### 3.356 $$\int x^3 \sinh (a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=195 $\frac{3 i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{6 i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{6 x \sinh (a+b x)}{b^3}-\frac{6 \cosh (a+b x)}{b^4}-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \sinh (a+b x)}{b}$

[Out]

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

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Rubi [A]  time = 0.197753, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {5449, 3296, 2638, 4180, 2531, 6609, 2282, 6589} $\frac{3 i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{6 i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{6 x \sinh (a+b x)}{b^3}-\frac{6 \cosh (a+b x)}{b^4}-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \sinh (a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*x^3*ArcTan[E^(a + b*x)])/b - (6*Cosh[a + b*x])/b^4 - (3*x^2*Cosh[a + b*x])/b^2 + ((3*I)*x^2*PolyLog[2, (-I
)*E^(a + b*x)])/b^2 - ((3*I)*x^2*PolyLog[2, I*E^(a + b*x)])/b^2 - ((6*I)*x*PolyLog[3, (-I)*E^(a + b*x)])/b^3 +
((6*I)*x*PolyLog[3, I*E^(a + b*x)])/b^3 + ((6*I)*PolyLog[4, (-I)*E^(a + b*x)])/b^4 - ((6*I)*PolyLog[4, I*E^(a
+ b*x)])/b^4 + (6*x*Sinh[a + b*x])/b^3 + (x^3*Sinh[a + b*x])/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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

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

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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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 \sinh (a+b x) \tanh (a+b x) \, dx &=\int x^3 \cosh (a+b x) \, dx-\int x^3 \text{sech}(a+b x) \, dx\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \sinh (a+b x)}{b}+\frac{(3 i) \int x^2 \log \left (1-i e^{a+b x}\right ) \, dx}{b}-\frac{(3 i) \int x^2 \log \left (1+i e^{a+b x}\right ) \, dx}{b}-\frac{3 \int x^2 \sinh (a+b x) \, dx}{b}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{x^3 \sinh (a+b x)}{b}-\frac{(6 i) \int x \text{Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}+\frac{(6 i) \int x \text{Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}+\frac{6 \int x \cosh (a+b x) \, dx}{b^2}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}+\frac{6 x \sinh (a+b x)}{b^3}+\frac{x^3 \sinh (a+b x)}{b}+\frac{(6 i) \int \text{Li}_3\left (-i e^{a+b x}\right ) \, dx}{b^3}-\frac{(6 i) \int \text{Li}_3\left (i e^{a+b x}\right ) \, dx}{b^3}-\frac{6 \int \sinh (a+b x) \, dx}{b^3}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 \cosh (a+b x)}{b^4}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}+\frac{6 x \sinh (a+b x)}{b^3}+\frac{x^3 \sinh (a+b x)}{b}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 \cosh (a+b x)}{b^4}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}+\frac{6 i \text{Li}_4\left (-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{Li}_4\left (i e^{a+b x}\right )}{b^4}+\frac{6 x \sinh (a+b x)}{b^3}+\frac{x^3 \sinh (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 1.43698, size = 211, normalized size = 1.08 $-\frac{i \left (-3 b^2 x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )+3 b^2 x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )+6 b x \text{PolyLog}\left (3,-i e^{a+b x}\right )-6 b x \text{PolyLog}\left (3,i e^{a+b x}\right )-6 \text{PolyLog}\left (4,-i e^{a+b x}\right )+6 \text{PolyLog}\left (4,i e^{a+b x}\right )+b^3 x^3 \log \left (1-i e^{a+b x}\right )-b^3 x^3 \log \left (1+i e^{a+b x}\right )+i b^3 x^3 \sinh (a+b x)-3 i b^2 x^2 \cosh (a+b x)+6 i b x \sinh (a+b x)-6 i \cosh (a+b x)\right )}{b^4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 0.125, size = 0, normalized size = 0. \begin{align*} \int{x}^{3}{\rm sech} \left (bx+a\right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left ({\left (b^{3} x^{3} e^{\left (2 \, a\right )} - 3 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 6 \, e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} -{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x\right )}\right )} e^{\left (-a\right )}}{2 \, b^{4}} - 2 \, \int \frac{x^{3} e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.38427, size = 1704, normalized size = 8.74 \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)*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/2*(b^3*x^3 + 3*b^2*x^2 - (b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*cosh(b*x + a)^2 - 2*(b^3*x^3 - 3*b^2*x^2 + 6*b*x
- 6)*cosh(b*x + a)*sinh(b*x + a) - (b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*sinh(b*x + a)^2 + 6*b*x - (-6*I*b^2*x^2*
cosh(b*x + a) - 6*I*b^2*x^2*sinh(b*x + a))*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (6*I*b^2*x^2*cosh(b*x +
a) + 6*I*b^2*x^2*sinh(b*x + a))*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) - (2*I*a^3*cosh(b*x + a) + 2*I*a^3*s
inh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + I) - (-2*I*a^3*cosh(b*x + a) - 2*I*a^3*sinh(b*x + a))*log(co
sh(b*x + a) + sinh(b*x + a) - I) - ((2*I*b^3*x^3 + 2*I*a^3)*cosh(b*x + a) + (2*I*b^3*x^3 + 2*I*a^3)*sinh(b*x +
a))*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) - ((-2*I*b^3*x^3 - 2*I*a^3)*cosh(b*x + a) + (-2*I*b^3*x^3 - 2*
I*a^3)*sinh(b*x + a))*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - (-12*I*cosh(b*x + a) - 12*I*sinh(b*x + a))
*polylog(4, I*cosh(b*x + a) + I*sinh(b*x + a)) - (12*I*cosh(b*x + a) + 12*I*sinh(b*x + a))*polylog(4, -I*cosh(
b*x + a) - I*sinh(b*x + a)) - (12*I*b*x*cosh(b*x + a) + 12*I*b*x*sinh(b*x + a))*polylog(3, I*cosh(b*x + a) + I
*sinh(b*x + a)) - (-12*I*b*x*cosh(b*x + a) - 12*I*b*x*sinh(b*x + a))*polylog(3, -I*cosh(b*x + a) - I*sinh(b*x
+ a)) + 6)/(b^4*cosh(b*x + a) + b^4*sinh(b*x + a))

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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**3*sech(b*x+a)*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^{3} \operatorname{sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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