### 3.495 $$\int x^3 \text{csch}^2(a+b x) \text{sech}^2(a+b x) \, dx$$

Optimal. Leaf size=85 $\frac{3 x \text{PolyLog}\left (2,e^{4 (a+b x)}\right )}{2 b^3}-\frac{3 \text{PolyLog}\left (3,e^{4 (a+b x)}\right )}{8 b^4}+\frac{3 x^2 \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac{2 x^3 \coth (2 a+2 b x)}{b}-\frac{2 x^3}{b}$

[Out]

(-2*x^3)/b - (2*x^3*Coth[2*a + 2*b*x])/b + (3*x^2*Log[1 - E^(4*(a + b*x))])/b^2 + (3*x*PolyLog[2, E^(4*(a + b*
x))])/(2*b^3) - (3*PolyLog[3, E^(4*(a + b*x))])/(8*b^4)

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Rubi [A]  time = 0.238234, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.35, Rules used = {5461, 4184, 3716, 2190, 2531, 2282, 6589} $\frac{3 x \text{PolyLog}\left (2,e^{4 (a+b x)}\right )}{2 b^3}-\frac{3 \text{PolyLog}\left (3,e^{4 (a+b x)}\right )}{8 b^4}+\frac{3 x^2 \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac{2 x^3 \coth (2 a+2 b x)}{b}-\frac{2 x^3}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Csch[a + b*x]^2*Sech[a + b*x]^2,x]

[Out]

(-2*x^3)/b - (2*x^3*Coth[2*a + 2*b*x])/b + (3*x^2*Log[1 - E^(4*(a + b*x))])/b^2 + (3*x*PolyLog[2, E^(4*(a + b*
x))])/(2*b^3) - (3*PolyLog[3, E^(4*(a + b*x))])/(8*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 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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 \text{csch}^2(a+b x) \text{sech}^2(a+b x) \, dx &=4 \int x^3 \text{csch}^2(2 a+2 b x) \, dx\\ &=-\frac{2 x^3 \coth (2 a+2 b x)}{b}+\frac{6 \int x^2 \coth (2 a+2 b x) \, dx}{b}\\ &=-\frac{2 x^3}{b}-\frac{2 x^3 \coth (2 a+2 b x)}{b}-\frac{12 \int \frac{e^{2 (2 a+2 b x)} x^2}{1-e^{2 (2 a+2 b x)}} \, dx}{b}\\ &=-\frac{2 x^3}{b}-\frac{2 x^3 \coth (2 a+2 b x)}{b}+\frac{3 x^2 \log \left (1-e^{4 (a+b x)}\right )}{b^2}-\frac{6 \int x \log \left (1-e^{2 (2 a+2 b x)}\right ) \, dx}{b^2}\\ &=-\frac{2 x^3}{b}-\frac{2 x^3 \coth (2 a+2 b x)}{b}+\frac{3 x^2 \log \left (1-e^{4 (a+b x)}\right )}{b^2}+\frac{3 x \text{Li}_2\left (e^{4 (a+b x)}\right )}{2 b^3}-\frac{3 \int \text{Li}_2\left (e^{2 (2 a+2 b x)}\right ) \, dx}{2 b^3}\\ &=-\frac{2 x^3}{b}-\frac{2 x^3 \coth (2 a+2 b x)}{b}+\frac{3 x^2 \log \left (1-e^{4 (a+b x)}\right )}{b^2}+\frac{3 x \text{Li}_2\left (e^{4 (a+b x)}\right )}{2 b^3}-\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 (2 a+2 b x)}\right )}{8 b^4}\\ &=-\frac{2 x^3}{b}-\frac{2 x^3 \coth (2 a+2 b x)}{b}+\frac{3 x^2 \log \left (1-e^{4 (a+b x)}\right )}{b^2}+\frac{3 x \text{Li}_2\left (e^{4 (a+b x)}\right )}{2 b^3}-\frac{3 \text{Li}_3\left (e^{4 (a+b x)}\right )}{8 b^4}\\ \end{align*}

