3.4.63 \(\int x^3 \tanh ^2(a+b x) \, dx\) [363]

Optimal. Leaf size=89 \[ -\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}-\frac {3 \text {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^4}-\frac {x^3 \tanh (a+b x)}{b} \]

[Out]

-x^3/b+1/4*x^4+3*x^2*ln(1+exp(2*b*x+2*a))/b^2+3*x*polylog(2,-exp(2*b*x+2*a))/b^3-3/2*polylog(3,-exp(2*b*x+2*a)
)/b^4-x^3*tanh(b*x+a)/b

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Rubi [A]
time = 0.12, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3801, 3799, 2221, 2611, 2320, 6724, 30} \begin {gather*} -\frac {3 \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {3 x^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}-\frac {x^3 \tanh (a+b x)}{b}-\frac {x^3}{b}+\frac {x^4}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Tanh[a + b*x]^2,x]

[Out]

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

Rule 30

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3799

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 3801

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 6724

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 \tanh ^2(a+b x) \, dx &=-\frac {x^3 \tanh (a+b x)}{b}+\frac {3 \int x^2 \tanh (a+b x) \, dx}{b}+\int x^3 \, dx\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \tanh (a+b x)}{b}+\frac {6 \int \frac {e^{2 (a+b x)} x^2}{1+e^{2 (a+b x)}} \, dx}{b}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac {x^3 \tanh (a+b x)}{b}-\frac {6 \int x \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {x^3 \tanh (a+b x)}{b}-\frac {3 \int \text {Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {x^3 \tanh (a+b x)}{b}-\frac {3 \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {3 \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}-\frac {x^3 \tanh (a+b x)}{b}\\ \end {align*}

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Mathematica [A]
time = 3.40, size = 109, normalized size = 1.22 \begin {gather*} \frac {x^4}{4}+\frac {2 b^2 x^2 \left (-\frac {2 b e^{2 a} x}{1+e^{2 a}}+3 \log \left (1+e^{2 (a+b x)}\right )\right )+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^4}-\frac {x^3 \text {sech}(a) \text {sech}(a+b x) \sinh (b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Tanh[a + b*x]^2,x]

[Out]

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

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Maple [A]
time = 2.14, size = 125, normalized size = 1.40

method	result	size
risch	\(\frac {x^{4}}{4}+\frac {2 x^{3}}{\left ({\mathrm e}^{2 b x +2 a}+1\right ) b}-\frac {6 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 x^{3}}{b}+\frac {6 x \,a^{2}}{b^{3}}+\frac {4 a^{3}}{b^{4}}+\frac {3 x^{2} \ln \left ({\mathrm e}^{2 b x +2 a}+1\right )}{b^{2}}+\frac {3 x \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}-\frac {3 \polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{4}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4+2*x^3/(exp(2*b*x+2*a)+1)/b-6/b^4*a^2*ln(exp(b*x+a))-2/b*x^3+6/b^3*x*a^2+4/b^4*a^3+3*x^2*ln(exp(2*b*x+2
*a)+1)/b^2+3*x*polylog(2,-exp(2*b*x+2*a))/b^3-3/2*polylog(3,-exp(2*b*x+2*a))/b^4

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Maxima [A]
time = 0.32, size = 108, normalized size = 1.21 \begin {gather*} -\frac {2 \, x^{3}}{b} + \frac {b x^{4} e^{\left (2 \, b x + 2 \, a\right )} + b x^{4} + 8 \, x^{3}}{4 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}} + \frac {3 \, {\left (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 )})\right )}}{2 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^3/b + 1/4*(b*x^4*e^(2*b*x + 2*a) + b*x^4 + 8*x^3)/(b*e^(2*b*x + 2*a) + b) + 3/2*(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^4

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Fricas [C] Result contains complex when optimal does not.
time = 0.38, size = 721, normalized size = 8.10 \begin {gather*} \frac {b^{4} x^{4} - 8 \, a^{3} + {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \sinh \left (b x + a\right )^{2} + 24 \, {\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} + b x\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 24 \, {\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} + b x\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 12 \, {\left (a^{2} \cosh \left (b x + a\right )^{2} + 2 \, a^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a^{2} \sinh \left (b x + a\right )^{2} + a^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + 12 \, {\left (a^{2} \cosh \left (b x + a\right )^{2} + 2 \, a^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a^{2} \sinh \left (b x + a\right )^{2} + a^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + 12 \, {\left (b^{2} x^{2} + {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \sinh \left (b x + a\right )^{2} - a^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + 12 \, {\left (b^{2} x^{2} + {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \sinh \left (b x + a\right )^{2} - a^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 24 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 24 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1\right )} {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{4 \, {\left (b^{4} \cosh \left (b x + a\right )^{2} + 2 \, b^{4} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{4} \sinh \left (b x + a\right )^{2} + b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^4*x^4 - 8*a^3 + (b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*x + a)^2 + 2*(b^4*x^4 - 8*b^3*x^3 - 8*a^3)*cosh(b*
x + a)*sinh(b*x + a) + (b^4*x^4 - 8*b^3*x^3 - 8*a^3)*sinh(b*x + a)^2 + 24*(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 + b*x)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + 24*(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 + b*x)*dilog(-I*cosh(b*x + a) - I*sinh(b*x +
a)) + 12*(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 + a^2)*log(cosh(b*x +
a) + sinh(b*x + a) + I) + 12*(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 +
a^2)*log(cosh(b*x + a) + sinh(b*x + a) - I) + 12*(b^2*x^2 + (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 - a^2)*log(I*cosh(b*x + a) + I*sinh(b*x + a) +
1) + 12*(b^2*x^2 + (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 - a^2)*log(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) - 24*(cosh(b*x + a)^2 + 2*cosh(b*x
+ a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*polylog(3, I*cosh(b*x + a) + I*sinh(b*x + a)) - 24*(cosh(b*x + a)^2
+ 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)*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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \sinh ^{2}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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