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

Optimal. Leaf size=130 \[ -\frac{x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}+\frac{\text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{\sinh ^2(a+b x)}{4 b^3}-\frac{x \sinh (a+b x) \cosh (a+b x)}{2 b^2}-\frac{x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{x^2}{4 b}+\frac{x^3}{3} \]

[Out]

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

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Rubi [A]  time = 0.18999, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5449, 5372, 3310, 30, 3718, 2190, 2531, 2282, 6589} \[ -\frac{x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}+\frac{\text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{\sinh ^2(a+b x)}{4 b^3}-\frac{x \sinh (a+b x) \cosh (a+b x)}{2 b^2}-\frac{x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{x^2}{4 b}+\frac{x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sinh[a + b*x]^2*Tanh[a + b*x],x]

[Out]

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

Rule 5449

Int[((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int
[(c + d*x)^m*Sinh[a + b*x]^n*Tanh[a + b*x]^(p - 2), x] - Int[(c + d*x)^m*Sinh[a + b*x]^(n - 2)*Tanh[a + b*x]^p
, x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 3310

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

Rule 30

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

Rubi steps

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

Mathematica [A]  time = 2.63895, size = 154, normalized size = 1.18 \[ \frac{1}{24} \left (\frac{4 \left (6 b x \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )+3 \text{PolyLog}\left (3,-e^{-2 (a+b x)}\right )+2 b^2 x^2 \left (-\frac{2 b x}{e^{2 a}+1}-3 \log \left (e^{-2 (a+b x)}+1\right )\right )\right )}{b^3}+\frac{3 \cosh (2 b x) \left (\cosh (2 a) \left (2 b^2 x^2+1\right )-2 b x \sinh (2 a)\right )}{b^3}+\frac{3 \sinh (2 b x) \left (\sinh (2 a) \left (2 b^2 x^2+1\right )-2 b x \cosh (2 a)\right )}{b^3}-8 x^3 \tanh (a)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sinh[a + b*x]^2*Tanh[a + b*x],x]

[Out]

((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))]))/b^3 + (3*Cosh[2*b*x]*((1 + 2*b^2*x^2)*Cosh[2*a] - 2*b*x*Sinh[2*a]))/b^3 + (3
*(-2*b*x*Cosh[2*a] + (1 + 2*b^2*x^2)*Sinh[2*a])*Sinh[2*b*x])/b^3 - 8*x^3*Tanh[a])/24

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Maple [A]  time = 0.069, size = 152, normalized size = 1.2 \begin{align*}{\frac{{x}^{3}}{3}}+{\frac{ \left ( 2\,{x}^{2}{b}^{2}-2\,bx+1 \right ){{\rm e}^{2\,bx+2\,a}}}{16\,{b}^{3}}}+{\frac{ \left ( 2\,{x}^{2}{b}^{2}+2\,bx+1 \right ){{\rm e}^{-2\,bx-2\,a}}}{16\,{b}^{3}}}+2\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{{a}^{2}x}{{b}^{2}}}-{\frac{4\,{a}^{3}}{3\,{b}^{3}}}-{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}-{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3+1/16*(2*b^2*x^2-2*b*x+1)/b^3*exp(2*b*x+2*a)+1/16*(2*b^2*x^2+2*b*x+1)/b^3*exp(-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 = 1.38972, size = 186, normalized size = 1.43 \begin{align*} \frac{2}{3} \, x^{3} - \frac{{\left (16 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 3 \,{\left (2 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} - 3 \,{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{48 \, b^{3}} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^3 - 1/48*(16*b^3*x^3*e^(2*a) - 3*(2*b^2*x^2*e^(4*a) - 2*b*x*e^(4*a) + e^(4*a))*e^(2*b*x) - 3*(2*b^2*x^2
+ 2*b*x + 1)*e^(-2*b*x))*e^(-2*a)/b^3 - 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

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Fricas [C]  time = 2.31393, size = 2090, normalized size = 16.08 \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)*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/48*(3*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)^4 + 12*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)*sinh(b*x + a)^3 + 3
*(2*b^2*x^2 - 2*b*x + 1)*sinh(b*x + a)^4 + 6*b^2*x^2 + 16*(b^3*x^3 + 2*a^3)*cosh(b*x + a)^2 + 2*(8*b^3*x^3 + 1
6*a^3 + 9*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 6*b*x - 96*(b*x*cosh(b*x + a)^2 + 2*b*x*c
osh(b*x + a)*sinh(b*x + a) + b*x*sinh(b*x + a)^2)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 96*(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)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) -
48*(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)*log(cosh(b*x + a) + sinh(b*
x + a) + I) - 48*(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)*log(cosh(b*x
+ a) + sinh(b*x + a) - I) - 48*((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)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - 48*((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)*log(-I*cosh(b*x + a
) - I*sinh(b*x + a) + 1) + 96*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*polylog(3, I
*cosh(b*x + a) + I*sinh(b*x + a)) + 96*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2)*pol
ylog(3, -I*cosh(b*x + a) - I*sinh(b*x + a)) + 4*(3*(2*b^2*x^2 - 2*b*x + 1)*cosh(b*x + a)^3 + 8*(b^3*x^3 + 2*a^
3)*cosh(b*x + a))*sinh(b*x + a) + 3)/(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)

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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)*sinh(b*x+a)**3,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 ) \sinh \left (b x + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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