3.517 \(\int x \text{csch}^3(a+b x) \text{sech}^2(a+b x) \, dx\)
Optimal. Leaf size=109 \[ \frac{3 \text{PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}-\frac{3 \text{PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}-\frac{\text{csch}(a+b x)}{2 b^2}+\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac{3 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b} \]
[Out]
ArcTan[Sinh[a + b*x]]/b^2 + (3*x*ArcTanh[E^(a + b*x)])/b - Csch[a + b*x]/(2*b^2) + (3*PolyLog[2, -E^(a + b*x)]
)/(2*b^2) - (3*PolyLog[2, E^(a + b*x)])/(2*b^2) - (3*x*Sech[a + b*x])/(2*b) - (x*Csch[a + b*x]^2*Sech[a + b*x]
)/(2*b)
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Rubi [A] time = 0.165331, antiderivative size = 109, normalized size of antiderivative =
1., number of steps used = 13, number of rules used = 12, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) =
0.667, Rules used = {2622, 288, 321, 207, 5462, 6271, 12, 4182, 2279, 2391, 3770, 2621}
\[ \frac{3 \text{PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}-\frac{3 \text{PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}-\frac{\text{csch}(a+b x)}{2 b^2}+\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac{3 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[x*Csch[a + b*x]^3*Sech[a + b*x]^2,x]
[Out]
ArcTan[Sinh[a + b*x]]/b^2 + (3*x*ArcTanh[E^(a + b*x)])/b - Csch[a + b*x]/(2*b^2) + (3*PolyLog[2, -E^(a + b*x)]
)/(2*b^2) - (3*PolyLog[2, E^(a + b*x)])/(2*b^2) - (3*x*Sech[a + b*x])/(2*b) - (x*Csch[a + b*x]^2*Sech[a + b*x]
)/(2*b)
Rule 2622
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Rule 288
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]
Rule 321
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rule 5462
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 6271
Int[ArcTanh[u_], x_Symbol] :> Simp[x*ArcTanh[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 - u^2), x], x] /; I
nverseFunctionFreeQ[u, x]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 2279
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Rule 2391
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rule 2621
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Rubi steps
\begin{align*} \int x \text{csch}^3(a+b x) \text{sech}^2(a+b x) \, dx &=\frac{3 x \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b}-\int \left (\frac{3 \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac{3 \text{sech}(a+b x)}{2 b}-\frac{\text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b}\right ) \, dx\\ &=\frac{3 x \tanh ^{-1}(\cosh (a+b x))}{2 b}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b}+\frac{\int \text{csch}^2(a+b x) \text{sech}(a+b x) \, dx}{2 b}-\frac{3 \int \tanh ^{-1}(\cosh (a+b x)) \, dx}{2 b}+\frac{3 \int \text{sech}(a+b x) \, dx}{2 b}\\ &=\frac{3 \tan ^{-1}(\sinh (a+b x))}{2 b^2}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b}-\frac{i \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,-i \text{csch}(a+b x)\right )}{2 b^2}-\frac{3 \int b x \text{csch}(a+b x) \, dx}{2 b}\\ &=\frac{3 \tan ^{-1}(\sinh (a+b x))}{2 b^2}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b}-\frac{3}{2} \int x \text{csch}(a+b x) \, dx-\frac{i \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,-i \text{csch}(a+b x)\right )}{2 b^2}\\ &=\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac{3 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b}+\frac{3 \int \log \left (1-e^{a+b x}\right ) \, dx}{2 b}-\frac{3 \int \log \left (1+e^{a+b x}\right ) \, dx}{2 b}\\ &=\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac{3 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\text{csch}(a+b x)}{2 b^2}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b}+\frac{3 \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}-\frac{3 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{2 b^2}\\ &=\frac{\tan ^{-1}(\sinh (a+b x))}{b^2}+\frac{3 x \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\text{csch}(a+b x)}{2 b^2}+\frac{3 \text{Li}_2\left (-e^{a+b x}\right )}{2 b^2}-\frac{3 \text{Li}_2\left (e^{a+b x}\right )}{2 b^2}-\frac{3 x \text{sech}(a+b x)}{2 b}-\frac{x \text{csch}^2(a+b x) \text{sech}(a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 3.0529, size = 168, normalized size = 1.54 \[ -\frac{12 \left (\text{PolyLog}\left (2,-e^{-a-b x}\right )-\text{PolyLog}\left (2,e^{-a-b x}\right )\right )+12 (a+b x) \left (\log \left (1-e^{-a-b x}\right )-\log \left (e^{-a-b x}+1\right )\right )-2 \tanh \left (\frac{1}{2} (a+b x)\right )+2 \coth \left (\frac{1}{2} (a+b x)\right )+b x \text{csch}^2\left (\frac{1}{2} (a+b x)\right )+b x \text{sech}^2\left (\frac{1}{2} (a+b x)\right )+8 b x \text{sech}(a+b x)-12 a \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )-16 \tan ^{-1}\left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{8 b^2} \]
Antiderivative was successfully verified.
