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

Optimal. Leaf size=155 \[ -\frac{3 b^2 \text{PolyLog}\left (3,e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{2 d e}-\frac{3 b \text{PolyLog}\left (2,e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e}-\frac{3 b^3 \text{PolyLog}\left (4,e^{-2 \sinh ^{-1}(c+d x)}\right )}{4 d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\log \left (1-e^{-2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e} \]

[Out]

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

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Rubi [A]  time = 0.223581, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5865, 12, 5659, 3716, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 b^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{2 d e}+\frac{3 b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d e}+\frac{3 b^3 \text{PolyLog}\left (4,e^{2 \sinh ^{-1}(c+d x)}\right )}{4 d e}-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{d e} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcSinh[c + d*x])^3/(c*e + d*e*x),x]

[Out]

-(a + b*ArcSinh[c + d*x])^4/(4*b*d*e) + ((a + b*ArcSinh[c + d*x])^3*Log[1 - E^(2*ArcSinh[c + d*x])])/(d*e) + (
3*b*(a + b*ArcSinh[c + d*x])^2*PolyLog[2, E^(2*ArcSinh[c + d*x])])/(2*d*e) - (3*b^2*(a + b*ArcSinh[c + d*x])*P
olyLog[3, E^(2*ArcSinh[c + d*x])])/(2*d*e) + (3*b^3*PolyLog[4, E^(2*ArcSinh[c + d*x])])/(4*d*e)

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5659

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tanh[x], x], x, ArcSinh
[c*x]] /; FreeQ[{a, b, c}, x] && 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 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{\left (a+b \sinh ^{-1}(c+d x)\right )^3}{c e+d e x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^3}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^3}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac{\operatorname{Subst}\left (\int (a+b x)^3 \coth (x) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{4 b d e}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)^3}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}-\frac{(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{3 b \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{3 b \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}-\frac{3 b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{3 b \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}-\frac{3 b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c+d x)}\right )}{4 d e}\\ &=-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{4 b d e}+\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right )}{d e}+\frac{3 b \left (a+b \sinh ^{-1}(c+d x)\right )^2 \text{Li}_2\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}-\frac{3 b^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Li}_3\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{2 d e}+\frac{3 b^3 \text{Li}_4\left (e^{2 \sinh ^{-1}(c+d x)}\right )}{4 d e}\\ \end{align*}

Mathematica [A]  time = 0.0467424, size = 128, normalized size = 0.83 \[ \frac{-6 b^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )+6 b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^2+3 b^3 \text{PolyLog}\left (4,e^{2 \sinh ^{-1}(c+d x)}\right )-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^4}{b}+4 \log \left (1-e^{2 \sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d e} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSinh[c + d*x])^3/(c*e + d*e*x),x]

[Out]

(-((a + b*ArcSinh[c + d*x])^4/b) + 4*(a + b*ArcSinh[c + d*x])^3*Log[1 - E^(2*ArcSinh[c + d*x])] + 6*b*(a + b*A
rcSinh[c + d*x])^2*PolyLog[2, E^(2*ArcSinh[c + d*x])] - 6*b^2*(a + b*ArcSinh[c + d*x])*PolyLog[3, E^(2*ArcSinh
[c + d*x])] + 3*b^3*PolyLog[4, E^(2*ArcSinh[c + d*x])])/(4*d*e)

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Maple [B]  time = 0.053, size = 736, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^3/e*ln(d*x+c)-1/4/d*b^3/e*arcsinh(d*x+c)^4+1/d*b^3/e*arcsinh(d*x+c)^3*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+3/
d*b^3/e*arcsinh(d*x+c)^2*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))-6/d*b^3/e*arcsinh(d*x+c)*polylog(3,-d*x-c-(1+(d
*x+c)^2)^(1/2))+6/d*b^3/e*polylog(4,-d*x-c-(1+(d*x+c)^2)^(1/2))+1/d*b^3/e*arcsinh(d*x+c)^3*ln(1-d*x-c-(1+(d*x+
c)^2)^(1/2))+3/d*b^3/e*arcsinh(d*x+c)^2*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))-6/d*b^3/e*arcsinh(d*x+c)*polylog(
3,d*x+c+(1+(d*x+c)^2)^(1/2))+6/d*b^3/e*polylog(4,d*x+c+(1+(d*x+c)^2)^(1/2))-1/d*a*b^2/e*arcsinh(d*x+c)^3+3/d*a
*b^2/e*arcsinh(d*x+c)^2*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+6/d*a*b^2/e*arcsinh(d*x+c)*polylog(2,-d*x-c-(1+(d*x+c)
^2)^(1/2))-6/d*a*b^2/e*polylog(3,-d*x-c-(1+(d*x+c)^2)^(1/2))+3/d*a*b^2/e*arcsinh(d*x+c)^2*ln(1-d*x-c-(1+(d*x+c
)^2)^(1/2))+6/d*a*b^2/e*arcsinh(d*x+c)*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))-6/d*a*b^2/e*polylog(3,d*x+c+(1+(d*
x+c)^2)^(1/2))-3/2/d*a^2*b/e*arcsinh(d*x+c)^2+3/d*a^2*b/e*arcsinh(d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+3/d*a
^2*b/e*polylog(2,-d*x-c-(1+(d*x+c)^2)^(1/2))+3/d*a^2*b/e*arcsinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))+3/d*a^
2*b/e*polylog(2,d*x+c+(1+(d*x+c)^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^3*arcsinh(d*x + c)^3 + 3*a*b^2*arcsinh(d*x + c)^2 + 3*a^2*b*arcsinh(d*x + c) + a^3)/(d*e*x + c*e),
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**3/(c + d*x), x) + Integral(b**3*asinh(c + d*x)**3/(c + d*x), x) + Integral(3*a*b**2*asinh(c + d*x
)**2/(c + d*x), x) + Integral(3*a**2*b*asinh(c + d*x)/(c + d*x), x))/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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