3.5.82 \(\int x^2 \text {csch}(a+b x) \text {sech}^3(a+b x) \, dx\) [482]

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

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2700, 14, 5570, 2631, 12, 4267, 2611, 2320, 6724, 3801, 3556, 30} \begin {gather*} \frac {\text {Li}_3\left (-e^{2 a+2 b x}\right )}{2 b^3}-\frac {\text {Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}+\frac {\log (\cosh (a+b x))}{b^3}-\frac {x \text {Li}_2\left (-e^{2 a+2 b x}\right )}{b^2}+\frac {x \text {Li}_2\left (e^{2 a+2 b x}\right )}{b^2}-\frac {x \tanh (a+b x)}{b^2}-\frac {2 x^2 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac {x^2 \tanh ^2(a+b x)}{2 b}+\frac {x^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2320

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 2611

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 2631

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1)*(Log[u]/(b*(m + 1))), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[(a + b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 2700

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3801

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

Rule 4267

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 5570

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Wit
h[{u = IntHide[Csch[a + b*x]^n*Sech[a + b*x]^p, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[(c + d*x)^(m - 1)
*u, x], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, p]

Rule 6724

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

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Mathematica [A]
time = 2.44, size = 284, normalized size = 1.92 \begin {gather*} \frac {1}{6} \left (\frac {-12 b e^{2 a} x+4 b^3 e^{2 a} x^3+6 \log \left (1+e^{2 (a+b x)}\right )+6 e^{2 a} \log \left (1+e^{2 (a+b x)}\right )-6 b^2 x^2 \log \left (1+e^{2 (a+b x)}\right )-6 b^2 e^{2 a} x^2 \log \left (1+e^{2 (a+b x)}\right )-6 b \left (1+e^{2 a}\right ) x \text {PolyLog}\left (2,-e^{2 (a+b x)}\right )+3 \left (1+e^{2 a}\right ) \text {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{b^3 \left (1+e^{2 a}\right )}+\frac {2 b^2 x^2 \left (-\frac {2 b e^{2 a} x}{-1+e^{2 a}}+3 \log \left (1-e^{2 (a+b x)}\right )\right )+6 b x \text {PolyLog}\left (2,e^{2 (a+b x)}\right )-3 \text {PolyLog}\left (3,e^{2 (a+b x)}\right )}{b^3}+2 x^3 \text {csch}(a) \text {sech}(a)+\frac {3 x^2 \text {sech}^2(a+b x)}{b}-\frac {6 x \text {sech}(a) \text {sech}(a+b x) \sinh (b x)}{b^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.83, size = 256, normalized size = 1.73

method	result	size
risch	\(\frac {2 x \left (b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 b x +2 a}+1\right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}+1\right )^{2}}+\frac {2 x \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b}+\frac {2 x \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {x^{2} \ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b}-\frac {x \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{2}}{b}-\frac {2 \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}}-\frac {2 \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b^{3}}-\frac {2 \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{3}}-\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{3}}\)	\(256\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 229, normalized size = 1.55 \begin {gather*} \frac {2 \, {\left ({\left (b x^{2} e^{\left (2 \, a\right )} + x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} - \frac {2 \, x}{b^{2}} - \frac {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 )})}{2 \, b^{3}} + \frac {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 )})}{b^{3}} + \frac {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 )})}{b^{3}} + \frac {\log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((b*x^2*e^(2*a) + x*e^(2*a))*e^(2*b*x) + x)/(b^2*e^(4*b*x + 4*a) + 2*b^2*e^(2*b*x + 2*a) + b^2) - 2*x/b^2 -
1/2*(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^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^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^3 + log(e^(2*b*x + 2*a) + 1)/b^3

