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

Optimal. Leaf size=120 \[ \frac{b \text{PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )}{2 c d^2}-\frac{b \text{PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{2 c d^2}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac{\tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c d^2}-\frac{b}{2 c d^2 \sqrt{c x-1} \sqrt{c x+1}} \]

[Out]

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

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Rubi [A]  time = 0.0946673, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {5689, 74, 5694, 4182, 2279, 2391} \[ \frac{b \text{PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )}{2 c d^2}-\frac{b \text{PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{2 c d^2}+\frac{x \left (a+b \cosh ^{-1}(c x)\right )}{2 d^2 \left (1-c^2 x^2\right )}+\frac{\tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c d^2}-\frac{b}{2 c d^2 \sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5689

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

Rule 74

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

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 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]

Rubi steps

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

Mathematica [A]  time = 1.34477, size = 189, normalized size = 1.58 \[ \frac{2 b \text{PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )-2 b \text{PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )+\frac{a c^2 x^2 \log (c x+1)+\left (a-a c^2 x^2\right ) \log (1-c x)-2 a c x-a \log (c x+1)-2 b \cosh ^{-1}(c x) \left (\left (c^2 x^2-1\right ) \log \left (1-e^{\cosh ^{-1}(c x)}\right )+\left (1-c^2 x^2\right ) \log \left (e^{\cosh ^{-1}(c x)}+1\right )+c x\right )-2 b c x \sqrt{\frac{c x-1}{c x+1}}-2 b \sqrt{\frac{c x-1}{c x+1}}}{c^2 x^2-1}}{4 c d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*a*c*x - 2*b*Sqrt[(-1 + c*x)/(1 + c*x)] - 2*b*c*x*Sqrt[(-1 + c*x)/(1 + c*x)] - 2*b*ArcCosh[c*x]*(c*x + (-1
+ c^2*x^2)*Log[1 - E^ArcCosh[c*x]] + (1 - c^2*x^2)*Log[1 + E^ArcCosh[c*x]]) + (a - a*c^2*x^2)*Log[1 - c*x] -
a*Log[1 + c*x] + a*c^2*x^2*Log[1 + c*x])/(-1 + c^2*x^2) + 2*b*PolyLog[2, -E^ArcCosh[c*x]] - 2*b*PolyLog[2, E^A
rcCosh[c*x]])/(4*c*d^2)

