3.3.17 \(\int \sin ^3(\coth (a+b x)) \, dx\) [217]

Optimal. Leaf size=157 \[ -\frac {3 \text {CosIntegral}(1-\coth (a+b x)) \sin (1)}{8 b}-\frac {3 \text {CosIntegral}(1+\coth (a+b x)) \sin (1)}{8 b}+\frac {\text {CosIntegral}(3-3 \coth (a+b x)) \sin (3)}{8 b}+\frac {\text {CosIntegral}(3+3 \coth (a+b x)) \sin (3)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \coth (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\coth (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1+\coth (a+b x))}{8 b}-\frac {\cos (3) \text {Si}(3+3 \coth (a+b x))}{8 b} \]

[Out]

1/8*cos(3)*Si(-3+3*coth(b*x+a))/b-3/8*cos(1)*Si(-1+coth(b*x+a))/b+3/8*cos(1)*Si(1+coth(b*x+a))/b-1/8*cos(3)*Si
(3+3*coth(b*x+a))/b-3/8*Ci(1-coth(b*x+a))*sin(1)/b-3/8*Ci(1+coth(b*x+a))*sin(1)/b+1/8*Ci(3-3*coth(b*x+a))*sin(
3)/b+1/8*Ci(3+3*coth(b*x+a))*sin(3)/b

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Rubi [A]
time = 0.28, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 5, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {6857, 3393, 3384, 3380, 3383} \begin {gather*} \frac {\sin (3) \text {CosIntegral}(3-3 \coth (a+b x))}{8 b}+\frac {\sin (3) \text {CosIntegral}(3 \coth (a+b x)+3)}{8 b}-\frac {3 \sin (1) \text {CosIntegral}(1-\coth (a+b x))}{8 b}-\frac {3 \sin (1) \text {CosIntegral}(\coth (a+b x)+1)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \coth (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\coth (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(\coth (a+b x)+1)}{8 b}-\frac {\cos (3) \text {Si}(3 \coth (a+b x)+3)}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[Coth[a + b*x]]^3,x]

[Out]

(-3*CosIntegral[1 - Coth[a + b*x]]*Sin[1])/(8*b) - (3*CosIntegral[1 + Coth[a + b*x]]*Sin[1])/(8*b) + (CosInteg
ral[3 - 3*Coth[a + b*x]]*Sin[3])/(8*b) + (CosIntegral[3 + 3*Coth[a + b*x]]*Sin[3])/(8*b) - (Cos[3]*SinIntegral
[3 - 3*Coth[a + b*x]])/(8*b) + (3*Cos[1]*SinIntegral[1 - Coth[a + b*x]])/(8*b) + (3*Cos[1]*SinIntegral[1 + Cot
h[a + b*x]])/(8*b) - (Cos[3]*SinIntegral[3 + 3*Coth[a + b*x]])/(8*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 3393

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 6857

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]

