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

Optimal. Leaf size=124 \[ -\frac {\tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c^3 d^2}+\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \text {Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac {b \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )}{2 c^3 d^2}-\frac {b}{2 c^3 d^2 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

1/2*x*(a+b*arccosh(c*x))/c^2/d^2/(-c^2*x^2+1)-(a+b*arccosh(c*x))*arctanh(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))/c^3/
d^2-1/2*b*polylog(2,-c*x-(c*x-1)^(1/2)*(c*x+1)^(1/2))/c^3/d^2+1/2*b*polylog(2,c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))
/c^3/d^2-1/2*b/c^3/d^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {5750, 74, 5694, 4182, 2279, 2391} \[ -\frac {b \text {PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac {b \text {PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {\tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c^3 d^2}-\frac {b}{2 c^3 d^2 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-b/(2*c^3*d^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (x*(a + b*ArcCosh[c*x]))/(2*c^2*d^2*(1 - c^2*x^2)) - ((a + b*Arc
Cosh[c*x])*ArcTanh[E^ArcCosh[c*x]])/(c^3*d^2) - (b*PolyLog[2, -E^ArcCosh[c*x]])/(2*c^3*d^2) + (b*PolyLog[2, E^
ArcCosh[c*x]])/(2*c^3*d^2)

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 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]

Rule 5750

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

Rubi steps

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

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Mathematica [A]  time = 0.76, size = 206, normalized size = 1.66 \[ \frac {-\frac {2 a c x}{c^2 x^2-1}+a \log (1-c x)-a \log (c x+1)-2 b \text {Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )+2 b \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )+\frac {b c x \sqrt {\frac {c x-1}{c x+1}}}{1-c x}+\frac {b \sqrt {\frac {c x-1}{c x+1}}}{1-c x}+b \sqrt {\frac {c x-1}{c x+1}}+\frac {b \cosh ^{-1}(c x)}{1-c x}-\frac {b \cosh ^{-1}(c x)}{c x+1}+2 b \cosh ^{-1}(c x) \log \left (1-e^{\cosh ^{-1}(c x)}\right )-2 b \cosh ^{-1}(c x) \log \left (e^{\cosh ^{-1}(c x)}+1\right )}{4 c^3 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b*Sqrt[(-1 + c*x)/(1 + c*x)] + (b*Sqrt[(-1 + c*x)/(1 + c*x)])/(1 - c*x) + (b*c*x*Sqrt[(-1 + c*x)/(1 + c*x)])/
(1 - c*x) - (2*a*c*x)/(-1 + c^2*x^2) + (b*ArcCosh[c*x])/(1 - c*x) - (b*ArcCosh[c*x])/(1 + c*x) + 2*b*ArcCosh[c
*x]*Log[1 - E^ArcCosh[c*x]] - 2*b*ArcCosh[c*x]*Log[1 + E^ArcCosh[c*x]] + a*Log[1 - c*x] - a*Log[1 + c*x] - 2*b
*PolyLog[2, -E^ArcCosh[c*x]] + 2*b*PolyLog[2, E^ArcCosh[c*x]])/(4*c^3*d^2)

