3.46 \(\int \frac{(a+b \text{sech}^{-1}(c x))^3}{x} \, dx\)

Optimal. Leaf size=114 \[ \frac{3}{2} b^2 \text{PolyLog}\left (3,-e^{2 \text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )-\frac{3}{2} b \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{3}{4} b^3 \text{PolyLog}\left (4,-e^{2 \text{sech}^{-1}(c x)}\right )+\frac{\left (a+b \text{sech}^{-1}(c x)\right )^4}{4 b}-\log \left (e^{2 \text{sech}^{-1}(c x)}+1\right ) \left (a+b \text{sech}^{-1}(c x)\right )^3 \]

[Out]

(a + b*ArcSech[c*x])^4/(4*b) - (a + b*ArcSech[c*x])^3*Log[1 + E^(2*ArcSech[c*x])] - (3*b*(a + b*ArcSech[c*x])^
2*PolyLog[2, -E^(2*ArcSech[c*x])])/2 + (3*b^2*(a + b*ArcSech[c*x])*PolyLog[3, -E^(2*ArcSech[c*x])])/2 - (3*b^3
*PolyLog[4, -E^(2*ArcSech[c*x])])/4

________________________________________________________________________________________

Rubi [A]  time = 0.14476, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6285, 3718, 2190, 2531, 6609, 2282, 6589} \[ \frac{3}{2} b^2 \text{PolyLog}\left (3,-e^{2 \text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )-\frac{3}{2} b \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )^2-\frac{3}{4} b^3 \text{PolyLog}\left (4,-e^{2 \text{sech}^{-1}(c x)}\right )+\frac{\left (a+b \text{sech}^{-1}(c x)\right )^4}{4 b}-\log \left (e^{2 \text{sech}^{-1}(c x)}+1\right ) \left (a+b \text{sech}^{-1}(c x)\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^3/x,x]

[Out]

(a + b*ArcSech[c*x])^4/(4*b) - (a + b*ArcSech[c*x])^3*Log[1 + E^(2*ArcSech[c*x])] - (3*b*(a + b*ArcSech[c*x])^
2*PolyLog[2, -E^(2*ArcSech[c*x])])/2 + (3*b^2*(a + b*ArcSech[c*x])*PolyLog[3, -E^(2*ArcSech[c*x])])/2 - (3*b^3
*PolyLog[4, -E^(2*ArcSech[c*x])])/4

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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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 )^3}{x} \, dx &=-\operatorname{Subst}\left (\int (a+b x)^3 \tanh (x) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=\frac{\left (a+b \text{sech}^{-1}(c x)\right )^4}{4 b}-2 \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)^3}{1+e^{2 x}} \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=\frac{\left (a+b \text{sech}^{-1}(c x)\right )^4}{4 b}-\left (a+b \text{sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text{sech}^{-1}(c x)}\right )+(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \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 )^4}{4 b}-\left (a+b \text{sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text{sech}^{-1}(c x)}\right )-\frac{3}{2} b \left (a+b \text{sech}^{-1}(c x)\right )^2 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(c x)}\right )+\left (3 b^2\right ) \operatorname{Subst}\left (\int (a+b x) \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 )^4}{4 b}-\left (a+b \text{sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text{sech}^{-1}(c x)}\right )-\frac{3}{2} b \left (a+b \text{sech}^{-1}(c x)\right )^2 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(c x)}\right )+\frac{3}{2} b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_3\left (-e^{2 \text{sech}^{-1}(c x)}\right )-\frac{1}{2} \left (3 b^3\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=\frac{\left (a+b \text{sech}^{-1}(c x)\right )^4}{4 b}-\left (a+b \text{sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text{sech}^{-1}(c x)}\right )-\frac{3}{2} b \left (a+b \text{sech}^{-1}(c x)\right )^2 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(c x)}\right )+\frac{3}{2} b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_3\left (-e^{2 \text{sech}^{-1}(c x)}\right )-\frac{1}{4} \left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 \text{sech}^{-1}(c x)}\right )\\ &=\frac{\left (a+b \text{sech}^{-1}(c x)\right )^4}{4 b}-\left (a+b \text{sech}^{-1}(c x)\right )^3 \log \left (1+e^{2 \text{sech}^{-1}(c x)}\right )-\frac{3}{2} b \left (a+b \text{sech}^{-1}(c x)\right )^2 \text{Li}_2\left (-e^{2 \text{sech}^{-1}(c x)}\right )+\frac{3}{2} b^2 \left (a+b \text{sech}^{-1}(c x)\right ) \text{Li}_3\left (-e^{2 \text{sech}^{-1}(c x)}\right )-\frac{3}{4} b^3 \text{Li}_4\left (-e^{2 \text{sech}^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.229202, size = 182, normalized size = 1.6 \[ \frac{1}{4} \left (6 b^2 \text{PolyLog}\left (3,-e^{-2 \text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )+6 b \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(c x)}\right ) \left (a+b \text{sech}^{-1}(c x)\right )^2+3 b^3 \text{PolyLog}\left (4,-e^{-2 \text{sech}^{-1}(c x)}\right )-6 a^2 b \text{sech}^{-1}(c x)^2-12 a^2 b \text{sech}^{-1}(c x) \log \left (e^{-2 \text{sech}^{-1}(c x)}+1\right )+4 a^3 \log (c x)-4 a b^2 \text{sech}^{-1}(c x)^3-12 a b^2 \text{sech}^{-1}(c x)^2 \log \left (e^{-2 \text{sech}^{-1}(c x)}+1\right )-b^3 \text{sech}^{-1}(c x)^4-4 b^3 \text{sech}^{-1}(c x)^3 \log \left (e^{-2 \text{sech}^{-1}(c x)}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSech[c*x])^3/x,x]

