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

Optimal. Leaf size=131 \[ \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+\text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+\sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+\sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )-\frac{1}{2} \sinh ^{-1}(a+b x)^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.243267, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5865, 5799, 5561, 2190, 2279, 2391} \[ \text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+\text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+\sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{a-\sqrt{a^2+1}}\right )+\sinh ^{-1}(a+b x) \log \left (1-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )-\frac{1}{2} \sinh ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]/x,x]

[Out]

-ArcSinh[a + b*x]^2/2 + ArcSinh[a + b*x]*Log[1 - E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] + ArcSinh[a + b*x]*Lo
g[1 - E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^2])] + PolyLog[2, E^ArcSinh[a + b*x]/(a - Sqrt[1 + a^2])] + PolyLog[2
, E^ArcSinh[a + b*x]/(a + Sqrt[1 + a^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 5799

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

Rule 5561

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

Rubi steps

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

Mathematica [A]  time = 0.0121277, size = 153, normalized size = 1.17 \[ \text{PolyLog}\left (2,-\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}-a}\right )+\text{PolyLog}\left (2,\frac{e^{\sinh ^{-1}(a+b x)}}{\sqrt{a^2+1}+a}\right )+\sinh ^{-1}(a+b x) \log \left (\frac{e^{\sinh ^{-1}(a+b x)}}{b \left (-\frac{\sqrt{a^2+1}}{b}-\frac{a}{b}\right )}+1\right )+\sinh ^{-1}(a+b x) \log \left (\frac{e^{\sinh ^{-1}(a+b x)}}{b \left (\frac{\sqrt{a^2+1}}{b}-\frac{a}{b}\right )}+1\right )-\frac{1}{2} \sinh ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]/x,x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.082, size = 388, normalized size = 3. \begin{align*} -{\frac{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}{2}}+{\frac{{\it Arcsinh} \left ( bx+a \right ) }{{a}^{2}+1} \left ({a}^{2}+1+\sqrt{{a}^{2}+1}a \right ) \left ( 2\,\ln \left ({\frac{\sqrt{{a}^{2}+1}-bx-\sqrt{1+ \left ( bx+a \right ) ^{2}}}{a+\sqrt{{a}^{2}+1}}} \right ){a}^{2}+\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 ) +\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 ) -2\,\sqrt{{a}^{2}+1}\ln \left ({\frac{\sqrt{{a}^{2}+1}-bx-\sqrt{1+ \left ( bx+a \right ) ^{2}}}{a+\sqrt{{a}^{2}+1}}} \right ) a \right ) }+{\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 ) +{\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 ) +{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 ){\frac{1}{\sqrt{{a}^{2}+1}}}}-{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 ){\frac{1}{\sqrt{{a}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(b*x + a)/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a + b*x)/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(b*x + a)/x, x)