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

Optimal. Leaf size=84 \[ \frac{\text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^2}-\frac{a \text{Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac{a \sqrt{(a+b x)^2+1}}{b^2 \sinh ^{-1}(a+b x)}-\frac{(a+b x) \sqrt{(a+b x)^2+1}}{b^2 \sinh ^{-1}(a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.131585, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5865, 5803, 5655, 5779, 3298, 5665, 3301} \[ \frac{\text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b^2}-\frac{a \text{Shi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac{a \sqrt{(a+b x)^2+1}}{b^2 \sinh ^{-1}(a+b x)}-\frac{(a+b x) \sqrt{(a+b x)^2+1}}{b^2 \sinh ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[x/ArcSinh[a + b*x]^2,x]

[Out]

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

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

Mathematica [A]  time = 0.162088, size = 62, normalized size = 0.74 \[ -\frac{-\sinh ^{-1}(a+b x) \text{Chi}\left (2 \sinh ^{-1}(a+b x)\right )+a \sinh ^{-1}(a+b x) \text{Shi}\left (\sinh ^{-1}(a+b x)\right )+b x \sqrt{(a+b x)^2+1}}{b^2 \sinh ^{-1}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[x/ArcSinh[a + b*x]^2,x]

[Out]

-((b*x*Sqrt[1 + (a + b*x)^2] - ArcSinh[a + b*x]*CoshIntegral[2*ArcSinh[a + b*x]] + a*ArcSinh[a + b*x]*SinhInte
gral[ArcSinh[a + b*x]])/(b^2*ArcSinh[a + b*x]))

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Maple [A]  time = 0.036, size = 73, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{2}} \left ( -{\frac{\sinh \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) }{2\,{\it Arcsinh} \left ( bx+a \right ) }}+{\it Chi} \left ( 2\,{\it Arcsinh} \left ( bx+a \right ) \right ) -{\frac{a}{{\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/arcsinh(b*x+a)^2,x)

[Out]

1/b^2*(-1/2/arcsinh(b*x+a)*sinh(2*arcsinh(b*x+a))+Chi(2*arcsinh(b*x+a))-a*(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^{4} + 3 \, a b^{2} x^{3} +{\left (3 \, a^{2} b + b\right )} x^{2} +{\left (a^{3} + a\right )} x +{\left (b^{2} x^{3} + 2 \, a b x^{2} +{\left (a^{2} + 1\right )} x\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{2 \, b^{5} x^{5} + 9 \, a b^{4} x^{4} + a^{5} + 4 \,{\left (4 \, a^{2} b^{3} + b^{3}\right )} x^{3} + 2 \, a^{3} + 2 \,{\left (7 \, a^{3} b^{2} + 5 \, a b^{2}\right )} x^{2} +{\left (2 \, b^{3} x^{3} + 5 \, a b^{2} x^{2} + 4 \, a^{2} b x + a^{3} + a\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )} + 2 \,{\left (3 \, a^{4} b + 4 \, a^{2} b + b\right )} x +{\left (4 \, b^{4} x^{4} + 14 \, a b^{3} x^{3} + 2 \, a^{4} + 2 \,{\left (9 \, a^{2} b^{2} + 2 \, b^{2}\right )} x^{2} + 3 \, a^{2} +{\left (10 \, a^{3} b + 7 \, a b\right )} x + 1\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + a}{{\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/arcsinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-(b^3*x^4 + 3*a*b^2*x^3 + (3*a^2*b + b)*x^2 + (a^3 + a)*x + (b^2*x^3 + 2*a*b*x^2 + (a^2 + 1)*x)*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((2*b^5*x^5 + 9*a*b^4*x^4 + a^5 + 4*(4*a^2*b^3 + b^3)
*x^3 + 2*a^3 + 2*(7*a^3*b^2 + 5*a*b^2)*x^2 + (2*b^3*x^3 + 5*a*b^2*x^2 + 4*a^2*b*x + a^3 + a)*(b^2*x^2 + 2*a*b*
x + a^2 + 1) + 2*(3*a^4*b + 4*a^2*b + b)*x + (4*b^4*x^4 + 14*a*b^3*x^3 + 2*a^4 + 2*(9*a^2*b^2 + 2*b^2)*x^2 + 3
*a^2 + (10*a^3*b + 7*a*b)*x + 1)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + a)/((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)*l
og(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}{\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/arcsinh(b*x+a)^2,x, algorithm="fricas")

[Out]

integral(x/arcsinh(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/asinh(a + b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\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/arcsinh(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x/arcsinh(b*x + a)^2, x)