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Fricas [C] Result contains complex when optimal does not.
time = 0.44, size = 2523, normalized size = 17.05 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(b*x + a)*cosh(b*x + a)^4 + 8*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + 2*(b*x + a)*sinh(b*x + a)^4 - 2*(b
^2*x^2 - b*x - 2*a)*cosh(b*x + a)^2 - 2*(b^2*x^2 - 6*(b*x + a)*cosh(b*x + a)^2 - b*x - 2*a)*sinh(b*x + a)^2 -
2*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 + 2*b*x*cosh(b*x + a)^2 + 2
*(3*b*x*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(b*x + a)^3 + b*x*cosh(b*x + a))*sinh(b*x +
a))*dilog(cosh(b*x + a) + sinh(b*x + a)) + 2*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*
sinh(b*x + a)^4 + 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(
b*x + a)^3 + b*x*cosh(b*x + a))*sinh(b*x + a))*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + 2*(b*x*cosh(b*x + a)
^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 + 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a
)^2 + b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(b*x + a)^3 + b*x*cosh(b*x + a))*sinh(b*x + a))*dilog(-I*cosh(b*
x + a) - I*sinh(b*x + a)) - 2*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4
+ 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(b*x + a)^3 + b*
x*cosh(b*x + a))*sinh(b*x + a))*dilog(-cosh(b*x + a) - sinh(b*x + a)) - (b^2*x^2*cosh(b*x + a)^4 + 4*b^2*x^2*c
osh(b*x + a)*sinh(b*x + a)^3 + b^2*x^2*sinh(b*x + a)^4 + 2*b^2*x^2*cosh(b*x + a)^2 + b^2*x^2 + 2*(3*b^2*x^2*co
sh(b*x + a)^2 + b^2*x^2)*sinh(b*x + a)^2 + 4*(b^2*x^2*cosh(b*x + a)^3 + b^2*x^2*cosh(b*x + a))*sinh(b*x + a))*
log(cosh(b*x + a) + sinh(b*x + a) + 1) + ((a^2 - 1)*cosh(b*x + a)^4 + 4*(a^2 - 1)*cosh(b*x + a)*sinh(b*x + a)^
3 + (a^2 - 1)*sinh(b*x + a)^4 + 2*(a^2 - 1)*cosh(b*x + a)^2 + 2*(3*(a^2 - 1)*cosh(b*x + a)^2 + a^2 - 1)*sinh(b
*x + a)^2 + a^2 + 4*((a^2 - 1)*cosh(b*x + a)^3 + (a^2 - 1)*cosh(b*x + a))*sinh(b*x + a) - 1)*log(cosh(b*x + a)
+ sinh(b*x + a) + I) + ((a^2 - 1)*cosh(b*x + a)^4 + 4*(a^2 - 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (a^2 - 1)*sin
h(b*x + a)^4 + 2*(a^2 - 1)*cosh(b*x + a)^2 + 2*(3*(a^2 - 1)*cosh(b*x + a)^2 + a^2 - 1)*sinh(b*x + a)^2 + a^2 +
4*((a^2 - 1)*cosh(b*x + a)^3 + (a^2 - 1)*cosh(b*x + a))*sinh(b*x + a) - 1)*log(cosh(b*x + a) + sinh(b*x + a)
- I) - (a^2*cosh(b*x + a)^4 + 4*a^2*cosh(b*x + a)*sinh(b*x + a)^3 + a^2*sinh(b*x + a)^4 + 2*a^2*cosh(b*x + a)^
2 + 2*(3*a^2*cosh(b*x + a)^2 + a^2)*sinh(b*x + a)^2 + a^2 + 4*(a^2*cosh(b*x + a)^3 + a^2*cosh(b*x + a))*sinh(b
*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - 1) + ((b^2*x^2 - a^2)*cosh(b*x + a)^4 + 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 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^
2*x^2 + 3*(b^2*x^2 - a^2)*cosh(b*x + a)^2 - a^2)*sinh(b*x + a)^2 - a^2 + 4*((b^2*x^2 - a^2)*cosh(b*x + a)^3 +
(b^2*x^2 - a^2)*cosh(b*x + a))*sinh(b*x + a))*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + ((b^2*x^2 - a^2)*co
sh(b*x + a)^4 + 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 +
2*(b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 + 3*(b^2*x^2 - a^2)*cosh(b*x + a)^2 - a^2)*sinh(b*x + a)^2 - a^
2 + 4*((b^2*x^2 - a^2)*cosh(b*x + a)^3 + (b^2*x^2 - a^2)*cosh(b*x + a))*sinh(b*x + a))*log(-I*cosh(b*x + a) -
I*sinh(b*x + a) + 1) - ((b^2*x^2 - a^2)*cosh(b*x + a)^4 + 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 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 + 3*(b^2*x^2 - a^2)*c
osh(b*x + a)^2 - a^2)*sinh(b*x + a)^2 - a^2 + 4*((b^2*x^2 - a^2)*cosh(b*x + a)^3 + (b^2*x^2 - a^2)*cosh(b*x +
a))*sinh(b*x + a))*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 2*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a
)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + c
osh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, cosh(b*x + a) + sinh(b*x + a)) - 2*(cosh(b*x + a)^4 + 4*cosh(b*x +
a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cos
h(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) - 2*(cosh(b*x +
a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh
(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, -I*cosh(b*x + a) - I*sinh(b*x
+ a)) + 2*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sin
h(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, -cosh(b*x
+ a) - sinh(b*x + a)) + 4*(2*(b*x + a)*cosh(b*x + a)^3 - (b^2*x^2 - b*x - 2*a)*cosh(b*x + a))*sinh(b*x + a) +
2*a)/(b^3*cosh(b*x + a)^4 + 4*b^3*cosh(b*x + a)*sinh(b*x + a)^3 + b^3*sinh(b*x + a)^4 + 2*b^3*cosh(b*x + a)^2
+ b^3 + 2*(3*b^3*cosh(b*x + a)^2 + b^3)*sinh(b*x + a)^2 + 4*(b^3*cosh(b*x + a)^3 + b^3*cosh(b*x + a))*sinh(b*
x + a))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {csch}{\left (a + b x \right )} \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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