3.1.26 \(\int \frac {a+b \text {sech}^{-1}(c x)}{x} \, dx\) [26]

Optimal. Leaf size=56 \[ -\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{2 b}-\left (a+b \text {sech}^{-1}(c x)\right ) \log \left (1+e^{-2 \text {sech}^{-1}(c x)}\right )+\frac {1}{2} b \text {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}(c x)}\right ) \]

[Out]

-1/2*(a+b*arcsech(c*x))^2/b-(a+b*arcsech(c*x))*ln(1+1/(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)+1/2*b*polylo
g(2,-1/(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)

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Rubi [A]
time = 0.09, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6416, 5882, 3799, 2221, 2317, 2438} \begin {gather*} -\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{2 b}-\log \left (e^{-2 \text {sech}^{-1}(c x)}+1\right ) \left (a+b \text {sech}^{-1}(c x)\right )+\frac {1}{2} b \text {Li}_2\left (-e^{-2 \text {sech}^{-1}(c x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])/x,x]

[Out]

-1/2*(a + b*ArcSech[c*x])^2/b - (a + b*ArcSech[c*x])*Log[1 + E^(-2*ArcSech[c*x])] + (b*PolyLog[2, -E^(-2*ArcSe
ch[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 2317

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 2438

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 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 5882

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Tanh[-a/b + x/b], x],
x, a + b*ArcCosh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 6416

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCosh[x/c])/x, x], x, 1/x] /; F
reeQ[{a, b, c}, x]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 47, normalized size = 0.84 \begin {gather*} a \log (x)+\frac {1}{2} 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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])/x,x]

[Out]

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

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Maple [A]
time = 0.17, size = 100, normalized size = 1.79

method	result	size
derivativedivides	\(a \ln \left (c x \right )+\frac {b \mathrm {arcsech}\left (c x \right )^{2}}{2}-b \,\mathrm {arcsech}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-\frac {b \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2}\)	\(100\)
default	\(a \ln \left (c x \right )+\frac {b \mathrm {arcsech}\left (c x \right )^{2}}{2}-b \,\mathrm {arcsech}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )-\frac {b \polylog \left (2, -\left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )^{2}\right )}{2}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*ln(c*x)+1/2*b*arcsech(c*x)^2-b*arcsech(c*x)*ln(1+(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)-1/2*b*polylog(2
,-(1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))^2)

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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))/x,x, algorithm="maxima")

[Out]

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

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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))/x,x, algorithm="fricas")

[Out]

integral((b*arcsech(c*x) + a)/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asech}{\left (c x \right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asech(c*x))/x, x)

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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))/x,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))/x,x)

[Out]

int((a + b*acosh(1/(c*x)))/x, x)

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