3.1.5 \(\int \text {sech}^{-1}(a x)^2 \, dx\) [5]

Optimal. Leaf size=63 \[ x \text {sech}^{-1}(a x)^2-\frac {4 \text {sech}^{-1}(a x) \text {ArcTan}\left (e^{\text {sech}^{-1}(a x)}\right )}{a}+\frac {2 i \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(a x)}\right )}{a}-\frac {2 i \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(a x)}\right )}{a} \]

[Out]

x*arcsech(a*x)^2-4*arcsech(a*x)*arctan(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))/a+2*I*polylog(2,-I*(1/a/x+(1/a/x
-1)^(1/2)*(1+1/a/x)^(1/2)))/a-2*I*polylog(2,I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))/a

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Rubi [A]
time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {6414, 5526, 4265, 2317, 2438} \begin {gather*} -\frac {4 \text {sech}^{-1}(a x) \text {ArcTan}\left (e^{\text {sech}^{-1}(a x)}\right )}{a}+\frac {2 i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a x)}\right )}{a}-\frac {2 i \text {Li}_2\left (i e^{\text {sech}^{-1}(a x)}\right )}{a}+x \text {sech}^{-1}(a x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSech[a*x]^2,x]

[Out]

x*ArcSech[a*x]^2 - (4*ArcSech[a*x]*ArcTan[E^ArcSech[a*x]])/a + ((2*I)*PolyLog[2, (-I)*E^ArcSech[a*x]])/a - ((2
*I)*PolyLog[2, I*E^ArcSech[a*x]])/a

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 4265

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 5526

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> Simp[(-
x^(m - n + 1))*(Sech[a + b*x^n]^p/(b*n*p)), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x]
, x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 1]

Rule 6414

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

Rubi steps

\begin {align*} \int \text {sech}^{-1}(a x)^2 \, dx &=-\frac {\text {Subst}\left (\int x^2 \text {sech}(x) \tanh (x) \, dx,x,\text {sech}^{-1}(a x)\right )}{a}\\ &=x \text {sech}^{-1}(a x)^2-\frac {2 \text {Subst}\left (\int x \text {sech}(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{a}\\ &=x \text {sech}^{-1}(a x)^2-\frac {4 \text {sech}^{-1}(a x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{a}+\frac {(2 i) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{a}-\frac {(2 i) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {sech}^{-1}(a x)\right )}{a}\\ &=x \text {sech}^{-1}(a x)^2-\frac {4 \text {sech}^{-1}(a x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{a}+\frac {(2 i) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a x)}\right )}{a}-\frac {(2 i) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {sech}^{-1}(a x)}\right )}{a}\\ &=x \text {sech}^{-1}(a x)^2-\frac {4 \text {sech}^{-1}(a x) \tan ^{-1}\left (e^{\text {sech}^{-1}(a x)}\right )}{a}+\frac {2 i \text {Li}_2\left (-i e^{\text {sech}^{-1}(a x)}\right )}{a}-\frac {2 i \text {Li}_2\left (i e^{\text {sech}^{-1}(a x)}\right )}{a}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 90, normalized size = 1.43 \begin {gather*} \frac {i \left (\text {sech}^{-1}(a x) \left (-i a x \text {sech}^{-1}(a x)+2 \log \left (1-i e^{-\text {sech}^{-1}(a x)}\right )-2 \log \left (1+i e^{-\text {sech}^{-1}(a x)}\right )\right )+2 \text {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(a x)}\right )-2 \text {PolyLog}\left (2,i e^{-\text {sech}^{-1}(a x)}\right )\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[a*x]^2,x]

[Out]

(I*(ArcSech[a*x]*((-I)*a*x*ArcSech[a*x] + 2*Log[1 - I/E^ArcSech[a*x]] - 2*Log[1 + I/E^ArcSech[a*x]]) + 2*PolyL
og[2, (-I)/E^ArcSech[a*x]] - 2*PolyLog[2, I/E^ArcSech[a*x]]))/a

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Maple [A]
time = 0.29, size = 183, normalized size = 2.90

method	result	size
derivativedivides	\(\frac {\mathrm {arcsech}\left (a x \right )^{2} a x +2 i \mathrm {arcsech}\left (a x \right ) \ln \left (1+i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )-2 i \mathrm {arcsech}\left (a x \right ) \ln \left (1-i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )+2 i \dilog \left (1+i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )-2 i \dilog \left (1-i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )}{a}\)	\(183\)
default	\(\frac {\mathrm {arcsech}\left (a x \right )^{2} a x +2 i \mathrm {arcsech}\left (a x \right ) \ln \left (1+i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )-2 i \mathrm {arcsech}\left (a x \right ) \ln \left (1-i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )+2 i \dilog \left (1+i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )-2 i \dilog \left (1-i \left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )\right )}{a}\)	\(183\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/a*(arcsech(a*x)^2*a*x+2*I*arcsech(a*x)*ln(1+I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))-2*I*arcsech(a*x)*ln(1
-I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))+2*I*dilog(1+I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2)))-2*I*dilog(1
-I*(1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))))

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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(arcsech(a*x)^2,x, algorithm="maxima")

[Out]

x*log(sqrt(a*x + 1)*sqrt(-a*x + 1) + 1)^2 - integrate(-(a^2*x^2*log(a)^2 + (a^2*x^2 - 1)*log(x)^2 + (a^2*x^2*l
og(a)^2 + (a^2*x^2 - 1)*log(x)^2 - log(a)^2 + 2*(a^2*x^2*log(a) - log(a))*log(x))*sqrt(a*x + 1)*sqrt(-a*x + 1)
- 2*(a^2*x^2*log(a) + (a^2*x^2*(log(a) + 1) + (a^2*x^2 - 1)*log(x) - log(a))*sqrt(a*x + 1)*sqrt(-a*x + 1) + (
a^2*x^2 - 1)*log(x) - log(a))*log(sqrt(a*x + 1)*sqrt(-a*x + 1) + 1) - log(a)^2 + 2*(a^2*x^2*log(a) - log(a))*l
og(x))/(a^2*x^2 + (a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(-a*x + 1) - 1), 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(arcsech(a*x)^2,x, algorithm="fricas")

[Out]

integral(arcsech(a*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \operatorname {asech}^{2}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asech(a*x)**2,x)

[Out]

Integral(asech(a*x)**2, 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(arcsech(a*x)^2,x, algorithm="giac")

[Out]

integrate(arcsech(a*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\mathrm {acosh}\left (\frac {1}{a\,x}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a*x))^2,x)

[Out]

int(acosh(1/(a*x))^2, x)

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