3.16 \(\int \frac{(a+b \sinh ^{-1}(c x))^2}{d+e x} \, dx\)

Optimal. Leaf size=291 \[ \frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}-\frac{2 b^2 \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{2 b^2 \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e} \]

[Out]

-(a + b*ArcSinh[c*x])^3/(3*b*e) + ((a + b*ArcSinh[c*x])^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2
])])/e + ((a + b*ArcSinh[c*x])^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/e + (2*b*(a + b*ArcS
inh[c*x])*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2]))])/e + (2*b*(a + b*ArcSinh[c*x])*PolyLog
[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/e - (2*b^2*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d - Sqrt
[c^2*d^2 + e^2]))])/e - (2*b^2*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/e

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Rubi [A]  time = 0.469464, antiderivative size = 291, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5799, 5561, 2190, 2531, 2282, 6589} \[ \frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}-\frac{2 b^2 \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{2 b^2 \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*ArcSinh[c*x])^3/(3*b*e) + ((a + b*ArcSinh[c*x])^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2
])])/e + ((a + b*ArcSinh[c*x])^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/e + (2*b*(a + b*ArcS
inh[c*x])*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2]))])/e + (2*b*(a + b*ArcSinh[c*x])*PolyLog
[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/e - (2*b^2*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d - Sqrt
[c^2*d^2 + e^2]))])/e - (2*b^2*PolyLog[3, -((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 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{\left (a+b \sinh ^{-1}(c x)\right )^2}{d+e x} \, dx &=\operatorname{Subst}\left (\int \frac{(a+b x)^2 \cosh (x)}{c d+e \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\operatorname{Subst}\left (\int \frac{e^x (a+b x)^2}{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 (a+b x)^2}{c d+\sqrt{c^2 d^2+e^2}+e e^x} \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \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{(2 b) \operatorname{Subst}\left (\int (a+b x) \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{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \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{\left (2 b^2\right ) \operatorname{Subst}\left (\int \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{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\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{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\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{\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}+\frac{2 b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{2 b^2 \text{Li}_3\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e}-\frac{2 b^2 \text{Li}_3\left (-\frac{e e^{\sinh ^{-1}(c x)}}{c d+\sqrt{c^2 d^2+e^2}}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.201623, size = 273, normalized size = 0.94 \[ \frac{6 b \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}-c d}\right )+6 b \left (a+b \sinh ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )-6 b^2 \text{PolyLog}\left (3,\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}-c d}\right )-6 b^2 \text{PolyLog}\left (3,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )+3 \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )+3 \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )-\frac{\left (a+b \sinh ^{-1}(c x)\right )^3}{b}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*ArcSinh[c*x])^3/b) + 3*(a + b*ArcSinh[c*x])^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])
] + 3*(a + b*ArcSinh[c*x])^2*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])] + 6*b*(a + b*ArcSinh[c*x]
)*PolyLog[2, (e*E^ArcSinh[c*x])/(-(c*d) + Sqrt[c^2*d^2 + e^2])] + 6*b*(a + b*ArcSinh[c*x])*PolyLog[2, -((e*E^A
rcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))] - 6*b^2*PolyLog[3, (e*E^ArcSinh[c*x])/(-(c*d) + Sqrt[c^2*d^2 + e^2]
)] - 6*b^2*PolyLog[3, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))])/(3*e)

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/(e*x+d),x)

[Out]

int((a+b*arcsinh(c*x))^2/(e*x+d),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(e*x + d)/e + integrate(b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(e*x + d) + 2*a*b*log(c*x + sqrt(c^2*x^2 + 1
))/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/(e*x+d),x)

[Out]

Integral((a + b*asinh(c*x))**2/(d + e*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(c*x) + a)^2/(e*x + d), x)