Mathematica [B]  time = 5.73757, size = 284, normalized size = 3.34 $4 \left (\frac{x^3 \text{csch}(2 a) \sinh (2 b x) \text{csch}(2 a+2 b x)}{2 b}-\frac{e^{4 a} \left (12 \left (1-e^{-4 a}\right ) \left (b x \text{PolyLog}\left (2,-e^{-a-b x}\right )+\text{PolyLog}\left (3,-e^{-a-b x}\right )\right )+12 \left (1-e^{-4 a}\right ) \left (b x \text{PolyLog}\left (2,e^{-a-b x}\right )+\text{PolyLog}\left (3,e^{-a-b x}\right )\right )+3 e^{-4 a} \left (e^{4 a}-1\right ) \left (2 b x \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )+\text{PolyLog}\left (3,-e^{-2 (a+b x)}\right )\right )+8 e^{-4 a} b^3 x^3-6 \left (1-e^{-4 a}\right ) b^2 x^2 \log \left (1-e^{-a-b x}\right )-6 \left (1-e^{-4 a}\right ) b^2 x^2 \log \left (e^{-a-b x}+1\right )-6 \left (1-e^{-4 a}\right ) b^2 x^2 \log \left (e^{-2 (a+b x)}+1\right )\right )}{8 \left (e^{4 a}-1\right ) b^4}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*Csch[a + b*x]^2*Sech[a + b*x]^2,x]

[Out]

4*(-(E^(4*a)*((8*b^3*x^3)/E^(4*a) - 6*b^2*(1 - E^(-4*a))*x^2*Log[1 - E^(-a - b*x)] - 6*b^2*(1 - E^(-4*a))*x^2*
Log[1 + E^(-a - b*x)] - 6*b^2*(1 - E^(-4*a))*x^2*Log[1 + E^(-2*(a + b*x))] + 12*(1 - E^(-4*a))*(b*x*PolyLog[2,
-E^(-a - b*x)] + PolyLog[3, -E^(-a - b*x)]) + 12*(1 - E^(-4*a))*(b*x*PolyLog[2, E^(-a - b*x)] + PolyLog[3, E^
(-a - b*x)]) + (3*(-1 + E^(4*a))*(2*b*x*PolyLog[2, -E^(-2*(a + b*x))] + PolyLog[3, -E^(-2*(a + b*x))]))/E^(4*a
)))/(8*b^4*(-1 + E^(4*a))) + (x^3*Csch[2*a]*Csch[2*a + 2*b*x]*Sinh[2*b*x])/(2*b))

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Maple [B]  time = 0.052, size = 263, normalized size = 3.1 \begin{align*} -4\,{\frac{{x}^{3}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}+8\,{\frac{{a}^{3}}{{b}^{4}}}-6\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-{\frac{3\,{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{4}}}+12\,{\frac{{a}^{2}x}{{b}^{3}}}+3\,{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}+3\,{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{3}}}+6\,{\frac{x{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+3\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}+6\,{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}-6\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-4\,{\frac{{x}^{3}}{b}}+3\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{4}}}-12\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{2}}{{b}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x^3/b/(exp(2*b*x+2*a)-1)/(1+exp(2*b*x+2*a))+8/b^4*a^3-6*polylog(3,exp(b*x+a))/b^4-3/2*polylog(3,-exp(2*b*x+
2*a))/b^4+12/b^3*a^2*x+3*x^2*ln(1+exp(2*b*x+2*a))/b^2+3*x*polylog(2,-exp(2*b*x+2*a))/b^3+6*x*polylog(2,exp(b*x
+a))/b^3+3/b^2*ln(1+exp(b*x+a))*x^2+6*x*polylog(2,-exp(b*x+a))/b^3+3/b^2*ln(1-exp(b*x+a))*x^2-6*polylog(3,-exp
(b*x+a))/b^4-4*x^3/b+3/b^4*a^2*ln(exp(b*x+a)-1)-12/b^4*a^2*ln(exp(b*x+a))-3/b^4*ln(1-exp(b*x+a))*a^2

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Maxima [B]  time = 1.17352, size = 243, normalized size = 2.86 \begin{align*} -\frac{4 \, x^{3}}{b e^{\left (4 \, b x + 4 \, a\right )} - b} - \frac{4 \, x^{3}}{b} + \frac{3 \,{\left (2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) -{\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})\right )}}{2 \, 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}} + \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*csch(b*x+a)^2*sech(b*x+a)^2,x, algorithm="maxima")

[Out]

