3.1.36 \(\int (a+b \text {sech}^{-1}(c x))^2 \, dx\) [36]

Optimal. Leaf size=78 \[ x \left (a+b \text {sech}^{-1}(c x)\right )^2-\frac {4 b \left (a+b \text {sech}^{-1}(c x)\right ) \text {ArcTan}\left (e^{\text {sech}^{-1}(c x)}\right )}{c}+\frac {2 i b^2 \text {PolyLog}\left (2,-i e^{\text {sech}^{-1}(c x)}\right )}{c}-\frac {2 i b^2 \text {PolyLog}\left (2,i e^{\text {sech}^{-1}(c x)}\right )}{c} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6414, 5559, 4265, 2317, 2438} \begin {gather*} -\frac {4 b \text {ArcTan}\left (e^{\text {sech}^{-1}(c x)}\right ) \left (a+b \text {sech}^{-1}(c x)\right )}{c}+x \left (a+b \text {sech}^{-1}(c x)\right )^2+\frac {2 i b^2 \text {Li}_2\left (-i e^{\text {sech}^{-1}(c x)}\right )}{c}-\frac {2 i b^2 \text {Li}_2\left (i e^{\text {sech}^{-1}(c x)}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2,x]

[Out]

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

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 5559

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Sim
p[(-(c + d*x)^m)*(Sech[a + b*x]^n/(b*n)), x] + Dist[d*(m/(b*n)), Int[(c + d*x)^(m - 1)*Sech[a + b*x]^n, x], x]
/; FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

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

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Mathematica [A]
time = 0.16, size = 126, normalized size = 1.62 \begin {gather*} a^2 x+\frac {2 a b \left (c x \text {sech}^{-1}(c x)-2 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \text {sech}^{-1}(c x)\right )\right )\right )}{c}+\frac {i b^2 \left (\text {sech}^{-1}(c x) \left (-i c x \text {sech}^{-1}(c x)+2 \log \left (1-i e^{-\text {sech}^{-1}(c x)}\right )-2 \log \left (1+i e^{-\text {sech}^{-1}(c x)}\right )\right )+2 \text {PolyLog}\left (2,-i e^{-\text {sech}^{-1}(c x)}\right )-2 \text {PolyLog}\left (2,i e^{-\text {sech}^{-1}(c x)}\right )\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])^2,x]

[Out]

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

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Maple [A]
time = 0.35, size = 242, normalized size = 3.10

method	result	size
derivativedivides	\(\frac {c \,a^{2} x -2 i b^{2} \mathrm {arcsech}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )+2 i b^{2} \mathrm {arcsech}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )+\mathrm {arcsech}\left (c x \right )^{2} b^{2} c x -2 i b^{2} \dilog \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )+2 i b^{2} \dilog \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )+2 \,\mathrm {arcsech}\left (c x \right ) a b c x -2 \arctan \left (\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right ) a b}{c}\)	\(242\)
default	\(\frac {c \,a^{2} x -2 i b^{2} \mathrm {arcsech}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )+2 i b^{2} \mathrm {arcsech}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )+\mathrm {arcsech}\left (c x \right )^{2} b^{2} c x -2 i b^{2} \dilog \left (1-i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )+2 i b^{2} \dilog \left (1+i \left (\frac {1}{c x}+\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right )\right )+2 \,\mathrm {arcsech}\left (c x \right ) a b c x -2 \arctan \left (\sqrt {-1+\frac {1}{c x}}\, \sqrt {1+\frac {1}{c x}}\right ) a b}{c}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^2,x, algorithm="maxima")

[Out]

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

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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))^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*arcsech(c*x) + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))^2,x)

[Out]

int((a + b*acosh(1/(c*x)))^2, x)

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