3.73 \(\int \frac{\sinh ^{-1}(a+b x)^2}{x^3} \, dx\)

Optimal. Leaf size=235 \[ \frac{a b^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\left (a^2+1\right )^{3/2}}-\frac{a b^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\left (a^2+1\right )^{3/2}}+\frac{b^2 \log (x)}{a^2+1}+\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\left (a^2+1\right )^{3/2}}-\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\left (a^2+1\right )^{3/2}}-\frac{b \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{\left (a^2+1\right ) x}-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2} \]

[Out]

-((b*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/((1 + a^2)*x)) - ArcSinh[a + b*x]^2/(2*x^2) + (a*b^2*ArcSinh[a +
b*x]*Log[1 - E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])/(1 + a^2)^(3/2) - (a*b^2*ArcSinh[a + b*x]*Log[1 - E^ArcS
inh[a + b*x]/(a + Sqrt[1 + a^2])])/(1 + a^2)^(3/2) + (b^2*Log[x])/(1 + a^2) + (a*b^2*PolyLog[2, E^ArcSinh[a +
b*x]/(a - Sqrt[1 + a^2])])/(1 + a^2)^(3/2) - (a*b^2*PolyLog[2, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])])/(1 + a
^2)^(3/2)

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Rubi [A]  time = 0.485757, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.917, Rules used = {5865, 5801, 5831, 3324, 3322, 2264, 2190, 2279, 2391, 2668, 31} \[ \frac{a b^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\left (a^2+1\right )^{3/2}}-\frac{a b^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\left (a^2+1\right )^{3/2}}+\frac{b^2 \log (x)}{a^2+1}+\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )}{\left (a^2+1\right )^{3/2}}-\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )}{\left (a^2+1\right )^{3/2}}-\frac{b \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)}{\left (a^2+1\right ) x}-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]^2/x^3,x]

[Out]

-((b*Sqrt[1 + (a + b*x)^2]*ArcSinh[a + b*x])/((1 + a^2)*x)) - ArcSinh[a + b*x]^2/(2*x^2) + (a*b^2*ArcSinh[a +
b*x]*Log[1 - E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])])/(1 + a^2)^(3/2) - (a*b^2*ArcSinh[a + b*x]*Log[1 - E^ArcS
inh[a + b*x]/(a + Sqrt[1 + a^2])])/(1 + a^2)^(3/2) + (b^2*Log[x])/(1 + a^2) + (a*b^2*PolyLog[2, E^ArcSinh[a +
b*x]/(a - Sqrt[1 + a^2])])/(1 + a^2)^(3/2) - (a*b^2*PolyLog[2, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^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 5831

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> Dist[1/(c^(m + 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*(c*f + g*Sinh[x])^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{
a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

