3.1.33 \(\int \frac {a+b \cosh ^{-1}(c x)}{x (d-c^2 d x^2)} \, dx\) [33]

Optimal. Leaf size=61 \[ \frac {2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d}+\frac {b \text {PolyLog}\left (2,-e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac {b \text {PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5916, 5569, 4267, 2317, 2438} \begin {gather*} \frac {2 \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d}+\frac {b \text {Li}_2\left (-e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac {b \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c*x])/(x*(d - c^2*d*x^2)),x]

[Out]

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

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 4267

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 5569

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 5916

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[-d^(-1), Subst[I
nt[(a + b*x)^n/(Cosh[x]*Sinh[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 {a+b \cosh ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx &=-\frac {\text {Subst}\left (\int (a+b x) \text {csch}(x) \text {sech}(x) \, dx,x,\cosh ^{-1}(c x)\right )}{d}\\ &=-\frac {2 \text {Subst}\left (\int (a+b x) \text {csch}(2 x) \, dx,x,\cosh ^{-1}(c x)\right )}{d}\\ &=\frac {2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d}+\frac {b \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{d}-\frac {b \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{d}\\ &=\frac {2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d}+\frac {b \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac {b \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 d}\\ &=\frac {2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d}+\frac {b \text {Li}_2\left (-e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac {b \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 93, normalized size = 1.52 \begin {gather*} \frac {a \log (x)}{d}-\frac {a \log \left (1-c^2 x^2\right )}{2 d}-\frac {b \left (2 \cosh ^{-1}(c x) \left (\log \left (1-e^{-2 \cosh ^{-1}(c x)}\right )-\log \left (1+e^{-2 \cosh ^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \cosh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,e^{-2 \cosh ^{-1}(c x)}\right )\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 5.27, size = 88, normalized size = 1.44

method	result	size
derivativedivides	\(-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a \ln \left (c x +1\right )}{2 d}+\frac {a \ln \left (c x \right )}{d}-\frac {b \left (\dilog \left (\frac {1}{\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}}\right )-\frac {\dilog \left (\frac {1}{\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{4}}\right )}{4}\right )}{d}\)	\(88\)
default	\(-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a \ln \left (c x +1\right )}{2 d}+\frac {a \ln \left (c x \right )}{d}-\frac {b \left (\dilog \left (\frac {1}{\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}}\right )-\frac {\dilog \left (\frac {1}{\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{4}}\right )}{4}\right )}{d}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d*ln(c*x-1)-1/2*a/d*ln(c*x+1)+a/d*ln(c*x)-b/d*(dilog(1/(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)-1/4*dilog(1
/(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^4))

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

[Out]

-1/2*a*(log(c*x + 1)/d + log(c*x - 1)/d - 2*log(x)/d) - b*integrate(log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))/(c^
2*d*x^3 - d*x), 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*arccosh(c*x))/x/(-c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral(-(b*arccosh(c*x) + a)/(c^2*d*x^3 - d*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {a}{c^{2} x^{3} - x}\, dx + \int \frac {b \operatorname {acosh}{\left (c x \right )}}{c^{2} x^{3} - x}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a/(c**2*x**3 - x), x) + Integral(b*acosh(c*x)/(c**2*x**3 - x), x))/d

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

[Out]

integrate(-(b*arccosh(c*x) + a)/((c^2*d*x^2 - d)*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 (c\,x\right )}{x\,\left (d-c^2\,d\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))/(x*(d - c^2*d*x^2)),x)

[Out]

int((a + b*acosh(c*x))/(x*(d - c^2*d*x^2)), x)

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