3.4.43 \(\int \frac {(a+b \sinh ^{-1}(\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))^3}{1-c^2 x^2} \, dx\) [343]

Optimal. Leaf size=261 \[ -\frac {\left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{4 b c}-\frac {\left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {3 b \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}+\frac {3 b^2 \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (3,e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c}+\frac {3 b^3 \text {PolyLog}\left (4,e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{4 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.20, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6813, 5775, 3797, 2221, 2611, 6744, 2320, 6724} \begin {gather*} \frac {3 b^2 \text {Li}_3\left (e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{2 c}+\frac {3 b \text {Li}_2\left (e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 c}-\frac {\left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^4}{4 b c}-\frac {\log \left (1-e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{c}+\frac {3 b^3 \text {Li}_4\left (e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2),x]

[Out]

-1/4*(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^4/(b*c) - ((a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3*Lo
g[1 - E^(-2*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c + (3*b*(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*Po
lyLog[2, E^(-2*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/(2*c) + (3*b^2*(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*
x]])*PolyLog[3, E^(-2*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/(2*c) + (3*b^3*PolyLog[4, E^(-2*ArcSinh[Sqrt[1 -
c*x]/Sqrt[1 + c*x]])])/(4*c)

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3797

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))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 6724

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]

Rule 6744

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 6813

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[2*e*(g/(C*(e*f - d*g))), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 244, normalized size = 0.93 \begin {gather*} \frac {\frac {\left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^4}{b}-4 \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )-6 b \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )+6 b^2 \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (3,e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )-3 b^3 \text {PolyLog}\left (4,e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3/(1 - c^2*x^2),x]

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1174\) vs. \(2(285)=570\).
time = 4.03, size = 1175, normalized size = 4.50

