### 3.358 $$\int x \sinh (a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=77 $\frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{\cosh (a+b x)}{b^2}-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \sinh (a+b x)}{b}$

[Out]

(-2*x*ArcTan[E^(a + b*x)])/b - Cosh[a + b*x]/b^2 + (I*PolyLog[2, (-I)*E^(a + b*x)])/b^2 - (I*PolyLog[2, I*E^(a
+ b*x)])/b^2 + (x*Sinh[a + b*x])/b

________________________________________________________________________________________

Rubi [A]  time = 0.0637488, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.429, Rules used = {5449, 3296, 2638, 4180, 2279, 2391} $\frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{\cosh (a+b x)}{b^2}-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \sinh (a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Sinh[a + b*x]*Tanh[a + b*x],x]

[Out]

(-2*x*ArcTan[E^(a + b*x)])/b - Cosh[a + b*x]/b^2 + (I*PolyLog[2, (-I)*E^(a + b*x)])/b^2 - (I*PolyLog[2, I*E^(a
+ b*x)])/b^2 + (x*Sinh[a + b*x])/b

Rule 5449

Int[((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int
[(c + d*x)^m*Sinh[a + b*x]^n*Tanh[a + b*x]^(p - 2), x] - Int[(c + d*x)^m*Sinh[a + b*x]^(n - 2)*Tanh[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 3296

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

Rule 2638

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

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 \sinh (a+b x) \tanh (a+b x) \, dx &=\int x \cosh (a+b x) \, dx-\int x \text{sech}(a+b x) \, dx\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \sinh (a+b x)}{b}+\frac{i \int \log \left (1-i e^{a+b x}\right ) \, dx}{b}-\frac{i \int \log \left (1+i e^{a+b x}\right ) \, dx}{b}-\frac{\int \sinh (a+b x) \, dx}{b}\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\cosh (a+b x)}{b^2}+\frac{x \sinh (a+b x)}{b}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\cosh (a+b x)}{b^2}+\frac{i \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{i \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{x \sinh (a+b x)}{b}\\ \end{align*}

Mathematica [B]  time = 0.109484, size = 212, normalized size = 2.75 $\frac{i \left (\text{PolyLog}\left (2,-e^{i \left (-i a-i b x+\frac{\pi }{2}\right )}\right )-\text{PolyLog}\left (2,e^{i \left (-i a-i b x+\frac{\pi }{2}\right )}\right )\right )+\left (-i a-i b x+\frac{\pi }{2}\right ) \left (\log \left (1-e^{i \left (-i a-i b x+\frac{\pi }{2}\right )}\right )-\log \left (1+e^{i \left (-i a-i b x+\frac{\pi }{2}\right )}\right )\right )-\left (\frac{\pi }{2}-i a\right ) \log \left (\tan \left (\frac{1}{2} \left (-i a-i b x+\frac{\pi }{2}\right )\right )\right )}{b^2}+\frac{\cosh (b x) (b x \sinh (a)-\cosh (a))}{b^2}+\frac{\sinh (b x) (b x \cosh (a)-\sinh (a))}{b^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Sinh[a + b*x]*Tanh[a + b*x],x]

[Out]

(((-I)*a + Pi/2 - I*b*x)*(Log[1 - E^(I*((-I)*a + Pi/2 - I*b*x))] - Log[1 + E^(I*((-I)*a + Pi/2 - I*b*x))]) - (
(-I)*a + Pi/2)*Log[Tan[((-I)*a + Pi/2 - I*b*x)/2]] + I*(PolyLog[2, -E^(I*((-I)*a + Pi/2 - I*b*x))] - PolyLog[2
, E^(I*((-I)*a + Pi/2 - I*b*x))]))/b^2 + (Cosh[b*x]*(-Cosh[a] + b*x*Sinh[a]))/b^2 + ((b*x*Cosh[a] - Sinh[a])*S
inh[b*x])/b^2

