3.510 \(\int x \text{csch}^3(a+b x) \text{sech}(a+b x) \, dx\)
Optimal. Leaf size=95 \[ \frac{\text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}-\frac{\text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac{\coth (a+b x)}{2 b^2}+\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{x \coth ^2(a+b x)}{2 b}+\frac{x}{2 b} \]
[Out]
x/(2*b) + (2*x*ArcTanh[E^(2*a + 2*b*x)])/b - Coth[a + b*x]/(2*b^2) - (x*Coth[a + b*x]^2)/(2*b) + PolyLog[2, -E
^(2*a + 2*b*x)]/(2*b^2) - PolyLog[2, E^(2*a + 2*b*x)]/(2*b^2)
________________________________________________________________________________________
Rubi [A] time = 0.120756, antiderivative size = 95, normalized size of antiderivative = 1.,
number of steps used = 11, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used
= {2620, 14, 5462, 3473, 8, 2548, 12, 4182, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}-\frac{\text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac{\coth (a+b x)}{2 b^2}+\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{x \coth ^2(a+b x)}{2 b}+\frac{x}{2 b} \]
Antiderivative was successfully verified.
[In]
Int[x*Csch[a + b*x]^3*Sech[a + b*x],x]
[Out]
x/(2*b) + (2*x*ArcTanh[E^(2*a + 2*b*x)])/b - Coth[a + b*x]/(2*b^2) - (x*Coth[a + b*x]^2)/(2*b) + PolyLog[2, -E
^(2*a + 2*b*x)]/(2*b^2) - PolyLog[2, E^(2*a + 2*b*x)]/(2*b^2)
Rule 2620
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 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 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 3473
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 2548
Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[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]
Rubi steps
\begin{align*} \int x \text{csch}^3(a+b x) \text{sech}(a+b x) \, dx &=-\frac{x \coth ^2(a+b x)}{2 b}-\frac{x \log (\tanh (a+b x))}{b}-\int \left (-\frac{\coth ^2(a+b x)}{2 b}-\frac{\log (\tanh (a+b x))}{b}\right ) \, dx\\ &=-\frac{x \coth ^2(a+b x)}{2 b}-\frac{x \log (\tanh (a+b x))}{b}+\frac{\int \coth ^2(a+b x) \, dx}{2 b}+\frac{\int \log (\tanh (a+b x)) \, dx}{b}\\ &=-\frac{\coth (a+b x)}{2 b^2}-\frac{x \coth ^2(a+b x)}{2 b}+\frac{\int 1 \, dx}{2 b}-\frac{\int 2 b x \text{csch}(2 a+2 b x) \, dx}{b}\\ &=\frac{x}{2 b}-\frac{\coth (a+b x)}{2 b^2}-\frac{x \coth ^2(a+b x)}{2 b}-2 \int x \text{csch}(2 a+2 b x) \, dx\\ &=\frac{x}{2 b}+\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{\coth (a+b x)}{2 b^2}-\frac{x \coth ^2(a+b x)}{2 b}+\frac{\int \log \left (1-e^{2 a+2 b x}\right ) \, dx}{b}-\frac{\int \log \left (1+e^{2 a+2 b x}\right ) \, dx}{b}\\ &=\frac{x}{2 b}+\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{\coth (a+b x)}{2 b^2}-\frac{x \coth ^2(a+b x)}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{2 b^2}-\frac{\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{2 b^2}\\ &=\frac{x}{2 b}+\frac{2 x \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac{\coth (a+b x)}{2 b^2}-\frac{x \coth ^2(a+b x)}{2 b}+\frac{\text{Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}-\frac{\text{Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}\\ \end{align*}
Mathematica [A] time = 0.656768, size = 137, normalized size = 1.44 \[ -\frac{\text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )-\text{PolyLog}\left (2,e^{-2 (a+b x)}\right )+2 a \log \left (1-e^{-2 (a+b x)}\right )+2 b x \log \left (1-e^{-2 (a+b x)}\right )-2 a \log \left (e^{-2 (a+b x)}+1\right )-2 b x \log \left (e^{-2 (a+b x)}+1\right )+\coth (a+b x)+b x \text{csch}^2(a+b x)-2 a \log (\sinh (a+b x))+2 a \log (\cosh (a+b x))}{2 b^2} \]
Antiderivative was successfully verified.
