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

Optimal. Leaf size=154 \[ \frac{a^2 \text{Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}-\frac{a^2 \sqrt{(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac{2 a \text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3}-\frac{\text{Shi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac{3 \text{Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac{2 a (a+b x) \sqrt{(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac{(a+b x)^2 \sqrt{(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.216496, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {5865, 5803, 5655, 5779, 3298, 5665, 3301} \[ \frac{a^2 \text{Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^3}-\frac{a^2 \sqrt{(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac{2 a \text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^3}-\frac{\text{Shi}\left (\sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac{3 \text{Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3}+\frac{2 a (a+b x) \sqrt{(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)}-\frac{(a+b x)^2 \sqrt{(a+b x)^2+1}}{b^3 \sinh ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSinh[a + b*x]^2,x]

[Out]

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

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 5803

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*ArcSinh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[m, 0] && LtQ[n, -1]

Rule 5655

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(n + 1
))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c^2*x^2], x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

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 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 5665

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

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

Mathematica [A]  time = 0.43585, size = 83, normalized size = 0.54 \[ \frac{-\frac{4 b^2 x^2 \sqrt{a^2+2 a b x+b^2 x^2+1}}{\sinh ^{-1}(a+b x)}+\left (4 a^2-1\right ) \text{Shi}\left (\sinh ^{-1}(a+b x)\right )-8 a \text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )+3 \text{Shi}\left (3 \sinh ^{-1}(a+b x)\right )}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSinh[a + b*x]^2,x]

[Out]

((-4*b^2*x^2*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/ArcSinh[a + b*x] - 8*a*CoshIntegral[2*ArcSinh[a + b*x]] + (-1
+ 4*a^2)*SinhIntegral[ArcSinh[a + b*x]] + 3*SinhIntegral[3*ArcSinh[a + b*x]])/(4*b^3)

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Maple [A]  time = 0.051, size = 146, normalized size = 1. \begin{align*}{\frac{1}{{b}^{3}} \left ( -{\frac{a \left ( 2\,{\it Chi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ){\it Arcsinh} \left ( bx+a \right ) -\sinh \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) \right ) }{{\it Arcsinh} \left ( bx+a \right ) }}+{\frac{1}{4\,{\it Arcsinh} \left ( bx+a \right ) }\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{{\it Shi} \left ({\it Arcsinh} \left ( bx+a \right ) \right ) }{4}}-{\frac{\cosh \left ( 3\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{4\,{\it Arcsinh} \left ( bx+a \right ) }}+{\frac{3\,{\it Shi} \left ( 3\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{4}}+{\frac{{a}^{2}}{{\it Arcsinh} \left ( bx+a \right ) } \left ({\it Shi} \left ({\it Arcsinh} \left ( bx+a \right ) \right ){\it Arcsinh} \left ( bx+a \right ) -\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(-a*(2*Chi(2*arcsinh(b*x+a))*arcsinh(b*x+a)-sinh(2*arcsinh(b*x+a)))/arcsinh(b*x+a)+1/4/arcsinh(b*x+a)*(1
+(b*x+a)^2)^(1/2)-1/4*Shi(arcsinh(b*x+a))-1/4/arcsinh(b*x+a)*cosh(3*arcsinh(b*x+a))+3/4*Shi(3*arcsinh(b*x+a))+
a^2*(Shi(arcsinh(b*x+a))*arcsinh(b*x+a)-(1+(b*x+a)^2)^(1/2))/arcsinh(b*x+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b^{3} x^{5} + 3 \, a b^{2} x^{4} +{\left (3 \, a^{2} b + b\right )} x^{3} +{\left (a^{3} + a\right )} x^{2} +{\left (b^{2} x^{4} + 2 \, a b x^{3} +{\left (a^{2} + 1\right )} x^{2}\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (b^{2} x + a b\right )} + b\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} + \int \frac{3 \, b^{5} x^{6} + 14 \, a b^{4} x^{5} + 2 \,{\left (13 \, a^{2} b^{3} + 3 \, b^{3}\right )} x^{4} + 8 \,{\left (3 \, a^{3} b^{2} + 2 \, a b^{2}\right )} x^{3} +{\left (11 \, a^{4} b + 14 \, a^{2} b + 3 \, b\right )} x^{2} +{\left (3 \, b^{3} x^{4} + 8 \, a b^{2} x^{3} +{\left (7 \, a^{2} b + b\right )} x^{2} + 2 \,{\left (a^{3} + a\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} + 2 \,{\left (a^{5} + 2 \, a^{3} + a\right )} x +{\left (6 \, b^{4} x^{5} + 22 \, a b^{3} x^{4} +{\left (30 \, a^{2} b^{2} + 7 \, b^{2}\right )} x^{3} +{\left (18 \, a^{3} b + 13 \, a b\right )} x^{2} + 2 \,{\left (2 \, a^{4} + 3 \, a^{2} + 1\right )} x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (b^{5} x^{4} + 4 \, a b^{4} x^{3} + a^{4} b + 2 \, a^{2} b + 2 \,{\left (3 \, a^{2} b^{3} + b^{3}\right )} x^{2} +{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} + 4 \,{\left (a^{3} b^{2} + a b^{2}\right )} x + 2 \,{\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + a^{3} b + a b +{\left (3 \, a^{2} b^{2} + b^{2}\right )} x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + b\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^3*x^5 + 3*a*b^2*x^4 + (3*a^2*b + b)*x^3 + (a^3 + a)*x^2 + (b^2*x^4 + 2*a*b*x^3 + (a^2 + 1)*x^2)*sqrt(b^2*x
^2 + 2*a*b*x + a^2 + 1))/((b^3*x^2 + 2*a*b^2*x + a^2*b + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(b^2*x + a*b) + b)*
log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))) + integrate((3*b^5*x^6 + 14*a*b^4*x^5 + 2*(13*a^2*b^3 + 3*b^
3)*x^4 + 8*(3*a^3*b^2 + 2*a*b^2)*x^3 + (11*a^4*b + 14*a^2*b + 3*b)*x^2 + (3*b^3*x^4 + 8*a*b^2*x^3 + (7*a^2*b +
b)*x^2 + 2*(a^3 + a)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + 2*(a^5 + 2*a^3 + a)*x + (6*b^4*x^5 + 22*a*b^3*x^4 + (
30*a^2*b^2 + 7*b^2)*x^3 + (18*a^3*b + 13*a*b)*x^2 + 2*(2*a^4 + 3*a^2 + 1)*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)
)/((b^5*x^4 + 4*a*b^4*x^3 + a^4*b + 2*a^2*b + 2*(3*a^2*b^3 + b^3)*x^2 + (b^3*x^2 + 2*a*b^2*x + a^2*b)*(b^2*x^2
+ 2*a*b*x + a^2 + 1) + 4*(a^3*b^2 + a*b^2)*x + 2*(b^4*x^3 + 3*a*b^3*x^2 + a^3*b + a*b + (3*a^2*b^2 + b^2)*x)*
sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + b)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2/arcsinh(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/asinh(a + b*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/arcsinh(b*x + a)^2, x)