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

Optimal. Leaf size=102 \[ \frac {2 \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c^3 d}-\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d}+\frac {b \text {Li}_2\left (-e^{\cosh ^{-1}(c x)}\right )}{c^3 d}-\frac {b \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )}{c^3 d}+\frac {b \sqrt {c x-1} \sqrt {c x+1}}{c^3 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 102, 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 = {5766, 74, 5694, 4182, 2279, 2391} \[ \frac {b \text {PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )}{c^3 d}-\frac {b \text {PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{c^3 d}-\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d}+\frac {2 \tanh ^{-1}\left (e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c^3 d}+\frac {b \sqrt {c x-1} \sqrt {c x+1}}{c^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5766

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)/(e*(m + 2*p + 1)), x] + (-Dist[(b*f*n*(-d)^p)/(c*
(m + 2*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))/(c^2*(m + 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcCosh[c*x])^n, x], x]) /;
FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && Inte
gerQ[p] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 155, normalized size = 1.52 \[ \frac {-2 a c x-a \log (1-c x)+a \log (c x+1)-2 b \text {Li}_2\left (-e^{-\cosh ^{-1}(c x)}\right )-2 b \text {Li}_2\left (e^{\cosh ^{-1}(c x)}\right )+2 b \sqrt {\frac {c x-1}{c x+1}}+2 b c x \sqrt {\frac {c x-1}{c x+1}}+b \cosh ^{-1}(c x)^2-2 b c x \cosh ^{-1}(c x)+2 b \cosh ^{-1}(c x) \log \left (e^{-\cosh ^{-1}(c x)}+1\right )-2 b \cosh ^{-1}(c x) \log \left (1-e^{\cosh ^{-1}(c x)}\right )}{2 c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^2*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^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*c*x*ArcCosh[c*x] + b*Arc
Cosh[c*x]^2 + 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]])/(2*c^3*d)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x^{2} \operatorname {arcosh}\left (c x\right ) + a x^{2}}{c^{2} d x^{2} - d}, x\right ) \]

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),x, algorithm="fricas")

[Out]

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

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

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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.27, size = 208, normalized size = 2.04 \[ -\frac {a x}{c^{2} d}-\frac {a \ln \left (c x -1\right )}{2 c^{3} d}+\frac {a \ln \left (c x +1\right )}{2 c^{3} d}+\frac {b \sqrt {c x -1}\, \sqrt {c x +1}}{c^{3} d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1-c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{c^{3} d}+\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{c^{3} d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) x}{c^{2} d}-\frac {b \polylog \left (2, c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{c^{3} d}+\frac {b \polylog \left (2, -c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{c^{3} d} \]

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),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, {\left (4 \, c^{2} {\left (\frac {2 \, x}{c^{4} d} - \frac {\log \left (c x + 1\right )}{c^{5} d} + \frac {\log \left (c x - 1\right )}{c^{5} d}\right )} + 24 \, c \int \frac {x \log \left (c x - 1\right )}{4 \, {\left (c^{4} d x^{2} - c^{2} d\right )}}\,{d x} - \frac {4 \, {\left (2 \, c x - \log \left (c x + 1\right ) + \log \left (c x - 1\right )\right )} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right ) + \log \left (c x + 1\right )^{2} + 2 \, \log \left (c x + 1\right ) \log \left (c x - 1\right )}{c^{3} d} + 8 \, \int -\frac {2 \, c x - \log \left (c x + 1\right ) + \log \left (c x - 1\right )}{2 \, {\left (c^{5} d x^{3} - c^{3} d x + {\left (c^{4} d x^{2} - c^{2} d\right )} \sqrt {c x + 1} \sqrt {c x - 1}\right )}}\,{d x} - 8 \, \int \frac {\log \left (c x - 1\right )}{4 \, {\left (c^{4} d x^{2} - c^{2} d\right )}}\,{d x}\right )} b - \frac {1}{2} \, a {\left (\frac {2 \, x}{c^{2} d} - \frac {\log \left (c x + 1\right )}{c^{3} d} + \frac {\log \left (c x - 1\right )}{c^{3} d}\right )} \]

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),x, algorithm="maxima")

[Out]

1/8*(4*c^2*(2*x/(c^4*d) - log(c*x + 1)/(c^5*d) + log(c*x - 1)/(c^5*d)) + 24*c*integrate(1/4*x*log(c*x - 1)/(c^
4*d*x^2 - c^2*d), x) - (4*(2*c*x - log(c*x + 1) + log(c*x - 1))*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1)) + log(c
*x + 1)^2 + 2*log(c*x + 1)*log(c*x - 1))/(c^3*d) + 8*integrate(-1/2*(2*c*x - log(c*x + 1) + log(c*x - 1))/(c^5
*d*x^3 - c^3*d*x + (c^4*d*x^2 - c^2*d)*sqrt(c*x + 1)*sqrt(c*x - 1)), x) - 8*integrate(1/4*log(c*x - 1)/(c^4*d*
x^2 - c^2*d), x))*b - 1/2*a*(2*x/(c^2*d) - log(c*x + 1)/(c^3*d) + log(c*x - 1)/(c^3*d))

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

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),x)

[Out]

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

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

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),x)

[Out]

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

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