### 3.364 $$\int x^2 \tanh ^2(a+b x) \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.118195, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {3720, 3718, 2190, 2279, 2391, 30} $\frac{\text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}+\frac{2 x \log \left (e^{2 (a+b x)}+1\right )}{b^2}-\frac{x^2 \tanh (a+b x)}{b}-\frac{x^2}{b}+\frac{x^3}{3}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Tanh[a + b*x]^2,x]

[Out]

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

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

Mathematica [C]  time = 3.20437, size = 168, normalized size = 2.58 $\frac{-3 \text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )-3 b^2 x^2 \text{sech}(a) \sinh (b x) \text{sech}(a+b x)-3 b^2 x^2 \tanh (a) \sqrt{-\text{csch}^2(a)} e^{-\tanh ^{-1}(\coth (a))}+6 b x \log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )+6 \tanh ^{-1}(\coth (a)) \left (\log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )+b x\right )+b^3 x^3+3 i \pi b x-3 i \pi \log \left (e^{2 b x}+1\right )+3 i \pi \log (\cosh (b x))}{3 b^3}$

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*Tanh[a + b*x]^2,x]

[Out]

((3*I)*b*Pi*x + b^3*x^3 - (3*I)*Pi*Log[1 + E^(2*b*x)] + 6*b*x*Log[1 - E^(-2*(b*x + ArcTanh[Coth[a]]))] + (3*I)
*Pi*Log[Cosh[b*x]] + 6*ArcTanh[Coth[a]]*(b*x + Log[1 - E^(-2*(b*x + ArcTanh[Coth[a]]))] - Log[I*Sinh[b*x + Arc
Tanh[Coth[a]]]]) - 3*PolyLog[2, E^(-2*(b*x + ArcTanh[Coth[a]]))] - 3*b^2*x^2*Sech[a]*Sech[a + b*x]*Sinh[b*x] -
(3*b^2*x^2*Sqrt[-Csch[a]^2]*Tanh[a])/E^ArcTanh[Coth[a]])/(3*b^3)

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Maple [A]  time = 0.055, size = 99, normalized size = 1.5 \begin{align*}{\frac{{x}^{3}}{3}}+2\,{\frac{{x}^{2}}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}-2\,{\frac{{x}^{2}}{b}}-4\,{\frac{ax}{{b}^{2}}}-2\,{\frac{{a}^{2}}{{b}^{3}}}+2\,{\frac{x\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{3}}}+4\,{\frac{a\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.3793, size = 113, normalized size = 1.74 \begin{align*} -\frac{2 \, x^{2}}{b} + \frac{b x^{3} e^{\left (2 \, b x + 2 \, a\right )} + b x^{3} + 6 \, x^{2}}{3 \,{\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}} + \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 )}{b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.25351, size = 1424, normalized size = 21.91 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^3*x^3 + (b^3*x^3 - 6*b^2*x^2 + 6*a^2)*cosh(b*x + a)^2 + 2*(b^3*x^3 - 6*b^2*x^2 + 6*a^2)*cosh(b*x + a)*s
inh(b*x + a) + (b^3*x^3 - 6*b^2*x^2 + 6*a^2)*sinh(b*x + a)^2 + 6*a^2 + 6*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*si
nh(b*x + a) + sinh(b*x + a)^2 + 1)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + 6*(cosh(b*x + a)^2 + 2*cosh(b*x
+ a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) - 6*(a*cosh(b*x + a)^2 + 2
*a*cosh(b*x + a)*sinh(b*x + a) + a*sinh(b*x + a)^2 + a)*log(cosh(b*x + a) + sinh(b*x + a) + I) - 6*(a*cosh(b*x
+ a)^2 + 2*a*cosh(b*x + a)*sinh(b*x + a) + a*sinh(b*x + a)^2 + a)*log(cosh(b*x + a) + sinh(b*x + a) - I) + 6*
((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 + b*x + a)*lo
g(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + 6*((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 + b*x + a)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1))/(b^3*cosh(b*x + a)^2 +
2*b^3*cosh(b*x + a)*sinh(b*x + a) + b^3*sinh(b*x + a)^2 + b^3)

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

[Out]

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