[In]
Integrate[x*Csch[a + b*x]^3*Sech[a + b*x]^2,x]
[Out]
-(-16*ArcTan[Tanh[(a + b*x)/2]] + 2*Coth[(a + b*x)/2] + b*x*Csch[(a + b*x)/2]^2 + 12*(a + b*x)*(Log[1 - E^(-a
- b*x)] - Log[1 + E^(-a - b*x)]) - 12*a*Log[Tanh[(a + b*x)/2]] + 12*(PolyLog[2, -E^(-a - b*x)] - PolyLog[2, E^
(-a - b*x)]) + b*x*Sech[(a + b*x)/2]^2 + 8*b*x*Sech[a + b*x] - 2*Tanh[(a + b*x)/2])/(8*b^2)
________________________________________________________________________________________
Maple [A] time = 0.05, size = 148, normalized size = 1.4 \begin{align*} -{\frac{{{\rm e}^{bx+a}} \left ( 3\,bx{{\rm e}^{4\,bx+4\,a}}-2\,bx{{\rm e}^{2\,bx+2\,a}}+{{\rm e}^{4\,bx+4\,a}}+3\,bx-1 \right ) }{{b}^{2} \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2} \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}+2\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{3\,a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{2\,{b}^{2}}}+{\frac{3\,{\it dilog} \left ({{\rm e}^{bx+a}} \right ) }{2\,{b}^{2}}}+{\frac{3\,{\it dilog} \left ( 1+{{\rm e}^{bx+a}} \right ) }{2\,{b}^{2}}}+{\frac{3\,\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{2\,b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*csch(b*x+a)^3*sech(b*x+a)^2,x)
[Out]
-exp(b*x+a)*(3*b*x*exp(4*b*x+4*a)-2*b*x*exp(2*b*x+2*a)+exp(4*b*x+4*a)+3*b*x-1)/b^2/(exp(2*b*x+2*a)-1)^2/(1+exp
(2*b*x+2*a))+2/b^2*arctan(exp(b*x+a))+3/2/b^2*a*ln(exp(b*x+a)-1)+3/2/b^2*dilog(exp(b*x+a))+3/2/b^2*dilog(1+exp
(b*x+a))+3/2/b*ln(1+exp(b*x+a))*x
________________________________________________________________________________________
Maxima [A] time = 1.71509, size = 224, normalized size = 2.06 \begin{align*} \frac{2 \, b x e^{\left (3 \, b x + 3 \, a\right )} -{\left (3 \, b x e^{\left (5 \, a\right )} + e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )} -{\left (3 \, b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (6 \, b x + 6 \, a\right )} - b^{2} e^{\left (4 \, b x + 4 \, a\right )} - b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \frac{3 \,{\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{2 \, b^{2}} - \frac{3 \,{\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{2 \, b^{2}} + \frac{2 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*csch(b*x+a)^3*sech(b*x+a)^2,x, algorithm="maxima")
[Out]
(2*b*x*e^(3*b*x + 3*a) - (3*b*x*e^(5*a) + e^(5*a))*e^(5*b*x) - (3*b*x*e^a - e^a)*e^(b*x))/(b^2*e^(6*b*x + 6*a)
- b^2*e^(4*b*x + 4*a) - b^2*e^(2*b*x + 2*a) + b^2) + 3/2*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^2
- 3/2*(b*x*log(-e^(b*x + a) + 1) + dilog(e^(b*x + a)))/b^2 + 2*arctan(e^(b*x + a))/b^2
________________________________________________________________________________________
Fricas [B] time = 2.04225, size = 4531, normalized size = 41.57 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*csch(b*x+a)^3*sech(b*x+a)^2,x, algorithm="fricas")
[Out]
-1/2*(2*(3*b*x + 1)*cosh(b*x + a)^5 + 10*(3*b*x + 1)*cosh(b*x + a)*sinh(b*x + a)^4 + 2*(3*b*x + 1)*sinh(b*x +
a)^5 - 4*b*x*cosh(b*x + a)^3 + 4*(5*(3*b*x + 1)*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^3 + 4*(5*(3*b*x + 1)*cosh
(b*x + a)^3 - 3*b*x*cosh(b*x + a))*sinh(b*x + a)^2 - 4*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + si
nh(b*x + a)^6 + (15*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - cosh(b*x +
