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

Optimal. Leaf size=93 \[ -\frac{\sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^3}-\frac{x^2 \left (c^2 x^2+1\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-((x^2*(1 + c^2*x^2))/(b*c*(a + b*ArcSinh[c*x]))) - (CoshIntegral[(4*(a + b*ArcSinh[c*x]))/b]*Sinh[(4*a)/b])/(
2*b^2*c^3) + (Cosh[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(2*b^2*c^3)

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Rubi [A]  time = 0.561542, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {5777, 5669, 5448, 12, 3303, 3298, 3301} \[ -\frac{\sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{2 b^2 c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{2 b^2 c^3}-\frac{x^2 \left (c^2 x^2+1\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^2*(1 + c^2*x^2))/(b*c*(a + b*ArcSinh[c*x]))) - (CoshIntegral[(4*a)/b + 4*ArcSinh[c*x]]*Sinh[(4*a)/b])/(2*
b^2*c^3) + (Cosh[(4*a)/b]*SinhIntegral[(4*a)/b + 4*ArcSinh[c*x]])/(2*b^2*c^3)

Rule 5777

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[((f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(f*m*d^IntP
art[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p -
1/2)*(a + b*ArcSinh[c*x])^(n + 1), x], x] - Dist[(c*(m + 2*p + 1)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(
n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1), x],
x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && LtQ[n, -1] && IGtQ[m, -3] && IGtQ[2*p, 0]

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.309546, size = 82, normalized size = 0.88 \[ \frac{-\frac{2 b c^2 x^2 \left (c^2 x^2+1\right )}{a+b \sinh ^{-1}(c x)}-\sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )}{2 b^2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*c^2*x^2*(1 + c^2*x^2))/(a + b*ArcSinh[c*x]) - CoshIntegral[4*(a/b + ArcSinh[c*x])]*Sinh[(4*a)/b] + Cosh
[(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[c*x])])/(2*b^2*c^3)

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Maple [B]  time = 0.145, size = 248, normalized size = 2.7 \begin{align*}{\frac{1}{8\,{c}^{3} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) b}}-{\frac{1}{16\,{c}^{3} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) b} \left ( 8\,{c}^{4}{x}^{4}-8\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}+8\,{c}^{2}{x}^{2}-4\,cx\sqrt{{c}^{2}{x}^{2}+1}+1 \right ) }+{\frac{1}{4\,{c}^{3}{b}^{2}}{{\rm e}^{4\,{\frac{a}{b}}}}{\it Ei} \left ( 1,4\,{\it Arcsinh} \left ( cx \right ) +4\,{\frac{a}{b}} \right ) }-{\frac{1}{16\,{c}^{3}{b}^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) } \left ( 8\,{x}^{4}b{c}^{4}+8\,\sqrt{{c}^{2}{x}^{2}+1}{x}^{3}b{c}^{3}+8\,{x}^{2}b{c}^{2}+4\,bc\sqrt{{c}^{2}{x}^{2}+1}x+4\,{\it Arcsinh} \left ( cx \right ){\it Ei} \left ( 1,-4\,{\it Arcsinh} \left ( cx \right ) -4\,{\frac{a}{b}} \right ){{\rm e}^{-4\,{\frac{a}{b}}}}b+4\,{\it Ei} \left ( 1,-4\,{\it Arcsinh} \left ( cx \right ) -4\,{\frac{a}{b}} \right ){{\rm e}^{-4\,{\frac{a}{b}}}}a+b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (c^{2} x^{4} + x^{2}\right )}{\left (c^{2} x^{2} + 1\right )} +{\left (c^{3} x^{5} + c x^{3}\right )} \sqrt{c^{2} x^{2} + 1}}{a b c^{3} x^{2} + \sqrt{c^{2} x^{2} + 1} a b c^{2} x + a b c +{\left (b^{2} c^{3} x^{2} + \sqrt{c^{2} x^{2} + 1} b^{2} c^{2} x + b^{2} c\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )} + \int \frac{{\left (4 \, c^{3} x^{4} + c x^{2}\right )}{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 2 \,{\left (4 \, c^{4} x^{5} + 4 \, c^{2} x^{3} + x\right )}{\left (c^{2} x^{2} + 1\right )} +{\left (4 \, c^{5} x^{6} + 7 \, c^{3} x^{4} + 3 \, c x^{2}\right )} \sqrt{c^{2} x^{2} + 1}}{a b c^{5} x^{4} +{\left (c^{2} x^{2} + 1\right )} a b c^{3} x^{2} + 2 \, a b c^{3} x^{2} + a b c +{\left (b^{2} c^{5} x^{4} +{\left (c^{2} x^{2} + 1\right )} b^{2} c^{3} x^{2} + 2 \, b^{2} c^{3} x^{2} + b^{2} c + 2 \,{\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (a b c^{4} x^{3} + a b c^{2} x\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c^2*x^4 + x^2)*(c^2*x^2 + 1) + (c^3*x^5 + c*x^3)*sqrt(c^2*x^2 + 1))/(a*b*c^3*x^2 + sqrt(c^2*x^2 + 1)*a*b*c^
2*x + a*b*c + (b^2*c^3*x^2 + sqrt(c^2*x^2 + 1)*b^2*c^2*x + b^2*c)*log(c*x + sqrt(c^2*x^2 + 1))) + integrate(((
4*c^3*x^4 + c*x^2)*(c^2*x^2 + 1)^(3/2) + 2*(4*c^4*x^5 + 4*c^2*x^3 + x)*(c^2*x^2 + 1) + (4*c^5*x^6 + 7*c^3*x^4
+ 3*c*x^2)*sqrt(c^2*x^2 + 1))/(a*b*c^5*x^4 + (c^2*x^2 + 1)*a*b*c^3*x^2 + 2*a*b*c^3*x^2 + a*b*c + (b^2*c^5*x^4
+ (c^2*x^2 + 1)*b^2*c^3*x^2 + 2*b^2*c^3*x^2 + b^2*c + 2*(b^2*c^4*x^3 + b^2*c^2*x)*sqrt(c^2*x^2 + 1))*log(c*x +
sqrt(c^2*x^2 + 1)) + 2*(a*b*c^4*x^3 + a*b*c^2*x)*sqrt(c^2*x^2 + 1)), 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}}{b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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