3.3 \(\int \frac{\sinh ^{-1}(c x)^3}{d+e x} \, dx\)

Optimal. Leaf size=348 \[ \frac{3 \sinh ^{-1}(c x)^2 \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{6 \text{PolyLog}\left (4,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{6 \text{PolyLog}\left (4,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e}-\frac{\sinh ^{-1}(c x)^4}{4 e} \]

[Out]

-ArcSinh[c*x]^4/(4*e) + (ArcSinh[c*x]^3*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])])/e + (ArcSinh[
c*x]^3*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/e + (3*ArcSinh[c*x]^2*PolyLog[2, -((e*E^ArcSin
h[c*x])/(c*d - Sqrt[c^2*d^2 + e^2]))])/e + (3*ArcSinh[c*x]^2*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d
^2 + e^2]))])/e - (6*ArcSinh[c*x]*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2]))])/e - (6*ArcSin
h[c*x]*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/e + (6*PolyLog[4, -((e*E^ArcSinh[c*x])/(
c*d - Sqrt[c^2*d^2 + e^2]))])/e + (6*PolyLog[4, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/e

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Rubi [A]  time = 0.433145, antiderivative size = 348, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5799, 5561, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 \sinh ^{-1}(c x)^2 \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{6 \text{PolyLog}\left (4,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{6 \text{PolyLog}\left (4,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e}-\frac{\sinh ^{-1}(c x)^4}{4 e} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[c*x]^3/(d + e*x),x]

[Out]

-ArcSinh[c*x]^4/(4*e) + (ArcSinh[c*x]^3*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])])/e + (ArcSinh[
c*x]^3*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/e + (3*ArcSinh[c*x]^2*PolyLog[2, -((e*E^ArcSin
h[c*x])/(c*d - Sqrt[c^2*d^2 + e^2]))])/e + (3*ArcSinh[c*x]^2*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d
^2 + e^2]))])/e - (6*ArcSinh[c*x]*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2]))])/e - (6*ArcSin
h[c*x]*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/e + (6*PolyLog[4, -((e*E^ArcSinh[c*x])/(
c*d - Sqrt[c^2*d^2 + e^2]))])/e + (6*PolyLog[4, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/e

Rule 5799

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

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

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

Rubi steps

\begin{align*} \int \frac{\sinh ^{-1}(c x)^3}{d+e x} \, dx &=\operatorname{Subst}\left (\int \frac{x^3 \cosh (x)}{c d+e \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac{\sinh ^{-1}(c x)^4}{4 e}+\operatorname{Subst}\left (\int \frac{e^x x^3}{c d-\sqrt{c^2 d^2+e^2}+e e^x} \, dx,x,\sinh ^{-1}(c x)\right )+\operatorname{Subst}\left (\int \frac{e^x x^3}{c d+\sqrt{c^2 d^2+e^2}+e e^x} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac{\sinh ^{-1}(c x)^4}{4 e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{3 \operatorname{Subst}\left (\int x^2 \log \left (1+\frac{e e^x}{c d-\sqrt{c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}-\frac{3 \operatorname{Subst}\left (\int x^2 \log \left (1+\frac{e e^x}{c d+\sqrt{c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}\\ &=-\frac{\sinh ^{-1}(c x)^4}{4 e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \operatorname{Subst}\left (\int x \text{Li}_2\left (-\frac{e e^x}{c d-\sqrt{c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}-\frac{6 \operatorname{Subst}\left (\int x \text{Li}_2\left (-\frac{e e^x}{c d+\sqrt{c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}\\ &=-\frac{\sinh ^{-1}(c x)^4}{4 e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{Li}_3\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{Li}_3\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{6 \operatorname{Subst}\left (\int \text{Li}_3\left (-\frac{e e^x}{c d-\sqrt{c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}+\frac{6 \operatorname{Subst}\left (\int \text{Li}_3\left (-\frac{e e^x}{c d+\sqrt{c^2 d^2+e^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{e}\\ &=-\frac{\sinh ^{-1}(c x)^4}{4 e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{Li}_3\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{Li}_3\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{e x}{-c d+\sqrt{c^2 d^2+e^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{e}+\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{e x}{c d+\sqrt{c^2 d^2+e^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{e}\\ &=-\frac{\sinh ^{-1}(c x)^4}{4 e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\sinh ^{-1}(c x)^3 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{3 \sinh ^{-1}(c x)^2 \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{Li}_3\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{6 \sinh ^{-1}(c x) \text{Li}_3\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{6 \text{Li}_4\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{6 \text{Li}_4\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.0433466, size = 322, normalized size = 0.93 \[ \frac{12 \sinh ^{-1}(c x)^2 \text{PolyLog}\left (2,\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}-c d}\right )+12 \sinh ^{-1}(c x)^2 \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )-24 \sinh ^{-1}(c x) \text{PolyLog}\left (3,\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}-c d}\right )-24 \sinh ^{-1}(c x) \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )+24 \text{PolyLog}\left (4,\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}-c d}\right )+24 \text{PolyLog}\left (4,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )+4 \sinh ^{-1}(c x)^3 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )+4 \sinh ^{-1}(c x)^3 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )-\sinh ^{-1}(c x)^4}{4 e} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[c*x]^3/(d + e*x),x]

[Out]

(-ArcSinh[c*x]^4 + 4*ArcSinh[c*x]^3*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])] + 4*ArcSinh[c*x]^3
*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])] + 12*ArcSinh[c*x]^2*PolyLog[2, (e*E^ArcSinh[c*x])/(-(
c*d) + Sqrt[c^2*d^2 + e^2])] + 12*ArcSinh[c*x]^2*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))]
- 24*ArcSinh[c*x]*PolyLog[3, (e*E^ArcSinh[c*x])/(-(c*d) + Sqrt[c^2*d^2 + e^2])] - 24*ArcSinh[c*x]*PolyLog[3,
-((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))] + 24*PolyLog[4, (e*E^ArcSinh[c*x])/(-(c*d) + Sqrt[c^2*d^2 +
e^2])] + 24*PolyLog[4, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/(4*e)

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Maple [F]  time = 0.113, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{3}}{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(c*x)^3/(e*x+d),x)

[Out]

int(arcsinh(c*x)^3/(e*x+d),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsinh}\left (c x\right )^{3}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(c*x)^3/(e*x+d),x, algorithm="maxima")

[Out]

integrate(arcsinh(c*x)^3/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(c*x)^3/(e*x+d),x, algorithm="fricas")

[Out]

integral(arcsinh(c*x)^3/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(c*x)**3/(e*x+d),x)

[Out]

Integral(asinh(c*x)**3/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(c*x)^3/(e*x+d),x, algorithm="giac")

[Out]

integrate(arcsinh(c*x)^3/(e*x + d), x)