### 3.363 $$\int x^3 \tanh ^2(a+b x) \, dx$$

Optimal. Leaf size=89 $\frac{3 x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}-\frac{3 \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^4}+\frac{3 x^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}-\frac{x^3 \tanh (a+b x)}{b}-\frac{x^3}{b}+\frac{x^4}{4}$

[Out]

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

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Rubi [A]  time = 0.17831, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.583, Rules used = {3720, 3718, 2190, 2531, 2282, 6589, 30} $\frac{3 x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}-\frac{3 \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^4}+\frac{3 x^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}-\frac{x^3 \tanh (a+b x)}{b}-\frac{x^3}{b}+\frac{x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Tanh[a + b*x]^2,x]

[Out]

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

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

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]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^3 \tanh ^2(a+b x) \, dx &=-\frac{x^3 \tanh (a+b x)}{b}+\frac{3 \int x^2 \tanh (a+b x) \, dx}{b}+\int x^3 \, dx\\ &=-\frac{x^3}{b}+\frac{x^4}{4}-\frac{x^3 \tanh (a+b x)}{b}+\frac{6 \int \frac{e^{2 (a+b x)} x^2}{1+e^{2 (a+b x)}} \, dx}{b}\\ &=-\frac{x^3}{b}+\frac{x^4}{4}+\frac{3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac{x^3 \tanh (a+b x)}{b}-\frac{6 \int x \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac{x^3}{b}+\frac{x^4}{4}+\frac{3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac{3 x \text{Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac{x^3 \tanh (a+b x)}{b}-\frac{3 \int \text{Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac{x^3}{b}+\frac{x^4}{4}+\frac{3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac{3 x \text{Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac{x^3 \tanh (a+b x)}{b}-\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=-\frac{x^3}{b}+\frac{x^4}{4}+\frac{3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac{3 x \text{Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac{3 \text{Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}-\frac{x^3 \tanh (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 4.53327, size = 104, normalized size = 1.17 $\frac{-6 b x \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )-3 \text{PolyLog}\left (3,-e^{-2 (a+b x)}\right )+2 b^2 x^2 \left (\frac{2 b x}{e^{2 a}+1}+3 \log \left (e^{-2 (a+b x)}+1\right )\right )}{2 b^4}-\frac{x^3 \text{sech}(a) \sinh (b x) \text{sech}(a+b x)}{b}+\frac{x^4}{4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*Tanh[a + b*x]^2,x]

[Out]

x^4/4 + (2*b^2*x^2*((2*b*x)/(1 + E^(2*a)) + 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))])/(2*b^4) - (x^3*Sech[a]*Sech[a + b*x]*Sinh[b*x])/b

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Maple [A]  time = 0.056, size = 125, normalized size = 1.4 \begin{align*}{\frac{{x}^{4}}{4}}+2\,{\frac{{x}^{3}}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}-6\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-2\,{\frac{{x}^{3}}{b}}+6\,{\frac{{a}^{2}x}{{b}^{3}}}+4\,{\frac{{a}^{3}}{{b}^{4}}}+3\,{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}+3\,{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{3}}}-{\frac{3\,{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{4}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sech(b*x+a)^2*sinh(b*x+a)^2,x)

[Out]

1/4*x^4+2*x^3/b/(1+exp(2*b*x+2*a))-6/b^4*a^2*ln(exp(b*x+a))-2*x^3/b+6/b^3*a^2*x+4/b^4*a^3+3*x^2*ln(1+exp(2*b*x
+2*a))/b^2+3*x*polylog(2,-exp(2*b*x+2*a))/b^3-3/2*polylog(3,-exp(2*b*x+2*a))/b^4

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Maxima [A]  time = 1.34843, size = 146, normalized size = 1.64 \begin{align*} -\frac{2 \, x^{3}}{b} + \frac{b x^{4} e^{\left (2 \, b x + 2 \, a\right )} + b x^{4} + 8 \, x^{3}}{4 \,{\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}} + \frac{3 \,{\left (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 )})\right )}}{2 \, b^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sech(b*x+a)^2*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-2*x^3/b + 1/4*(b*x^4*e^(2*b*x + 2*a) + b*x^4 + 8*x^3)/(b*e^(2*b*x + 2*a) + b) + 3/2*(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^4

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Fricas [C]  time = 2.31947, size = 1890, normalized size = 21.24 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sech(b*x+a)^2*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/4*(b^4*x^4 - 8*a^3 + (b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*x + a)^2 + 2*(b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*
x + a)*sinh(b*x + a) + (b^4*x^4 - 8*b^3*x^3 - 8*a^3)*sinh(b*x + a)^2 + 24*(b*x*cosh(b*x + a)^2 + 2*b*x*cosh(b*
x + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2 + b*x)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + 24*(b*x*cosh(b*x
+ a)^2 + 2*b*x*cosh(b*x + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2 + b*x)*dilog(-I*cosh(b*x + a) - I*sinh(b*x +
a)) + 12*(a^2*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2*sinh(b*x + a)^2 + a^2)*log(cosh(b*x +
a) + sinh(b*x + a) + I) + 12*(a^2*cosh(b*x + a)^2 + 2*a^2*cosh(b*x + a)*sinh(b*x + a) + a^2*sinh(b*x + a)^2 +
a^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) + 12*(b^2*x^2 + (b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 - a^2
)*cosh(b*x + a)*sinh(b*x + a) + (b^2*x^2 - a^2)*sinh(b*x + a)^2 - a^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) +
1) + 12*(b^2*x^2 + (b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a) + (b^2*x^2
- a^2)*sinh(b*x + a)^2 - a^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - 24*(cosh(b*x + a)^2 + 2*cosh(b*x
+ a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) - 24*(cosh(b*x + a)^2
+ 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*polylog(3, -I*cosh(b*x + a) - I*sinh(b*x + a)))/(b^4*co
sh(b*x + a)^2 + 2*b^4*cosh(b*x + a)*sinh(b*x + a) + b^4*sinh(b*x + a)^2 + b^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*sech(b*x+a)**2*sinh(b*x+a)**2,x)

[Out]

Timed out

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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 )^{2} \sinh \left (b x + a\right )^{2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sech(b*x+a)^2*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^3*sech(b*x + a)^2*sinh(b*x + a)^2, x)