3.392 \(\int x^2 \tanh ^3(a+b x) \, dx\)

Optimal. Leaf size=116 \[ -\frac {\text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac {\log (\cosh (a+b x))}{b^3}+\frac {x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}-\frac {x \tanh (a+b x)}{b^2}+\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x^2 \tanh ^2(a+b x)}{2 b}+\frac {x^2}{2 b}-\frac {x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3720, 3475, 30, 3718, 2190, 2531, 2282, 6589} \[ \frac {x \text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}-\frac {\text {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}-\frac {x \tanh (a+b x)}{b^2}+\frac {\log (\cosh (a+b x))}{b^3}+\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x^2 \tanh ^2(a+b x)}{2 b}+\frac {x^2}{2 b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Tanh[a + b*x]^3,x]

[Out]

x^2/(2*b) - x^3/3 + (x^2*Log[1 + E^(2*(a + b*x))])/b + Log[Cosh[a + b*x]]/b^3 + (x*PolyLog[2, -E^(2*(a + b*x))
])/b^2 - PolyLog[3, -E^(2*(a + b*x))]/(2*b^3) - (x*Tanh[a + b*x])/b^2 - (x^2*Tanh[a + b*x]^2)/(2*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 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 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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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]

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

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Mathematica [A]  time = 2.38, size = 185, normalized size = 1.59 \[ \frac {1}{6} \left (\frac {e^{2 a} \left (4 e^{-2 a} b^2 x^3-3 \left (e^{-2 a}+1\right ) \left (2 x \text {Li}_2\left (-e^{-2 (a+b x)}\right )+\frac {\text {Li}_3\left (-e^{-2 (a+b x)}\right )}{b}\right )+6 \left (e^{-2 a}+1\right ) b x^2 \log \left (e^{-2 (a+b x)}+1\right )-\frac {6 \left (e^{-2 a}+1\right ) \left (2 b x-\log \left (e^{2 (a+b x)}+1\right )\right )}{b}+12 e^{-2 a} x\right )}{\left (e^{2 a}+1\right ) b^2}-\frac {6 x \text {sech}(a) \sinh (b x) \text {sech}(a+b x)}{b^2}+\frac {3 x^2 \text {sech}^2(a+b x)}{b}+2 x^3 \tanh (a)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Tanh[a + b*x]^3,x]

[Out]

((E^(2*a)*((12*x)/E^(2*a) + (4*b^2*x^3)/E^(2*a) + 6*b*(1 + E^(-2*a))*x^2*Log[1 + E^(-2*(a + b*x))] - (6*(1 + E
^(-2*a))*(2*b*x - Log[1 + E^(2*(a + b*x))]))/b - 3*(1 + E^(-2*a))*(2*x*PolyLog[2, -E^(-2*(a + b*x))] + PolyLog
[3, -E^(-2*(a + b*x))]/b)))/(b^2*(1 + E^(2*a))) + (3*x^2*Sech[a + b*x]^2)/b - (6*x*Sech[a]*Sech[a + b*x]*Sinh[
b*x])/b^2 + 2*x^3*Tanh[a])/6