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{2} \operatorname {arcosh}\left (c x\right ) + a x^{2}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.31, size = 255, normalized size = 2.06 \[ -\frac {a}{4 c^{3} d^{2} \left (c x -1\right )}+\frac {a \ln \left (c x -1\right )}{4 c^{3} d^{2}}-\frac {a}{4 c^{3} d^{2} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{4 c^{3} d^{2}}-\frac {b \,\mathrm {arccosh}\left (c x \right ) x}{2 c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {c x +1}\, \sqrt {c x -1}}{2 c^{3} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{2 c^{3} d^{2}}-\frac {b \polylog \left (2, -c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{2 c^{3} d^{2}}+\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1-c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{2 c^{3} d^{2}}+\frac {b \polylog \left (2, c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{2 c^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/c^3*a/d^2/(c*x-1)+1/4/c^3*a/d^2*ln(c*x-1)-1/4/c^3*a/d^2/(c*x+1)-1/4/c^3*a/d^2*ln(c*x+1)-1/2/c^2*b/d^2/(c^
2*x^2-1)*arccosh(c*x)*x-1/2/c^3*b/d^2/(c^2*x^2-1)*(c*x+1)^(1/2)*(c*x-1)^(1/2)-1/2/c^3*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^3/d^2+1/2/c^3*b/d^2*arcco
sh(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^3/d^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{64} \, {\left (192 \, c^{3} \int \frac {x^{3} \log \left (c x - 1\right )}{8 \, {\left (c^{6} d^{2} x^{4} - 2 \, c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}}\,{d x} + 8 \, c^{2} {\left (\frac {2 \, x}{c^{6} d^{2} x^{2} - c^{4} d^{2}} + \frac {\log \left (c x + 1\right )}{c^{5} d^{2}} - \frac {\log \left (c x - 1\right )}{c^{5} d^{2}}\right )} - 64 \, c^{2} \int \frac {x^{2} \log \left (c x - 1\right )}{8 \, {\left (c^{6} d^{2} x^{4} - 2 \, c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}}\,{d x} + 3 \, {\left (c {\left (\frac {2}{c^{6} d^{2} x - c^{5} d^{2}} - \frac {\log \left (c x + 1\right )}{c^{5} d^{2}} + \frac {\log \left (c x - 1\right )}{c^{5} d^{2}}\right )} + \frac {4 \, \log \left (c x - 1\right )}{c^{6} d^{2} x^{2} - c^{4} d^{2}}\right )} c - \frac {4 \, {\left ({\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right )^{2} + 2 \, {\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) \log \left (c x - 1\right ) - 4 \, {\left (2 \, c x + {\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) - {\left (c^{2} x^{2} - 1\right )} \log \left (c x - 1\right )\right )} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )\right )}}{c^{5} d^{2} x^{2} - c^{3} d^{2}} + 64 \, \int \frac {2 \, c x + {\left (c^{2} x^{2} - 1\right )} \log \left (c x + 1\right ) - {\left (c^{2} x^{2} - 1\right )} \log \left (c x - 1\right )}{4 \, {\left (c^{7} d^{2} x^{5} - 2 \, c^{5} d^{2} x^{3} + c^{3} d^{2} x + {\left (c^{6} d^{2} x^{4} - 2 \, c^{4} d^{2} x^{2} + c^{2} d^{2}\right )} \sqrt {c x + 1} \sqrt {c x - 1}\right )}}\,{d x} + 64 \, \int \frac {\log \left (c x - 1\right )}{8 \, {\left (c^{6} d^{2} x^{4} - 2 \, c^{4} d^{2} x^{2} + c^{2} d^{2}\right )}}\,{d x}\right )} b - \frac {1}{4} \, a {\left (\frac {2 \, x}{c^{4} d^{2} x^{2} - c^{2} d^{2}} + \frac {\log \left (c x + 1\right )}{c^{3} d^{2}} - \frac {\log \left (c x - 1\right )}{c^{3} d^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(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^6*d^2*x^4 - 2*c^4*d^2*x^2 + c^2*d^2), x) + 8*c^2*(2*x/(c^6*d^
2*x^2 - c^4*d^2) + log(c*x + 1)/(c^5*d^2) - log(c*x - 1)/(c^5*d^2)) - 64*c^2*integrate(1/8*x^2*log(c*x - 1)/(c
^6*d^2*x^4 - 2*c^4*d^2*x^2 + c^2*d^2), x) + 3*(c*(2/(c^6*d^2*x - c^5*d^2) - log(c*x + 1)/(c^5*d^2) + log(c*x -
1)/(c^5*d^2)) + 4*log(c*x - 1)/(c^6*d^2*x^2 - c^4*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 + sqr
t(c*x + 1)*sqrt(c*x - 1)))/(c^5*d^2*x^2 - c^3*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^7*d^2*x^5 - 2*c^5*d^2*x^3 + c^3*d^2*x + (c^6*d^2*x^4 - 2*c^4*d^2*x^2 + c^2*d^2)*s
qrt(c*x + 1)*sqrt(c*x - 1)), x) + 64*integrate(1/8*log(c*x - 1)/(c^6*d^2*x^4 - 2*c^4*d^2*x^2 + c^2*d^2), x))*b
- 1/4*a*(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x^{2}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{2} \operatorname {acosh}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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