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

Optimal. Leaf size=82 \[ \frac{\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b c^3}-\frac{\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b c^3}-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c^3} \]

[Out]

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

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Rubi [A]  time = 0.291869, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {5779, 3312, 3303, 3298, 3301} \[ \frac{\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{2 b c^3}-\frac{\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{2 b c^3}-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 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 \sinh ^{-1}(c x)\right )} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{1}{2 (a+b x)}-\frac{\cosh (2 x)}{2 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3}\\ &=-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c^3}+\frac{\operatorname{Subst}\left (\int \frac{\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^3}\\ &=-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c^3}+\frac{\cosh \left (\frac{2 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^3}-\frac{\sinh \left (\frac{2 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^3}\\ &=\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{2 b c^3}-\frac{\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c^3}-\frac{\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{2 b c^3}\\ \end{align*}

Mathematica [A]  time = 0.180165, size = 65, normalized size = 0.79 \[ \frac{\cosh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-\log \left (a+b \sinh ^{-1}(c x)\right )}{2 b c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.083, size = 79, normalized size = 1. \begin{align*} -{\frac{\ln \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) }{2\,{c}^{3}b}}-{\frac{1}{4\,{c}^{3}b}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\it Arcsinh} \left ( cx \right ) +2\,{\frac{a}{b}} \right ) }-{\frac{1}{4\,{c}^{3}b}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\it Arcsinh} \left ( cx \right ) -2\,{\frac{a}{b}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*ln(a+b*arcsinh(c*x))/b/c^3-1/4/c^3/b*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)-1/4/c^3/b*exp(-2*a/b)*Ei(1,-2*
arcsinh(c*x)-2*a/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{arsinh}\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*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/(sqrt(c^2*x^2 + 1)*(b*arcsinh(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{arsinh}\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*arcsinh(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)*arcsinh(c*x) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/((a + b*asinh(c*x))*sqrt(c**2*x**2 + 1)), 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{arsinh}\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*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

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