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

Optimal. Leaf size=116 \[ \frac {a+b \cosh ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}+\frac {2 \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d^2}+\frac {b \text {Li}_2\left (-e^{2 \cosh ^{-1}(c x)}\right )}{2 d^2}-\frac {b \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 d^2}-\frac {b c x}{2 d^2 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {5754, 5721, 5461, 4182, 2279, 2391, 39} \[ \frac {b \text {PolyLog}\left (2,-e^{2 \cosh ^{-1}(c x)}\right )}{2 d^2}-\frac {b \text {PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 d^2}+\frac {a+b \cosh ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}+\frac {2 \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d^2}-\frac {b c x}{2 d^2 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*x)/(2*d^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (a + b*ArcCosh[c*x])/(2*d^2*(1 - c^2*x^2)) + (2*(a + b*ArcCosh
[c*x])*ArcTanh[E^(2*ArcCosh[c*x])])/d^2 + (b*PolyLog[2, -E^(2*ArcCosh[c*x])])/(2*d^2) - (b*PolyLog[2, E^(2*Arc
Cosh[c*x])])/(2*d^2)

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

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 5461

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 5721

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]

Rule 5754

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp
[((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcCosh[c*x])^n, x], x] - Dist[(b*c*n*(-d)^p)/(2*f*(p + 1)), Int
[(f*x)^(m + 1)*(1 + c*x)^(p + 1/2)*(-1 + c*x)^(p + 1/2)*(a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b,
c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {a+b \cosh ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^2} \, dx &=\frac {a+b \cosh ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}+\frac {(b c) \int \frac {1}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{2 d^2}+\frac {\int \frac {a+b \cosh ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx}{d}\\ &=-\frac {b c x}{2 d^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {a+b \cosh ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac {\operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \text {sech}(x) \, dx,x,\cosh ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c x}{2 d^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {a+b \cosh ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac {2 \operatorname {Subst}\left (\int (a+b x) \text {csch}(2 x) \, dx,x,\cosh ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c x}{2 d^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {a+b \cosh ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}+\frac {2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d^2}+\frac {b \operatorname {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{d^2}-\frac {b \operatorname {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c x}{2 d^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {a+b \cosh ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}+\frac {2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d^2}+\frac {b \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 d^2}-\frac {b \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 d^2}\\ &=-\frac {b c x}{2 d^2 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {a+b \cosh ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}+\frac {2 \left (a+b \cosh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right )}{d^2}+\frac {b \text {Li}_2\left (-e^{2 \cosh ^{-1}(c x)}\right )}{2 d^2}-\frac {b \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 d^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.81, size = 149, normalized size = 1.28 \[ \frac {\frac {a}{1-c^2 x^2}-a \log \left (1-c^2 x^2\right )+2 a \log (x)+b \left (\frac {\cosh ^{-1}(c x)}{1-c^2 x^2}-\text {Li}_2\left (-e^{-2 \cosh ^{-1}(c x)}\right )+\text {Li}_2\left (e^{-2 \cosh ^{-1}(c x)}\right )+\frac {c x \sqrt {\frac {c x-1}{c x+1}}}{1-c x}-2 \cosh ^{-1}(c x) \log \left (1-e^{-2 \cosh ^{-1}(c x)}\right )+2 \cosh ^{-1}(c x) \log \left (e^{-2 \cosh ^{-1}(c x)}+1\right )\right )}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arcosh}\left (c x\right ) + a}{c^{4} d^{2} x^{5} - 2 \, c^{2} d^{2} x^{3} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arccosh(c*x) + a)/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.28, size = 339, normalized size = 2.92 \[ \frac {a \ln \left (c x \right )}{d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}-\frac {a \ln \left (c x -1\right )}{2 d^{2}}+\frac {a}{4 d^{2} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{2 d^{2}}-\frac {b \sqrt {c x +1}\, \sqrt {c x -1}\, c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \,c^{2} x^{2}}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \,\mathrm {arccosh}\left (c x \right )}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{d^{2}}+\frac {b \polylog \left (2, -\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}\right )}{2 d^{2}}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{2}}-\frac {b \polylog \left (2, -c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{2}}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1-c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{2}}-\frac {b \polylog \left (2, c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {1}{c^{2} d^{2} x^{2} - d^{2}} + \frac {\log \left (c x + 1\right )}{d^{2}} + \frac {\log \left (c x - 1\right )}{d^{2}} - \frac {2 \, \log \relax (x)}{d^{2}}\right )} + b \int \frac {\log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{c^{4} d^{2} x^{5} - 2 \, c^{2} d^{2} x^{3} + d^{2} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(1/(c^2*d^2*x^2 - d^2) + log(c*x + 1)/d^2 + log(c*x - 1)/d^2 - 2*log(x)/d^2) + b*integrate(log(c*x + sq
rt(c*x + 1)*sqrt(c*x - 1))/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2*x), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{4} x^{5} - 2 c^{2} x^{3} + x}\, dx + \int \frac {b \operatorname {acosh}{\left (c x \right )}}{c^{4} x^{5} - 2 c^{2} x^{3} + x}\, dx}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**4*x**5 - 2*c**2*x**3 + x), x) + Integral(b*acosh(c*x)/(c**4*x**5 - 2*c**2*x**3 + x), x))/d**2

________________________________________________________________________________________