3.3.19 \(\int \sin (\coth (a+b x)) \, dx\) [219]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 = {3414, 3384, 3380, 3383} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[Coth[a + b*x]],x]

[Out]

-1/2*(CosIntegral[1 - Coth[a + b*x]]*Sin[1])/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 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 3414

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])

Rubi steps

\begin {align*} \int \sin (\coth (a+b x)) \, dx &=\frac {\text {Subst}\left (\int \frac {\sin (x)}{1-x^2} \, dx,x,\coth (a+b x)\right )}{b}\\ &=\frac {\text {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 {\text {Subst}\left (\int \frac {\sin (x)}{1-x} \, dx,x,\coth (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\sin (x)}{1+x} \, dx,x,\coth (a+b x)\right )}{2 b}\\ &=-\frac {\cos (1) \text {Subst}\left (\int \frac {\sin (1-x)}{1-x} \, dx,x,\coth (a+b x)\right )}{2 b}+\frac {\cos (1) \text {Subst}\left (\int \frac {\sin (1+x)}{1+x} \, dx,x,\coth (a+b x)\right )}{2 b}+\frac {\sin (1) \text {Subst}\left (\int \frac {\cos (1-x)}{1-x} \, dx,x,\coth (a+b x)\right )}{2 b}-\frac {\sin (1) \text {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 = 0.10, size = 59, normalized size = 0.77 \begin {gather*} -\frac {\text {CosIntegral}(1-\coth (a+b x)) \sin (1)+\text {CosIntegral}(1+\coth (a+b x)) \sin (1)-\cos (1) (\text {Si}(1-\coth (a+b x))+\text {Si}(1+\coth (a+b x)))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[Coth[a + b*x]],x]

[Out]

-1/2*(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]]))/b

________________________________________________________________________________________

Maple [A]
time = 3.32, size = 58, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {\frac {\sinIntegral \left (\coth \left (b x +a \right )+1\right ) \cos \left (1\right )}{2}-\frac {\cosineIntegral \left (\coth \left (b x +a \right )+1\right ) \sin \left (1\right )}{2}-\frac {\sinIntegral \left (-1+\coth \left (b x +a \right )\right ) \cos \left (1\right )}{2}-\frac {\cosineIntegral \left (-1+\coth \left (b x +a \right )\right ) \sin \left (1\right )}{2}}{b}\)	\(58\)
default	\(\frac {\frac {\sinIntegral \left (\coth \left (b x +a \right )+1\right ) \cos \left (1\right )}{2}-\frac {\cosineIntegral \left (\coth \left (b x +a \right )+1\right ) \sin \left (1\right )}{2}-\frac {\sinIntegral \left (-1+\coth \left (b x +a \right )\right ) \cos \left (1\right )}{2}-\frac {\cosineIntegral \left (-1+\coth \left (b x +a \right )\right ) \sin \left (1\right )}{2}}{b}\)	\(58\)
risch	\(\frac {i {\mathrm e}^{-i} \expIntegral \left (1, -\frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}-2 i\right )}{4 b}-\frac {\pi \,\mathrm {csgn}\left (\frac {{\mathrm e}^{-a}}{-{\mathrm e}^{2 b x +a}+{\mathrm e}^{-a}}\right ) {\mathrm e}^{i}}{4 b}-\frac {\sinIntegral \left (\frac {2 \,{\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right ) {\mathrm e}^{i}}{2 b}-\frac {i \expIntegral \left (1, \frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right ) {\mathrm e}^{i}}{4 b}-\frac {i {\mathrm e}^{i} \expIntegral \left (1, \frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}+2 i\right )}{4 b}+\frac {i {\mathrm e}^{-i} \expIntegral \left (1, \frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{4 b}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(coth(b*x+a)),x,method=_RETURNVERBOSE)

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (67) = 134\).
time = 0.39, size = 149, normalized size = 1.94 \begin {gather*} -\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 {gather*}

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin {\left (\coth {\left (a + b x \right )} \right )}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(coth(a + b*x)),x)

[Out]

int(sin(coth(a + b*x)), x)

________________________________________________________________________________________