{\sin (2) \operatorname{Subst}\left (\int \frac{\sin (2+2 x)}{1+x} \, dx,x,\coth (a+b x)\right )}{4 b}\\ &=\frac{\cos (2) \text{Ci}(2-2 \coth (a+b x))}{4 b}-\frac{\cos (2) \text{Ci}(2+2 \coth (a+b x))}{4 b}-\frac{\log (1-\coth (a+b x))}{4 b}+\frac{\log (1+\coth (a+b x))}{4 b}+\frac{\sin (2) \text{Si}(2-2 \coth (a+b x))}{4 b}-\frac{\sin (2) \text{Si}(2+2 \coth (a+b x))}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.166621, size = 88, normalized size = 0.77 \[ \frac{\cos (2) \text{CosIntegral}(2-2 \coth (a+b x))-\cos (2) \text{CosIntegral}(2 (\coth (a+b x)+1))+\sin (2) \text{Si}(2-2 \coth (a+b x))-\sin (2) \text{Si}(2 (\coth (a+b x)+1))-\log (1-\coth (a+b x))+\log (\coth (a+b x)+1)}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[Coth[a + b*x]]^2,x]

[Out]

(Cos[2]*CosIntegral[2 - 2*Coth[a + b*x]] - Cos[2]*CosIntegral[2*(1 + Coth[a + b*x])] - Log[1 - Coth[a + b*x]]
+ Log[1 + Coth[a + b*x]] + Sin[2]*SinIntegral[2 - 2*Coth[a + b*x]] - Sin[2]*SinIntegral[2*(1 + Coth[a + b*x])]
)/(4*b)

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Maple [A]  time = 0.02, size = 102, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ({\rm coth} \left (bx+a\right )-1 \right ) }{4\,b}}+{\frac{\ln \left ({\rm coth} \left (bx+a\right )+1 \right ) }{4\,b}}-{\frac{{\it Si} \left ( 2+2\,{\rm coth} \left (bx+a\right ) \right ) \sin \left ( 2 \right ) }{4\,b}}-{\frac{{\it Ci} \left ( 2+2\,{\rm coth} \left (bx+a\right ) \right ) \cos \left ( 2 \right ) }{4\,b}}-{\frac{{\it Si} \left ( -2+2\,{\rm coth} \left (bx+a\right ) \right ) \sin \left ( 2 \right ) }{4\,b}}+{\frac{{\it Ci} \left ( -2+2\,{\rm coth} \left (bx+a\right ) \right ) \cos \left ( 2 \right ) }{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(coth(b*x+a))^2,x)

[Out]

-1/4/b*ln(coth(b*x+a)-1)+1/4/b*ln(coth(b*x+a)+1)-1/4*Si(2+2*coth(b*x+a))*sin(2)/b-1/4*Ci(2+2*coth(b*x+a))*cos(
2)/b-1/4*Si(-2+2*coth(b*x+a))*sin(2)/b+1/4/b*Ci(-2+2*coth(b*x+a))*cos(2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x - \frac{1}{2} \, \int \cos \left (\frac{2 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x - 1/2*integrate(cos(2*(e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1)), x)

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Fricas [A]  time = 2.52604, size = 474, normalized size = 4.12 \begin{align*} \frac{4 \, b x - \cos \left (2\right ) \operatorname{Ci}\left (\frac{4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) - \cos \left (2\right ) \operatorname{Ci}\left (-\frac{4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) + \cos \left (2\right ) \operatorname{Ci}\left (\frac{4}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) + \cos \left (2\right ) \operatorname{Ci}\left (-\frac{4}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) - 2 \, \sin \left (2\right ) \operatorname{Si}\left (\frac{4 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) - 2 \, \sin \left (2\right ) \operatorname{Si}\left (\frac{4}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*b*x - cos(2)*cos_integral(4*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) - 1)) - cos(2)*cos_integral(-4*e^(2*b*x +
2*a)/(e^(2*b*x + 2*a) - 1)) + cos(2)*cos_integral(4/(e^(2*b*x + 2*a) - 1)) + cos(2)*cos_integral(-4/(e^(2*b*x
+ 2*a) - 1)) - 2*sin(2)*sin_integral(4*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) - 1)) - 2*sin(2)*sin_integral(4/(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 ^{2}{\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))**2,x)

[Out]

Integral(sin(coth(a + b*x))**2, 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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(coth(b*x + a))^2, x)