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

Optimal. Leaf size=162 \[ \frac{6 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}-\frac{6 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{3 x^2 \sinh (a+b x)}{b^2}-\frac{6 \sinh (a+b x)}{b^4}+\frac{6 x \cosh (a+b x)}{b^3}+\frac{x^3 \cosh (a+b x)}{b}+\frac{x^3 \text{sech}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.18291, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {5449, 3296, 2637, 5418, 4180, 2531, 2282, 6589} \[ \frac{6 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}-\frac{6 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{3 x^2 \sinh (a+b x)}{b^2}-\frac{6 \sinh (a+b x)}{b^4}+\frac{6 x \cosh (a+b x)}{b^3}+\frac{x^3 \cosh (a+b x)}{b}+\frac{x^3 \text{sech}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sinh[a + b*x]*Tanh[a + b*x]^2,x]

[Out]

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

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 \sinh (a+b x) \tanh ^2(a+b x) \, dx &=\int x^3 \sinh (a+b x) \, dx-\int x^3 \text{sech}(a+b x) \tanh (a+b x) \, dx\\ &=\frac{x^3 \cosh (a+b x)}{b}+\frac{x^3 \text{sech}(a+b x)}{b}-\frac{3 \int x^2 \cosh (a+b x) \, dx}{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 \cosh (a+b x)}{b}+\frac{x^3 \text{sech}(a+b x)}{b}-\frac{3 x^2 \sinh (a+b x)}{b^2}+\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 \int x \sinh (a+b x) \, dx}{b^2}\\ &=-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac{6 x \cosh (a+b x)}{b^3}+\frac{x^3 \cosh (a+b x)}{b}+\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{3 x^2 \sinh (a+b x)}{b^2}-\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 \int \cosh (a+b x) \, dx}{b^3}\\ &=-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}+\frac{6 x \cosh (a+b x)}{b^3}+\frac{x^3 \cosh (a+b x)}{b}+\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 \sinh (a+b x)}{b^4}-\frac{3 x^2 \sinh (a+b x)}{b^2}-\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 x \cosh (a+b x)}{b^3}+\frac{x^3 \cosh (a+b x)}{b}+\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}-\frac{6 \sinh (a+b x)}{b^4}-\frac{3 x^2 \sinh (a+b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 3.25533, size = 196, normalized size = 1.21 \[ \frac{-3 i \left (-2 b x \text{PolyLog}\left (2,-i e^{a+b x}\right )+2 b x \text{PolyLog}\left (2,i e^{a+b x}\right )+2 \text{PolyLog}\left (3,-i e^{a+b x}\right )-2 \text{PolyLog}\left (3,i e^{a+b x}\right )+b^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 x^2 \log \left (1+i e^{a+b x}\right )\right )+b^3 x^3 \text{sech}(a+b x)+\cosh (b x) \left (b x \cosh (a) \left (b^2 x^2+6\right )-3 \sinh (a) \left (b^2 x^2+2\right )\right )+\sinh (b x) \left (b x \sinh (a) \left (b^2 x^2+6\right )-3 \cosh (a) \left (b^2 x^2+2\right )\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sinh[a + b*x]*Tanh[a + b*x]^2,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^3*x^3*Se
ch[a + b*x] + Cosh[b*x]*(b*x*(6 + b^2*x^2)*Cosh[a] - 3*(2 + b^2*x^2)*Sinh[a]) + (-3*(2 + b^2*x^2)*Cosh[a] + b*
x*(6 + b^2*x^2)*Sinh[a])*Sinh[b*x])/b^4

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Maple [F]  time = 0.103, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ({\rm sech} \left (bx+a\right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{3}\, 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)^3,x)

[Out]

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

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

[Out]

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

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Fricas [C]  time = 2.50269, size = 3351, normalized size = 20.69 \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)^3,x, algorithm="fricas")

[Out]