a))*sinh(b*x + a)^3 + (15*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - cosh(b*x + a)^2 + 2*(3*c
osh(b*x + a)^5 - 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) +
2*(3*b*x - 1)*cosh(b*x + a) + 3*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + (15*co
sh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^3 +
(15*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - cosh(b*x + a)^2 + 2*(3*cosh(b*x + a)^5 - 2*cos
h(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*dilog(cosh(b*x + a) + sinh(b*x + a)) - 3*(cosh(b*x + a)^6 + 6
*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + (15*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - cosh(b*x + a)^4
+ 4*(5*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a)^3 + (15*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 - 1)*sinh(b*
x + a)^2 - cosh(b*x + a)^2 + 2*(3*cosh(b*x + a)^5 - 2*cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*dilo
g(-cosh(b*x + a) - sinh(b*x + a)) - 3*(b*x*cosh(b*x + a)^6 + 6*b*x*cosh(b*x + a)*sinh(b*x + a)^5 + b*x*sinh(b*
x + a)^6 - b*x*cosh(b*x + a)^4 + (15*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^4 - b*x*cosh(b*x + a)^2 + 4*(5*b
*x*cosh(b*x + a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a)^3 + (15*b*x*cosh(b*x + a)^4 - 6*b*x*cosh(b*x + a)^2 - b*
x)*sinh(b*x + a)^2 + b*x + 2*(3*b*x*cosh(b*x + a)^5 - 2*b*x*cosh(b*x + a)^3 - b*x*cosh(b*x + a))*sinh(b*x + a)
)*log(cosh(b*x + a) + sinh(b*x + a) + 1) - 3*(a*cosh(b*x + a)^6 + 6*a*cosh(b*x + a)*sinh(b*x + a)^5 + a*sinh(b
*x + a)^6 - a*cosh(b*x + a)^4 + (15*a*cosh(b*x + a)^2 - a)*sinh(b*x + a)^4 + 4*(5*a*cosh(b*x + a)^3 - a*cosh(b
*x + a))*sinh(b*x + a)^3 - a*cosh(b*x + a)^2 + (15*a*cosh(b*x + a)^4 - 6*a*cosh(b*x + a)^2 - a)*sinh(b*x + a)^
2 + 2*(3*a*cosh(b*x + a)^5 - 2*a*cosh(b*x + a)^3 - a*cosh(b*x + a))*sinh(b*x + a) + a)*log(cosh(b*x + a) + sin
h(b*x + a) - 1) + 3*((b*x + a)*cosh(b*x + a)^6 + 6*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^5 + (b*x + a)*sinh(b*
x + a)^6 - (b*x + a)*cosh(b*x + a)^4 + (15*(b*x + a)*cosh(b*x + a)^2 - b*x - a)*sinh(b*x + a)^4 + 4*(5*(b*x +
a)*cosh(b*x + a)^3 - (b*x + a)*cosh(b*x + a))*sinh(b*x + a)^3 - (b*x + a)*cosh(b*x + a)^2 + (15*(b*x + a)*cosh
(b*x + a)^4 - 6*(b*x + a)*cosh(b*x + a)^2 - b*x - a)*sinh(b*x + a)^2 + b*x + 2*(3*(b*x + a)*cosh(b*x + a)^5 -
2*(b*x + a)*cosh(b*x + a)^3 - (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(-cosh(b*x + a) - sinh(b*x + a) +
1) + 2*(5*(3*b*x + 1)*cosh(b*x + a)^4 - 6*b*x*cosh(b*x + a)^2 + 3*b*x - 1)*sinh(b*x + a))/(b^2*cosh(b*x + a)^
6 + 6*b^2*cosh(b*x + a)*sinh(b*x + a)^5 + b^2*sinh(b*x + a)^6 - b^2*cosh(b*x + a)^4 + (15*b^2*cosh(b*x + a)^2
- b^2)*sinh(b*x + a)^4 - b^2*cosh(b*x + a)^2 + 4*(5*b^2*cosh(b*x + a)^3 - b^2*cosh(b*x + a))*sinh(b*x + a)^3 +
(15*b^2*cosh(b*x + a)^4 - 6*b^2*cosh(b*x + a)^2 - b^2)*sinh(b*x + a)^2 + b^2 + 2*(3*b^2*cosh(b*x + a)^5 - 2*b
^2*cosh(b*x + a)^3 - b^2*cosh(b*x + a))*sinh(b*x + a))
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{csch}^{3}{\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*csch(b*x+a)**3*sech(b*x+a)**2,x)
[Out]
Integral(x*csch(a + b*x)**3*sech(a + b*x)**2, x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{csch}\left (b x + a\right )^{3} \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*csch(b*x+a)^3*sech(b*x+a)^2,x, algorithm="giac")
[Out]
integrate(x*csch(b*x + a)^3*sech(b*x + a)^2, x)