3.337 \(\int x \tanh (a+b x) \, dx\)

Optimal. Leaf size=45 \[ \frac {\text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}+\frac {x \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x^2}{2} \]

[Out]

-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

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Rubi [A]  time = 0.08, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3718, 2190, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}+\frac {x \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[x*Tanh[a + b*x],x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [C]  time = 3.24, size = 149, normalized size = 3.31 \[ \frac {1}{2} \left (x^2 \tanh (a)+\frac {-b^2 x^2 \tanh (a) \sqrt {-\text {csch}^2(a)} e^{-\tanh ^{-1}(\coth (a))}-\text {Li}_2\left (e^{-2 \left (b x+\tanh ^{-1}(\coth (a))\right )}\right )+2 b x \log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )+2 \tanh ^{-1}(\coth (a)) \left (\log \left (1-e^{-2 \left (\tanh ^{-1}(\coth (a))+b x\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}(\coth (a))+b x\right )\right )+b x\right )+i \pi b x-i \pi \log \left (e^{2 b x}+1\right )+i \pi \log (\cosh (b x))}{b^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*Tanh[a + b*x],x]

[Out]

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

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fricas [C]  time = 0.69, size = 141, normalized size = 3.13 \[ -\frac {b^{2} x^{2} + 2 \, a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + 2 \, a \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - 2 \, {\left (b x + a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - 2 \, {\left (b x + a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 2 \, {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 2 \, {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.31, size = 70, normalized size = 1.56 \[ -\frac {x^{2}}{2}-\frac {2 a x}{b}-\frac {a^{2}}{b^{2}}+\frac {x \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}+\frac {\polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^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 = 0.50, size = 40, normalized size = 0.89 \[ -\frac {1}{2} \, x^{2} + \frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^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 = -0.02 \[ \int \frac {x\,\mathrm {sinh}\left (a+b\,x\right )}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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