### 3.336 $$\int x^2 \tanh (a+b x) \, dx$$

Optimal. Leaf size=65 $\frac{x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}-\frac{\text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac{x^3}{3}$

[Out]

-x^3/3 + (x^2*Log[1 + E^(2*(a + b*x))])/b + (x*PolyLog[2, -E^(2*(a + b*x))])/b^2 - PolyLog[3, -E^(2*(a + b*x))
]/(2*b^3)

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Rubi [A]  time = 0.135225, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {3718, 2190, 2531, 2282, 6589} $\frac{x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}-\frac{\text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac{x^3}{3}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Tanh[a + b*x],x]

[Out]

-x^3/3 + (x^2*Log[1 + E^(2*(a + b*x))])/b + (x*PolyLog[2, -E^(2*(a + b*x))])/b^2 - PolyLog[3, -E^(2*(a + b*x))
]/(2*b^3)

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 2531

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 2282

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 6589

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

Mathematica [A]  time = 2.05807, size = 66, normalized size = 1.02 $\frac{-6 b x \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )-3 \text{PolyLog}\left (3,-e^{-2 (a+b x)}\right )+2 b^2 x^2 \left (3 \log \left (e^{-2 (a+b x)}+1\right )+b x\right )}{6 b^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*Tanh[a + b*x],x]

[Out]

(2*b^2*x^2*(b*x + 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))])/(6*b^3)

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Maple [A]  time = 0.026, size = 94, normalized size = 1.5 \begin{align*} -{\frac{{x}^{3}}{3}}-2\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{{a}^{2}x}{{b}^{2}}}+{\frac{4\,{a}^{3}}{3\,{b}^{3}}}+{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}+{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}-{\frac{{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3-2/b^3*a^2*ln(exp(b*x+a))+2/b^2*a^2*x+4/3/b^3*a^3+x^2*ln(1+exp(2*b*x+2*a))/b+x*polylog(2,-exp(2*b*x+2*
a))/b^2-1/2*polylog(3,-exp(2*b*x+2*a))/b^3

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Maxima [A]  time = 1.20669, size = 85, normalized size = 1.31 \begin{align*} -\frac{1}{3} \, x^{3} + \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}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3 + 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

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Fricas [C]  time = 1.96507, size = 595, normalized size = 9.15 \begin{align*} -\frac{b^{3} x^{3} - 6 \, b x{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 6 \, b x{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 3 \, a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) - 3 \, a^{2} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - 3 \,{\left (b^{2} x^{2} - a^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - 3 \,{\left (b^{2} x^{2} - a^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) + 6 \,{\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 6 \,{\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{3 \, b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 - 6*b*x*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 6*b*x*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a
)) - 3*a^2*log(cosh(b*x + a) + sinh(b*x + a) + I) - 3*a^2*log(cosh(b*x + a) + sinh(b*x + a) - I) - 3*(b^2*x^2
- a^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - 3*(b^2*x^2 - a^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) +
1) + 6*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + 6*polylog(3, -I*cosh(b*x + a) - I*sinh(b*x + a)))/b^3

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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