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fricas [C]  time = 0.94, size = 1649, normalized size = 14.22 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 + (b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*cosh(b*x + a)^4 + 4*(b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*cosh(b*x +
a)*sinh(b*x + a)^3 + (b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*sinh(b*x + a)^4 + 2*a^3 + 2*(b^3*x^3 - 3*b^2*x^2 + 2*a^3
+ 3*b*x + 6*a)*cosh(b*x + a)^2 + 2*(b^3*x^3 - 3*b^2*x^2 + 2*a^3 + 3*(b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*cosh(b*x
+ a)^2 + 3*b*x + 6*a)*sinh(b*x + a)^2 - 6*(b*x*cosh(b*x + a)^4 + 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sin
h(b*x + a)^4 + 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2 + b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(b*x
+ a)^3 + b*x*cosh(b*x + a))*sinh(b*x + a))*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 6*(b*x*cosh(b*x + a)^4
+ 4*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + b*x*sinh(b*x + a)^4 + 2*b*x*cosh(b*x + a)^2 + 2*(3*b*x*cosh(b*x + a)^2
+ b*x)*sinh(b*x + a)^2 + b*x + 4*(b*x*cosh(b*x + a)^3 + b*x*cosh(b*x + a))*sinh(b*x + a))*dilog(-I*cosh(b*x +
a) - I*sinh(b*x + a)) - 3*((a^2 + 1)*cosh(b*x + a)^4 + 4*(a^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (a^2 + 1)*
sinh(b*x + a)^4 + 2*(a^2 + 1)*cosh(b*x + a)^2 + 2*(3*(a^2 + 1)*cosh(b*x + a)^2 + a^2 + 1)*sinh(b*x + a)^2 + a^
2 + 4*((a^2 + 1)*cosh(b*x + a)^3 + (a^2 + 1)*cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x +
a) + I) - 3*((a^2 + 1)*cosh(b*x + a)^4 + 4*(a^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + (a^2 + 1)*sinh(b*x + a)^4
+ 2*(a^2 + 1)*cosh(b*x + a)^2 + 2*(3*(a^2 + 1)*cosh(b*x + a)^2 + a^2 + 1)*sinh(b*x + a)^2 + a^2 + 4*((a^2 + 1
)*cosh(b*x + a)^3 + (a^2 + 1)*cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh(b*x + a) - I) - 3*((b
^2*x^2 - a^2)*cosh(b*x + a)^4 + 4*(b^2*x^2 - a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2 - a^2)*sinh(b*x + a
)^4 + b^2*x^2 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 + 3*(b^2*x^2 - a^2)*cosh(b*x + a)^2 - a^2)*sinh
(b*x + a)^2 - a^2 + 4*((b^2*x^2 - a^2)*cosh(b*x + a)^3 + (b^2*x^2 - a^2)*cosh(b*x + a))*sinh(b*x + a))*log(I*c
osh(b*x + a) + I*sinh(b*x + a) + 1) - 3*((b^2*x^2 - a^2)*cosh(b*x + a)^4 + 4*(b^2*x^2 - a^2)*cosh(b*x + a)*sin
h(b*x + a)^3 + (b^2*x^2 - a^2)*sinh(b*x + a)^4 + b^2*x^2 + 2*(b^2*x^2 - a^2)*cosh(b*x + a)^2 + 2*(b^2*x^2 + 3*
(b^2*x^2 - a^2)*cosh(b*x + a)^2 - a^2)*sinh(b*x + a)^2 - a^2 + 4*((b^2*x^2 - a^2)*cosh(b*x + a)^3 + (b^2*x^2 -
a^2)*cosh(b*x + a))*sinh(b*x + a))*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + 6*(cosh(b*x + a)^4 + 4*cosh(
b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4
*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + 6*(cosh(
b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2
*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*polylog(3, -I*cosh(b*x + a) - I*sinh
(b*x + a)) + 4*((b^3*x^3 + 2*a^3 + 6*b*x + 6*a)*cosh(b*x + a)^3 + (b^3*x^3 - 3*b^2*x^2 + 2*a^3 + 3*b*x + 6*a)*
cosh(b*x + a))*sinh(b*x + a) + 6*a)/(b^3*cosh(b*x + a)^4 + 4*b^3*cosh(b*x + a)*sinh(b*x + a)^3 + b^3*sinh(b*x
+ a)^4 + 2*b^3*cosh(b*x + a)^2 + b^3 + 2*(3*b^3*cosh(b*x + a)^2 + b^3)*sinh(b*x + a)^2 + 4*(b^3*cosh(b*x + a)^
3 + b^3*cosh(b*x + a))*sinh(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {sech}\left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.61, size = 164, normalized size = 1.41 \[ -\frac {x^{3}}{3}+\frac {2 x \left (b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 b x +2 a}+1\right )}{b^{2} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}-\frac {2 \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}-\frac {2 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {2 a^{2} x}{b^{2}}+\frac {4 a^{3}}{3 b^{3}}+\frac {x^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}+\frac {x \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}-\frac {\polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3+2*x*(b*x*exp(2*b*x+2*a)+exp(2*b*x+2*a)+1)/b^2/(1+exp(2*b*x+2*a))^2-2/b^3*ln(exp(b*x+a))+1/b^3*ln(1+ex
p(2*b*x+2*a))-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 = 0.97, size = 183, normalized size = 1.58 \[ -\frac {2}{3} \, x^{3} + \frac {b^{2} x^{3} e^{\left (4 \, b x + 4 \, a\right )} + b^{2} x^{3} + 2 \, {\left (b^{2} x^{3} e^{\left (2 \, a\right )} + 3 \, b x^{2} e^{\left (2 \, a\right )} + 3 \, x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 6 \, x}{3 \, {\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} - \frac {2 \, x}{b^{2}} + \frac {2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})}{2 \, b^{3}} + \frac {\log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{{\mathrm {cosh}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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