3.344 \(\int \frac {(a+b \sinh ^{-1}(\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))^2}{1-c^2 x^2} \, dx\)

Optimal. Leaf size=194 \[ \frac {b \text {Li}_2\left (e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}-\frac {\left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{3 b c}-\frac {\log \left (1-e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{c}+\frac {b^2 \text {Li}_3\left (e^{-2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 195, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6681, 5659, 3716, 2190, 2531, 2282, 6589} \[ -\frac {b \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}+\frac {b^2 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{2 c}+\frac {\left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^3}{3 b c}-\frac {\log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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 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 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 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 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 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]

Rule 6681

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

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Mathematica [A]  time = 0.06, size = 187, normalized size = 0.96 \[ \frac {-6 b^2 \text {Li}_2\left (e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )+2 \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2 \left (a+b \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )-3 b \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )\right )+3 b^3 \text {Li}_3\left (e^{2 \sinh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{6 b c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \operatorname {arsinh}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + 2 \, a b \operatorname {arsinh}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a^{2}}{c^{2} x^{2} - 1}, x\right ) \]

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)))^2/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \operatorname {arsinh}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{2}}{c^{2} x^{2} - 1}\,{d x} \]

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)))^2/(-c^2*x^2+1),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 649, normalized size = 3.35 \[ \frac {a^{2} \ln \left (c x +1\right )}{2 c}-\frac {a^{2} \ln \left (c x -1\right )}{2 c}+\frac {b^{2} \arcsinh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{3}}{3 c}-\frac {b^{2} \arcsinh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1+\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}-\frac {2 b^{2} \arcsinh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}+\frac {2 b^{2} \polylog \left (3, -\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}-\frac {b^{2} \arcsinh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1-\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}-\frac {2 b^{2} \arcsinh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, \frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}+\frac {2 b^{2} \polylog \left (3, \frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}+\frac {a b \arcsinh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{c}-\frac {2 a b \arcsinh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}-\frac {2 a b \polylog \left (2, -\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}-\frac {2 a b \arcsinh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}-\frac {2 a b \polylog \left (2, \frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c} \]

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)))^2/(-c^2*x^2+1),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} + \frac {{\left (b^{2} \log \left (c x + 1\right ) - b^{2} \log \left (-c x + 1\right )\right )} \log \left (\sqrt {2} + \sqrt {-c x + 1}\right )^{2}}{2 \, c} + \int -\frac {{\left (\sqrt {2} b^{2} + \sqrt {-c x + 1} b^{2}\right )} \log \left (c x + 1\right )^{2} - 4 \, {\left (\sqrt {2} a b + \sqrt {-c x + 1} a b\right )} \log \left (c x + 1\right ) + 2 \, {\left (4 \, \sqrt {2} a b - 2 \, {\left (\sqrt {2} b^{2} + \sqrt {-c x + 1} b^{2}\right )} \log \left (c x + 1\right ) + {\left (4 \, a b + {\left (b^{2} c x + b^{2}\right )} \log \left (c x + 1\right ) - {\left (b^{2} c x + b^{2}\right )} \log \left (-c x + 1\right )\right )} \sqrt {-c x + 1}\right )} \log \left (\sqrt {2} + \sqrt {-c x + 1}\right )}{4 \, {\left (\sqrt {2} c^{2} x^{2} + {\left (c^{2} x^{2} - 1\right )} \sqrt {-c x + 1} - \sqrt {2}\right )}}\,{d x} \]

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)))^2/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*a^2*(log(c*x + 1)/c - log(c*x - 1)/c) + 1/2*(b^2*log(c*x + 1) - b^2*log(-c*x + 1))*log(sqrt(2) + sqrt(-c*x
+ 1))^2/c + integrate(-1/4*((sqrt(2)*b^2 + sqrt(-c*x + 1)*b^2)*log(c*x + 1)^2 - 4*(sqrt(2)*a*b + sqrt(-c*x +
1)*a*b)*log(c*x + 1) + 2*(4*sqrt(2)*a*b - 2*(sqrt(2)*b^2 + sqrt(-c*x + 1)*b^2)*log(c*x + 1) + (4*a*b + (b^2*c*
x + b^2)*log(c*x + 1) - (b^2*c*x + b^2)*log(-c*x + 1))*sqrt(-c*x + 1))*log(sqrt(2) + sqrt(-c*x + 1)))/(sqrt(2)
*c^2*x^2 + (c^2*x^2 - 1)*sqrt(-c*x + 1) - sqrt(2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {{\left (a+b\,\mathrm {asinh}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^2}{c^2\,x^2-1} \,d x \]

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)))^2/(c^2*x^2 - 1),x)

[Out]

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

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

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)))**2/(-c**2*x**2+1),x)

[Out]

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

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