3.2.15 \(\int \frac {1}{\sinh ^{-1}(a x)^{7/2}} \, dx\) [115]

Optimal. Leaf size=112 \[ -\frac {2 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {8 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}-\frac {4 \sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{15 a}+\frac {4 \sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{15 a} \]

[Out]

-4/15*x/arcsinh(a*x)^(3/2)-4/15*erf(arcsinh(a*x)^(1/2))*Pi^(1/2)/a+4/15*erfi(arcsinh(a*x)^(1/2))*Pi^(1/2)/a-2/
5*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^(5/2)-8/15*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 112, 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 = {5773, 5818, 5819, 3389, 2211, 2235, 2236} \begin {gather*} -\frac {8 \sqrt {a^2 x^2+1}}{15 a \sqrt {\sinh ^{-1}(a x)}}-\frac {2 \sqrt {a^2 x^2+1}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 \sqrt {\pi } \text {Erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{15 a}+\frac {4 \sqrt {\pi } \text {Erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{15 a}-\frac {4 x}{15 \sinh ^{-1}(a x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^(-7/2),x]

[Out]

(-2*Sqrt[1 + a^2*x^2])/(5*a*ArcSinh[a*x]^(5/2)) - (4*x)/(15*ArcSinh[a*x]^(3/2)) - (8*Sqrt[1 + a^2*x^2])/(15*a*
Sqrt[ArcSinh[a*x]]) - (4*Sqrt[Pi]*Erf[Sqrt[ArcSinh[a*x]]])/(15*a) + (4*Sqrt[Pi]*Erfi[Sqrt[ArcSinh[a*x]]])/(15*
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 5773

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

Rule 5818

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Dist[f*(m/
(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[(f*x)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x]
/; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -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 \frac {1}{\sinh ^{-1}(a x)^{7/2}} \, dx &=-\frac {2 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}+\frac {1}{5} (2 a) \int \frac {x}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{5/2}} \, dx\\ &=-\frac {2 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x}{15 \sinh ^{-1}(a x)^{3/2}}+\frac {4}{15} \int \frac {1}{\sinh ^{-1}(a x)^{3/2}} \, dx\\ &=-\frac {2 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {8 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}+\frac {1}{15} (8 a) \int \frac {x}{\sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}} \, dx\\ &=-\frac {2 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {8 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}+\frac {8 \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a}\\ &=-\frac {2 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {8 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}-\frac {4 \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a}+\frac {4 \text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a}\\ &=-\frac {2 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {8 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}-\frac {8 \text {Subst}\left (\int e^{-x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{15 a}+\frac {8 \text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{15 a}\\ &=-\frac {2 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {8 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}-\frac {4 \sqrt {\pi } \text {erf}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{15 a}+\frac {4 \sqrt {\pi } \text {erfi}\left (\sqrt {\sinh ^{-1}(a x)}\right )}{15 a}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 111, normalized size = 0.99 \begin {gather*} \frac {-2 e^{\sinh ^{-1}(a x)} \left (3+2 \sinh ^{-1}(a x)+4 \sinh ^{-1}(a x)^2\right )+8 \left (-\sinh ^{-1}(a x)\right )^{5/2} \Gamma \left (\frac {1}{2},-\sinh ^{-1}(a x)\right )+e^{-\sinh ^{-1}(a x)} \left (-6+4 \sinh ^{-1}(a x)-8 \sinh ^{-1}(a x)^2+8 e^{\sinh ^{-1}(a x)} \sinh ^{-1}(a x)^{5/2} \Gamma \left (\frac {1}{2},\sinh ^{-1}(a x)\right )\right )}{30 a \sinh ^{-1}(a x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]^(-7/2),x]

[Out]

(-2*E^ArcSinh[a*x]*(3 + 2*ArcSinh[a*x] + 4*ArcSinh[a*x]^2) + 8*(-ArcSinh[a*x])^(5/2)*Gamma[1/2, -ArcSinh[a*x]]
+ (-6 + 4*ArcSinh[a*x] - 8*ArcSinh[a*x]^2 + 8*E^ArcSinh[a*x]*ArcSinh[a*x]^(5/2)*Gamma[1/2, ArcSinh[a*x]])/E^A
rcSinh[a*x])/(30*a*ArcSinh[a*x]^(5/2))

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Maple [A]
time = 2.73, size = 105, normalized size = 0.94

method	result	size
default	\(-\frac {2 \left (2 \arcsinh \left (a x \right )^{3} \pi \erf \left (\sqrt {\arcsinh \left (a x \right )}\right )-2 \arcsinh \left (a x \right )^{3} \pi \erfi \left (\sqrt {\arcsinh \left (a x \right )}\right )+4 \arcsinh \left (a x \right )^{\frac {5}{2}} \sqrt {a^{2} x^{2}+1}\, \sqrt {\pi }+2 \arcsinh \left (a x \right )^{\frac {3}{2}} \sqrt {\pi }\, a x +3 \sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}\right )}{15 \sqrt {\pi }\, a \arcsinh \left (a x \right )^{3}}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arcsinh(a*x)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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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(1/arcsinh(a*x)^(7/2),x, algorithm="maxima")

[Out]

integrate(arcsinh(a*x)^(-7/2), 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(1/arcsinh(a*x)^(7/2),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 \frac {1}{\operatorname {asinh}^{\frac {7}{2}}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/asinh(a*x)**(7/2),x)

[Out]

Integral(asinh(a*x)**(-7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arcsinh(a*x)^(7/2),x, algorithm="giac")

[Out]

integrate(arcsinh(a*x)^(-7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {asinh}\left (a\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/asinh(a*x)^(7/2),x)

[Out]

int(1/asinh(a*x)^(7/2), x)

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