\(\int \frac {a+b \text {arcsinh}(\frac {\sqrt {1-c x}}{\sqrt {1+c x}})}{1-c^2 x^2} \, dx\) [345]

Optimal result

Integrand size = 38, antiderivative size = 133 \[ \int \frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=-\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{2 b c}-\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \log \left (1-e^{-2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {b \operatorname {PolyLog}\left (2,e^{-2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {212, 6813, 5775, 3797, 2221, 2317, 2438} \[ \int \frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=-\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 b c}-\frac {\log \left (1-e^{-2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}+\frac {b \operatorname {PolyLog}\left (2,e^{-2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{2 c} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 2317

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

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 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*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {a+b \text {arcsinh}(x)}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c} \\ & = \frac {\text {Subst}\left (\int x \coth \left (\frac {a}{b}-\frac {x}{b}\right ) \, dx,x,a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{b c} \\ & = -\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{2 b c}-\frac {2 \text {Subst}\left (\int \frac {e^{2 \left (\frac {a}{b}-\frac {x}{b}\right )} x}{1-e^{2 \left (\frac {a}{b}-\frac {x}{b}\right )}} \, dx,x,a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{b c} \\ & = -\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{2 b c}-\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \log \left (1-e^{-2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {\text {Subst}\left (\int \log \left (1-e^{2 \left (\frac {a}{b}-\frac {x}{b}\right )}\right ) \, dx,x,a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )}{c} \\ & = -\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{2 b c}-\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \log \left (1-e^{-2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}-\frac {b \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \left (\frac {a}{b}-\frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{b}\right )}\right )}{2 c} \\ & = -\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{2 b c}-\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \log \left (1-e^{-2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{c}+\frac {b \operatorname {PolyLog}\left (2,e^{2 \left (\frac {a}{b}-\frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{b}\right )}\right )}{2 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\frac {\left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \left (a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )-2 b \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )\right )-b^2 \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}\right )}{2 b c} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.94

method	result	size
default	\(\frac {a \ln \left (c x +1\right )}{2 c}-\frac {a \ln \left (c x -1\right )}{2 c}-b \left (-\frac {\operatorname {arcsinh}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{2 c}+\frac {\operatorname {arcsinh}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}+1\right )}{c}+\frac {\operatorname {polylog}\left (2, -\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}+\frac {\operatorname {arcsinh}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}+\frac {\operatorname {polylog}\left (2, \frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}\right )\)	\(258\)
parts	\(\frac {a \ln \left (c x +1\right )}{2 c}-\frac {a \ln \left (c x -1\right )}{2 c}-b \left (-\frac {\operatorname {arcsinh}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{2 c}+\frac {\operatorname {arcsinh}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}+1\right )}{c}+\frac {\operatorname {polylog}\left (2, -\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}+\frac {\operatorname {arcsinh}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}+\frac {\operatorname {polylog}\left (2, \frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+\sqrt {1+\frac {-c x +1}{c x +1}}\right )}{c}\right )\)	\(258\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \operatorname {arsinh}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=- \int \frac {a}{c^{2} x^{2} - 1}\, dx - \int \frac {b \operatorname {asinh}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \operatorname {arsinh}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \]

[In]

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

[Out]

-1/8*b*((2*(log(c*x + 1) - log(-c*x + 1))*log(c*x + 1) - log(c*x + 1)^2 + 2*log(c*x + 1)*log(-c*x + 1) - log(-
c*x + 1)^2 - 4*(log(c*x + 1) - log(-c*x + 1))*log(sqrt(2) + sqrt(-c*x + 1)))/c + 8*integrate(-1/4*(sqrt(2)*log
(c*x + 1) - sqrt(2)*log(-c*x + 1))/(sqrt(2)*c*x + (c*x - 1)*sqrt(-c*x + 1) - sqrt(2)), x)) + 1/2*a*(log(c*x +
1)/c - log(c*x - 1)/c)

Giac [F]

\[ \int \frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int { -\frac {b \operatorname {arsinh}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx=\int -\frac {a+b\,\mathrm {asinh}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )}{c^2\,x^2-1} \,d x \]

[In]

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

[Out]

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