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Maple [A]  time = 0.062, size = 252, normalized size = 2.1 \begin{align*} -{\frac{a}{4\,c{d}^{2} \left ( cx-1 \right ) }}-{\frac{a\ln \left ( cx-1 \right ) }{4\,c{d}^{2}}}-{\frac{a}{4\,c{d}^{2} \left ( cx+1 \right ) }}+{\frac{a\ln \left ( cx+1 \right ) }{4\,c{d}^{2}}}-{\frac{b{\rm arccosh} \left (cx\right )x}{2\,{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{b}{2\,c{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{cx-1}\sqrt{cx+1}}+{\frac{b{\rm arccosh} \left (cx\right )}{2\,c{d}^{2}}\ln \left ( 1+cx+\sqrt{cx-1}\sqrt{cx+1} \right ) }+{\frac{b}{2\,c{d}^{2}}{\it polylog} \left ( 2,-cx-\sqrt{cx-1}\sqrt{cx+1} \right ) }-{\frac{b{\rm arccosh} \left (cx\right )}{2\,c{d}^{2}}\ln \left ( 1-cx-\sqrt{cx-1}\sqrt{cx+1} \right ) }-{\frac{b}{2\,c{d}^{2}}{\it polylog} \left ( 2,cx+\sqrt{cx-1}\sqrt{cx+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{64} \,{\left (192 \, c^{3} \int \frac{x^{3} \log \left (c x - 1\right )}{8 \,{\left (c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}\right )}}\,{d x} - 8 \, c^{2}{\left (\frac{2 \, x}{c^{4} d^{2} x^{2} - c^{2} d^{2}} + \frac{\log \left (c x + 1\right )}{c^{3} d^{2}} - \frac{\log \left (c x - 1\right )}{c^{3} d^{2}}\right )} - 64 \, c^{2} \int \frac{x^{2} \log \left (c x - 1\right )}{8 \,{\left (c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}\right )}}\,{d x} + 3 \,{\left (c{\left (\frac{2}{c^{4} d^{2} x - c^{3} d^{2}} - \frac{\log \left (c x + 1\right )}{c^{3} d^{2}} + \frac{\log \left (c x - 1\right )}{c^{3} d^{2}}\right )} + \frac{4 \, \log \left (c x - 1\right )}{c^{4} d^{2} x^{2} - c^{2} d^{2}}\right )} c - \frac{4 \,{\left ({\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right )^{2} + 2 \,{\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) \log \left (c x - 1\right ) + 4 \,{\left (2 \, c x -{\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) +{\left (c^{2} x^{2} - 1\right )} \log \left (c x - 1\right )\right )} \log \left (c x + \sqrt{c x + 1} \sqrt{c x - 1}\right )\right )}}{c^{3} d^{2} x^{2} - c d^{2}} + 64 \, \int -\frac{2 \, c x -{\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) +{\left (c^{2} x^{2} - 1\right )} \log \left (c x - 1\right )}{4 \,{\left (c^{5} d^{2} x^{5} - 2 \, c^{3} d^{2} x^{3} + c d^{2} x +{\left (c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}\right )} \sqrt{c x + 1} \sqrt{c x - 1}\right )}}\,{d x} + 64 \, \int \frac{\log \left (c x - 1\right )}{8 \,{\left (c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}\right )}}\,{d x}\right )} b - \frac{1}{4} \, a{\left (\frac{2 \, x}{c^{2} d^{2} x^{2} - d^{2}} - \frac{\log \left (c x + 1\right )}{c d^{2}} + \frac{\log \left (c x - 1\right )}{c d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(192*c^3*integrate(1/8*x^3*log(c*x - 1)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x) - 8*c^2*(2*x/(c^4*d^2*x^2
- c^2*d^2) + log(c*x + 1)/(c^3*d^2) - log(c*x - 1)/(c^3*d^2)) - 64*c^2*integrate(1/8*x^2*log(c*x - 1)/(c^4*d^
2*x^4 - 2*c^2*d^2*x^2 + d^2), x) + 3*(c*(2/(c^4*d^2*x - c^3*d^2) - log(c*x + 1)/(c^3*d^2) + log(c*x - 1)/(c^3*
d^2)) + 4*log(c*x - 1)/(c^4*d^2*x^2 - c^2*d^2))*c - 4*((c^2*x^2 - 1)*log(c*x + 1)^2 + 2*(c^2*x^2 - 1)*log(c*x
+ 1)*log(c*x - 1) + 4*(2*c*x - (c^2*x^2 - 1)*log(c*x + 1) + (c^2*x^2 - 1)*log(c*x - 1))*log(c*x + sqrt(c*x + 1
)*sqrt(c*x - 1)))/(c^3*d^2*x^2 - c*d^2) + 64*integrate(-1/4*(2*c*x - (c^2*x^2 - 1)*log(c*x + 1) + (c^2*x^2 - 1
)*log(c*x - 1))/(c^5*d^2*x^5 - 2*c^3*d^2*x^3 + c*d^2*x + (c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2)*sqrt(c*x + 1)*sqr
t(c*x - 1)), x) + 64*integrate(1/8*log(c*x - 1)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x))*b - 1/4*a*(2*x/(c^2*d
^2*x^2 - d^2) - log(c*x + 1)/(c*d^2) + log(c*x - 1)/(c*d^2))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arcosh}\left (c x\right ) + a}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac{b \operatorname{acosh}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b*acosh(c*x)/(c**4*x**4 - 2*c**2*x**2 + 1), x))/d**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcosh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*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)