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

Optimal. Leaf size=65 \[ \frac {\text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {2 x \log \left (e^{2 (a+b x)}+1\right )}{b^2}-\frac {x^2 \tanh (a+b x)}{b}-\frac {x^2}{b}+\frac {x^3}{3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 30

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

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 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 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]

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 = 3.34, size = 168, normalized size = 2.58 \[ \frac {-3 b^2 x^2 \text {sech}(a) \sinh (b x) \text {sech}(a+b x)-3 b^2 x^2 \tanh (a) \sqrt {-\text {csch}^2(a)} e^{-\tanh ^{-1}(\coth (a))}-3 \text {Li}_2\left (e^{-2 \left (b x+\tanh ^{-1}(\coth (a))\right )}\right )+6 b x \log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )+6 \tanh ^{-1}(\coth (a)) \left (\log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )+b x\right )+b^3 x^3+3 i \pi b x-3 i \pi \log \left (e^{2 b x}+1\right )+3 i \pi \log (\cosh (b x))}{3 b^3} \]

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)

________________________________________________________________________________________

fricas [C]  time = 0.83, size = 515, normalized size = 7.92 \[ \frac {b^{3} x^{3} + {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 6 \, a^{2}\right )} \sinh \left (b x + a\right )^{2} + 6 \, a^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 6 \, {\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} + a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) - 6 \, {\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} + a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + 6 \, {\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x + a\right )} \sinh \left (b x + a\right )^{2} + b x + a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x + a\right )} \sinh \left (b x + a\right )^{2} + b x + a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{3 \, {\left (b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} + b^{3}\right )}} \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2}\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.57, size = 99, normalized size = 1.52 \[ \frac {x^{3}}{3}+\frac {2 x^{2}}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {2 x^{2}}{b}-\frac {4 a x}{b^{2}}-\frac {2 a^{2}}{b^{3}}+\frac {2 x \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {\polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}+\frac {4 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}} \]

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*a*x/b^2-2/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))

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 84, normalized size = 1.29 \[ -\frac {2 \, x^{2}}{b} + \frac {b x^{3} e^{\left (2 \, b x + 2 \, a\right )} + b x^{3} + 6 \, x^{2}}{3 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}} + \frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{b^{3}} \]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sinh ^{2}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]

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]

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

________________________________________________________________________________________