\(\int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx\) [483]

Optimal result

Integrand size = 15, antiderivative size = 46 \[ \int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx=-\frac {1}{8} \text {arcsinh}(2 x)^2+\frac {1}{4} \text {arcsinh}(2 x) \log \left (1+e^{2 \text {arcsinh}(2 x)}\right )+\frac {1}{8} \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(2 x)}\right ) \] Output:

-1/8*arcsinh(2*x)^2+1/4*arcsinh(2*x)*ln(1+(2*x+(4*x^2+1)^(1/2))^2)+1/8*pol 
ylog(2,-(2*x+(4*x^2+1)^(1/2))^2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.85 \[ \int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx=-\frac {1}{8} \text {arcsinh}(2 x)^2+\frac {1}{4} \text {arcsinh}(2 x) \log \left (1-i e^{\text {arcsinh}(2 x)}\right )+\frac {1}{4} \text {arcsinh}(2 x) \log \left (1+i e^{\text {arcsinh}(2 x)}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(2 x)}\right )+\frac {1}{4} \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(2 x)}\right ) \] Input:

Integrate[(x*ArcSinh[2*x])/(1 + 4*x^2),x]
 

Output:

-1/8*ArcSinh[2*x]^2 + (ArcSinh[2*x]*Log[1 - I*E^ArcSinh[2*x]])/4 + (ArcSin 
h[2*x]*Log[1 + I*E^ArcSinh[2*x]])/4 + PolyLog[2, (-I)*E^ArcSinh[2*x]]/4 + 
PolyLog[2, I*E^ArcSinh[2*x]]/4
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.28, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {6212, 3042, 26, 4201, 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.

\(\displaystyle \int \frac {x \text {arcsinh}(2 x)}{4 x^2+1} \, dx\)

\(\Big \downarrow \) 6212

\(\displaystyle \frac {1}{4} \int \frac {2 x \text {arcsinh}(2 x)}{\sqrt {4 x^2+1}}d\text {arcsinh}(2 x)\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int -i \text {arcsinh}(2 x) \tan (i \text {arcsinh}(2 x))d\text {arcsinh}(2 x)\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {1}{4} i \int \text {arcsinh}(2 x) \tan (i \text {arcsinh}(2 x))d\text {arcsinh}(2 x)\)

\(\Big \downarrow \) 4201

\(\displaystyle -\frac {1}{4} i \left (2 i \int \frac {e^{2 \text {arcsinh}(2 x)} \text {arcsinh}(2 x)}{1+e^{2 \text {arcsinh}(2 x)}}d\text {arcsinh}(2 x)-\frac {1}{2} i \text {arcsinh}(2 x)^2\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {1}{4} i \left (2 i \left (\frac {1}{2} \text {arcsinh}(2 x) \log \left (e^{2 \text {arcsinh}(2 x)}+1\right )-\frac {1}{2} \int \log \left (1+e^{2 \text {arcsinh}(2 x)}\right )d\text {arcsinh}(2 x)\right )-\frac {1}{2} i \text {arcsinh}(2 x)^2\right )\)

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {1}{4} i \left (2 i \left (\frac {1}{2} \text {arcsinh}(2 x) \log \left (e^{2 \text {arcsinh}(2 x)}+1\right )-\frac {1}{4} \int e^{-2 \text {arcsinh}(2 x)} \log \left (1+e^{2 \text {arcsinh}(2 x)}\right )de^{2 \text {arcsinh}(2 x)}\right )-\frac {1}{2} i \text {arcsinh}(2 x)^2\right )\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {1}{4} i \left (2 i \left (\frac {1}{4} \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(2 x)}\right )+\frac {1}{2} \text {arcsinh}(2 x) \log \left (e^{2 \text {arcsinh}(2 x)}+1\right )\right )-\frac {1}{2} i \text {arcsinh}(2 x)^2\right )\)

Input:

Int[(x*ArcSinh[2*x])/(1 + 4*x^2),x]
 

