Integrand size = 10, antiderivative size = 46 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=-\text {arcsinh}\left (\sqrt {x}\right )^2+2 \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )+\operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right ) \]
-arcsinh(x^(1/2))^2+2*arcsinh(x^(1/2))*ln(1-(x^(1/2)+(1+x)^(1/2))^2)+polyl og(2,(x^(1/2)+(1+x)^(1/2))^2)
Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=-\text {arcsinh}\left (\sqrt {x}\right )^2+2 \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )+\operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right ) \]
-ArcSinh[Sqrt[x]]^2 + 2*ArcSinh[Sqrt[x]]*Log[1 - E^(2*ArcSinh[Sqrt[x]])] + PolyLog[2, E^(2*ArcSinh[Sqrt[x]])]
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.41, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {6284, 3042, 26, 4199, 25, 2620, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx\) |
\(\Big \downarrow \) 6284 |
\(\displaystyle 2 \int \frac {\sqrt {x+1} \text {arcsinh}\left (\sqrt {x}\right )}{\sqrt {x}}d\text {arcsinh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int -i \text {arcsinh}\left (\sqrt {x}\right ) \tan \left (i \text {arcsinh}\left (\sqrt {x}\right )+\frac {\pi }{2}\right )d\text {arcsinh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -2 i \int \text {arcsinh}\left (\sqrt {x}\right ) \tan \left (i \text {arcsinh}\left (\sqrt {x}\right )+\frac {\pi }{2}\right )d\text {arcsinh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 4199 |
\(\displaystyle -2 i \left (2 i \int -\frac {e^{2 \text {arcsinh}\left (\sqrt {x}\right )} \text {arcsinh}\left (\sqrt {x}\right )}{1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}}d\text {arcsinh}\left (\sqrt {x}\right )-\frac {1}{2} i \text {arcsinh}\left (\sqrt {x}\right )^2\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 i \left (-2 i \int \frac {e^{2 \text {arcsinh}\left (\sqrt {x}\right )} \text {arcsinh}\left (\sqrt {x}\right )}{1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}}d\text {arcsinh}\left (\sqrt {x}\right )-\frac {1}{2} i \text {arcsinh}\left (\sqrt {x}\right )^2\right )\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -2 i \left (-2 i \left (\frac {1}{2} \int \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )d\text {arcsinh}\left (\sqrt {x}\right )-\frac {1}{2} \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )\right )-\frac {1}{2} i \text {arcsinh}\left (\sqrt {x}\right )^2\right )\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -2 i \left (-2 i \left (\frac {1}{4} \int e^{-2 \text {arcsinh}\left (\sqrt {x}\right )} \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )de^{2 \text {arcsinh}\left (\sqrt {x}\right )}-\frac {1}{2} \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )\right )-\frac {1}{2} i \text {arcsinh}\left (\sqrt {x}\right )^2\right )\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -2 i \left (-2 i \left (-\frac {1}{4} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )-\frac {1}{2} \text {arcsinh}\left (\sqrt {x}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\sqrt {x}\right )}\right )\right )-\frac {1}{2} i \text {arcsinh}\left (\sqrt {x}\right )^2\right )\) |
(-2*I)*((-1/2*I)*ArcSinh[Sqrt[x]]^2 - (2*I)*(-1/2*(ArcSinh[Sqrt[x]]*Log[1 - E^(2*ArcSinh[Sqrt[x]])]) - PolyLog[2, E^(2*ArcSinh[Sqrt[x]])]/4))
3.3.95.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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] - Si mp[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]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[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]
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]
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] + Simp [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] && In tegerQ[4*k] && IGtQ[m, 0]
Int[ArcSinh[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> Simp[1/p Subst[Int[ x^n*Coth[x], x], x, ArcSinh[a*x^p]], x] /; FreeQ[{a, p}, x] && IGtQ[n, 0]
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 | \(-\operatorname {arcsinh}\left (\sqrt {x}\right )^{2}+2 \,\operatorname {arcsinh}\left (\sqrt {x}\right ) \ln \left (1+\sqrt {x}+\sqrt {1+x}\right )+2 \operatorname {polylog}\left (2, -\sqrt {x}-\sqrt {1+x}\right )+2 \,\operatorname {arcsinh}\left (\sqrt {x}\right ) \ln \left (1-\sqrt {x}-\sqrt {1+x}\right )+2 \operatorname {polylog}\left (2, \sqrt {x}+\sqrt {1+x}\right )\) | \(78\) |
-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*po lylog(2,x^(1/2)+(1+x)^(1/2))
\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{x} \,d x } \]
\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int \frac {\operatorname {asinh}{\left (\sqrt {x} \right )}}{x}\, dx \]
\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{x} \,d x } \]
\[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int { \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{x} \,d x } \]
Timed out. \[ \int \frac {\text {arcsinh}\left (\sqrt {x}\right )}{x} \, dx=\int \frac {\mathrm {asinh}\left (\sqrt {x}\right )}{x} \,d x \]