Rubi steps

\begin {align*} \int \sin ^3(\coth (a+b x)) \, dx &=\frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{1-x^2} \, dx,x,\coth (a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {\sin ^3(x)}{2 (-1+x)}+\frac {\sin ^3(x)}{2 (1+x)}\right ) \, dx,x,\coth (a+b x)\right )}{b}\\ &=-\frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \frac {\sin ^3(x)}{1+x} \, dx,x,\coth (a+b x)\right )}{2 b}\\ &=-\frac {\text {Subst}\left (\int \left (\frac {3 \sin (x)}{4 (-1+x)}-\frac {\sin (3 x)}{4 (-1+x)}\right ) \, dx,x,\coth (a+b x)\right )}{2 b}+\frac {\text {Subst}\left (\int \left (\frac {3 \sin (x)}{4 (1+x)}-\frac {\sin (3 x)}{4 (1+x)}\right ) \, dx,x,\coth (a+b x)\right )}{2 b}\\ &=\frac {\text {Subst}\left (\int \frac {\sin (3 x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{8 b}-\frac {\text {Subst}\left (\int \frac {\sin (3 x)}{1+x} \, dx,x,\coth (a+b x)\right )}{8 b}-\frac {3 \text {Subst}\left (\int \frac {\sin (x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{8 b}+\frac {3 \text {Subst}\left (\int \frac {\sin (x)}{1+x} \, dx,x,\coth (a+b x)\right )}{8 b}\\ &=\frac {(3 \cos (1)) \text {Subst}\left (\int \frac {\sin (1-x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{8 b}+\frac {(3 \cos (1)) \text {Subst}\left (\int \frac {\sin (1+x)}{1+x} \, dx,x,\coth (a+b x)\right )}{8 b}-\frac {\cos (3) \text {Subst}\left (\int \frac {\sin (3-3 x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{8 b}-\frac {\cos (3) \text {Subst}\left (\int \frac {\sin (3+3 x)}{1+x} \, dx,x,\coth (a+b x)\right )}{8 b}-\frac {(3 \sin (1)) \text {Subst}\left (\int \frac {\cos (1-x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{8 b}-\frac {(3 \sin (1)) \text {Subst}\left (\int \frac {\cos (1+x)}{1+x} \, dx,x,\coth (a+b x)\right )}{8 b}+\frac {\sin (3) \text {Subst}\left (\int \frac {\cos (3-3 x)}{-1+x} \, dx,x,\coth (a+b x)\right )}{8 b}+\frac {\sin (3) \text {Subst}\left (\int \frac {\cos (3+3 x)}{1+x} \, dx,x,\coth (a+b x)\right )}{8 b}\\ &=-\frac {3 \text {Ci}(1-\coth (a+b x)) \sin (1)}{8 b}-\frac {3 \text {Ci}(1+\coth (a+b x)) \sin (1)}{8 b}+\frac {\text {Ci}(3-3 \coth (a+b x)) \sin (3)}{8 b}+\frac {\text {Ci}(3+3 \coth (a+b x)) \sin (3)}{8 b}-\frac {\cos (3) \text {Si}(3-3 \coth (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1-\coth (a+b x))}{8 b}+\frac {3 \cos (1) \text {Si}(1+\coth (a+b x))}{8 b}-\frac {\cos (3) \text {Si}(3+3 \coth (a+b x))}{8 b}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 124, normalized size = 0.79 \begin {gather*} \frac {-6 \text {CosIntegral}(1-\coth (a+b x)) \sin (1)-6 \text {CosIntegral}(1+\coth (a+b x)) \sin (1)+2 \text {CosIntegral}(3-3 \coth (a+b x)) \sin (3)+2 \text {CosIntegral}(3+3 \coth (a+b x)) \sin (3)-2 \cos (3) \text {Si}(3-3 \coth (a+b x))+6 \cos (1) \text {Si}(1-\coth (a+b x))+6 \cos (1) \text {Si}(1+\coth (a+b x))-2 \cos (3) \text {Si}(3+3 \coth (a+b x))}{16 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[Coth[a + b*x]]^3,x]

[Out]

(-6*CosIntegral[1 - Coth[a + b*x]]*Sin[1] - 6*CosIntegral[1 + Coth[a + b*x]]*Sin[1] + 2*CosIntegral[3 - 3*Coth
[a + b*x]]*Sin[3] + 2*CosIntegral[3 + 3*Coth[a + b*x]]*Sin[3] - 2*Cos[3]*SinIntegral[3 - 3*Coth[a + b*x]] + 6*
Cos[1]*SinIntegral[1 - Coth[a + b*x]] + 6*Cos[1]*SinIntegral[1 + Coth[a + b*x]] - 2*Cos[3]*SinIntegral[3 + 3*C
oth[a + b*x]])/(16*b)

