3.3.69 \(\int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \cosh ^{-1}(c x)} \, dx\) [269]

Optimal. Leaf size=139 \[ \frac {\sqrt {1-c x} \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \cosh ^{-1}(c x)\right )}{b}\right )}{8 b c^3 \sqrt {-1+c x}}-\frac {\sqrt {1-c x} \log \left (a+b \cosh ^{-1}(c x)\right )}{8 b c^3 \sqrt {-1+c x}}-\frac {\sqrt {1-c x} \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \cosh ^{-1}(c x)\right )}{b}\right )}{8 b c^3 \sqrt {-1+c x}} \]

[Out]

1/8*Chi(4*(a+b*arccosh(c*x))/b)*cosh(4*a/b)*(-c*x+1)^(1/2)/b/c^3/(c*x-1)^(1/2)-1/8*ln(a+b*arccosh(c*x))*(-c*x+
1)^(1/2)/b/c^3/(c*x-1)^(1/2)-1/8*Shi(4*(a+b*arccosh(c*x))/b)*sinh(4*a/b)*(-c*x+1)^(1/2)/b/c^3/(c*x-1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {5952, 5556, 3384, 3379, 3382} \begin {gather*} \frac {\sqrt {1-c x} \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \cosh ^{-1}(c x)\right )}{b}\right )}{8 b c^3 \sqrt {c x-1}}-\frac {\sqrt {1-c x} \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \cosh ^{-1}(c x)\right )}{b}\right )}{8 b c^3 \sqrt {c x-1}}-\frac {\sqrt {1-c x} \log \left (a+b \cosh ^{-1}(c x)\right )}{8 b c^3 \sqrt {c x-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - c*x]*Cosh[(4*a)/b]*CoshIntegral[(4*(a + b*ArcCosh[c*x]))/b])/(8*b*c^3*Sqrt[-1 + c*x]) - (Sqrt[1 - c*
x]*Log[a + b*ArcCosh[c*x]])/(8*b*c^3*Sqrt[-1 + c*x]) - (Sqrt[1 - c*x]*Sinh[(4*a)/b]*SinhIntegral[(4*(a + b*Arc
Cosh[c*x]))/b])/(8*b*c^3*Sqrt[-1 + c*x])

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5952

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/((1 + c*x)^p*(-1 + c*x)^p)], Subst[Int[x^n*Cosh[-a/b + x/b]^m*Sinh[-a/b + x/b]^
(2*p + 1), x], x, a + b*ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p + 2
, 0] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \sqrt {1-c^2 x^2}}{a+b \cosh ^{-1}(c x)} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {x^2 \sqrt {-1+c x} \sqrt {1+c x}}{a+b \cosh ^{-1}(c x)} \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {\sqrt {1-c^2 x^2} \text {Subst}\left (\int \frac {\cosh ^2(x) \sinh ^2(x)}{a+b x} \, dx,x,\cosh ^{-1}(c x)\right )}{c^3 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {\sqrt {1-c^2 x^2} \text {Subst}\left (\int \left (-\frac {1}{8 (a+b x)}+\frac {\cosh (4 x)}{8 (a+b x)}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{c^3 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {\sqrt {1-c^2 x^2} \log \left (a+b \cosh ^{-1}(c x)\right )}{8 b c^3 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {\sqrt {1-c^2 x^2} \text {Subst}\left (\int \frac {\cosh (4 x)}{a+b x} \, dx,x,\cosh ^{-1}(c x)\right )}{8 c^3 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {\sqrt {1-c^2 x^2} \log \left (a+b \cosh ^{-1}(c x)\right )}{8 b c^3 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {\left (\sqrt {1-c^2 x^2} \cosh \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c x)\right )}{8 c^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (\sqrt {1-c^2 x^2} \sinh \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\cosh ^{-1}(c x)\right )}{8 c^3 \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {\sqrt {1-c^2 x^2} \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \cosh ^{-1}(c x)\right )}{8 b c^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\sqrt {1-c^2 x^2} \log \left (a+b \cosh ^{-1}(c x)\right )}{8 b c^3 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\sqrt {1-c^2 x^2} \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \cosh ^{-1}(c x)\right )}{8 b c^3 \sqrt {-1+c x} \sqrt {1+c x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.22, size = 103, normalized size = 0.74 \begin {gather*} -\frac {\sqrt {-((-1+c x) (1+c x))} \left (-\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\cosh ^{-1}(c x)\right )\right )+\log \left (a+b \cosh ^{-1}(c x)\right )+\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\cosh ^{-1}(c x)\right )\right )\right )}{8 b c^3 \sqrt {\frac {-1+c x}{1+c x}} (1+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(Sqrt[-((-1 + c*x)*(1 + c*x))]*(-(Cosh[(4*a)/b]*CoshIntegral[4*(a/b + ArcCosh[c*x])]) + Log[a + b*ArcCosh
[c*x]] + Sinh[(4*a)/b]*SinhIntegral[4*(a/b + ArcCosh[c*x])]))/(b*c^3*Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x))

________________________________________________________________________________________

Maple [A]
time = 5.30, size = 229, normalized size = 1.65

method	result	size
default	\(\frac {\sqrt {-c^{2} x^{2}+1}\, \left (-\sqrt {c x +1}\, \sqrt {c x -1}\, x c +c^{2} x^{2}-1\right ) \expIntegral \left (1, 4 \,\mathrm {arccosh}\left (c x \right )+\frac {4 a}{b}\right ) {\mathrm e}^{\frac {b \,\mathrm {arccosh}\left (c x \right )+4 a}{b}}}{16 \left (c x +1\right ) c^{3} \left (c x -1\right ) b}+\frac {\sqrt {-c^{2} x^{2}+1}\, \left (-\sqrt {c x +1}\, \sqrt {c x -1}\, x c +c^{2} x^{2}-1\right ) \expIntegral \left (1, -4 \,\mathrm {arccosh}\left (c x \right )-\frac {4 a}{b}\right ) {\mathrm e}^{-\frac {-b \,\mathrm {arccosh}\left (c x \right )+4 a}{b}}}{16 \left (c x +1\right ) c^{3} \left (c x -1\right ) b}-\frac {\sqrt {-c^{2} x^{2}+1}\, \ln \left (a +b \,\mathrm {arccosh}\left (c x \right )\right )}{8 \sqrt {c x -1}\, \sqrt {c x +1}\, c^{3} b}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(-c^2*x^2+1)^(1/2)*(-(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+c^2*x^2-1)*Ei(1,4*arccosh(c*x)+4*a/b)*exp((b*arccosh
(c*x)+4*a)/b)/(c*x+1)/c^3/(c*x-1)/b+1/16*(-c^2*x^2+1)^(1/2)*(-(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+c^2*x^2-1)*Ei(1,
-4*arccosh(c*x)-4*a/b)*exp(-(-b*arccosh(c*x)+4*a)/b)/(c*x+1)/c^3/(c*x-1)/b-1/8*(-c^2*x^2+1)^(1/2)/(c*x-1)^(1/2
)/(c*x+1)^(1/2)/c^3*ln(a+b*arccosh(c*x))/b

________________________________________________________________________________________

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

[Out]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sqrt(-(c*x - 1)*(c*x + 1))/(a + b*acosh(c*x)), 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(x^2*(-c^2*x^2+1)^(1/2)/(a+b*arccosh(c*x)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\sqrt {1-c^2\,x^2}}{a+b\,\mathrm {acosh}\left (c\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________