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

Optimal. Leaf size=91 \[ -\frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}+\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}-\frac{\text{sech}(a+b x)}{2 b^2}+\frac{x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x \tanh (a+b x) \text{sech}(a+b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.101526, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5455, 4180, 2279, 2391, 4185} \[ -\frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}+\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}-\frac{\text{sech}(a+b x)}{2 b^2}+\frac{x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x \tanh (a+b x) \text{sech}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5455

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]*Tanh[(a_.) + (b_.)*(x_)]^(p_), x_Symbol] :> Int[(c + d
*x)^m*Sech[a + b*x]*Tanh[a + b*x]^(p - 2), x] - Int[(c + d*x)^m*Sech[a + b*x]^3*Tanh[a + b*x]^(p - 2), x] /; F
reeQ[{a, b, c, d, m}, x] && IGtQ[p/2, 0]

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]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rubi steps

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

Mathematica [A]  time = 0.752659, size = 93, normalized size = 1.02 \[ -\frac{i \text{PolyLog}(2,-i (\sinh (a+b x)+\cosh (a+b x)))-i \text{PolyLog}(2,i (\sinh (a+b x)+\cosh (a+b x)))+\text{sech}(a+b x)+b x \tanh (a+b x) \text{sech}(a+b x)-2 b x \tan ^{-1}(\sinh (a+b x)+\cosh (a+b x))}{2 b^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(-2*b*x*ArcTan[Cosh[a + b*x] + Sinh[a + b*x]] + I*PolyLog[2, (-I)*(Cosh[a + b*x] + Sinh[a + b*x])] - I*PolyLo
g[2, I*(Cosh[a + b*x] + Sinh[a + b*x])] + Sech[a + b*x] + b*x*Sech[a + b*x]*Tanh[a + b*x])/(2*b^2)

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Maple [B]  time = 0.068, size = 178, normalized size = 2. \begin{align*} -{\frac{{{\rm e}^{bx+a}} \left ( bx{{\rm e}^{2\,bx+2\,a}}-bx+{{\rm e}^{2\,bx+2\,a}}+1 \right ) }{{b}^{2} \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) ^{2}}}-{\frac{{\frac{i}{2}}\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{{\frac{i}{2}}\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\frac{i}{2}}\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{{\frac{i}{2}}\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}-{\frac{{\frac{i}{2}}{\it dilog} \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{{\frac{i}{2}}{\it dilog} \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{a\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (b x e^{\left (3 \, a\right )} + e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} -{\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + 2 \, \int \frac{x e^{\left (b x + a\right )}}{2 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.35389, size = 2963, normalized size = 32.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(b*x + 1)*cosh(b*x + a)^3 + 6*(b*x + 1)*cosh(b*x + a)*sinh(b*x + a)^2 + 2*(b*x + 1)*sinh(b*x + a)^3 -
2*(b*x - 1)*cosh(b*x + a) - (I*cosh(b*x + a)^4 + 4*I*cosh(b*x + a)*sinh(b*x + a)^3 + I*sinh(b*x + a)^4 + (6*I*
cosh(b*x + a)^2 + 2*I)*sinh(b*x + a)^2 + 2*I*cosh(b*x + a)^2 + (4*I*cosh(b*x + a)^3 + 4*I*cosh(b*x + a))*sinh(
b*x + a) + I)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - (-I*cosh(b*x + a)^4 - 4*I*cosh(b*x + a)*sinh(b*x + a)
^3 - I*sinh(b*x + a)^4 + (-6*I*cosh(b*x + a)^2 - 2*I)*sinh(b*x + a)^2 - 2*I*cosh(b*x + a)^2 + (-4*I*cosh(b*x +
a)^3 - 4*I*cosh(b*x + a))*sinh(b*x + a) - I)*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) - (-I*a*cosh(b*x + a)^
4 - 4*I*a*cosh(b*x + a)*sinh(b*x + a)^3 - I*a*sinh(b*x + a)^4 - 2*I*a*cosh(b*x + a)^2 + (-6*I*a*cosh(b*x + a)^
2 - 2*I*a)*sinh(b*x + a)^2 + (-4*I*a*cosh(b*x + a)^3 - 4*I*a*cosh(b*x + a))*sinh(b*x + a) - I*a)*log(cosh(b*x
+ a) + sinh(b*x + a) + I) - (I*a*cosh(b*x + a)^4 + 4*I*a*cosh(b*x + a)*sinh(b*x + a)^3 + I*a*sinh(b*x + a)^4 +
2*I*a*cosh(b*x + a)^2 + (6*I*a*cosh(b*x + a)^2 + 2*I*a)*sinh(b*x + a)^2 + (4*I*a*cosh(b*x + a)^3 + 4*I*a*cosh
(b*x + a))*sinh(b*x + a) + I*a)*log(cosh(b*x + a) + sinh(b*x + a) - I) - ((-I*b*x - I*a)*cosh(b*x + a)^4 + (-4
*I*b*x - 4*I*a)*cosh(b*x + a)*sinh(b*x + a)^3 + (-I*b*x - I*a)*sinh(b*x + a)^4 + (-2*I*b*x - 2*I*a)*cosh(b*x +
a)^2 + ((-6*I*b*x - 6*I*a)*cosh(b*x + a)^2 - 2*I*b*x - 2*I*a)*sinh(b*x + a)^2 - I*b*x + ((-4*I*b*x - 4*I*a)*c
osh(b*x + a)^3 + (-4*I*b*x - 4*I*a)*cosh(b*x + a))*sinh(b*x + a) - I*a)*log(I*cosh(b*x + a) + I*sinh(b*x + a)
+ 1) - ((I*b*x + I*a)*cosh(b*x + a)^4 + (4*I*b*x + 4*I*a)*cosh(b*x + a)*sinh(b*x + a)^3 + (I*b*x + I*a)*sinh(b
*x + a)^4 + (2*I*b*x + 2*I*a)*cosh(b*x + a)^2 + ((6*I*b*x + 6*I*a)*cosh(b*x + a)^2 + 2*I*b*x + 2*I*a)*sinh(b*x
+ a)^2 + I*b*x + ((4*I*b*x + 4*I*a)*cosh(b*x + a)^3 + (4*I*b*x + 4*I*a)*cosh(b*x + a))*sinh(b*x + a) + I*a)*l
og(-I*cosh(b*x + a) - I*sinh(b*x + a) + 1) + 2*(3*(b*x + 1)*cosh(b*x + a)^2 - b*x + 1)*sinh(b*x + a))/(b^2*cos
h(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*co
sh(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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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*sech(b*x+a)**3*sinh(b*x+a)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{sech}\left (b x + a\right )^{3} \sinh \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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