\(\int \frac {\text {arcsinh}(a x)^2}{x \sqrt {1+a^2 x^2}} \, dx\) [293]

Optimal result

Integrand size = 23, antiderivative size = 68 \[ \int \frac {\text {arcsinh}(a x)^2}{x \sqrt {1+a^2 x^2}} \, dx=-2 \text {arcsinh}(a x)^2 \text {arctanh}\left (e^{\text {arcsinh}(a x)}\right )-2 \text {arcsinh}(a x) \operatorname {PolyLog}\left (2,-e^{\text {arcsinh}(a x)}\right )+2 \text {arcsinh}(a x) \operatorname {PolyLog}\left (2,e^{\text {arcsinh}(a x)}\right )+2 \operatorname {PolyLog}\left (3,-e^{\text {arcsinh}(a x)}\right )-2 \operatorname {PolyLog}\left (3,e^{\text {arcsinh}(a x)}\right ) \] Output:

-2*arcsinh(a*x)^2*arctanh(a*x+(a^2*x^2+1)^(1/2))-2*arcsinh(a*x)*polylog(2, 
-a*x-(a^2*x^2+1)^(1/2))+2*arcsinh(a*x)*polylog(2,a*x+(a^2*x^2+1)^(1/2))+2* 
polylog(3,-a*x-(a^2*x^2+1)^(1/2))-2*polylog(3,a*x+(a^2*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.47 \[ \int \frac {\text {arcsinh}(a x)^2}{x \sqrt {1+a^2 x^2}} \, dx=\text {arcsinh}(a x)^2 \log \left (1-e^{-\text {arcsinh}(a x)}\right )-\text {arcsinh}(a x)^2 \log \left (1+e^{-\text {arcsinh}(a x)}\right )+2 \text {arcsinh}(a x) \operatorname {PolyLog}\left (2,-e^{-\text {arcsinh}(a x)}\right )-2 \text {arcsinh}(a x) \operatorname {PolyLog}\left (2,e^{-\text {arcsinh}(a x)}\right )+2 \operatorname {PolyLog}\left (3,-e^{-\text {arcsinh}(a x)}\right )-2 \operatorname {PolyLog}\left (3,e^{-\text {arcsinh}(a x)}\right ) \] Input:

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

Output:

ArcSinh[a*x]^2*Log[1 - E^(-ArcSinh[a*x])] - ArcSinh[a*x]^2*Log[1 + E^(-Arc 
Sinh[a*x])] + 2*ArcSinh[a*x]*PolyLog[2, -E^(-ArcSinh[a*x])] - 2*ArcSinh[a* 
x]*PolyLog[2, E^(-ArcSinh[a*x])] + 2*PolyLog[3, -E^(-ArcSinh[a*x])] - 2*Po 
lyLog[3, E^(-ArcSinh[a*x])]
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.85 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6231, 3042, 26, 4670, 3011, 2720, 7143}

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.

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

\(\Big \downarrow \) 6231

\(\displaystyle \int \frac {\text {arcsinh}(a x)^2}{a x}d\text {arcsinh}(a x)\)

\(\Big \downarrow \) 3042

\(\displaystyle \int i \text {arcsinh}(a x)^2 \csc (i \text {arcsinh}(a x))d\text {arcsinh}(a x)\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \text {arcsinh}(a x)^2 \csc (i \text {arcsinh}(a x))d\text {arcsinh}(a x)\)

\(\Big \downarrow \) 4670

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

\(\Big \downarrow \) 3011

\(\displaystyle i \left (-2 i \left (\int \operatorname {PolyLog}\left (2,-e^{\text {arcsinh}(a x)}\right )d\text {arcsinh}(a x)-\text {arcsinh}(a x) \operatorname {PolyLog}\left (2,-e^{\text {arcsinh}(a x)}\right )\right )+2 i \left (\int \operatorname {PolyLog}\left (2,e^{\text {arcsinh}(a x)}\right )d\text {arcsinh}(a x)-\text {arcsinh}(a x) \operatorname {PolyLog}\left (2,e^{\text {arcsinh}(a x)}\right )\right )+2 i \text {arcsinh}(a x)^2 \text {arctanh}\left (e^{\text {arcsinh}(a x)}\right )\right )\)

