3.64 \(\int \frac{\sinh ^{-1}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=92 \[ \frac{a b^2 \tanh ^{-1}\left (\frac{a (a+b x)+1}{\sqrt{a^2+1} \sqrt{(a+b x)^2+1}}\right )}{2 \left (a^2+1\right )^{3/2}}-\frac{b \sqrt{(a+b x)^2+1}}{2 \left (a^2+1\right ) x}-\frac{\sinh ^{-1}(a+b x)}{2 x^2} \]

[Out]

-(b*Sqrt[1 + (a + b*x)^2])/(2*(1 + a^2)*x) - ArcSinh[a + b*x]/(2*x^2) + (a*b^2*ArcTanh[(1 + a*(a + b*x))/(Sqrt
[1 + a^2]*Sqrt[1 + (a + b*x)^2])])/(2*(1 + a^2)^(3/2))

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Rubi [A]  time = 0.102091, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5865, 5801, 731, 725, 206} \[ \frac{a b^2 \tanh ^{-1}\left (\frac{a (a+b x)+1}{\sqrt{a^2+1} \sqrt{(a+b x)^2+1}}\right )}{2 \left (a^2+1\right )^{3/2}}-\frac{b \sqrt{(a+b x)^2+1}}{2 \left (a^2+1\right ) x}-\frac{\sinh ^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]/x^3,x]

[Out]

-(b*Sqrt[1 + (a + b*x)^2])/(2*(1 + a^2)*x) - ArcSinh[a + b*x]/(2*x^2) + (a*b^2*ArcTanh[(1 + a*(a + b*x))/(Sqrt
[1 + a^2]*Sqrt[1 + (a + b*x)^2])])/(2*(1 + a^2)^(3/2))

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 5801

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)
*(a + b*ArcSinh[c*x])^n)/(e*(m + 1)), x] - Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*x)^(m + 1)*(a + b*ArcSinh[c*x
])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sinh ^{-1}(a+b x)}{x^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\left (-\frac{a}{b}+\frac{x}{b}\right )^3} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\sinh ^{-1}(a+b x)}{2 x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{2 \left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)}{2 x^2}-\frac{(a b) \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right ) \sqrt{1+x^2}} \, dx,x,a+b x\right )}{2 \left (1+a^2\right )}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{2 \left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)}{2 x^2}+\frac{(a b) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{b^2}+\frac{a^2}{b^2}-x^2} \, dx,x,\frac{\frac{1}{b}+\frac{a (a+b x)}{b}}{\sqrt{1+(a+b x)^2}}\right )}{2 \left (1+a^2\right )}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{2 \left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)}{2 x^2}+\frac{a b^2 \tanh ^{-1}\left (\frac{1+a (a+b x)}{\sqrt{1+a^2} \sqrt{1+(a+b x)^2}}\right )}{2 \left (1+a^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.166296, size = 110, normalized size = 1.2 \[ -\frac{\frac{b x \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}-a b x \log \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )+a b x \log (x)\right )}{\left (a^2+1\right )^{3/2}}+\sinh ^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]/x^3,x]

[Out]

-(ArcSinh[a + b*x] + (b*x*(Sqrt[1 + a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + a*b*x*Log[x] - a*b*x*Log[1 + a^2
+ a*b*x + Sqrt[1 + a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]]))/(1 + a^2)^(3/2))/(2*x^2)

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Maple [A]  time = 0.008, size = 106, normalized size = 1.2 \begin{align*} -{\frac{{\it Arcsinh} \left ( bx+a \right ) }{2\,{x}^{2}}}-{\frac{b}{ \left ( 2\,{a}^{2}+2 \right ) x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{b}^{2}a}{2}\ln \left ({\frac{1}{bx} \left ( 2\,{a}^{2}+2+2\,xab+2\,\sqrt{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ) } \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)/x^3,x)

[Out]

-1/2*arcsinh(b*x+a)/x^2-1/2*b/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2*b^2*a/(a^2+1)^(3/2)*ln((2*a^2+2+2*x*
a*b+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/b/x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.708, size = 564, normalized size = 6.13 \begin{align*} \frac{\sqrt{a^{2} + 1} a b^{2} x^{2} \log \left (-\frac{a^{2} b x + a^{3} + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (a^{2} + \sqrt{a^{2} + 1} a + 1\right )} +{\left (a b x + a^{2} + 1\right )} \sqrt{a^{2} + 1} + a}{x}\right ) -{\left (a^{2} + 1\right )} b^{2} x^{2} +{\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (a^{2} + 1\right )} b x -{\left (a^{4} -{\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2} + 2 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{2 \,{\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)/x^3,x, algorithm="fricas")

[Out]

1/2*(sqrt(a^2 + 1)*a*b^2*x^2*log(-(a^2*b*x + a^3 + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(a^2 + sqrt(a^2 + 1)*a +
1) + (a*b*x + a^2 + 1)*sqrt(a^2 + 1) + a)/x) - (a^2 + 1)*b^2*x^2 + (a^4 + 2*a^2 + 1)*x^2*log(-b*x - a + sqrt(b
^2*x^2 + 2*a*b*x + a^2 + 1)) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(a^2 + 1)*b*x - (a^4 - (a^4 + 2*a^2 + 1)*x^2
+ 2*a^2 + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)))/((a^4 + 2*a^2 + 1)*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}{\left (a + b x \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(b*x+a)/x**3,x)

[Out]

Integral(asinh(a + b*x)/x**3, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)/x^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError