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

Optimal. Leaf size=139 \[ \frac{\sqrt{c x-1} \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \cosh ^{-1}(c x)\right )}{b}\right )}{2 b c^3 \sqrt{1-c x}}-\frac{\sqrt{c x-1} \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \cosh ^{-1}(c x)\right )}{b}\right )}{2 b c^3 \sqrt{1-c x}}+\frac{\sqrt{c x-1} \log \left (a+b \cosh ^{-1}(c x)\right )}{2 b c^3 \sqrt{1-c x}} \]

[Out]

(Sqrt[-1 + c*x]*Cosh[(2*a)/b]*CoshIntegral[(2*(a + b*ArcCosh[c*x]))/b])/(2*b*c^3*Sqrt[1 - c*x]) + (Sqrt[-1 + c
*x]*Log[a + b*ArcCosh[c*x]])/(2*b*c^3*Sqrt[1 - c*x]) - (Sqrt[-1 + c*x]*Sinh[(2*a)/b]*SinhIntegral[(2*(a + b*Ar
cCosh[c*x]))/b])/(2*b*c^3*Sqrt[1 - c*x])

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Rubi [A]  time = 0.631826, antiderivative size = 178, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {5798, 5781, 3312, 3303, 3298, 3301} \[ \frac{\sqrt{c x-1} \sqrt{c x+1} \cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \cosh ^{-1}(c x)\right )}{2 b c^3 \sqrt{1-c^2 x^2}}-\frac{\sqrt{c x-1} \sqrt{c x+1} \sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \cosh ^{-1}(c x)\right )}{2 b c^3 \sqrt{1-c^2 x^2}}+\frac{\sqrt{c x-1} \sqrt{c x+1} \log \left (a+b \cosh ^{-1}(c x)\right )}{2 b c^3 \sqrt{1-c^2 x^2}} \]

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]*Sqrt[1 + c*x]*Cosh[(2*a)/b]*CoshIntegral[(2*a)/b + 2*ArcCosh[c*x]])/(2*b*c^3*Sqrt[1 - c^2*x^2]
) + (Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Log[a + b*ArcCosh[c*x]])/(2*b*c^3*Sqrt[1 - c^2*x^2]) - (Sqrt[-1 + c*x]*Sqrt[
1 + c*x]*Sinh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcCosh[c*x]])/(2*b*c^3*Sqrt[1 - c^2*x^2])

Rule 5798

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

Rule 5781

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x_))^(p_
.), x_Symbol] :> Dist[(-(d1*d2))^p/c^(m + 1), Subst[Int[(a + b*x)^n*Cosh[x]^m*Sinh[x]^(2*p + 1), x], x, ArcCos
h[c*x]], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && IntegerQ[p
+ 1/2] && GtQ[p, -1] && IGtQ[m, 0] && (GtQ[d1, 0] && LtQ[d2, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3303

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 3298

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 3301

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]

Rubi steps

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

Mathematica [A]  time = 0.261181, size = 99, normalized size = 0.71 \[ -\frac{\sqrt{1-c^2 x^2} \left (\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\cosh ^{-1}(c x)\right )\right )-\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\cosh ^{-1}(c x)\right )\right )+\log \left (a+b \cosh ^{-1}(c x)\right )\right )}{2 c^3 \sqrt{\frac{c x-1}{c x+1}} (b c x+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - c^2*x^2]*(Cosh[(2*a)/b]*CoshIntegral[2*(a/b + ArcCosh[c*x])] + Log[a + b*ArcCosh[c*x]] - Sinh[(2*a)
/b]*SinhIntegral[2*(a/b + ArcCosh[c*x])]))/(2*c^3*Sqrt[(-1 + c*x)/(1 + c*x)]*(b + b*c*x))

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Maple [A]  time = 0.166, size = 232, normalized size = 1.7 \begin{align*}{\frac{1}{4\,{c}^{3} \left ({c}^{2}{x}^{2}-1 \right ) b}\sqrt{-{c}^{2}{x}^{2}+1} \left ( \sqrt{cx+1}\sqrt{cx-1}xc+{c}^{2}{x}^{2}-1 \right ){\it Ei} \left ( 1,2\,{\rm arccosh} \left (cx\right )+2\,{\frac{a}{b}} \right ){{\rm e}^{-{\frac{b{\rm arccosh} \left (cx\right )-2\,a}{b}}}}}+{\frac{1}{4\,{c}^{3} \left ({c}^{2}{x}^{2}-1 \right ) b}\sqrt{-{c}^{2}{x}^{2}+1} \left ( \sqrt{cx+1}\sqrt{cx-1}xc+{c}^{2}{x}^{2}-1 \right ){\it Ei} \left ( 1,-2\,{\rm arccosh} \left (cx\right )-2\,{\frac{a}{b}} \right ){{\rm e}^{-{\frac{b{\rm arccosh} \left (cx\right )+2\,a}{b}}}}}-{\frac{\ln \left ( a+b{\rm arccosh} \left (cx\right ) \right ) }{2\,{c}^{3} \left ({c}^{2}{x}^{2}-1 \right ) b}\sqrt{-{c}^{2}{x}^{2}+1}\sqrt{cx-1}\sqrt{cx+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-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(1,2*arccosh(c*x)+2*a/b)*exp(-(b*arccosh(
c*x)-2*a)/b)/c^3/(c^2*x^2-1)/b+1/4*(-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,-2*arcc
osh(c*x)-2*a/b)*exp(-(b*arccosh(c*x)+2*a)/b)/c^3/(c^2*x^2-1)/b-1/2*(-c^2*x^2+1)^(1/2)*(c*x-1)^(1/2)*(c*x+1)^(1
/2)/c^3/(c^2*x^2-1)*ln(a+b*arccosh(c*x))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{-c^{2} x^{2} + 1}{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} x^{2} + 1} x^{2}}{a c^{2} x^{2} +{\left (b c^{2} x^{2} - b\right )} \operatorname{arcosh}\left (c x\right ) - a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{-c^{2} x^{2} + 1}{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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