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

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

[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

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Rubi [A]  time = 0.104649, antiderivative size = 93, normalized size of antiderivative = 1., 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 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 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 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 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 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*}

Mathematica [A]  time = 6.86935, size = 167, normalized size = 1.8 \[ -\frac{\text{csch}\left (\frac{1}{2} (a+b x)\right ) \text{sech}\left (\frac{1}{2} (a+b x)\right ) \left (6 b x \sinh (a+b x) \text{PolyLog}(2,-\sinh (a+b x)-\cosh (a+b x))-6 b x \sinh (a+b x) \text{PolyLog}(2,\sinh (a+b x)+\cosh (a+b x))-6 \sinh (a+b x) \text{PolyLog}(3,-\sinh (a+b x)-\cosh (a+b x))+6 \sinh (a+b x) \text{PolyLog}(3,\sinh (a+b x)+\cosh (a+b x))+6 b^2 x^2 \sinh (a+b x) \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))+b^3 x^3\right )}{2 b^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(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]))/(2*b^4)

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Maple [A]  time = 0.028, size = 174, normalized size = 1.9 \begin{align*} -2\,{\frac{{x}^{3}{{\rm e}^{bx+a}}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-6\,{\frac{{a}^{2}{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-3\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}+3\,{\frac{{a}^{2}\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-6\,{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+6\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}-3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{4}}}+6\,{\frac{x{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-6\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}} \end{align*}

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 = 1.38132, size = 163, normalized size = 1.75 \begin{align*} -\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}} \end{align*}

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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Fricas [C]  time = 2.11485, size = 1461, normalized size = 15.71 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \cosh \left (b x + a\right ) \operatorname{csch}\left (b x + a\right )^{2}\,{d x} \end{align*}

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)