### 3.335 $$\int x^3 \tanh (a+b x) \, dx$$

Optimal. Leaf size=91 $\frac{3 x^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 x \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{3 \text{PolyLog}\left (4,-e^{2 (a+b x)}\right )}{4 b^4}+\frac{x^3 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac{x^4}{4}$

[Out]

-x^4/4 + (x^3*Log[1 + E^(2*(a + b*x))])/b + (3*x^2*PolyLog[2, -E^(2*(a + b*x))])/(2*b^2) - (3*x*PolyLog[3, -E^
(2*(a + b*x))])/(2*b^3) + (3*PolyLog[4, -E^(2*(a + b*x))])/(4*b^4)

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Rubi [A]  time = 0.159999, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.6, Rules used = {3718, 2190, 2531, 6609, 2282, 6589} $\frac{3 x^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 x \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{3 \text{PolyLog}\left (4,-e^{2 (a+b x)}\right )}{4 b^4}+\frac{x^3 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac{x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Tanh[a + b*x],x]

[Out]

-x^4/4 + (x^3*Log[1 + E^(2*(a + b*x))])/b + (3*x^2*PolyLog[2, -E^(2*(a + b*x))])/(2*b^2) - (3*x*PolyLog[3, -E^
(2*(a + b*x))])/(2*b^3) + (3*PolyLog[4, -E^(2*(a + b*x))])/(4*b^4)

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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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^3 \tanh (a+b x) \, dx &=-\frac{x^4}{4}+2 \int \frac{e^{2 (a+b x)} x^3}{1+e^{2 (a+b x)}} \, dx\\ &=-\frac{x^4}{4}+\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{3 \int x^2 \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b}\\ &=-\frac{x^4}{4}+\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac{3 x^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 \int x \text{Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{x^4}{4}+\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac{3 x^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac{3 \int \text{Li}_3\left (-e^{2 (a+b x)}\right ) \, dx}{2 b^3}\\ &=-\frac{x^4}{4}+\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac{3 x^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{4 b^4}\\ &=-\frac{x^4}{4}+\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac{3 x^2 \text{Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac{3 \text{Li}_4\left (-e^{2 (a+b x)}\right )}{4 b^4}\\ \end{align*}

Mathematica [A]  time = 2.23159, size = 88, normalized size = 0.97 $\frac{-6 b^2 x^2 \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )-6 b x \text{PolyLog}\left (3,-e^{-2 (a+b x)}\right )-3 \text{PolyLog}\left (4,-e^{-2 (a+b x)}\right )+4 b^3 x^3 \log \left (e^{-2 (a+b x)}+1\right )+b^4 x^4}{4 b^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*Tanh[a + b*x],x]

[Out]

(b^4*x^4 + 4*b^3*x^3*Log[1 + E^(-2*(a + b*x))] - 6*b^2*x^2*PolyLog[2, -E^(-2*(a + b*x))] - 6*b*x*PolyLog[3, -E
^(-2*(a + b*x))] - 3*PolyLog[4, -E^(-2*(a + b*x))])/(4*b^4)

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Maple [A]  time = 0.098, size = 116, normalized size = 1.3 \begin{align*} -{\frac{{x}^{4}}{4}}-2\,{\frac{{a}^{3}x}{{b}^{3}}}-{\frac{3\,{a}^{4}}{2\,{b}^{4}}}+{\frac{{x}^{3}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}+{\frac{3\,{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{2}}}-{\frac{3\,x{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{3}}}+{\frac{3\,{\it polylog} \left ( 4,-{{\rm e}^{2\,bx+2\,a}} \right ) }{4\,{b}^{4}}}+2\,{\frac{{a}^{3}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.20261, size = 113, normalized size = 1.24 \begin{align*} -\frac{1}{4} \, x^{4} + \frac{4 \, b^{3} x^{3} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x{\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )}) + 3 \,{\rm Li}_{4}(-e^{\left (2 \, b x + 2 \, a\right )})}{3 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^4 + 1/3*(4*b^3*x^3*log(e^(2*b*x + 2*a) + 1) + 6*b^2*x^2*dilog(-e^(2*b*x + 2*a)) - 6*b*x*polylog(3, -e^(
2*b*x + 2*a)) + 3*polylog(4, -e^(2*b*x + 2*a)))/b^4

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Fricas [C]  time = 2.08174, size = 761, normalized size = 8.36 \begin{align*} -\frac{b^{4} x^{4} - 12 \, b^{2} x^{2}{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 12 \, b^{2} x^{2}{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 4 \, a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + 4 \, a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + 24 \, b x{\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 24 \, b x{\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 4 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - 4 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 24 \,{\rm polylog}\left (4, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 24 \,{\rm polylog}\left (4, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{4 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^4*x^4 - 12*b^2*x^2*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 12*b^2*x^2*dilog(-I*cosh(b*x + a) - I*si
nh(b*x + a)) + 4*a^3*log(cosh(b*x + a) + sinh(b*x + a) + I) + 4*a^3*log(cosh(b*x + a) + sinh(b*x + a) - I) + 2
4*b*x*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + 24*b*x*polylog(3, -I*cosh(b*x + a) - I*sinh(b*x + a)) -
4*(b^3*x^3 + a^3)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - 4*(b^3*x^3 + a^3)*log(-I*cosh(b*x + a) - I*sinh
(b*x + a) + 1) - 24*polylog(4, I*cosh(b*x + a) + I*sinh(b*x + a)) - 24*polylog(4, -I*cosh(b*x + a) - I*sinh(b*
x + a)))/b^4

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

[Out]

Integral(x**3*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^{3} \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^3*sech(b*x+a)*sinh(b*x+a),x, algorithm="giac")

[Out]

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