\(\int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx\) [381]

Optimal result

Integrand size = 23, antiderivative size = 27 \[ \int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx=-\frac {3 \text {Shi}(\text {arcsinh}(a x))}{4 a^4}+\frac {\text {Shi}(3 \text {arcsinh}(a x))}{4 a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5819, 3393, 3379} \[ \int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx=\frac {\text {Shi}(3 \text {arcsinh}(a x))}{4 a^4}-\frac {3 \text {Shi}(\text {arcsinh}(a x))}{4 a^4} \]

[In]

Int[x^3/(Sqrt[1 + a^2*x^2]*ArcSinh[a*x]),x]

[Out]

(-3*SinhIntegral[ArcSinh[a*x]])/(4*a^4) + SinhIntegral[3*ArcSinh[a*x]]/(4*a^4)

Rule 3379

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

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5819

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sinh ^3(x)}{x} \, dx,x,\text {arcsinh}(a x)\right )}{a^4} \\ & = \frac {i \text {Subst}\left (\int \left (\frac {3 i \sinh (x)}{4 x}-\frac {i \sinh (3 x)}{4 x}\right ) \, dx,x,\text {arcsinh}(a x)\right )}{a^4} \\ & = \frac {\text {Subst}\left (\int \frac {\sinh (3 x)}{x} \, dx,x,\text {arcsinh}(a x)\right )}{4 a^4}-\frac {3 \text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\text {arcsinh}(a x)\right )}{4 a^4} \\ & = -\frac {3 \text {Shi}(\text {arcsinh}(a x))}{4 a^4}+\frac {\text {Shi}(3 \text {arcsinh}(a x))}{4 a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx=\frac {-3 \text {Shi}(\text {arcsinh}(a x))+\text {Shi}(3 \text {arcsinh}(a x))}{4 a^4} \]

[In]

Integrate[x^3/(Sqrt[1 + a^2*x^2]*ArcSinh[a*x]),x]

[Out]

(-3*SinhIntegral[ArcSinh[a*x]] + SinhIntegral[3*ArcSinh[a*x]])/(4*a^4)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {3 \,\operatorname {Shi}\left (\operatorname {arcsinh}\left (a x \right )\right )-\operatorname {Shi}\left (3 \,\operatorname {arcsinh}\left (a x \right )\right )}{4 a^{4}}\)	\(23\)

[In]

int(x^3/arcsinh(a*x)/(a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(3*Shi(arcsinh(a*x))-Shi(3*arcsinh(a*x)))/a^4

Fricas [F]

\[ \int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx=\int { \frac {x^{3}}{\sqrt {a^{2} x^{2} + 1} \operatorname {arsinh}\left (a x\right )} \,d x } \]

[In]

integrate(x^3/arcsinh(a*x)/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx=\int \frac {x^{3}}{\sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}\, dx \]

[In]

integrate(x**3/asinh(a*x)/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(x**3/(sqrt(a**2*x**2 + 1)*asinh(a*x)), x)

Maxima [F]

\[ \int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx=\int { \frac {x^{3}}{\sqrt {a^{2} x^{2} + 1} \operatorname {arsinh}\left (a x\right )} \,d x } \]

[In]

integrate(x^3/arcsinh(a*x)/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^3/arcsinh(a*x)/(a^2*x^2+1)^(1/2),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. \[ \int \frac {x^3}{\sqrt {1+a^2 x^2} \text {arcsinh}(a x)} \, dx=\int \frac {x^3}{\mathrm {asinh}\left (a\,x\right )\,\sqrt {a^2\,x^2+1}} \,d x \]

[In]

int(x^3/(asinh(a*x)*(a^2*x^2 + 1)^(1/2)),x)

[Out]

int(x^3/(asinh(a*x)*(a^2*x^2 + 1)^(1/2)), x)