1/2*(b^3*x^3 + (b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*cosh(b*x + a)^4 + 4*(b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*cosh(b*
x + a)*sinh(b*x + a)^3 + (b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*sinh(b*x + a)^4 + 3*b^2*x^2 + 6*(b^3*x^3 + 2*b*x)*c
osh(b*x + a)^2 + 6*(b^3*x^3 + (b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*cosh(b*x + a)^2 + 2*b*x)*sinh(b*x + a)^2 + 6*b
*x + (-12*I*b*x*cosh(b*x + a)^3 - 36*I*b*x*cosh(b*x + a)*sinh(b*x + a)^2 - 12*I*b*x*sinh(b*x + a)^3 - 12*I*b*x
*cosh(b*x + a) + (-36*I*b*x*cosh(b*x + a)^2 - 12*I*b*x)*sinh(b*x + a))*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)
) + (12*I*b*x*cosh(b*x + a)^3 + 36*I*b*x*cosh(b*x + a)*sinh(b*x + a)^2 + 12*I*b*x*sinh(b*x + a)^3 + 12*I*b*x*c
osh(b*x + a) + (36*I*b*x*cosh(b*x + a)^2 + 12*I*b*x)*sinh(b*x + a))*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a))
+ (-6*I*a^2*cosh(b*x + a)^3 - 18*I*a^2*cosh(b*x + a)*sinh(b*x + a)^2 - 6*I*a^2*sinh(b*x + a)^3 - 6*I*a^2*cosh(
b*x + a) + (-18*I*a^2*cosh(b*x + a)^2 - 6*I*a^2)*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + I) + (6*I*
a^2*cosh(b*x + a)^3 + 18*I*a^2*cosh(b*x + a)*sinh(b*x + a)^2 + 6*I*a^2*sinh(b*x + a)^3 + 6*I*a^2*cosh(b*x + a)
+ (18*I*a^2*cosh(b*x + a)^2 + 6*I*a^2)*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - I) + ((6*I*b^2*x^2
- 6*I*a^2)*cosh(b*x + a)^3 + (18*I*b^2*x^2 - 18*I*a^2)*cosh(b*x + a)*sinh(b*x + a)^2 + (6*I*b^2*x^2 - 6*I*a^2)
*sinh(b*x + a)^3 + (6*I*b^2*x^2 - 6*I*a^2)*cosh(b*x + a) + (6*I*b^2*x^2 + (18*I*b^2*x^2 - 18*I*a^2)*cosh(b*x +
a)^2 - 6*I*a^2)*sinh(b*x + a))*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + ((-6*I*b^2*x^2 + 6*I*a^2)*cosh(b*
x + a)^3 + (-18*I*b^2*x^2 + 18*I*a^2)*cosh(b*x + a)*sinh(b*x + a)^2 + (-6*I*b^2*x^2 + 6*I*a^2)*sinh(b*x + a)^3
+ (-6*I*b^2*x^2 + 6*I*a^2)*cosh(b*x + a) + (-6*I*b^2*x^2 + (-18*I*b^2*x^2 + 18*I*a^2)*cosh(b*x + a)^2 + 6*I*a
^2)*sinh(b*x + a))*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + (12*I*cosh(b*x + a)^3 + 36*I*cosh(b*x + a)*si
nh(b*x + a)^2 + 12*I*sinh(b*x + a)^3 + (36*I*cosh(b*x + a)^2 + 12*I)*sinh(b*x + a) + 12*I*cosh(b*x + a))*polyl
og(3, I*cosh(b*x + a) + I*sinh(b*x + a)) + (-12*I*cosh(b*x + a)^3 - 36*I*cosh(b*x + a)*sinh(b*x + a)^2 - 12*I*
sinh(b*x + a)^3 + (-36*I*cosh(b*x + a)^2 - 12*I)*sinh(b*x + a) - 12*I*cosh(b*x + a))*polylog(3, -I*cosh(b*x +
a) - I*sinh(b*x + a)) + 4*((b^3*x^3 - 3*b^2*x^2 + 6*b*x - 6)*cosh(b*x + a)^3 + 3*(b^3*x^3 + 2*b*x)*cosh(b*x +
a))*sinh(b*x + a) + 6)/(b^4*cosh(b*x + a)^3 + 3*b^4*cosh(b*x + a)*sinh(b*x + a)^2 + b^4*sinh(b*x + a)^3 + b^4*
cosh(b*x + a) + (3*b^4*cosh(b*x + a)^2 + 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)**2*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^{3} \operatorname{sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )^{3}\,{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)^3,x, algorithm="giac")

[Out]

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