-4*x^3/(b*e^(4*b*x + 4*a) - b) - 4*x^3/b + 3/2*(2*b^2*x^2*log(e^(2*b*x + 2*a) + 1) + 2*b*x*dilog(-e^(2*b*x + 2
*a)) - polylog(3, -e^(2*b*x + 2*a)))/b^4 + 3*(b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*pol
ylog(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 = 2.82888, size = 5019, normalized size = 59.05 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*(b^3*x^3 + a^3)*cosh(b*x + a)^4 + 16*(b^3*x^3 + a^3)*cosh(b*x + a)^3*sinh(b*x + a) + 24*(b^3*x^3 + a^3)*co
sh(b*x + a)^2*sinh(b*x + a)^2 + 16*(b^3*x^3 + a^3)*cosh(b*x + a)*sinh(b*x + a)^3 + 4*(b^3*x^3 + a^3)*sinh(b*x
+ a)^4 - 4*a^3 - 6*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*x*cosh(b*x + a)^2*sinh(b*x
+ a)^2 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - b*x)*dilog(cosh(b*x + a) + sinh(b*x + a)
) - 6*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*x*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b
*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - b*x)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 6*(b*
x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*x*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b*x*cosh(b
*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - b*x)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) - 6*(b*x*cosh(b
*x + a)^4 + 4*b*x*cosh(b*x + a)^3*sinh(b*x + a) + 6*b*x*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b*x*cosh(b*x + a)*
sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 - b*x)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - 3*(b^2*x^2*cosh(b*x + a)^
4 + 4*b^2*x^2*cosh(b*x + a)^3*sinh(b*x + a) + 6*b^2*x^2*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*b^2*x^2*cosh(b*x +
a)*sinh(b*x + a)^3 + b^2*x^2*sinh(b*x + a)^4 - b^2*x^2)*log(cosh(b*x + a) + sinh(b*x + a) + 1) - 3*(a^2*cosh(
b*x + a)^4 + 4*a^2*cosh(b*x + a)^3*sinh(b*x + a) + 6*a^2*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*a^2*cosh(b*x + a)
*sinh(b*x + a)^3 + a^2*sinh(b*x + a)^4 - a^2)*log(cosh(b*x + a) + sinh(b*x + a) + I) - 3*(a^2*cosh(b*x + a)^4
+ 4*a^2*cosh(b*x + a)^3*sinh(b*x + a) + 6*a^2*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*a^2*cosh(b*x + a)*sinh(b*x +
a)^3 + a^2*sinh(b*x + a)^4 - a^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) - 3*(a^2*cosh(b*x + a)^4 + 4*a^2*cos
h(b*x + a)^3*sinh(b*x + a) + 6*a^2*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*a^2*cosh(b*x + a)*sinh(b*x + a)^3 + a^2
*sinh(b*x + a)^4 - a^2)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - 3*((b^2*x^2 - a^2)*cosh(b*x + a)^4 + 4*(b^2*x
^2 - a^2)*cosh(b*x + a)^3*sinh(b*x + a) + 6*(b^2*x^2 - a^2)*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*(b^2*x^2 - a^2
)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2 - a^2)*sinh(b*x + a)^4 - b^2*x^2 + a^2)*log(I*cosh(b*x + a) + I*sin
h(b*x + a) + 1) - 3*((b^2*x^2 - a^2)*cosh(b*x + a)^4 + 4*(b^2*x^2 - a^2)*cosh(b*x + a)^3*sinh(b*x + a) + 6*(b^
2*x^2 - a^2)*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2 - a^
2)*sinh(b*x + a)^4 - b^2*x^2 + a^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - 3*((b^2*x^2 - a^2)*cosh(b*x
+ a)^4 + 4*(b^2*x^2 - a^2)*cosh(b*x + a)^3*sinh(b*x + a) + 6*(b^2*x^2 - a^2)*cosh(b*x + a)^2*sinh(b*x + a)^2 +
4*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2 - a^2)*sinh(b*x + a)^4 - b^2*x^2 + a^2)*log(-cosh(
b*x + a) - sinh(b*x + a) + 1) + 6*(cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(
b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*polylog(3, cosh(b*x + a) + sinh(b*x + a))
+ 6*(cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*s
inh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + 6*(cosh(b*x + a)^4 + 4*c
osh(b*x + a)^3*sinh(b*x + a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x
+ a)^4 - 1)*polylog(3, -I*cosh(b*x + a) - I*sinh(b*x + a)) + 6*(cosh(b*x + a)^4 + 4*cosh(b*x + a)^3*sinh(b*x +
a) + 6*cosh(b*x + a)^2*sinh(b*x + a)^2 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 - 1)*polylog(3, -c
osh(b*x + a) - sinh(b*x + a)))/(b^4*cosh(b*x + a)^4 + 4*b^4*cosh(b*x + a)^3*sinh(b*x + a) + 6*b^4*cosh(b*x + a
)^2*sinh(b*x + a)^2 + 4*b^4*cosh(b*x + a)*sinh(b*x + a)^3 + b^4*sinh(b*x + a)^4 - b^4)

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

[Out]

Integral(x**3*csch(a + b*x)**2*sech(a + b*x)**2, 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 )^{2} \operatorname{sech}\left (b x + a\right )^{2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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