### 3.342 $$\int x^3 \text{sech}(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=113 $-\frac{6 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{6 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^4}+\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{sech}(a+b x)}{b}$

[Out]

(6*x^2*ArcTan[E^(a + b*x)])/b^2 - ((6*I)*x*PolyLog[2, (-I)*E^(a + b*x)])/b^3 + ((6*I)*x*PolyLog[2, I*E^(a + b*
x)])/b^3 + ((6*I)*PolyLog[3, (-I)*E^(a + b*x)])/b^4 - ((6*I)*PolyLog[3, I*E^(a + b*x)])/b^4 - (x^3*Sech[a + b*
x])/b

________________________________________________________________________________________

Rubi [A]  time = 0.105284, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.312, Rules used = {5418, 4180, 2531, 2282, 6589} $-\frac{6 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{6 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^4}+\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^3*Sech[a + b*x]*Tanh[a + b*x],x]

[Out]

(6*x^2*ArcTan[E^(a + b*x)])/b^2 - ((6*I)*x*PolyLog[2, (-I)*E^(a + b*x)])/b^3 + ((6*I)*x*PolyLog[2, I*E^(a + b*
x)])/b^3 + ((6*I)*PolyLog[3, (-I)*E^(a + b*x)])/b^4 - ((6*I)*PolyLog[3, I*E^(a + b*x)])/b^4 - (x^3*Sech[a + b*
x])/b

Rule 5418

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> -Simp[(
x^(m - n + 1)*Sech[a + b*x^n]^p)/(b*n*p), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x],
x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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^3 \text{sech}(a+b x) \tanh (a+b x) \, dx &=-\frac{x^3 \text{sech}(a+b x)}{b}+\frac{3 \int x^2 \text{sech}(a+b x) \, dx}{b}\\ &=\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{sech}(a+b x)}{b}-\frac{(6 i) \int x \log \left (1-i e^{a+b x}\right ) \, dx}{b^2}+\frac{(6 i) \int x \log \left (1+i e^{a+b x}\right ) \, dx}{b^2}\\ &=\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_2\left (i e^{a+b x}\right )}{b^3}-\frac{x^3 \text{sech}(a+b x)}{b}+\frac{(6 i) \int \text{Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^3}-\frac{(6 i) \int \text{Li}_2\left (i e^{a+b x}\right ) \, dx}{b^3}\\ &=\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_2\left (i e^{a+b x}\right )}{b^3}-\frac{x^3 \text{sech}(a+b x)}{b}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac{6 i \text{Li}_3\left (-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{Li}_3\left (i e^{a+b x}\right )}{b^4}-\frac{x^3 \text{sech}(a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 2.19236, size = 130, normalized size = 1.15 $-\frac{x^3 \text{sech}(a+b x)}{b}+\frac{3 i \left (-2 b x \text{PolyLog}\left (2,-i e^{a+b x}\right )+2 b x \text{PolyLog}\left (2,i e^{a+b x}\right )+2 \text{PolyLog}\left (3,-i e^{a+b x}\right )-2 \text{PolyLog}\left (3,i e^{a+b x}\right )+b^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 x^2 \log \left (1+i e^{a+b x}\right )\right )}{b^4}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^3*Sech[a + b*x]*Tanh[a + b*x],x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.102, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ({\rm sech} \left (bx+a\right ) \right ) ^{2}\sinh \left ( bx+a \right ) \, dx \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),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, x^{3} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} + 6 \, \int \frac{x^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b}\,{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),x, algorithm="maxima")

[Out]

-2*x^3*e^(b*x + a)/(b*e^(2*b*x + 2*a) + b) + 6*integrate(x^2*e^(b*x + a)/(b*e^(2*b*x + 2*a) + b), x)

________________________________________________________________________________________

Fricas [C]  time = 2.02914, size = 1890, normalized size = 16.73 \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),x, algorithm="fricas")

[Out]

-(2*b^3*x^3*cosh(b*x + a) + 2*b^3*x^3*sinh(b*x + a) - (6*I*b*x*cosh(b*x + a)^2 + 12*I*b*x*cosh(b*x + a)*sinh(b
*x + a) + 6*I*b*x*sinh(b*x + a)^2 + 6*I*b*x)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (-6*I*b*x*cosh(b*x + a
)^2 - 12*I*b*x*cosh(b*x + a)*sinh(b*x + a) - 6*I*b*x*sinh(b*x + a)^2 - 6*I*b*x)*dilog(-I*cosh(b*x + a) - I*sin
h(b*x + a)) - (3*I*a^2*cosh(b*x + a)^2 + 6*I*a^2*cosh(b*x + a)*sinh(b*x + a) + 3*I*a^2*sinh(b*x + a)^2 + 3*I*a
^2)*log(cosh(b*x + a) + sinh(b*x + a) + I) - (-3*I*a^2*cosh(b*x + a)^2 - 6*I*a^2*cosh(b*x + a)*sinh(b*x + a) -
3*I*a^2*sinh(b*x + a)^2 - 3*I*a^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) - (-3*I*b^2*x^2 + (-3*I*b^2*x^2 + 3
*I*a^2)*cosh(b*x + a)^2 + (-6*I*b^2*x^2 + 6*I*a^2)*cosh(b*x + a)*sinh(b*x + a) + (-3*I*b^2*x^2 + 3*I*a^2)*sinh
(b*x + a)^2 + 3*I*a^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - (3*I*b^2*x^2 + (3*I*b^2*x^2 - 3*I*a^2)*cos
h(b*x + a)^2 + (6*I*b^2*x^2 - 6*I*a^2)*cosh(b*x + a)*sinh(b*x + a) + (3*I*b^2*x^2 - 3*I*a^2)*sinh(b*x + a)^2 -
3*I*a^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - (-6*I*cosh(b*x + a)^2 - 12*I*cosh(b*x + a)*sinh(b*x +
a) - 6*I*sinh(b*x + a)^2 - 6*I)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) - (6*I*cosh(b*x + a)^2 + 12*I*co
sh(b*x + a)*sinh(b*x + a) + 6*I*sinh(b*x + a)^2 + 6*I)*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)

________________________________________________________________________________________

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),x)

[Out]

Timed out

________________________________________________________________________________________

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 )\,{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),x, algorithm="giac")

[Out]

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