\(\int \text {arcsinh}(a x)^{5/2} \, dx\) [90]

Optimal result

Integrand size = 8, antiderivative size = 94 \[ \int \text {arcsinh}(a x)^{5/2} \, dx=\frac {15}{4} x \sqrt {\text {arcsinh}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^{3/2}}{2 a}+x \text {arcsinh}(a x)^{5/2}+\frac {15 \sqrt {\pi } \text {erf}\left (\sqrt {\text {arcsinh}(a x)}\right )}{16 a}-\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {\text {arcsinh}(a x)}\right )}{16 a} \]

[Out]

x*arcsinh(a*x)^(5/2)+15/16*erf(arcsinh(a*x)^(1/2))*Pi^(1/2)/a-15/16*erfi(arcsinh(a*x)^(1/2))*Pi^(1/2)/a-5/2*ar
csinh(a*x)^(3/2)*(a^2*x^2+1)^(1/2)/a+15/4*x*arcsinh(a*x)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5772, 5798, 5819, 3389, 2211, 2235, 2236} \[ \int \text {arcsinh}(a x)^{5/2} \, dx=-\frac {5 \sqrt {a^2 x^2+1} \text {arcsinh}(a x)^{3/2}}{2 a}+\frac {15 \sqrt {\pi } \text {erf}\left (\sqrt {\text {arcsinh}(a x)}\right )}{16 a}-\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {\text {arcsinh}(a x)}\right )}{16 a}+x \text {arcsinh}(a x)^{5/2}+\frac {15}{4} x \sqrt {\text {arcsinh}(a x)} \]

[In]

Int[ArcSinh[a*x]^(5/2),x]

[Out]

(15*x*Sqrt[ArcSinh[a*x]])/4 - (5*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))/(2*a) + x*ArcSinh[a*x]^(5/2) + (15*Sqrt
[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(16*a) - (15*Sqrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]]])/(16*a)

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 5772

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[x*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5798

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^
(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] - Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^p], Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e
, c^2*d] && GtQ[n, 0] && NeQ[p, -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}& = x \text {arcsinh}(a x)^{5/2}-\frac {1}{2} (5 a) \int \frac {x \text {arcsinh}(a x)^{3/2}}{\sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {5 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^{3/2}}{2 a}+x \text {arcsinh}(a x)^{5/2}+\frac {15}{4} \int \sqrt {\text {arcsinh}(a x)} \, dx \\ & = \frac {15}{4} x \sqrt {\text {arcsinh}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^{3/2}}{2 a}+x \text {arcsinh}(a x)^{5/2}-\frac {1}{8} (15 a) \int \frac {x}{\sqrt {1+a^2 x^2} \sqrt {\text {arcsinh}(a x)}} \, dx \\ & = \frac {15}{4} x \sqrt {\text {arcsinh}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^{3/2}}{2 a}+x \text {arcsinh}(a x)^{5/2}-\frac {15 \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {x}} \, dx,x,\text {arcsinh}(a x)\right )}{8 a} \\ & = \frac {15}{4} x \sqrt {\text {arcsinh}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^{3/2}}{2 a}+x \text {arcsinh}(a x)^{5/2}+\frac {15 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {x}} \, dx,x,\text {arcsinh}(a x)\right )}{16 a}-\frac {15 \text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\text {arcsinh}(a x)\right )}{16 a} \\ & = \frac {15}{4} x \sqrt {\text {arcsinh}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^{3/2}}{2 a}+x \text {arcsinh}(a x)^{5/2}+\frac {15 \text {Subst}\left (\int e^{-x^2} \, dx,x,\sqrt {\text {arcsinh}(a x)}\right )}{8 a}-\frac {15 \text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\text {arcsinh}(a x)}\right )}{8 a} \\ & = \frac {15}{4} x \sqrt {\text {arcsinh}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)^{3/2}}{2 a}+x \text {arcsinh}(a x)^{5/2}+\frac {15 \sqrt {\pi } \text {erf}\left (\sqrt {\text {arcsinh}(a x)}\right )}{16 a}-\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {\text {arcsinh}(a x)}\right )}{16 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.48 \[ \int \text {arcsinh}(a x)^{5/2} \, dx=-\frac {\frac {\sqrt {-\text {arcsinh}(a x)} \Gamma \left (\frac {7}{2},-\text {arcsinh}(a x)\right )}{\sqrt {\text {arcsinh}(a x)}}+\Gamma \left (\frac {7}{2},\text {arcsinh}(a x)\right )}{2 a} \]

[In]

Integrate[ArcSinh[a*x]^(5/2),x]

[Out]

-1/2*((Sqrt[-ArcSinh[a*x]]*Gamma[7/2, -ArcSinh[a*x]])/Sqrt[ArcSinh[a*x]] + Gamma[7/2, ArcSinh[a*x]])/a

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {-16 \operatorname {arcsinh}\left (a x \right )^{\frac {5}{2}} \sqrt {\pi }\, a x +40 \operatorname {arcsinh}\left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}-60 \sqrt {\operatorname {arcsinh}\left (a x \right )}\, \sqrt {\pi }\, a x -15 \pi \,\operatorname {erf}\left (\sqrt {\operatorname {arcsinh}\left (a x \right )}\right )+15 \pi \,\operatorname {erfi}\left (\sqrt {\operatorname {arcsinh}\left (a x \right )}\right )}{16 \sqrt {\pi }\, a}\)	\(78\)

[In]

int(arcsinh(a*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/16*(-16*arcsinh(a*x)^(5/2)*Pi^(1/2)*a*x+40*arcsinh(a*x)^(3/2)*Pi^(1/2)*(a^2*x^2+1)^(1/2)-60*arcsinh(a*x)^(1
/2)*Pi^(1/2)*a*x-15*Pi*erf(arcsinh(a*x)^(1/2))+15*Pi*erfi(arcsinh(a*x)^(1/2)))/Pi^(1/2)/a

Fricas [F(-2)]

Exception generated. \[ \int \text {arcsinh}(a x)^{5/2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(arcsinh(a*x)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \text {arcsinh}(a x)^{5/2} \, dx=\int \operatorname {asinh}^{\frac {5}{2}}{\left (a x \right )}\, dx \]

[In]

integrate(asinh(a*x)**(5/2),x)

[Out]

Integral(asinh(a*x)**(5/2), x)

Maxima [F]

\[ \int \text {arcsinh}(a x)^{5/2} \, dx=\int { \operatorname {arsinh}\left (a x\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(arcsinh(a*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(arcsinh(a*x)^(5/2), x)

Giac [F(-2)]

Exception generated. \[ \int \text {arcsinh}(a x)^{5/2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(arcsinh(a*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> an error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \text {arcsinh}(a x)^{5/2} \, dx=\int {\mathrm {asinh}\left (a\,x\right )}^{5/2} \,d x \]

[In]

int(asinh(a*x)^(5/2),x)

[Out]

int(asinh(a*x)^(5/2), x)