[In]
Integrate[x*Csch[a + b*x]^3*Sech[a + b*x],x]
[Out]
-(Coth[a + b*x] + b*x*Csch[a + b*x]^2 + 2*a*Log[1 - E^(-2*(a + b*x))] + 2*b*x*Log[1 - E^(-2*(a + b*x))] - 2*a*
Log[1 + E^(-2*(a + b*x))] - 2*b*x*Log[1 + E^(-2*(a + b*x))] + 2*a*Log[Cosh[a + b*x]] - 2*a*Log[Sinh[a + b*x]]
+ PolyLog[2, -E^(-2*(a + b*x))] - PolyLog[2, E^(-2*(a + b*x))])/(2*b^2)
________________________________________________________________________________________
Maple [B] time = 0.043, size = 170, normalized size = 1.8 \begin{align*} -{\frac{2\,bx{{\rm e}^{2\,bx+2\,a}}+{{\rm e}^{2\,bx+2\,a}}-1}{{b}^{2} \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{x\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{2}}}-{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}-{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*csch(b*x+a)^3*sech(b*x+a),x)
[Out]
-(2*b*x*exp(2*b*x+2*a)+exp(2*b*x+2*a)-1)/b^2/(exp(2*b*x+2*a)-1)^2-1/b*ln(1+exp(b*x+a))*x-1/b^2*polylog(2,-exp(
b*x+a))+x*ln(1+exp(2*b*x+2*a))/b+1/2*polylog(2,-exp(2*b*x+2*a))/b^2-1/b*ln(1-exp(b*x+a))*x-1/b^2*ln(1-exp(b*x+
a))*a-1/b^2*polylog(2,exp(b*x+a))+1/b^2*a*ln(exp(b*x+a)-1)
________________________________________________________________________________________
Maxima [A] time = 1.12766, size = 196, normalized size = 2.06 \begin{align*} -\frac{{\left (2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} - 1}{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 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} - \frac{b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} - \frac{b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\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),x, algorithm="maxima")
[Out]
-((2*b*x*e^(2*a) + e^(2*a))*e^(2*b*x) - 1)/(b^2*e^(4*b*x + 4*a) - 2*b^2*e^(2*b*x + 2*a) + b^2) + 1/2*(2*b*x*lo
g(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b*x + 2*a)))/b^2 - (b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^2 -
(b*x*log(-e^(b*x + a) + 1) + dilog(e^(b*x + a)))/b^2
________________________________________________________________________________________
Fricas [C] time = 2.34645, size = 4327, normalized size = 45.55 \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),x, algorithm="fricas")
[Out]
-((2*b*x + 1)*cosh(b*x + a)^2 + 2*(2*b*x + 1)*cosh(b*x + a)*sinh(b*x + a) + (2*b*x + 1)*sinh(b*x + a)^2 + (cos
h(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)*dilog(cosh(b*x + a) + sinh(b*x + a
)) - (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)*dilog(I*cosh(b*x + a) + I
*sinh(b*x + a)) - (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)*dilog(-I*cos
h(b*x + a) - I*sinh(b*x + a)) + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*co
sh(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
)*dilog(-cosh(b*x + a) - sinh(b*x + a)) + (b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sin
h(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))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (a*cosh(b*x + a)^4 + 4*a*
cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - 2*a*cosh(b*x + a)^2 + 2*(3*a*cosh(b*x + a)^2 - a)*sinh(b*x
+ a)^2 + 4*(a*cosh(b*x + a)^3 - a*cosh(b*x + a))*sinh(b*x + a) + a)*log(cosh(b*x + a) + sinh(b*x + a) + I) +
(a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - 2*a*cosh(b*x + a)^2 + 2*(3*a*cosh
(b*x + a)^2 - a)*sinh(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 - a*cosh(b*x + a))*sinh(b*x + a) + a)*log(cosh(b*x + a
) + sinh(b*x + a) - I) - (a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 - 2*a*cosh
(b*x + a)^2 + 2*(3*a*cosh(b*x + a)^2 - a)*sinh(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 - a*cosh(b*x + a))*sinh(b*x +
a) + a)*log(cosh(b*x + a) + sinh(b*x + a) - 1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(
b*x + a)^3 + (b*x + a)*sinh(b*x + a)^4 - 2*(b*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 - b*x -
a)*sinh(b*x + a)^2 + b*x + 4*((b*x + a)*cosh(b*x + a)^3 - (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(I*co
sh(b*x + a) + I*sinh(b*x + a) + 1) - ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 +
(b*x + a)*sinh(b*x + a)^4 - 2*(b*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 - b*x - a)*sinh(b*x +
a)^2 + b*x + 4*((b*x + a)*cosh(b*x + a)^3 - (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(-I*cosh(b*x + a)
- I*sinh(b*x + a) + 1) + ((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x + a)*si
nh(b*x + a)^4 - 2*(b*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 - b*x - a)*sinh(b*x + a)^2 + b*x
+ 4*((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) - 1)/(b^2*cosh(b*x + a)^4 + 4*b^2*cosh(b*x + a)*sinh(b*x + a)^3 + b^2*sinh(b*x + a)^4 - 2*b^2*cosh(b*x
+ a)^2 + 2*(3*b^2*cosh(b*x + a)^2 - b^2)*sinh(b*x + a)^2 + b^2 + 4*(b^2*cosh(b*x + a)^3 - b^2*cosh(b*x + a))*s
inh(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}{\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),x)
[Out]
Integral(x*csch(a + b*x)**3*sech(a + b*x), 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 )\,{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),x, algorithm="giac")
[Out]
integrate(x*csch(b*x + a)^3*sech(b*x + a), x)