3.379 $$\int x \sinh ^2(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=89 $-\frac{\text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}-\frac{x \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}+\frac{x^2}{2}$

[Out]

x/(4*b) + x^2/2 - (x*Log[1 + E^(2*(a + b*x))])/b - PolyLog[2, -E^(2*(a + b*x))]/(2*b^2) - (Cosh[a + b*x]*Sinh[
a + b*x])/(4*b^2) + (x*Sinh[a + b*x]^2)/(2*b)

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Rubi [A]  time = 0.116242, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {5449, 5372, 2635, 8, 3718, 2190, 2279, 2391} $-\frac{\text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}-\frac{x \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}+\frac{x^2}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Sinh[a + b*x]^2*Tanh[a + b*x],x]

[Out]

x/(4*b) + x^2/2 - (x*Log[1 + E^(2*(a + b*x))])/b - PolyLog[2, -E^(2*(a + b*x))]/(2*b^2) - (Cosh[a + b*x]*Sinh[
a + b*x])/(4*b^2) + (x*Sinh[a + b*x]^2)/(2*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 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && 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 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 ^2(a+b x) \tanh (a+b x) \, dx &=\int x \cosh (a+b x) \sinh (a+b x) \, dx-\int x \tanh (a+b x) \, dx\\ &=\frac{x^2}{2}+\frac{x \sinh ^2(a+b x)}{2 b}-2 \int \frac{e^{2 (a+b x)} x}{1+e^{2 (a+b x)}} \, dx-\frac{\int \sinh ^2(a+b x) \, dx}{2 b}\\ &=\frac{x^2}{2}-\frac{x \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{\int 1 \, dx}{4 b}+\frac{\int \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b}\\ &=\frac{x}{4 b}+\frac{x^2}{2}-\frac{x \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^2}\\ &=\frac{x}{4 b}+\frac{x^2}{2}-\frac{x \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{\text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.221195, size = 102, normalized size = 1.15 $-\frac{-4 \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )+4 a^2+8 a b x+8 a \log \left (e^{-2 (a+b x)}+1\right )+8 b x \log \left (e^{-2 (a+b x)}+1\right )+\sinh (2 (a+b x))-2 b x \cosh (2 (a+b x))-8 a \log (\cosh (a+b x))+4 b^2 x^2}{8 b^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Sinh[a + b*x]^2*Tanh[a + b*x],x]

[Out]

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

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Maple [A]  time = 0.065, size = 110, normalized size = 1.2 \begin{align*}{\frac{{x}^{2}}{2}}+{\frac{ \left ( 2\,bx-1 \right ){{\rm e}^{2\,bx+2\,a}}}{16\,{b}^{2}}}+{\frac{ \left ( 2\,bx+1 \right ){{\rm e}^{-2\,bx-2\,a}}}{16\,{b}^{2}}}+2\,{\frac{ax}{b}}+{\frac{{a}^{2}}{{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}}}-2\,{\frac{a\ln \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)^3,x)

[Out]

1/2*x^2+1/16*(2*b*x-1)/b^2*exp(2*b*x+2*a)+1/16*(2*b*x+1)/b^2*exp(-2*b*x-2*a)+2/b*a*x+a^2/b^2-x*ln(1+exp(2*b*x+
2*a))/b-1/2*polylog(2,-exp(2*b*x+2*a))/b^2-2/b^2*a*ln(exp(b*x+a))

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Maxima [A]  time = 1.36182, size = 128, normalized size = 1.44 \begin{align*} x^{2} - \frac{{\left (8 \, b^{2} x^{2} e^{\left (2 \, a\right )} -{\left (2 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} -{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{16 \, 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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 - 1/16*(8*b^2*x^2*e^(2*a) - (2*b*x*e^(4*a) - e^(4*a))*e^(2*b*x) - (2*b*x + 1)*e^(-2*b*x))*e^(-2*a)/b^2 - 1
/2*(2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b*x + 2*a)))/b^2

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Fricas [C]  time = 2.28533, size = 1561, normalized size = 17.54 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((2*b*x - 1)*cosh(b*x + a)^4 + 4*(2*b*x - 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (2*b*x - 1)*sinh(b*x + a)^4
+ 8*(b^2*x^2 - 2*a^2)*cosh(b*x + a)^2 + 2*(4*b^2*x^2 + 3*(2*b*x - 1)*cosh(b*x + a)^2 - 8*a^2)*sinh(b*x + a)^2
+ 2*b*x - 16*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*dilog(I*cosh(b*x + a) + I*sin
h(b*x + a)) - 16*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*dilog(-I*cosh(b*x + a) -
I*sinh(b*x + a)) + 16*(a*cosh(b*x + a)^2 + 2*a*cosh(b*x + a)*sinh(b*x + a) + a*sinh(b*x + a)^2)*log(cosh(b*x +
a) + sinh(b*x + a) + I) + 16*(a*cosh(b*x + a)^2 + 2*a*cosh(b*x + a)*sinh(b*x + a) + a*sinh(b*x + a)^2)*log(co
sh(b*x + a) + sinh(b*x + a) - I) - 16*((b*x + a)*cosh(b*x + a)^2 + 2*(b*x + a)*cosh(b*x + a)*sinh(b*x + a) + (
b*x + a)*sinh(b*x + a)^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - 16*((b*x + a)*cosh(b*x + a)^2 + 2*(b*x
+ a)*cosh(b*x + a)*sinh(b*x + a) + (b*x + a)*sinh(b*x + a)^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + 4*
((2*b*x - 1)*cosh(b*x + a)^3 + 4*(b^2*x^2 - 2*a^2)*cosh(b*x + a))*sinh(b*x + a) + 1)/(b^2*cosh(b*x + a)^2 + 2*
b^2*cosh(b*x + a)*sinh(b*x + a) + b^2*sinh(b*x + a)^2)

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

[Out]

Integral(x*sinh(a + b*x)**3*sech(a + b*x), x)

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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 )^{3}\,{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)^3,x, algorithm="giac")

[Out]

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