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Maple [A]
time = 4.29, size = 118, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {-\frac {\sinIntegral \left (3+3 \coth \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\cosineIntegral \left (3+3 \coth \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {\sinIntegral \left (-3+3 \coth \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\cosineIntegral \left (-3+3 \coth \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {3 \sinIntegral \left (\coth \left (b x +a \right )+1\right ) \cos \left (1\right )}{8}-\frac {3 \cosineIntegral \left (\coth \left (b x +a \right )+1\right ) \sin \left (1\right )}{8}-\frac {3 \sinIntegral \left (-1+\coth \left (b x +a \right )\right ) \cos \left (1\right )}{8}-\frac {3 \cosineIntegral \left (-1+\coth \left (b x +a \right )\right ) \sin \left (1\right )}{8}}{b}\)	\(118\)
default	\(\frac {-\frac {\sinIntegral \left (3+3 \coth \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\cosineIntegral \left (3+3 \coth \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {\sinIntegral \left (-3+3 \coth \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\cosineIntegral \left (-3+3 \coth \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {3 \sinIntegral \left (\coth \left (b x +a \right )+1\right ) \cos \left (1\right )}{8}-\frac {3 \cosineIntegral \left (\coth \left (b x +a \right )+1\right ) \sin \left (1\right )}{8}-\frac {3 \sinIntegral \left (-1+\coth \left (b x +a \right )\right ) \cos \left (1\right )}{8}-\frac {3 \cosineIntegral \left (-1+\coth \left (b x +a \right )\right ) \sin \left (1\right )}{8}}{b}\)	\(118\)
risch	\(\frac {i {\mathrm e}^{3 i} \expIntegral \left (1, -\frac {6 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{16 b}-\frac {i {\mathrm e}^{-3 i} \expIntegral \left (1, -\frac {6 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}-6 i\right )}{16 b}+\frac {\pi \,{\mathrm e}^{-3 i} \mathrm {csgn}\left (\frac {{\mathrm e}^{-a}}{-{\mathrm e}^{2 b x +a}+{\mathrm e}^{-a}}\right )}{16 b}+\frac {{\mathrm e}^{-3 i} \sinIntegral \left (\frac {6 \,{\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{8 b}-\frac {i \expIntegral \left (1, -\frac {6 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right ) {\mathrm e}^{-3 i}}{16 b}+\frac {i {\mathrm e}^{3 i} \expIntegral \left (1, \frac {6 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}+6 i\right )}{16 b}-\frac {3 \pi \,\mathrm {csgn}\left (\frac {{\mathrm e}^{-a}}{-{\mathrm e}^{2 b x +a}+{\mathrm e}^{-a}}\right ) {\mathrm e}^{-i}}{16 b}-\frac {3 \,{\mathrm e}^{-i} \sinIntegral \left (\frac {2 \,{\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{8 b}+\frac {3 i {\mathrm e}^{-i} \expIntegral \left (1, -\frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{16 b}-\frac {3 i {\mathrm e}^{i} \expIntegral \left (1, \frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}+2 i\right )}{16 b}-\frac {3 i {\mathrm e}^{i} \expIntegral \left (1, -\frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{16 b}+\frac {3 i {\mathrm e}^{-i} \expIntegral \left (1, -\frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}-2 i\right )}{16 b}\)	\(410\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(coth(b*x+a))^3,x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/8*Si(3+3*coth(b*x+a))*cos(3)+1/8*Ci(3+3*coth(b*x+a))*sin(3)+1/8*Si(-3+3*coth(b*x+a))*cos(3)+1/8*Ci(-3+
3*coth(b*x+a))*sin(3)+3/8*Si(coth(b*x+a)+1)*cos(1)-3/8*Ci(coth(b*x+a)+1)*sin(1)-3/8*Si(-1+coth(b*x+a))*cos(1)-
3/8*Ci(-1+coth(b*x+a))*sin(1))

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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))^3,x, algorithm="maxima")

[Out]

integrate(sin(coth(b*x + a))^3, x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (139) = 278\).
time = 0.42, size = 296, normalized size = 1.89 \begin {gather*} \frac {\operatorname {Ci}\left (\frac {6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (3\right ) + \operatorname {Ci}\left (-\frac {6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (3\right ) + \operatorname {Ci}\left (\frac {6}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (3\right ) + \operatorname {Ci}\left (-\frac {6}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (3\right ) - 3 \, \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 ) - 3 \, \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 ) - 3 \, \operatorname {Ci}\left (\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (1\right ) - 3 \, \operatorname {Ci}\left (-\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) \sin \left (1\right ) - 2 \, \cos \left (3\right ) \operatorname {Si}\left (\frac {6 \, e^{\left (2 \, b x + 2 \, a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) + 6 \, \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 (3\right ) \operatorname {Si}\left (\frac {6}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right ) - 6 \, \cos \left (1\right ) \operatorname {Si}\left (\frac {2}{e^{\left (2 \, b x + 2 \, a\right )} - 1}\right )}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(coth(b*x+a))^3,x, algorithm="fricas")

[Out]

1/16*(cos_integral(6*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) - 1))*sin(3) + cos_integral(-6*e^(2*b*x + 2*a)/(e^(2*b*x
+ 2*a) - 1))*sin(3) + cos_integral(6/(e^(2*b*x + 2*a) - 1))*sin(3) + cos_integral(-6/(e^(2*b*x + 2*a) - 1))*s
in(3) - 3*cos_integral(2*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) - 1))*sin(1) - 3*cos_integral(-2*e^(2*b*x + 2*a)/(e^
(2*b*x + 2*a) - 1))*sin(1) - 3*cos_integral(2/(e^(2*b*x + 2*a) - 1))*sin(1) - 3*cos_integral(-2/(e^(2*b*x + 2*
a) - 1))*sin(1) - 2*cos(3)*sin_integral(6*e^(2*b*x + 2*a)/(e^(2*b*x + 2*a) - 1)) + 6*cos(1)*sin_integral(2*e^(
2*b*x + 2*a)/(e^(2*b*x + 2*a) - 1)) + 2*cos(3)*sin_integral(6/(e^(2*b*x + 2*a) - 1)) - 6*cos(1)*sin_integral(2
/(e^(2*b*x + 2*a) - 1)))/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{3}{\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))**3,x)

[Out]

Integral(sin(coth(a + b*x))**3, x)

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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))^3,x, algorithm="giac")

[Out]

integrate(sin(coth(b*x + a))^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (\mathrm {coth}\left (a+b\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(coth(a + b*x))^3,x)

[Out]

int(sin(coth(a + b*x))^3, x)

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