3.1.89 \(\int x \sinh ^{-1}(a x)^{5/2} \, dx\) [89]

Optimal. Leaf size=152 \[ \frac {15 \sqrt {\sinh ^{-1}(a x)}}{64 a^2}+\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac {15 \sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac {15 \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{256 a^2} \]

[Out]

1/4*arcsinh(a*x)^(5/2)/a^2+1/2*x^2*arcsinh(a*x)^(5/2)-15/512*erf(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/
a^2-15/512*erfi(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^2-5/8*x*arcsinh(a*x)^(3/2)*(a^2*x^2+1)^(1/2)/a+
15/64*arcsinh(a*x)^(1/2)/a^2+15/32*x^2*arcsinh(a*x)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {5777, 5812, 5783, 5819, 3393, 3388, 2211, 2235, 2236} \begin {gather*} -\frac {15 \sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac {15 \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac {5 x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {15 \sqrt {\sinh ^{-1}(a x)}}{64 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}+\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*Sqrt[ArcSinh[a*x]])/(64*a^2) + (15*x^2*Sqrt[ArcSinh[a*x]])/32 - (5*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^(3/2))
/(8*a) + ArcSinh[a*x]^(5/2)/(4*a^2) + (x^2*ArcSinh[a*x]^(5/2))/2 - (15*Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x
]]])/(256*a^2) - (15*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x]]])/(256*a^2)

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 3388

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

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 5777

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

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5812

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

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 x \sinh ^{-1}(a x)^{5/2} \, dx &=\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac {1}{4} (5 a) \int \frac {x^2 \sinh ^{-1}(a x)^{3/2}}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}+\frac {15}{16} \int x \sqrt {\sinh ^{-1}(a x)} \, dx+\frac {5 \int \frac {\sinh ^{-1}(a x)^{3/2}}{\sqrt {1+a^2 x^2}} \, dx}{8 a}\\ &=\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac {1}{64} (15 a) \int \frac {x^2}{\sqrt {1+a^2 x^2} \sqrt {\sinh ^{-1}(a x)}} \, dx\\ &=\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac {15 \text {Subst}\left (\int \frac {\sinh ^2(x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^2}\\ &=\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}+\frac {15 \text {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}-\frac {\cosh (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{64 a^2}\\ &=\frac {15 \sqrt {\sinh ^{-1}(a x)}}{64 a^2}+\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac {15 \text {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{128 a^2}\\ &=\frac {15 \sqrt {\sinh ^{-1}(a x)}}{64 a^2}+\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac {15 \text {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{256 a^2}-\frac {15 \text {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{256 a^2}\\ &=\frac {15 \sqrt {\sinh ^{-1}(a x)}}{64 a^2}+\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac {15 \text {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{128 a^2}-\frac {15 \text {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{128 a^2}\\ &=\frac {15 \sqrt {\sinh ^{-1}(a x)}}{64 a^2}+\frac {15}{32} x^2 \sqrt {\sinh ^{-1}(a x)}-\frac {5 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{3/2}}{8 a}+\frac {\sinh ^{-1}(a x)^{5/2}}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^{5/2}-\frac {15 \sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{256 a^2}-\frac {15 \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{256 a^2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 52, normalized size = 0.34 \begin {gather*} \frac {\frac {\sqrt {\sinh ^{-1}(a x)} \Gamma \left (\frac {7}{2},-2 \sinh ^{-1}(a x)\right )}{\sqrt {-\sinh ^{-1}(a x)}}+\Gamma \left (\frac {7}{2},2 \sinh ^{-1}(a x)\right )}{32 \sqrt {2} a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[ArcSinh[a*x]]*Gamma[7/2, -2*ArcSinh[a*x]])/Sqrt[-ArcSinh[a*x]] + Gamma[7/2, 2*ArcSinh[a*x]])/(32*Sqrt[2
]*a^2)

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Maple [A]
time = 2.46, size = 136, normalized size = 0.89

method	result	size
default	\(-\frac {\sqrt {2}\, \left (-128 \arcsinh \left (a x \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }\, a^{2} x^{2}+160 \arcsinh \left (a x \right )^{\frac {3}{2}} \sqrt {2}\, \sqrt {\pi }\, \sqrt {a^{2} x^{2}+1}\, a x -120 \sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }\, a^{2} x^{2}-64 \arcsinh \left (a x \right )^{\frac {5}{2}} \sqrt {2}\, \sqrt {\pi }-60 \sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\, \sqrt {\pi }+15 \pi \erf \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )+15 \pi \erfi \left (\sqrt {2}\, \sqrt {\arcsinh \left (a x \right )}\right )\right )}{512 \sqrt {\pi }\, a^{2}}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/512*2^(1/2)*(-128*arcsinh(a*x)^(5/2)*2^(1/2)*Pi^(1/2)*a^2*x^2+160*arcsinh(a*x)^(3/2)*2^(1/2)*Pi^(1/2)*(a^2*
x^2+1)^(1/2)*a*x-120*2^(1/2)*arcsinh(a*x)^(1/2)*Pi^(1/2)*a^2*x^2-64*arcsinh(a*x)^(5/2)*2^(1/2)*Pi^(1/2)-60*2^(
1/2)*arcsinh(a*x)^(1/2)*Pi^(1/2)+15*Pi*erf(2^(1/2)*arcsinh(a*x)^(1/2))+15*Pi*erfi(2^(1/2)*arcsinh(a*x)^(1/2)))
/Pi^(1/2)/a^2

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

[Out]

integrate(x*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(x*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 x \operatorname {asinh}^{\frac {5}{2}}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*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(x*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 x\,{\mathrm {asinh}\left (a\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*asinh(a*x)^(5/2), x)

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