3.1.90 \(\int \sinh ^{-1}(a x)^{5/2} \, dx\) [90]

Optimal. Leaf size=94 \[ \frac {15}{4} x \sqrt {\sinh ^{-1}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac {15 \sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{16 a}-\frac {15 \sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{16 a} \]

[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)

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Rubi [A]
time = 0.12, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5772, 5798, 5819, 3389, 2211, 2235, 2236} \begin {gather*} -\frac {5 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{2 a}+\frac {15 \sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{16 a}-\frac {15 \sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{16 a}+x \sinh ^{-1}(a x)^{5/2}+\frac {15}{4} x \sqrt {\sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \sinh ^{-1}(a x)^{5/2} \, dx &=x \sinh ^{-1}(a x)^{5/2}-\frac {1}{2} (5 a) \int \frac {x \sinh ^{-1}(a x)^{3/2}}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {5 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac {15}{4} \int \sqrt {\sinh ^{-1}(a x)} \, dx\\ &=\frac {15}{4} x \sqrt {\sinh ^{-1}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}-\frac {1}{8} (15 a) \int \frac {x}{\sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}} \, dx\\ &=\frac {15}{4} x \sqrt {\sinh ^{-1}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}-\frac {15 \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a}\\ &=\frac {15}{4} x \sqrt {\sinh ^{-1}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac {15 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a}-\frac {15 \text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{16 a}\\ &=\frac {15}{4} x \sqrt {\sinh ^{-1}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac {15 \text {Subst}\left (\int e^{-x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{8 a}-\frac {15 \text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{8 a}\\ &=\frac {15}{4} x \sqrt {\sinh ^{-1}(a x)}-\frac {5 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{2 a}+x \sinh ^{-1}(a x)^{5/2}+\frac {15 \sqrt {\pi } \text {erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{16 a}-\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{16 a}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 45, normalized size = 0.48 \begin {gather*} -\frac {\frac {\sqrt {-\sinh ^{-1}(a x)} \Gamma \left (\frac {7}{2},-\sinh ^{-1}(a x)\right )}{\sqrt {\sinh ^{-1}(a x)}}+\Gamma \left (\frac {7}{2},\sinh ^{-1}(a x)\right )}{2 a} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 2.53, size = 78, normalized size = 0.83

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x)^(5/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \operatorname {asinh}^{\frac {5}{2}}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**(5/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {asinh}\left (a\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(a*x)^(5/2), x)

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