3.425 \(\int x^3 \coth (a+b x) \text {csch}(a+b x) \, dx\)

Optimal. Leaf size=93 \[ \frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac {6 \text {Li}_3\left (e^{a+b x}\right )}{b^4}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b} \]

[Out]

-6*x^2*arctanh(exp(b*x+a))/b^2-x^3*csch(b*x+a)/b-6*x*polylog(2,-exp(b*x+a))/b^3+6*x*polylog(2,exp(b*x+a))/b^3+
6*polylog(3,-exp(b*x+a))/b^4-6*polylog(3,exp(b*x+a))/b^4

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Rubi [A]  time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5419, 4182, 2531, 2282, 6589} \[ -\frac {6 x \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac {6 x \text {PolyLog}\left (2,e^{a+b x}\right )}{b^3}+\frac {6 \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^4}-\frac {6 \text {PolyLog}\left (3,e^{a+b x}\right )}{b^4}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Coth[a + b*x]*Csch[a + b*x],x]

[Out]

(-6*x^2*ArcTanh[E^(a + b*x)])/b^2 - (x^3*Csch[a + b*x])/b - (6*x*PolyLog[2, -E^(a + b*x)])/b^3 + (6*x*PolyLog[
2, E^(a + b*x)])/b^3 + (6*PolyLog[3, -E^(a + b*x)])/b^4 - (6*PolyLog[3, E^(a + b*x)])/b^4

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5419

Int[Coth[(a_.) + (b_.)*(x_)^(n_.)]^(q_.)*Csch[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Csch[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Csch[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int x^3 \coth (a+b x) \text {csch}(a+b x) \, dx &=-\frac {x^3 \text {csch}(a+b x)}{b}+\frac {3 \int x^2 \text {csch}(a+b x) \, dx}{b}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 \int x \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac {6 \int x \log \left (1+e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \int \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^3}-\frac {6 \int \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^3}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac {6 \text {Li}_3\left (e^{a+b x}\right )}{b^4}\\ \end {align*}

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Mathematica [A]  time = 7.21, size = 167, normalized size = 1.80 \[ -\frac {\text {csch}\left (\frac {1}{2} (a+b x)\right ) \text {sech}\left (\frac {1}{2} (a+b x)\right ) \left (6 b^2 x^2 \sinh (a+b x) \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))+6 b x \sinh (a+b x) \text {Li}_2(-\cosh (a+b x)-\sinh (a+b x))-6 b x \sinh (a+b x) \text {Li}_2(\cosh (a+b x)+\sinh (a+b x))-6 \sinh (a+b x) \text {Li}_3(-\cosh (a+b x)-\sinh (a+b x))+6 \sinh (a+b x) \text {Li}_3(\cosh (a+b x)+\sinh (a+b x))+b^3 x^3\right )}{2 b^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3*Coth[a + b*x]*Csch[a + b*x],x]

[Out]

-1/2*(Csch[(a + b*x)/2]*Sech[(a + b*x)/2]*(b^3*x^3 + 6*b^2*x^2*ArcTanh[Cosh[a + b*x] + Sinh[a + b*x]]*Sinh[a +
b*x] + 6*b*x*PolyLog[2, -Cosh[a + b*x] - Sinh[a + b*x]]*Sinh[a + b*x] - 6*b*x*PolyLog[2, Cosh[a + b*x] + Sinh
[a + b*x]]*Sinh[a + b*x] - 6*PolyLog[3, -Cosh[a + b*x] - Sinh[a + b*x]]*Sinh[a + b*x] + 6*PolyLog[3, Cosh[a +
b*x] + Sinh[a + b*x]]*Sinh[a + b*x]))/b^4

