3.393 \(\int x \tanh ^3(a+b x) \, dx\)

Optimal. Leaf size=82 \[ \frac {\text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac {\tanh (a+b x)}{2 b^2}+\frac {x \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x \tanh ^2(a+b x)}{2 b}+\frac {x}{2 b}-\frac {x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3720, 3473, 8, 3718, 2190, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac {\tanh (a+b x)}{2 b^2}+\frac {x \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x \tanh ^2(a+b x)}{2 b}+\frac {x}{2 b}-\frac {x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*Tanh[a + b*x]^3,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 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 3720

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]

Rubi steps

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

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Mathematica [C]  time = 6.13, size = 231, normalized size = 2.82 \[ \frac {\text {csch}(a) \text {sech}(a) \left (b^2 x^2 e^{-\tanh ^{-1}(\coth (a))}-\frac {i \coth (a) \left (i \text {Li}_2\left (e^{2 i \left (i b x+i \tanh ^{-1}(\coth (a))\right )}\right )-b x \left (-\pi +2 i \tanh ^{-1}(\coth (a))\right )-2 \left (i \tanh ^{-1}(\coth (a))+i b x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (a))+i b x\right )}\right )+2 i \tanh ^{-1}(\coth (a)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )-\pi \log \left (e^{2 b x}+1\right )+\pi \log (\cosh (b x))\right )}{\sqrt {1-\coth ^2(a)}}\right )}{2 b^2 \sqrt {\text {csch}^2(a) \left (\sinh ^2(a)-\cosh ^2(a)\right )}}-\frac {\text {sech}(a) \sinh (b x) \text {sech}(a+b x)}{2 b^2}+\frac {x \text {sech}^2(a+b x)}{2 b}+\frac {1}{2} x^2 \tanh (a) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*Tanh[a + b*x]^3,x]

[Out]

(x*Sech[a + b*x]^2)/(2*b) + (Csch[a]*((b^2*x^2)/E^ArcTanh[Coth[a]] - (I*Coth[a]*(-(b*x*(-Pi + (2*I)*ArcTanh[Co
th[a]])) - Pi*Log[1 + E^(2*b*x)] - 2*(I*b*x + I*ArcTanh[Coth[a]])*Log[1 - E^((2*I)*(I*b*x + I*ArcTanh[Coth[a]]
))] + Pi*Log[Cosh[b*x]] + (2*I)*ArcTanh[Coth[a]]*Log[I*Sinh[b*x + ArcTanh[Coth[a]]]] + I*PolyLog[2, E^((2*I)*(
I*b*x + I*ArcTanh[Coth[a]]))]))/Sqrt[1 - Coth[a]^2])*Sech[a])/(2*b^2*Sqrt[Csch[a]^2*(-Cosh[a]^2 + Sinh[a]^2)])
- (Sech[a]*Sech[a + b*x]*Sinh[b*x])/(2*b^2) + (x^2*Tanh[a])/2

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fricas [C]  time = 0.57, size = 1106, normalized size = 13.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((b^2*x^2 - 2*a^2)*cosh(b*x + a)^4 + 4*(b^2*x^2 - 2*a^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*x^2 - 2*a^2
)*sinh(b*x + a)^4 + b^2*x^2 + 2*(b^2*x^2 - 2*a^2 - 2*b*x - 1)*cosh(b*x + a)^2 + 2*(b^2*x^2 + 3*(b^2*x^2 - 2*a^
2)*cosh(b*x + a)^2 - 2*a^2 - 2*b*x - 1)*sinh(b*x + a)^2 - 2*a^2 - 2*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*
x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^
3 + cosh(b*x + a))*sinh(b*x + a) + 1)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - 2*(cosh(b*x + a)^4 + 4*cosh(b
*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*
(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 1)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + 2*(a*cosh(b*x
+ a)^4 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 + 2*a*cosh(b*x + a)^2 + 2*(3*a*cosh(b*x + a)^2
+ a)*sinh(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 + a*cosh(b*x + a))*sinh(b*x + a) + a)*log(cosh(b*x + a) + sinh(b*
x + a) + I) + 2*(a*cosh(b*x + a)^4 + 4*a*cosh(b*x + a)*sinh(b*x + a)^3 + a*sinh(b*x + a)^4 + 2*a*cosh(b*x + a)
^2 + 2*(3*a*cosh(b*x + a)^2 + a)*sinh(b*x + a)^2 + 4*(a*cosh(b*x + a)^3 + a*cosh(b*x + a))*sinh(b*x + a) + a)*
log(cosh(b*x + a) + sinh(b*x + a) - I) - 2*((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a
)^3 + (b*x + a)*sinh(b*x + a)^4 + 2*(b*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 + b*x + a)*sinh
(b*x + a)^2 + b*x + 4*((b*x + a)*cosh(b*x + a)^3 + (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(I*cosh(b*x
+ a) + I*sinh(b*x + a) + 1) - 2*((b*x + a)*cosh(b*x + a)^4 + 4*(b*x + a)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*x
+ a)*sinh(b*x + a)^4 + 2*(b*x + a)*cosh(b*x + a)^2 + 2*(3*(b*x + a)*cosh(b*x + a)^2 + b*x + a)*sinh(b*x + a)^2
+ b*x + 4*((b*x + a)*cosh(b*x + a)^3 + (b*x + a)*cosh(b*x + a))*sinh(b*x + a) + a)*log(-I*cosh(b*x + a) - I*s
inh(b*x + a) + 1) + 4*((b^2*x^2 - 2*a^2)*cosh(b*x + a)^3 + (b^2*x^2 - 2*a^2 - 2*b*x - 1)*cosh(b*x + a))*sinh(b
*x + a) - 2)/(b^2*cosh(b*x + a)^4 + 4*b^2*cosh(b*x + a)*sinh(b*x + a)^3 + b^2*sinh(b*x + a)^4 + 2*b^2*cosh(b*x
+ a)^2 + 2*(3*b^2*cosh(b*x + a)^2 + b^2)*sinh(b*x + a)^2 + b^2 + 4*(b^2*cosh(b*x + a)^3 + b^2*cosh(b*x + a))*
sinh(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {sech}\left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.52, size = 111, normalized size = 1.35 \[ -\frac {x^{2}}{2}+\frac {2 b x \,{\mathrm e}^{2 b x +2 a}+{\mathrm e}^{2 b x +2 a}+1}{b^{2} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}-\frac {2 a x}{b}-\frac {a^{2}}{b^{2}}+\frac {x \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}+\frac {\polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 131, normalized size = 1.60 \[ -x^{2} + \frac {b^{2} x^{2} e^{\left (4 \, b x + 4 \, a\right )} + b^{2} x^{2} + 2 \, {\left (b^{2} x^{2} e^{\left (2 \, a\right )} + 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )} + 2}{2 \, {\left (b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}\right )}} + \frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 + 1/2*(b^2*x^2*e^(4*b*x + 4*a) + b^2*x^2 + 2*(b^2*x^2*e^(2*a) + 2*b*x*e^(2*a) + e^(2*a))*e^(2*b*x) + 2)/(
b^2*e^(4*b*x + 4*a) + 2*b^2*e^(2*b*x + 2*a) + b^2) + 1/2*(2*b*x*log(e^(2*b*x + 2*a) + 1) + dilog(-e^(2*b*x + 2
*a)))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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