3.1.37 \(\int \frac {(a+b \text {sech}^{-1}(c x))^2}{x} \, dx\) [37]

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

[Out]

1/3*(a+b*arcsech(c*x))^3/b-(a+b*arcsech(c*x))^2*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)-b*(a+b*arcsec
h(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)

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 = {6420, 3799, 2221, 2611, 2320, 6724} \begin {gather*} -b \text {Li}_2\left (-e^{2 \text {sech}^{-1}(c x)}\right ) \left (a+b \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+\frac {1}{2} b^2 \text {Li}_3\left (-e^{2 \text {sech}^{-1}(c x)}\right ) \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3799

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 6420

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 6724

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 &=-\text {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 \text {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) \text {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 \text {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 \text {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 = 0.10, size = 116, normalized size = 1.40 \begin {gather*} a^2 \log (c x)+a b \left (-\text {sech}^{-1}(c x) \left (\text {sech}^{-1}(c x)+2 \log \left (1+e^{-2 \text {sech}^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}(c x)}\right )\right )+b^2 \left (-\frac {1}{3} \text {sech}^{-1}(c x)^3-\text {sech}^{-1}(c x)^2 \log \left (1+e^{-2 \text {sech}^{-1}(c x)}\right )+\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 )\right ) \end {gather*}

Antiderivative was successfully verified.

[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*(-1/3*ArcSech[c*x]^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 = 0.22, size = 250, normalized size = 3.01

method	result	size
derivativedivides	\(a^{2} \ln \left (c x \right )+\frac {b^{2} \mathrm {arcsech}\left (c x \right )^{3}}{3}-b^{2} \mathrm {arcsech}\left (c x \right )^{2} \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-b^{2} \mathrm {arcsech}\left (c x \right ) \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )+\frac {b^{2} \polylog \left (3, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2}+a b \mathrm {arcsech}\left (c x \right )^{2}-2 a b \,\mathrm {arcsech}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-a b \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )\)	\(250\)
default	\(a^{2} \ln \left (c x \right )+\frac {b^{2} \mathrm {arcsech}\left (c x \right )^{3}}{3}-b^{2} \mathrm {arcsech}\left (c x \right )^{2} \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-b^{2} \mathrm {arcsech}\left (c x \right ) \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )+\frac {b^{2} \polylog \left (3, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2}+a b \mathrm {arcsech}\left (c x \right )^{2}-2 a b \,\mathrm {arcsech}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-a b \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )\)	\(250\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)+a*b*arcsech(c*x)^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*p
olylog(2,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right )^{2}}{x}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))^2/x,x)

[Out]

int((a + b*acosh(1/(c*x)))^2/x, x)

________________________________________________________________________________________