________________________________________________________________________________________

Maple [B]  time = 0.057, size = 162, normalized size = 2.1 \begin{align*}{\frac{ \left ( bx-1 \right ){{\rm e}^{bx+a}}}{2\,{b}^{2}}}-{\frac{ \left ( bx+1 \right ){{\rm e}^{-bx-a}}}{2\,{b}^{2}}}+{\frac{i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}-{\frac{i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{i{\it dilog} \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{i{\it dilog} \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+2\,{\frac{a\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x-1)/b^2*exp(b*x+a)-1/2*(b*x+1)/b^2*exp(-b*x-a)+I/b*ln(1+I*exp(b*x+a))*x+I/b^2*ln(1+I*exp(b*x+a))*a-I/b
*ln(1-I*exp(b*x+a))*x-I/b^2*ln(1-I*exp(b*x+a))*a+I/b^2*dilog(1+I*exp(b*x+a))-I/b^2*dilog(1-I*exp(b*x+a))+2/b^2
*a*arctan(exp(b*x+a))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left ({\left (b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} -{\left (b x + 1\right )} e^{\left (-b x\right )}\right )} e^{\left (-a\right )}}{2 \, b^{2}} - 2 \, \int \frac{x e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*x*e^(2*a) - e^(2*a))*e^(b*x) - (b*x + 1)*e^(-b*x))*e^(-a)/b^2 - 2*integrate(x*e^(b*x + a)/(e^(2*b*x +
2*a) + 1), x)

________________________________________________________________________________________

Fricas [B]  time = 2.23961, size = 973, normalized size = 12.64 \begin{align*} \frac{{\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b x - 1\right )} \sinh \left (b x + a\right )^{2} - b x +{\left (-2 i \, \cosh \left (b x + a\right ) - 2 i \, \sinh \left (b x + a\right )\right )}{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) +{\left (2 i \, \cosh \left (b x + a\right ) + 2 i \, \sinh \left (b x + a\right )\right )}{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) +{\left (2 i \, a \cosh \left (b x + a\right ) + 2 i \, a \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) +{\left (-2 i \, a \cosh \left (b x + a\right ) - 2 i \, a \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) +{\left ({\left (2 i \, b x + 2 i \, a\right )} \cosh \left (b x + a\right ) +{\left (2 i \, b x + 2 i \, a\right )} \sinh \left (b x + a\right )\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) +{\left ({\left (-2 i \, b x - 2 i \, a\right )} \cosh \left (b x + a\right ) +{\left (-2 i \, b x - 2 i \, a\right )} \sinh \left (b x + a\right )\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 1}{2 \,{\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*x - 1)*cosh(b*x + a)^2 + 2*(b*x - 1)*cosh(b*x + a)*sinh(b*x + a) + (b*x - 1)*sinh(b*x + a)^2 - b*x + (
-2*I*cosh(b*x + a) - 2*I*sinh(b*x + a))*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + (2*I*cosh(b*x + a) + 2*I*si
nh(b*x + a))*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + (2*I*a*cosh(b*x + a) + 2*I*a*sinh(b*x + a))*log(cosh(
b*x + a) + sinh(b*x + a) + I) + (-2*I*a*cosh(b*x + a) - 2*I*a*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a)
- I) + ((2*I*b*x + 2*I*a)*cosh(b*x + a) + (2*I*b*x + 2*I*a)*sinh(b*x + a))*log(I*cosh(b*x + a) + I*sinh(b*x +
a) + 1) + ((-2*I*b*x - 2*I*a)*cosh(b*x + a) + (-2*I*b*x - 2*I*a)*sinh(b*x + a))*log(-I*cosh(b*x + a) - I*sinh
(b*x + a) + 1) - 1)/(b^2*cosh(b*x + a) + b^2*sinh(b*x + a))

________________________________________________________________________________________

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

[Out]

Integral(x*sinh(a + b*x)**2*sech(a + b*x), x)

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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