\(\int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx\) [54]

Optimal result

Integrand size = 10, antiderivative size = 56 \[ \int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx=-\frac {x^3 \sqrt {1+a^2 x^2}}{a \text {arcsinh}(a x)}-\frac {\text {Chi}(2 \text {arcsinh}(a x))}{2 a^4}+\frac {\text {Chi}(4 \text {arcsinh}(a x))}{2 a^4} \]

[Out]

-1/2*Chi(2*arcsinh(a*x))/a^4+1/2*Chi(4*arcsinh(a*x))/a^4-x^3*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5778, 3382} \[ \int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx=-\frac {\text {Chi}(2 \text {arcsinh}(a x))}{2 a^4}+\frac {\text {Chi}(4 \text {arcsinh}(a x))}{2 a^4}-\frac {x^3 \sqrt {a^2 x^2+1}}{a \text {arcsinh}(a x)} \]

[In]

Int[x^3/ArcSinh[a*x]^2,x]

[Out]

-((x^3*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x])) - CoshIntegral[2*ArcSinh[a*x]]/(2*a^4) + CoshIntegral[4*ArcSinh[a*
x]]/(2*a^4)

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 5778

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^m*Sqrt[1 + c^2*x^2]*((a + b*ArcSi
nh[c*x])^(n + 1)/(b*c*(n + 1))), x] - Dist[1/(b^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}
, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^3 \sqrt {1+a^2 x^2}}{a \text {arcsinh}(a x)}+\frac {\text {Subst}\left (\int \left (-\frac {\cosh (2 x)}{2 x}+\frac {\cosh (4 x)}{2 x}\right ) \, dx,x,\text {arcsinh}(a x)\right )}{a^4} \\ & = -\frac {x^3 \sqrt {1+a^2 x^2}}{a \text {arcsinh}(a x)}-\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\text {arcsinh}(a x)\right )}{2 a^4}+\frac {\text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\text {arcsinh}(a x)\right )}{2 a^4} \\ & = -\frac {x^3 \sqrt {1+a^2 x^2}}{a \text {arcsinh}(a x)}-\frac {\text {Chi}(2 \text {arcsinh}(a x))}{2 a^4}+\frac {\text {Chi}(4 \text {arcsinh}(a x))}{2 a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx=-\frac {4 \text {arcsinh}(a x) \text {Chi}(2 \text {arcsinh}(a x))-4 \text {arcsinh}(a x) \text {Chi}(4 \text {arcsinh}(a x))-2 \sinh (2 \text {arcsinh}(a x))+\sinh (4 \text {arcsinh}(a x))}{8 a^4 \text {arcsinh}(a x)} \]

[In]

Integrate[x^3/ArcSinh[a*x]^2,x]

[Out]

-1/8*(4*ArcSinh[a*x]*CoshIntegral[2*ArcSinh[a*x]] - 4*ArcSinh[a*x]*CoshIntegral[4*ArcSinh[a*x]] - 2*Sinh[2*Arc
Sinh[a*x]] + Sinh[4*ArcSinh[a*x]])/(a^4*ArcSinh[a*x])

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {\sinh \left (2 \,\operatorname {arcsinh}\left (a x \right )\right )}{4 \,\operatorname {arcsinh}\left (a x \right )}-\frac {\operatorname {Chi}\left (2 \,\operatorname {arcsinh}\left (a x \right )\right )}{2}-\frac {\sinh \left (4 \,\operatorname {arcsinh}\left (a x \right )\right )}{8 \,\operatorname {arcsinh}\left (a x \right )}+\frac {\operatorname {Chi}\left (4 \,\operatorname {arcsinh}\left (a x \right )\right )}{2}}{a^{4}}\)	\(54\)
default	\(\frac {\frac {\sinh \left (2 \,\operatorname {arcsinh}\left (a x \right )\right )}{4 \,\operatorname {arcsinh}\left (a x \right )}-\frac {\operatorname {Chi}\left (2 \,\operatorname {arcsinh}\left (a x \right )\right )}{2}-\frac {\sinh \left (4 \,\operatorname {arcsinh}\left (a x \right )\right )}{8 \,\operatorname {arcsinh}\left (a x \right )}+\frac {\operatorname {Chi}\left (4 \,\operatorname {arcsinh}\left (a x \right )\right )}{2}}{a^{4}}\)	\(54\)

[In]

int(x^3/arcsinh(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/4/arcsinh(a*x)*sinh(2*arcsinh(a*x))-1/2*Chi(2*arcsinh(a*x))-1/8/arcsinh(a*x)*sinh(4*arcsinh(a*x))+1/2
*Chi(4*arcsinh(a*x)))

Fricas [F]

\[ \int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx=\int { \frac {x^{3}}{\operatorname {arsinh}\left (a x\right )^{2}} \,d x } \]

[In]

integrate(x^3/arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

integral(x^3/arcsinh(a*x)^2, x)

Sympy [F]

\[ \int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx=\int \frac {x^{3}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx \]

[In]

integrate(x**3/asinh(a*x)**2,x)

[Out]

Integral(x**3/asinh(a*x)**2, x)

Maxima [F]

\[ \int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx=\int { \frac {x^{3}}{\operatorname {arsinh}\left (a x\right )^{2}} \,d x } \]

[In]

integrate(x^3/arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

-(a^3*x^6 + a*x^4 + (a^2*x^5 + x^3)*sqrt(a^2*x^2 + 1))/((a^3*x^2 + sqrt(a^2*x^2 + 1)*a^2*x + a)*log(a*x + sqrt
(a^2*x^2 + 1))) + integrate((4*a^5*x^7 + 8*a^3*x^5 + 4*a*x^3 + 2*(2*a^3*x^5 + a*x^3)*(a^2*x^2 + 1) + (8*a^4*x^
6 + 10*a^2*x^4 + 3*x^2)*sqrt(a^2*x^2 + 1))/((a^5*x^4 + (a^2*x^2 + 1)*a^3*x^2 + 2*a^3*x^2 + 2*(a^4*x^3 + a^2*x)
*sqrt(a^2*x^2 + 1) + a)*log(a*x + sqrt(a^2*x^2 + 1))), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^3/arcsinh(a*x)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\text {arcsinh}(a x)^2} \, dx=\int \frac {x^3}{{\mathrm {asinh}\left (a\,x\right )}^2} \,d x \]

[In]

int(x^3/asinh(a*x)^2,x)

[Out]

int(x^3/asinh(a*x)^2, x)