3.55 \(\int \frac {\cosh ^{-1}(a x)}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=53 \[ \frac {\text {Li}_2\left (-e^{\cosh ^{-1}(a x)}\right )}{a c}-\frac {\text {Li}_2\left (e^{\cosh ^{-1}(a x)}\right )}{a c}+\frac {2 \cosh ^{-1}(a x) \tanh ^{-1}\left (e^{\cosh ^{-1}(a x)}\right )}{a c} \]

[Out]

2*arccosh(a*x)*arctanh(a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2))/a/c+polylog(2,-a*x-(a*x-1)^(1/2)*(a*x+1)^(1/2))/a/c-po
lylog(2,a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2))/a/c

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Rubi [A]  time = 0.05, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5694, 4182, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,-e^{\cosh ^{-1}(a x)}\right )}{a c}-\frac {\text {PolyLog}\left (2,e^{\cosh ^{-1}(a x)}\right )}{a c}+\frac {2 \cosh ^{-1}(a x) \tanh ^{-1}\left (e^{\cosh ^{-1}(a x)}\right )}{a c} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]/(c - a^2*c*x^2),x]

[Out]

(2*ArcCosh[a*x]*ArcTanh[E^ArcCosh[a*x]])/(a*c) + PolyLog[2, -E^ArcCosh[a*x]]/(a*c) - PolyLog[2, E^ArcCosh[a*x]
]/(a*c)

Rule 2279

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 2391

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 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5694

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(c*d)^(-1), Subst[Int[
(a + b*x)^n*Csch[x], x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 77, normalized size = 1.45 \[ \frac {\text {Li}_2\left (-e^{\cosh ^{-1}(a x)}\right )}{a c}-\frac {\text {Li}_2\left (e^{\cosh ^{-1}(a x)}\right )}{a c}-\frac {\cosh ^{-1}(a x) \log \left (1-e^{\cosh ^{-1}(a x)}\right )}{a c}+\frac {\cosh ^{-1}(a x) \log \left (e^{\cosh ^{-1}(a x)}+1\right )}{a c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCosh[a*x]/(c - a^2*c*x^2),x]

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {arcosh}\left (a x\right )}{a^{2} c x^{2} - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/(-a^2*c*x^2+c),x, algorithm="fricas")

[Out]

integral(-arccosh(a*x)/(a^2*c*x^2 - c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\operatorname {arcosh}\left (a x\right )}{a^{2} c x^{2} - c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/(-a^2*c*x^2+c),x, algorithm="giac")

[Out]

integrate(-arccosh(a*x)/(a^2*c*x^2 - c), x)

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maple [C]  time = 0.02, size = 309, normalized size = 5.83 \[ \frac {\arctanh \left (a x \right ) \mathrm {arccosh}\left (a x \right )}{a c}+\frac {2 i \sqrt {-a^{2} x^{2}+1}\, \sqrt {\frac {1}{2}+\frac {a x}{2}}\, \sqrt {-\frac {1}{2}+\frac {a x}{2}}\, \arctanh \left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a c \left (a^{2} x^{2}-1\right )}-\frac {2 i \sqrt {-a^{2} x^{2}+1}\, \sqrt {\frac {1}{2}+\frac {a x}{2}}\, \sqrt {-\frac {1}{2}+\frac {a x}{2}}\, \arctanh \left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a c \left (a^{2} x^{2}-1\right )}+\frac {2 i \sqrt {-a^{2} x^{2}+1}\, \sqrt {\frac {1}{2}+\frac {a x}{2}}\, \sqrt {-\frac {1}{2}+\frac {a x}{2}}\, \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a c \left (a^{2} x^{2}-1\right )}-\frac {2 i \sqrt {-a^{2} x^{2}+1}\, \sqrt {\frac {1}{2}+\frac {a x}{2}}\, \sqrt {-\frac {1}{2}+\frac {a x}{2}}\, \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{a c \left (a^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(a*x)/(-a^2*c*x^2+c),x)

[Out]

1/a/c*arctanh(a*x)*arccosh(a*x)+2*I/a/c*(-a^2*x^2+1)^(1/2)*(1/2+1/2*a*x)^(1/2)*(-1/2+1/2*a*x)^(1/2)/(a^2*x^2-1
)*arctanh(a*x)*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))-2*I/a/c*(-a^2*x^2+1)^(1/2)*(1/2+1/2*a*x)^(1/2)*(-1/2+1/2*a*x
)^(1/2)/(a^2*x^2-1)*arctanh(a*x)*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))+2*I/a/c*(-a^2*x^2+1)^(1/2)*(1/2+1/2*a*x)^(
1/2)*(-1/2+1/2*a*x)^(1/2)/(a^2*x^2-1)*dilog(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))-2*I/a/c*(-a^2*x^2+1)^(1/2)*(1/2+1/
2*a*x)^(1/2)*(-1/2+1/2*a*x)^(1/2)/(a^2*x^2-1)*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {4 \, {\left (\log \left (a x + 1\right ) - \log \left (a x - 1\right )\right )} \log \left (a x + \sqrt {a x + 1} \sqrt {a x - 1}\right ) - \log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (a x - 1\right ) + \log \left (a x - 1\right )^{2}}{8 \, a c} + \frac {\log \left (a x - 1\right ) \log \left (\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (-\frac {1}{2} \, a x + \frac {1}{2}\right )}{2 \, a c} + \int \frac {\log \left (a x + 1\right ) - \log \left (a x - 1\right )}{2 \, {\left (a^{3} c x^{3} - a c x + {\left (a^{2} c x^{2} - c\right )} \sqrt {a x + 1} \sqrt {a x - 1}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/(-a^2*c*x^2+c),x, algorithm="maxima")

[Out]

1/8*(4*(log(a*x + 1) - log(a*x - 1))*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1)) - log(a*x + 1)^2 - 2*log(a*x + 1)*
log(a*x - 1) + log(a*x - 1)^2)/(a*c) + 1/2*(log(a*x - 1)*log(1/2*a*x + 1/2) + dilog(-1/2*a*x + 1/2))/(a*c) + i
ntegrate(1/2*(log(a*x + 1) - log(a*x - 1))/(a^3*c*x^3 - a*c*x + (a^2*c*x^2 - c)*sqrt(a*x + 1)*sqrt(a*x - 1)),
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(a*x)/(c - a^2*c*x^2),x)

[Out]

int(acosh(a*x)/(c - a^2*c*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\operatorname {acosh}{\left (a x \right )}}{a^{2} x^{2} - 1}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(a*x)/(-a**2*c*x**2+c),x)

[Out]

-Integral(acosh(a*x)/(a**2*x**2 - 1), x)/c

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