3.37 \(\int \frac{(a+b \text{sech}^{-1}(c x))^2}{x} \, dx\)

Optimal. Leaf size=83 \[ -b \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,-e^{2 \text{sech}^{-1}(c x)}\right )+\frac{\left (a+b \text{sech}^{-1}(c x)\right )^3}{3 b}-\log \left (e^{2 \text{sech}^{-1}(c x)}+1\right ) \left (a+b \text{sech}^{-1}(c x)\right )^2 \]

[Out]

(a + b*ArcSech[c*x])^3/(3*b) - (a + b*ArcSech[c*x])^2*Log[1 + E^(2*ArcSech[c*x])] - b*(a + b*ArcSech[c*x])*Pol
yLog[2, -E^(2*ArcSech[c*x])] + (b^2*PolyLog[3, -E^(2*ArcSech[c*x])])/2

________________________________________________________________________________________

Rubi [A]  time = 0.125281, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6285, 3718, 2190, 2531, 2282, 6589} \[ -b \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,-e^{2 \text{sech}^{-1}(c x)}\right )+\frac{\left (a+b \text{sech}^{-1}(c x)\right )^3}{3 b}-\log \left (e^{2 \text{sech}^{-1}(c x)}+1\right ) \left (a+b \text{sech}^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2/x,x]

[Out]

(a + b*ArcSech[c*x])^3/(3*b) - (a + b*ArcSech[c*x])^2*Log[1 + E^(2*ArcSech[c*x])] - b*(a + b*ArcSech[c*x])*Pol
yLog[2, -E^(2*ArcSech[c*x])] + (b^2*PolyLog[3, -E^(2*ArcSech[c*x])])/2

Rule 6285

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -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 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 2531

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 2282

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 6589

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

Mathematica [A]  time = 0.156747, size = 116, normalized size = 1.4 \[ a b \left (\text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(c x)}\right )-\text{sech}^{-1}(c x) \left (\text{sech}^{-1}(c x)+2 \log \left (e^{-2 \text{sech}^{-1}(c x)}+1\right )\right )\right )+b^2 \left (\text{sech}^{-1}(c x) \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-2 \text{sech}^{-1}(c x)}\right )-\frac{1}{3} \text{sech}^{-1}(c x)^3-\text{sech}^{-1}(c x)^2 \log \left (e^{-2 \text{sech}^{-1}(c x)}+1\right )\right )+a^2 \log (c x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSech[c*x])^2/x,x]

[Out]

a^2*Log[c*x] + a*b*(-(ArcSech[c*x]*(ArcSech[c*x] + 2*Log[1 + E^(-2*ArcSech[c*x])])) + PolyLog[2, -E^(-2*ArcSec
h[c*x])]) + b^2*(-ArcSech[c*x]^3/3 - ArcSech[c*x]^2*Log[1 + E^(-2*ArcSech[c*x])] + ArcSech[c*x]*PolyLog[2, -E^
(-2*ArcSech[c*x])] + PolyLog[3, -E^(-2*ArcSech[c*x])]/2)

________________________________________________________________________________________

Maple [A]  time = 0.237, size = 250, normalized size = 3. \begin{align*}{a}^{2}\ln \left ( cx \right ) +{\frac{{b}^{2} \left ({\rm arcsech} \left (cx\right ) \right ) ^{3}}{3}}-{b}^{2} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}\ln \left ( 1+ \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) -{b}^{2}{\rm arcsech} \left (cx\right ){\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) +{\frac{{b}^{2}}{2}{\it polylog} \left ( 3,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }-2\,ab{\rm arcsech} \left (cx\right )\ln \left ( 1+ \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) +ab \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}-ab{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsech(c*x))^2/x,x)

[Out]

a^2*ln(c*x)+1/3*b^2*arcsech(c*x)^3-b^2*arcsech(c*x)^2*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)-b^2*arc
sech(c*x)*polylog(2,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+1/2*b^2*polylog(3,-(1/c/x+(-1+1/c/x)^(1/2)*(1
+1/c/x)^(1/2))^2)-2*a*b*arcsech(c*x)*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+a*b*arcsech(c*x)^2-a*b*p
olylog(2,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x,x, algorithm="maxima")

[Out]

a^2*log(x) + integrate(b^2*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2/x + 2*a*b*log(sqrt(1/(c*x) + 1
)*sqrt(1/(c*x) - 1) + 1/(c*x))/x, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{arsech}\left (c x\right )^{2} + 2 \, a b \operatorname{arsech}\left (c x\right ) + a^{2}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x,x, algorithm="fricas")

[Out]

integral((b^2*arcsech(c*x)^2 + 2*a*b*arcsech(c*x) + a^2)/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))**2/x,x)

[Out]

Integral((a + b*asech(c*x))**2/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)^2/x, x)