\(\Big \downarrow \) 2720

\(\displaystyle i \left (-2 i \left (\int e^{-\text {arcsinh}(a x)} \operatorname {PolyLog}\left (2,-e^{\text {arcsinh}(a x)}\right )de^{\text {arcsinh}(a x)}-\text {arcsinh}(a x) \operatorname {PolyLog}\left (2,-e^{\text {arcsinh}(a x)}\right )\right )+2 i \left (\int e^{-\text {arcsinh}(a x)} \operatorname {PolyLog}\left (2,e^{\text {arcsinh}(a x)}\right )de^{\text {arcsinh}(a x)}-\text {arcsinh}(a x) \operatorname {PolyLog}\left (2,e^{\text {arcsinh}(a x)}\right )\right )+2 i \text {arcsinh}(a x)^2 \text {arctanh}\left (e^{\text {arcsinh}(a x)}\right )\right )\)

\(\Big \downarrow \) 7143

\(\displaystyle i \left (2 i \text {arcsinh}(a x)^2 \text {arctanh}\left (e^{\text {arcsinh}(a x)}\right )-2 i \left (\operatorname {PolyLog}\left (3,-e^{\text {arcsinh}(a x)}\right )-\text {arcsinh}(a x) \operatorname {PolyLog}\left (2,-e^{\text {arcsinh}(a x)}\right )\right )+2 i \left (\operatorname {PolyLog}\left (3,e^{\text {arcsinh}(a x)}\right )-\text {arcsinh}(a x) \operatorname {PolyLog}\left (2,e^{\text {arcsinh}(a x)}\right )\right )\right )\)

Input:

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

Output:

I*((2*I)*ArcSinh[a*x]^2*ArcTanh[E^ArcSinh[a*x]] - (2*I)*(-(ArcSinh[a*x]*Po 
lyLog[2, -E^ArcSinh[a*x]]) + PolyLog[3, -E^ArcSinh[a*x]]) + (2*I)*(-(ArcSi 
nh[a*x]*PolyLog[2, E^ArcSinh[a*x]]) + PolyLog[3, E^ArcSinh[a*x]]))
 

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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4670

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x 
_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] 
 + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*fz*x 
)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e 
+ f*fz*x)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 6231

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

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [A] (verified)

Time = 1.39 (sec) , antiderivative size = 144, normalized size of antiderivative = 2.12

method	result	size
default	\(\operatorname {arcsinh}\left (x a \right )^{2} \ln \left (1-x a -\sqrt {a^{2} x^{2}+1}\right )+2 \,\operatorname {arcsinh}\left (x a \right ) \operatorname {polylog}\left (2, x a +\sqrt {a^{2} x^{2}+1}\right )-2 \operatorname {polylog}\left (3, x a +\sqrt {a^{2} x^{2}+1}\right )-\operatorname {arcsinh}\left (x a \right )^{2} \ln \left (1+x a +\sqrt {a^{2} x^{2}+1}\right )-2 \,\operatorname {arcsinh}\left (x a \right ) \operatorname {polylog}\left (2, -x a -\sqrt {a^{2} x^{2}+1}\right )+2 \operatorname {polylog}\left (3, -x a -\sqrt {a^{2} x^{2}+1}\right )\)	\(144\)

Input:

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

Output:

arcsinh(x*a)^2*ln(1-x*a-(a^2*x^2+1)^(1/2))+2*arcsinh(x*a)*polylog(2,x*a+(a 
^2*x^2+1)^(1/2))-2*polylog(3,x*a+(a^2*x^2+1)^(1/2))-arcsinh(x*a)^2*ln(1+x* 
a+(a^2*x^2+1)^(1/2))-2*arcsinh(x*a)*polylog(2,-x*a-(a^2*x^2+1)^(1/2))+2*po 
lylog(3,-x*a-(a^2*x^2+1)^(1/2))
 

Fricas [F]

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

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

Output:

integral(sqrt(a^2*x^2 + 1)*arcsinh(a*x)^2/(a^2*x^3 + x), x)
 

Sympy [F]

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

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

Output:

Integral(asinh(a*x)**2/(x*sqrt(a**2*x**2 + 1)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(asinh(a*x)**2/(sqrt(a**2*x**2 + 1)*x),x)