3.29 \(\int \frac{x^3 (a+b \cosh ^{-1}(c x))}{d-c^2 d x^2} \, dx\)

Optimal. Leaf size=140 \[ -\frac{b \text{PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 c^4 d}-\frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\log \left (1-e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d}+\frac{b x \sqrt{c x-1} \sqrt{c x+1}}{4 c^3 d}+\frac{b \cosh ^{-1}(c x)}{4 c^4 d} \]

[Out]

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

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Rubi [A]  time = 0.197875, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {5766, 90, 52, 5715, 3716, 2190, 2279, 2391} \[ -\frac{b \text{PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 c^4 d}-\frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\log \left (1-e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d}+\frac{b x \sqrt{c x-1} \sqrt{c x+1}}{4 c^3 d}+\frac{b \cosh ^{-1}(c x)}{4 c^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2),x]

[Out]

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

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]

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 52

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

Rule 5715

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/e, Subst[Int[(
a + b*x)^n*Coth[x], x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, 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{x^3 \left (a+b \cosh ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=-\frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\int \frac{x \left (a+b \cosh ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx}{c^2}+\frac{b \int \frac{x^2}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{2 c d}\\ &=\frac{b x \sqrt{-1+c x} \sqrt{1+c x}}{4 c^3 d}-\frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}-\frac{\operatorname{Subst}\left (\int (a+b x) \coth (x) \, dx,x,\cosh ^{-1}(c x)\right )}{c^4 d}+\frac{b \int \frac{1}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx}{4 c^3 d}\\ &=\frac{b x \sqrt{-1+c x} \sqrt{1+c x}}{4 c^3 d}+\frac{b \cosh ^{-1}(c x)}{4 c^4 d}-\frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}+\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1-e^{2 x}} \, dx,x,\cosh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac{b x \sqrt{-1+c x} \sqrt{1+c x}}{4 c^3 d}+\frac{b \cosh ^{-1}(c x)}{4 c^4 d}-\frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\left (a+b \cosh ^{-1}(c x)\right ) \log \left (1-e^{2 \cosh ^{-1}(c x)}\right )}{c^4 d}+\frac{b \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac{b x \sqrt{-1+c x} \sqrt{1+c x}}{4 c^3 d}+\frac{b \cosh ^{-1}(c x)}{4 c^4 d}-\frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\left (a+b \cosh ^{-1}(c x)\right ) \log \left (1-e^{2 \cosh ^{-1}(c x)}\right )}{c^4 d}+\frac{b \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 c^4 d}\\ &=\frac{b x \sqrt{-1+c x} \sqrt{1+c x}}{4 c^3 d}+\frac{b \cosh ^{-1}(c x)}{4 c^4 d}-\frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac{\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac{\left (a+b \cosh ^{-1}(c x)\right ) \log \left (1-e^{2 \cosh ^{-1}(c x)}\right )}{c^4 d}-\frac{b \text{Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 c^4 d}\\ \end{align*}

Mathematica [A]  time = 0.296799, size = 151, normalized size = 1.08 \[ -\frac{4 b \text{PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )+4 b \text{PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )+2 c^2 x^2 \left (a+b \cosh ^{-1}(c x)\right )-\frac{2 \left (a+b \cosh ^{-1}(c x)\right )^2}{b}+4 \log \left (1-e^{\cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )+4 \log \left (e^{\cosh ^{-1}(c x)}+1\right ) \left (a+b \cosh ^{-1}(c x)\right )-b \left (c x \sqrt{c x-1} \sqrt{c x+1}+2 \tanh ^{-1}\left (\sqrt{\frac{c x-1}{c x+1}}\right )\right )}{4 c^4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.087, size = 244, normalized size = 1.7 \begin{align*} -{\frac{a{x}^{2}}{2\,{c}^{2}d}}-{\frac{a\ln \left ( cx-1 \right ) }{2\,d{c}^{4}}}-{\frac{a\ln \left ( cx+1 \right ) }{2\,d{c}^{4}}}+{\frac{b \left ({\rm arccosh} \left (cx\right ) \right ) ^{2}}{2\,d{c}^{4}}}-{\frac{b{x}^{2}{\rm arccosh} \left (cx\right )}{2\,{c}^{2}d}}+{\frac{bx}{4\,{c}^{3}d}\sqrt{cx-1}\sqrt{cx+1}}+{\frac{b{\rm arccosh} \left (cx\right )}{4\,d{c}^{4}}}-{\frac{b{\rm arccosh} \left (cx\right )}{d{c}^{4}}\ln \left ( 1+cx+\sqrt{cx-1}\sqrt{cx+1} \right ) }-{\frac{b}{d{c}^{4}}{\it polylog} \left ( 2,-cx-\sqrt{cx-1}\sqrt{cx+1} \right ) }-{\frac{b{\rm arccosh} \left (cx\right )}{d{c}^{4}}\ln \left ( 1-cx-\sqrt{cx-1}\sqrt{cx+1} \right ) }-{\frac{b}{d{c}^{4}}{\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(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a{\left (\frac{x^{2}}{c^{2} d} + \frac{\log \left (c^{2} x^{2} - 1\right )}{c^{4} d}\right )} + \frac{1}{8} \, b{\left (\frac{2 \, c^{2} x^{2} - 4 \,{\left (c^{2} x^{2} + \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 ) + 2 \,{\left (\log \left (c x - 1\right ) + 1\right )} \log \left (c x + 1\right ) + \log \left (c x + 1\right )^{2} + \log \left (c x - 1\right )^{2} + 2 \, \log \left (c x - 1\right )}{c^{4} d} - 8 \, \int \frac{c^{2} x^{2} + \log \left (c x + 1\right ) + \log \left (c x - 1\right )}{2 \,{\left (c^{6} d x^{3} - c^{4} d x +{\left (c^{5} d x^{2} - c^{3} d\right )} e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (c x - 1\right )\right )}\right )}}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(x^2/(c^2*d) + log(c^2*x^2 - 1)/(c^4*d)) + 1/8*b*((2*c^2*x^2 - 4*(c^2*x^2 + log(c*x + 1) + log(c*x - 1)
)*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1)) + 2*(log(c*x - 1) + 1)*log(c*x + 1) + log(c*x + 1)^2 + log(c*x - 1)^2
+ 2*log(c*x - 1))/(c^4*d) - 8*integrate(1/2*(c^2*x^2 + log(c*x + 1) + log(c*x - 1))/(c^6*d*x^3 - c^4*d*x + (c
^5*d*x^2 - c^3*d)*e^(1/2*log(c*x + 1) + 1/2*log(c*x - 1))), x))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b x^{3} \operatorname{arcosh}\left (c x\right ) + a x^{3}}{c^{2} d x^{2} - d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a x^{3}}{c^{2} x^{2} - 1}\, dx + \int \frac{b x^{3} \operatorname{acosh}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a*x**3/(c**2*x**2 - 1), x) + Integral(b*x**3*acosh(c*x)/(c**2*x**2 - 1), x))/d

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )} x^{3}}{c^{2} d x^{2} - d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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