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fricas [C]  time = 0.43, size = 551, normalized size = 5.92 \[ -\frac {2 \, b^{3} x^{3} \cosh \left (b x + a\right ) + 2 \, b^{3} x^{3} \sinh \left (b x + a\right ) - 6 \, {\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} - b x\right )} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, {\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} - b x\right )} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + 3 \, {\left (b^{2} x^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} x^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} x^{2} \sinh \left (b x + a\right )^{2} - b^{2} x^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 3 \, {\left (a^{2} \cosh \left (b x + a\right )^{2} + 2 \, a^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a^{2} \sinh \left (b x + a\right )^{2} - a^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 3 \, {\left (b^{2} x^{2} - {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (b^{2} x^{2} - a^{2}\right )} \sinh \left (b x + a\right )^{2} - a^{2}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{b^{4} \cosh \left (b x + a\right )^{2} + 2 \, b^{4} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{4} \sinh \left (b x + a\right )^{2} - b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="fricas")

[Out]

-(2*b^3*x^3*cosh(b*x + a) + 2*b^3*x^3*sinh(b*x + a) - 6*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x + a)*sinh(b*x +
a) + b*x*sinh(b*x + a)^2 - b*x)*dilog(cosh(b*x + a) + sinh(b*x + a)) + 6*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*x
+ a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2 - b*x)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + 3*(b^2*x^2*cosh(b*x +
a)^2 + 2*b^2*x^2*cosh(b*x + a)*sinh(b*x + a) + b^2*x^2*sinh(b*x + a)^2 - b^2*x^2)*log(cosh(b*x + a) + sinh(b*
x + a) + 1) - 3*(a^2*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2*sinh(b*x + a)^2 - a^2)*log(cosh
(b*x + a) + sinh(b*x + a) - 1) + 3*(b^2*x^2 - (b^2*x^2 - a^2)*cosh(b*x + a)^2 - 2*(b^2*x^2 - a^2)*cosh(b*x + a
)*sinh(b*x + a) - (b^2*x^2 - a^2)*sinh(b*x + a)^2 - a^2)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 6*(cosh(b*x
+ a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*polylog(3, cosh(b*x + a) + sinh(b*x + a)) - 6*(
cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)*polylog(3, -cosh(b*x + a) - sinh(b*x +
a)))/(b^4*cosh(b*x + a)^2 + 2*b^4*cosh(b*x + a)*sinh(b*x + a) + b^4*sinh(b*x + a)^2 - b^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \cosh \left (b x + a\right ) \operatorname {csch}\left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^3*cosh(b*x + a)*csch(b*x + a)^2, x)

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maple [A]  time = 0.22, size = 174, normalized size = 1.87 \[ -\frac {2 x^{3} {\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}-\frac {6 a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {3 \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}+\frac {3 \ln \left (1+{\mathrm e}^{b x +a}\right ) a^{2}}{b^{4}}-\frac {6 x \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {6 \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}-\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{4}}+\frac {6 x \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {6 \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosh(b*x+a)*csch(b*x+a)^2,x)

[Out]

-2/b*x^3*exp(b*x+a)/(exp(2*b*x+2*a)-1)-6/b^4*a^2*arctanh(exp(b*x+a))-3/b^2*ln(1+exp(b*x+a))*x^2+3/b^4*ln(1+exp
(b*x+a))*a^2-6*x*polylog(2,-exp(b*x+a))/b^3+6*polylog(3,-exp(b*x+a))/b^4+3/b^2*ln(1-exp(b*x+a))*x^2-3/b^4*ln(1
-exp(b*x+a))*a^2+6*x*polylog(2,exp(b*x+a))/b^3-6*polylog(3,exp(b*x+a))/b^4

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maxima [A]  time = 0.54, size = 121, normalized size = 1.30 \[ -\frac {2 \, x^{3} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac {3 \, {\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x+a)*csch(b*x+a)^2,x, algorithm="maxima")

[Out]

-2*x^3*e^(b*x + a)/(b*e^(2*b*x + 2*a) - b) - 3*(b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*p
olylog(3, -e^(b*x + a)))/b^4 + 3*(b^2*x^2*log(-e^(b*x + a) + 1) + 2*b*x*dilog(e^(b*x + a)) - 2*polylog(3, e^(b
*x + a)))/b^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\mathrm {cosh}\left (a+b\,x\right )}{{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*cosh(a + b*x))/sinh(a + b*x)^2,x)

[Out]

int((x^3*cosh(a + b*x))/sinh(a + b*x)^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cosh(b*x+a)*csch(b*x+a)**2,x)

[Out]

Timed out

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