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

Optimal. Leaf size=148 \[ -\frac{3 x^2 \text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,-e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 x \text{PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 \text{PolyLog}\left (4,-e^{2 a+2 b x}\right )}{4 b^4}+\frac{3 \text{PolyLog}\left (4,e^{2 a+2 b x}\right )}{4 b^4}-\frac{2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.14994, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {5461, 4182, 2531, 6609, 2282, 6589} \[ -\frac{3 x^2 \text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,-e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 x \text{PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 \text{PolyLog}\left (4,-e^{2 a+2 b x}\right )}{4 b^4}+\frac{3 \text{PolyLog}\left (4,e^{2 a+2 b x}\right )}{4 b^4}-\frac{2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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 \text{csch}(a+b x) \text{sech}(a+b x) \, dx &=2 \int x^3 \text{csch}(2 a+2 b x) \, dx\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{3 \int x^2 \log \left (1-e^{2 a+2 b x}\right ) \, dx}{b}+\frac{3 \int x^2 \log \left (1+e^{2 a+2 b x}\right ) \, dx}{b}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{3 x^2 \text{Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x^2 \text{Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 \int x \text{Li}_2\left (-e^{2 a+2 b x}\right ) \, dx}{b^2}-\frac{3 \int x \text{Li}_2\left (e^{2 a+2 b x}\right ) \, dx}{b^2}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{3 x^2 \text{Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x^2 \text{Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 x \text{Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 \int \text{Li}_3\left (-e^{2 a+2 b x}\right ) \, dx}{2 b^3}+\frac{3 \int \text{Li}_3\left (e^{2 a+2 b x}\right ) \, dx}{2 b^3}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{3 x^2 \text{Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x^2 \text{Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 x \text{Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b^4}\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{3 x^2 \text{Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x^2 \text{Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 x \text{Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 \text{Li}_4\left (-e^{2 a+2 b x}\right )}{4 b^4}+\frac{3 \text{Li}_4\left (e^{2 a+2 b x}\right )}{4 b^4}\\ \end{align*}

Mathematica [A]  time = 4.00892, size = 150, normalized size = 1.01 \[ \frac{-6 b^2 x^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )+6 b^2 x^2 \text{PolyLog}\left (2,e^{2 (a+b x)}\right )+6 b x \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )-6 b x \text{PolyLog}\left (3,e^{2 (a+b x)}\right )-3 \text{PolyLog}\left (4,-e^{2 (a+b x)}\right )+3 \text{PolyLog}\left (4,e^{2 (a+b x)}\right )+4 b^3 x^3 \log \left (1-e^{2 (a+b x)}\right )-4 b^3 x^3 \log \left (e^{2 (a+b x)}+1\right )}{4 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^3*x^3*Log[1 - E^(2*(a + b*x))] - 4*b^3*x^3*Log[1 + E^(2*(a + b*x))] - 6*b^2*x^2*PolyLog[2, -E^(2*(a + b*x
))] + 6*b^2*x^2*PolyLog[2, E^(2*(a + b*x))] + 6*b*x*PolyLog[3, -E^(2*(a + b*x))] - 6*b*x*PolyLog[3, E^(2*(a +
b*x))] - 3*PolyLog[4, -E^(2*(a + b*x))] + 3*PolyLog[4, E^(2*(a + b*x))])/(4*b^4)

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Maple [A]  time = 0.082, size = 241, normalized size = 1.6 \begin{align*}{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{3}}{b}}-{\frac{{x}^{3}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}-{\frac{3\,{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{2}}}+{\frac{3\,x{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{3}}}+3\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}-6\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{3}}{{b}^{4}}}+6\,{\frac{{\it polylog} \left ( 4,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+6\,{\frac{{\it polylog} \left ( 4,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-{\frac{3\,{\it polylog} \left ( 4,-{{\rm e}^{2\,bx+2\,a}} \right ) }{4\,{b}^{4}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{3}}{b}}+3\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}-6\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}-{\frac{{a}^{3}\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16308, size = 274, normalized size = 1.85 \begin{align*} -\frac{4 \, b^{3} x^{3} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x{\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )}) + 3 \,{\rm Li}_{4}(-e^{\left (2 \, b x + 2 \, a\right )})}{3 \, b^{4}} + \frac{b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x{\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \,{\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} + \frac{b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x{\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \,{\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.63732, size = 1323, normalized size = 8.94 \begin{align*} \frac{b^{3} x^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2}{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2}{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 6 \, b x{\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, b x{\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 6 \, b x{\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 6 \, b x{\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) -{\left (b^{3} x^{3} + a^{3}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) -{\left (b^{3} x^{3} + a^{3}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) +{\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \,{\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \,{\rm polylog}\left (4, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 6 \,{\rm polylog}\left (4, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 6 \,{\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*csch(b*x+a)*sech(b*x+a),x)

[Out]

Integral(x**3*csch(a + b*x)*sech(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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