3.343 $$\int x^2 \text{sech}(a+b x) \tanh (a+b x) \, dx$$

Optimal. Leaf size=69 $-\frac{2 i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac{2 i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{sech}(a+b x)}{b}$

[Out]

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

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Rubi [A]  time = 0.0614134, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {5418, 4180, 2279, 2391} $-\frac{2 i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac{2 i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{4 x \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{sech}(a+b x)}{b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.435018, size = 125, normalized size = 1.81 $-\frac{2 i \left (\text{PolyLog}\left (2,-i e^{a+b x}\right )-\text{PolyLog}\left (2,i e^{a+b x}\right )\right )+b^2 x^2 \text{sech}(a+b x)+(-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-(\pi -2 i a) \log \left (\cot \left (\frac{1}{4} (2 i a+2 i b x+\pi )\right )\right )}{b^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*Sech[a + b*x]*Tanh[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.054, size = 154, normalized size = 2.2 \begin{align*} -2\,{\frac{{x}^{2}{{\rm e}^{bx+a}}}{b \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }}-{\frac{2\,i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-{\frac{2\,i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+{\frac{2\,i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+{\frac{2\,i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}-{\frac{2\,i{\it dilog} \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+{\frac{2\,i{\it dilog} \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-4\,{\frac{a\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^2*exp(b*x+a)/b/(1+exp(2*b*x+2*a))-2*I/b^2*ln(1+I*exp(b*x+a))*x-2*I/b^3*ln(1+I*exp(b*x+a))*a+2*I/b^2*ln(1-
I*exp(b*x+a))*x+2*I/b^3*ln(1-I*exp(b*x+a))*a-2*I/b^3*dilog(1+I*exp(b*x+a))+2*I/b^3*dilog(1-I*exp(b*x+a))-4/b^3
*a*arctan(exp(b*x+a))

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

[Out]

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

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Fricas [B]  time = 2.06587, size = 1395, normalized size = 20.22 \begin{align*} -\frac{2 \, b^{2} x^{2} \cosh \left (b x + a\right ) + 2 \, b^{2} x^{2} \sinh \left (b x + a\right ) -{\left (2 i \, \cosh \left (b x + a\right )^{2} + 4 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 i \, \sinh \left (b x + a\right )^{2} + 2 i\right )}{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) -{\left (-2 i \, \cosh \left (b x + a\right )^{2} - 4 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 2 i \, \sinh \left (b x + a\right )^{2} - 2 i\right )}{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) -{\left (-2 i \, a \cosh \left (b x + a\right )^{2} - 4 i \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 2 i \, a \sinh \left (b x + a\right )^{2} - 2 i \, a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) -{\left (2 i \, a \cosh \left (b x + a\right )^{2} + 4 i \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 2 i \, a \sinh \left (b x + a\right )^{2} + 2 i \, a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) -{\left ({\left (-2 i \, b x - 2 i \, a\right )} \cosh \left (b x + a\right )^{2} +{\left (-4 i \, b x - 4 i \, a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (-2 i \, b x - 2 i \, a\right )} \sinh \left (b x + a\right )^{2} - 2 i \, b x - 2 i \, a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) -{\left ({\left (2 i \, b x + 2 i \, a\right )} \cosh \left (b x + a\right )^{2} +{\left (4 i \, b x + 4 i \, a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (2 i \, b x + 2 i \, a\right )} \sinh \left (b x + a\right )^{2} + 2 i \, b x + 2 i \, a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} + b^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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