Output:

(-1/4*I)*((-1/2*I)*ArcSinh[2*x]^2 + (2*I)*((ArcSinh[2*x]*Log[1 + E^(2*ArcS 
inh[2*x])])/2 + PolyLog[2, -E^(2*ArcSinh[2*x])]/4))
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2620

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]
 

rule 2715

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]
 

rule 2838

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 6212

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

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(-\frac {\operatorname {arcsinh}\left (2 x \right )^{2}}{8}+\frac {\operatorname {arcsinh}\left (2 x \right ) \ln \left (1+\left (2 x +\sqrt {4 x^{2}+1}\right )^{2}\right )}{4}+\frac {\operatorname {polylog}\left (2, -\left (2 x +\sqrt {4 x^{2}+1}\right )^{2}\right )}{8}\)	\(55\)
default	\(-\frac {\operatorname {arcsinh}\left (2 x \right )^{2}}{8}+\frac {\operatorname {arcsinh}\left (2 x \right ) \ln \left (1+\left (2 x +\sqrt {4 x^{2}+1}\right )^{2}\right )}{4}+\frac {\operatorname {polylog}\left (2, -\left (2 x +\sqrt {4 x^{2}+1}\right )^{2}\right )}{8}\)	\(55\)

Input:

int(x*arcsinh(2*x)/(4*x^2+1),x,method=_RETURNVERBOSE)
 

Output:

-1/8*arcsinh(2*x)^2+1/4*arcsinh(2*x)*ln(1+(2*x+(4*x^2+1)^(1/2))^2)+1/8*pol 
ylog(2,-(2*x+(4*x^2+1)^(1/2))^2)
 

Fricas [F]

\[ \int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx=\int { \frac {x \operatorname {arsinh}\left (2 \, x\right )}{4 \, x^{2} + 1} \,d x } \] Input:

integrate(x*arcsinh(2*x)/(4*x^2+1),x, algorithm="fricas")
 

Output:

integral(x*arcsinh(2*x)/(4*x^2 + 1), x)
 

Sympy [F]

\[ \int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx=\int \frac {x \operatorname {asinh}{\left (2 x \right )}}{4 x^{2} + 1}\, dx \] Input:

integrate(x*asinh(2*x)/(4*x**2+1),x)
 

Output:

Integral(x*asinh(2*x)/(4*x**2 + 1), x)
 

Maxima [F]

\[ \int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx=\int { \frac {x \operatorname {arsinh}\left (2 \, x\right )}{4 \, x^{2} + 1} \,d x } \] Input:

integrate(x*arcsinh(2*x)/(4*x^2+1),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

-1/32*log(4*x^2 + 1)^2 + 1/8*log(4*x^2 + 1)*log(2*x + sqrt(4*x^2 + 1)) - i 
ntegrate(1/4*log(4*x^2 + 1)/(8*x^3 + (4*x^2 + 1)^(3/2) + 2*x), x)
 

Giac [F]

\[ \int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx=\int { \frac {x \operatorname {arsinh}\left (2 \, x\right )}{4 \, x^{2} + 1} \,d x } \] Input:

integrate(x*arcsinh(2*x)/(4*x^2+1),x, algorithm="giac")
 

Output:

integrate(x*arcsinh(2*x)/(4*x^2 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx=\int \frac {x\,\mathrm {asinh}\left (2\,x\right )}{4\,x^2+1} \,d x \] Input:

int((x*asinh(2*x))/(4*x^2 + 1),x)
 

Output:

int((x*asinh(2*x))/(4*x^2 + 1), x)
 

Reduce [F]

\[ \int \frac {x \text {arcsinh}(2 x)}{1+4 x^2} \, dx=\int \frac {\mathit {asinh} \left (2 x \right ) x}{4 x^{2}+1}d x \] Input:

int(x*asinh(2*x)/(4*x^2+1),x)
 

Output:

int((asinh(2*x)*x)/(4*x**2 + 1),x)