\(\int \frac {\text {arcsinh}(\sqrt {x})}{x} \, dx\) [295]

Optimal result

Integrand size = 10, antiderivative size = 46 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=-\text {arcsinh}\left (\sqrt {x}\right )^2+2 \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )+\operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right ) \]

[Out]

-arcsinh(x^(1/2))^2+2*arcsinh(x^(1/2))*ln(1-(x^(1/2)+(1+x)^(1/2))^2)+polylog(2,(x^(1/2)+(1+x)^(1/2))^2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5869, 3797, 2221, 2317, 2438} \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )-\text {arcsinh}\left (\sqrt {x}\right )^2+2 \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right ) \]

[In]

Int[ArcSinh[Sqrt[x]]/x,x]

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5869

Int[ArcSinh[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> Dist[1/p, Subst[Int[x^n*Coth[x], x], x, ArcSinh[a*x^p]],
x] /; FreeQ[{a, p}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int x \coth (x) \, dx,x,\text {arcsinh}\left (\sqrt {x}\right )\right ) \\ & = -\text {arcsinh}\left (\sqrt {x}\right )^2-4 \text {Subst}\left (\int \frac {e^{2 x} x}{1-e^{2 x}} \, dx,x,\text {arcsinh}\left (\sqrt {x}\right )\right ) \\ & = -\text {arcsinh}\left (\sqrt {x}\right )^2+2 \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )-2 \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\text {arcsinh}\left (\sqrt {x}\right )\right ) \\ & = -\text {arcsinh}\left (\sqrt {x}\right )^2+2 \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )-\text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right ) \\ & = -\text {arcsinh}\left (\sqrt {x}\right )^2+2 \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )+\operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=-\text {arcsinh}\left (\sqrt {x}\right )^2+2 \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )+\operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right ) \]

[In]

Integrate[ArcSinh[Sqrt[x]]/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.70

method	result	size
derivativedivides	\(-\operatorname {arcsinh}\left (\sqrt {x}\right )^{2}+2 \,\operatorname {arcsinh}\left (\sqrt {x}\right ) \ln \left (1+\sqrt {x}+\sqrt {1+x}\right )+2 \operatorname {polylog}\left (2, -\sqrt {x}-\sqrt {1+x}\right )+2 \,\operatorname {arcsinh}\left (\sqrt {x}\right ) \ln \left (1-\sqrt {x}-\sqrt {1+x}\right )+2 \operatorname {polylog}\left (2, \sqrt {x}+\sqrt {1+x}\right )\)	\(78\)
default	\(-\operatorname {arcsinh}\left (\sqrt {x}\right )^{2}+2 \,\operatorname {arcsinh}\left (\sqrt {x}\right ) \ln \left (1+\sqrt {x}+\sqrt {1+x}\right )+2 \operatorname {polylog}\left (2, -\sqrt {x}-\sqrt {1+x}\right )+2 \,\operatorname {arcsinh}\left (\sqrt {x}\right ) \ln \left (1-\sqrt {x}-\sqrt {1+x}\right )+2 \operatorname {polylog}\left (2, \sqrt {x}+\sqrt {1+x}\right )\)	\(78\)

[In]

int(arcsinh(x^(1/2))/x,x,method=_RETURNVERBOSE)

[Out]

-arcsinh(x^(1/2))^2+2*arcsinh(x^(1/2))*ln(1+x^(1/2)+(1+x)^(1/2))+2*polylog(2,-x^(1/2)-(1+x)^(1/2))+2*arcsinh(x
^(1/2))*ln(1-x^(1/2)-(1+x)^(1/2))+2*polylog(2,x^(1/2)+(1+x)^(1/2))

Fricas [F]

\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{x} \,d x } \]

[In]

integrate(arcsinh(x^(1/2))/x,x, algorithm="fricas")

[Out]

integral(arcsinh(sqrt(x))/x, x)

Sympy [F]

\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int \frac {\operatorname {asinh}{\left (\sqrt {x} \right )}}{x}\, dx \]

[In]

integrate(asinh(x**(1/2))/x,x)

[Out]

Integral(asinh(sqrt(x))/x, x)

Maxima [F]

\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{x} \,d x } \]

[In]

integrate(arcsinh(x^(1/2))/x,x, algorithm="maxima")

[Out]

integrate(arcsinh(sqrt(x))/x, x)

Giac [F]

\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{x} \,d x } \]

[In]

integrate(arcsinh(x^(1/2))/x,x, algorithm="giac")

[Out]

integrate(arcsinh(sqrt(x))/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int \frac {\mathrm {asinh}\left (\sqrt {x}\right )}{x} \,d x \]

[In]

int(asinh(x^(1/2))/x,x)

[Out]

int(asinh(x^(1/2))/x, x)