method	result	size
default	\(\text {Expression too large to display}\)	\(1175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/2*a^3/c*ln(c*x+1)-1/2*a^3/c*ln(c*x-1)+1/4*b^3/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^4-b^3/c*arcsinh((-c*x+
1)^(1/2)/(c*x+1)^(1/2))^3*ln(1+(-c*x+1)^(1/2)/(c*x+1)^(1/2)+((-c*x+1)/(c*x+1)+1)^(1/2))-3*b^3/c*arcsinh((-c*x+
1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,-(-c*x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))+6*b^3/c*arcsinh(
(-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,-(-c*x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))-6*b^3/c*poly
log(4,-(-c*x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))-b^3/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*
ln(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))-3*b^3/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*
polylog(2,(-c*x+1)^(1/2)/(c*x+1)^(1/2)+((-c*x+1)/(c*x+1)+1)^(1/2))+6*b^3/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2
))*polylog(3,(-c*x+1)^(1/2)/(c*x+1)^(1/2)+((-c*x+1)/(c*x+1)+1)^(1/2))-6*b^3/c*polylog(4,(-c*x+1)^(1/2)/(c*x+1)
^(1/2)+((-c*x+1)/(c*x+1)+1)^(1/2))+a*b^2*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3/c-3*a*b^2/c*arcsinh((-c*x+1)^
(1/2)/(c*x+1)^(1/2))^2*ln(1+(-c*x+1)^(1/2)/(c*x+1)^(1/2)+((-c*x+1)/(c*x+1)+1)^(1/2))-6*a*b^2/c*arcsinh((-c*x+1
)^(1/2)/(c*x+1)^(1/2))*polylog(2,-(-c*x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))+6*a*b^2/c*polylog(3
,-(-c*x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))-3*a*b^2/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*l
n(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))-6*a*b^2/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*p
olylog(2,(-c*x+1)^(1/2)/(c*x+1)^(1/2)+((-c*x+1)/(c*x+1)+1)^(1/2))+6*a*b^2/c*polylog(3,(-c*x+1)^(1/2)/(c*x+1)^(
1/2)+((-c*x+1)/(c*x+1)+1)^(1/2))+3/2*a^2*b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/c-3*a^2*b/c*arcsinh((-c*x+1
)^(1/2)/(c*x+1)^(1/2))*ln(1+(-c*x+1)^(1/2)/(c*x+1)^(1/2)+((-c*x+1)/(c*x+1)+1)^(1/2))-3*a^2*b/c*polylog(2,-(-c*
x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))-3*a^2*b/c*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1-(-c*
x+1)^(1/2)/(c*x+1)^(1/2)-((-c*x+1)/(c*x+1)+1)^(1/2))-3*a^2*b/c*polylog(2,(-c*x+1)^(1/2)/(c*x+1)^(1/2)+((-c*x+1
)/(c*x+1)+1)^(1/2))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*a^3*(log(c*x + 1)/c - log(c*x - 1)/c) + 1/2*(b^3*log(c*x + 1) - b^3*log(-c*x + 1))*log(sqrt(2) + sqrt(-c*x
+ 1))^3/c + integrate(1/8*((sqrt(2)*b^3 + sqrt(-c*x + 1)*b^3)*log(c*x + 1)^3 - 6*(sqrt(2)*a*b^2 + sqrt(-c*x +
1)*a*b^2)*log(c*x + 1)^2 - 6*(4*sqrt(2)*a*b^2 - 2*(sqrt(2)*b^3 + sqrt(-c*x + 1)*b^3)*log(c*x + 1) + (4*a*b^2
+ (b^3*c*x + b^3)*log(c*x + 1) - (b^3*c*x + b^3)*log(-c*x + 1))*sqrt(-c*x + 1))*log(sqrt(2) + sqrt(-c*x + 1))^
2 + 12*(sqrt(2)*a^2*b + sqrt(-c*x + 1)*a^2*b)*log(c*x + 1) - 6*(4*sqrt(2)*a^2*b + 4*sqrt(-c*x + 1)*a^2*b + (sq
rt(2)*b^3 + sqrt(-c*x + 1)*b^3)*log(c*x + 1)^2 - 4*(sqrt(2)*a*b^2 + sqrt(-c*x + 1)*a*b^2)*log(c*x + 1))*log(sq
rt(2) + sqrt(-c*x + 1)))/(sqrt(2)*c^2*x^2 + (c^2*x^2 - 1)*sqrt(-c*x + 1) - sqrt(2)), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-(b^3*arcsinh(sqrt(-c*x + 1)/sqrt(c*x + 1))^3 + 3*a*b^2*arcsinh(sqrt(-c*x + 1)/sqrt(c*x + 1))^2 + 3*a
^2*b*arcsinh(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a^3)/(c^2*x^2 - 1), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {a^{3}}{c^{2} x^{2} - 1}\, dx - \int \frac {b^{3} \operatorname {asinh}^{3}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx - \int \frac {3 a b^{2} \operatorname {asinh}^{2}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx - \int \frac {3 a^{2} b \operatorname {asinh}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh((-c*x+1)**(1/2)/(c*x+1)**(1/2)))**3/(-c**2*x**2+1),x)

[Out]

-Integral(a**3/(c**2*x**2 - 1), x) - Integral(b**3*asinh(sqrt(-c*x + 1)/sqrt(c*x + 1))**3/(c**2*x**2 - 1), x)
- Integral(3*a*b**2*asinh(sqrt(-c*x + 1)/sqrt(c*x + 1))**2/(c**2*x**2 - 1), x) - Integral(3*a**2*b*asinh(sqrt(
-c*x + 1)/sqrt(c*x + 1))/(c**2*x**2 - 1), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^3/(-c^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-(b*arcsinh(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)^3/(c^2*x^2 - 1), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {{\left (a+b\,\mathrm {asinh}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^3}{c^2\,x^2-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a + b*asinh((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1),x)

[Out]

int(-(a + b*asinh((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1), x)

________________________________________________________________________________________