Rule 3324

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

Rule 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{\sinh ^{-1}(a+b x)^2}{x^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)^2}{\left (-\frac{a}{b}+\frac{x}{b}\right )^3} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2}+\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2}+\operatorname{Subst}\left (\int \frac{x}{\left (-\frac{a}{b}+\frac{\sinh (x)}{b}\right )^2} \, dx,x,\sinh ^{-1}(a+b x)\right )\\ &=-\frac{b \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{\left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2}+\frac{b \operatorname{Subst}\left (\int \frac{\cosh (x)}{-\frac{a}{b}+\frac{\sinh (x)}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{1+a^2}-\frac{(a b) \operatorname{Subst}\left (\int \frac{x}{-\frac{a}{b}+\frac{\sinh (x)}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{1+a^2}\\ &=-\frac{b \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{\left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2}-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{e^x x}{-\frac{1}{b}-\frac{2 a e^x}{b}+\frac{e^{2 x}}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{1+a^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+x} \, dx,x,\frac{a}{b}+x\right )}{1+a^2}\\ &=-\frac{b \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{\left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2}+\frac{b^2 \log (x)}{1+a^2}-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{e^x x}{-\frac{2 a}{b}-\frac{2 \sqrt{1+a^2}}{b}+\frac{2 e^x}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{\left (1+a^2\right )^{3/2}}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{e^x x}{-\frac{2 a}{b}+\frac{2 \sqrt{1+a^2}}{b}+\frac{2 e^x}{b}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{\left (1+a^2\right )^{3/2}}\\ &=-\frac{b \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{\left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2}+\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{1+a^2}}\right )}{\left (1+a^2\right )^{3/2}}-\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a+\sqrt{1+a^2}}\right )}{\left (1+a^2\right )^{3/2}}+\frac{b^2 \log (x)}{1+a^2}+\frac{\left (a b^2\right ) \operatorname{Subst}\left (\int \log \left (1+\frac{2 e^x}{\left (-\frac{2 a}{b}-\frac{2 \sqrt{1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{\left (1+a^2\right )^{3/2}}-\frac{\left (a b^2\right ) \operatorname{Subst}\left (\int \log \left (1+\frac{2 e^x}{\left (-\frac{2 a}{b}+\frac{2 \sqrt{1+a^2}}{b}\right ) b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{\left (1+a^2\right )^{3/2}}\\ &=-\frac{b \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{\left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2}+\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{1+a^2}}\right )}{\left (1+a^2\right )^{3/2}}-\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a+\sqrt{1+a^2}}\right )}{\left (1+a^2\right )^{3/2}}+\frac{b^2 \log (x)}{1+a^2}+\frac{\left (a b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 x}{\left (-\frac{2 a}{b}-\frac{2 \sqrt{1+a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{\left (1+a^2\right )^{3/2}}-\frac{\left (a b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 x}{\left (-\frac{2 a}{b}+\frac{2 \sqrt{1+a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{\left (1+a^2\right )^{3/2}}\\ &=-\frac{b \sqrt{1+(a+b x)^2} \sinh ^{-1}(a+b x)}{\left (1+a^2\right ) x}-\frac{\sinh ^{-1}(a+b x)^2}{2 x^2}+\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{1+a^2}}\right )}{\left (1+a^2\right )^{3/2}}-\frac{a b^2 \sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a+\sqrt{1+a^2}}\right )}{\left (1+a^2\right )^{3/2}}+\frac{b^2 \log (x)}{1+a^2}+\frac{a b^2 \text{Li}_2\left (\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{1+a^2}}\right )}{\left (1+a^2\right )^{3/2}}-\frac{a b^2 \text{Li}_2\left (\frac{e^{\sinh ^{-1}(a+b x)}}{a+\sqrt{1+a^2}}\right )}{\left (1+a^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.123856, size = 279, normalized size = 1.19 \[ -\frac{-2 a b^2 x^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+2 a b^2 x^2 \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )-2 \sqrt{a^2+1} b^2 x^2 \log (x)+2 a b^2 x^2 \sinh ^{-1}(a+b x) \log \left (\frac{\sqrt{a^2+1}-e^{\sinh ^{-1}(a+b x)}+a}{\sqrt{a^2+1}+a}\right )-2 a b^2 x^2 \sinh ^{-1}(a+b x) \log \left (\frac{\sqrt{a^2+1}+e^{\sinh ^{-1}(a+b x)}-a}{\sqrt{a^2+1}-a}\right )+2 \sqrt{a^2+1} b x \sqrt{(a+b x)^2+1} \sinh ^{-1}(a+b x)+a^2 \sqrt{a^2+1} \sinh ^{-1}(a+b x)^2+\sqrt{a^2+1} \sinh ^{-1}(a+b x)^2}{2 \left (a^2+1\right )^{3/2} x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]^2/x^3,x]

[Out]

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

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Maple [A]  time = 0.224, size = 384, normalized size = 1.6 \begin{align*}{\frac{{b}^{2}{\it Arcsinh} \left ( bx+a \right ) }{{a}^{2}+1}}-{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}{a}^{2}}{ \left ( 2\,{a}^{2}+2 \right ){x}^{2}}}-{\frac{b{\it Arcsinh} \left ( bx+a \right ) }{ \left ({a}^{2}+1 \right ) x}\sqrt{1+ \left ( bx+a \right ) ^{2}}}-{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}{ \left ( 2\,{a}^{2}+2 \right ){x}^{2}}}-{{b}^{2}a{\it Arcsinh} \left ( bx+a \right ) \ln \left ({ \left ( \sqrt{{a}^{2}+1}-bx-\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) \left ( a+\sqrt{{a}^{2}+1} \right ) ^{-1}} \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{{b}^{2}a{\it Arcsinh} \left ( bx+a \right ) \ln \left ({ \left ( \sqrt{{a}^{2}+1}+bx+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) \left ( -a+\sqrt{{a}^{2}+1} \right ) ^{-1}} \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{{b}^{2}a{\it dilog} \left ({ \left ( \sqrt{{a}^{2}+1}-bx-\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) \left ( a+\sqrt{{a}^{2}+1} \right ) ^{-1}} \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}+{{b}^{2}a{\it dilog} \left ({ \left ( \sqrt{{a}^{2}+1}+bx+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) \left ( -a+\sqrt{{a}^{2}+1} \right ) ^{-1}} \right ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-2\,{\frac{{b}^{2}\ln \left ( bx+a+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) }{{a}^{2}+1}}+{\frac{{b}^{2}}{{a}^{2}+1}\ln \left ( \left ( bx+a+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) ^{2}-2\,a \left ( bx+a+\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)^2/x^3,x)

[Out]

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

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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)^2/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arsinh}\left (b x + a\right )^{2}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)^2/x^3,x, algorithm="fricas")

[Out]

integral(arcsinh(b*x + a)^2/x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}^{2}{\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)**2/x**3,x)

[Out]

Integral(asinh(a + b*x)**2/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arsinh}\left (b x + a\right )^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)^2/x^3,x, algorithm="giac")

[Out]

integrate(arcsinh(b*x + a)^2/x^3, x)