[Out]

(-6*a^2*b*ArcSech[c*x]^2 - 4*a*b^2*ArcSech[c*x]^3 - b^3*ArcSech[c*x]^4 - 12*a^2*b*ArcSech[c*x]*Log[1 + E^(-2*A
rcSech[c*x])] - 12*a*b^2*ArcSech[c*x]^2*Log[1 + E^(-2*ArcSech[c*x])] - 4*b^3*ArcSech[c*x]^3*Log[1 + E^(-2*ArcS
ech[c*x])] + 4*a^3*Log[c*x] + 6*b*(a + b*ArcSech[c*x])^2*PolyLog[2, -E^(-2*ArcSech[c*x])] + 6*b^2*(a + b*ArcSe
ch[c*x])*PolyLog[3, -E^(-2*ArcSech[c*x])] + 3*b^3*PolyLog[4, -E^(-2*ArcSech[c*x])])/4

________________________________________________________________________________________

Maple [B]  time = 0.272, size = 454, normalized size = 4. \begin{align*}{a}^{3}\ln \left ( cx \right ) +{\frac{{b}^{3} \left ({\rm arcsech} \left (cx\right ) \right ) ^{4}}{4}}-{b}^{3} \left ({\rm arcsech} \left (cx\right ) \right ) ^{3}\ln \left ( 1+ \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) -{\frac{3\,{b}^{3} \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{2}{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }+{\frac{3\,{b}^{3}{\rm arcsech} \left (cx\right )}{2}{\it polylog} \left ( 3,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }-{\frac{3\,{b}^{3}}{4}{\it polylog} \left ( 4,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }+a{b}^{2} \left ({\rm arcsech} \left (cx\right ) \right ) ^{3}-3\,a{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 ) -3\,a{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{3\,a{b}^{2}}{2}{\it polylog} \left ( 3,- \left ({\frac{1}{cx}}+\sqrt{-1+{\frac{1}{cx}}}\sqrt{1+{\frac{1}{cx}}} \right ) ^{2} \right ) }+{\frac{3\,{a}^{2}b \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}}{2}}-3\,{a}^{2}b{\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 ) -{\frac{3\,{a}^{2}b}{2}{\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))^3/x,x)

[Out]

a^3*ln(c*x)+1/4*b^3*arcsech(c*x)^4-b^3*arcsech(c*x)^3*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)-3/2*b^3
*arcsech(c*x)^2*polylog(2,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+3/2*b^3*arcsech(c*x)*polylog(3,-(1/c/x+
(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)-3/4*b^3*polylog(4,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+a*b^2*arcs
ech(c*x)^3-3*a*b^2*arcsech(c*x)^2*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)-3*a*b^2*arcsech(c*x)*polylo
g(2,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+3/2*a*b^2*polylog(3,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))
^2)+3/2*a^2*b*arcsech(c*x)^2-3*a^2*b*arcsech(c*x)*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)-3/2*a^2*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^{3} \log \left (x\right ) + \int \frac{b^{3} \log \left (\sqrt{\frac{1}{c x} + 1} \sqrt{\frac{1}{c x} - 1} + \frac{1}{c x}\right )^{3}}{x} + \frac{3 \, a b^{2} \log \left (\sqrt{\frac{1}{c x} + 1} \sqrt{\frac{1}{c x} - 1} + \frac{1}{c x}\right )^{2}}{x} + \frac{3 \, a^{2} 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))^3/x,x, algorithm="maxima")

[Out]

a^3*log(x) + integrate(b^3*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^3/x + 3*a*b^2*log(sqrt(1/(c*x) +
1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2/x + 3*a^2*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^{3} \operatorname{arsech}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname{arsech}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname{arsech}\left (c x\right ) + a^{3}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^3/x,x, algorithm="fricas")

[Out]

integral((b^3*arcsech(c*x)^3 + 3*a*b^2*arcsech(c*x)^2 + 3*a^2*b*arcsech(c*x) + a^3)/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 )^{3}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))**3/x,x)

[Out]

Integral((a + b*asech(c*x))**3/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 )}^{3}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^3/x,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)^3/x, x)