3.112 \(\int \frac {x^3}{\sinh ^{-1}(a x)^{7/2}} \, dx\)

Optimal. Leaf size=229 \[ \frac {16 \sqrt {\pi } \text {erf}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {4 \sqrt {2 \pi } \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}+\frac {16 \sqrt {\pi } \text {erfi}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {4 \sqrt {2 \pi } \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {4 x^2}{5 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {128 x^3 \sqrt {a^2 x^2+1}}{15 a \sqrt {\sinh ^{-1}(a x)}}-\frac {2 x^3 \sqrt {a^2 x^2+1}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {16 x \sqrt {a^2 x^2+1}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{15 \sinh ^{-1}(a x)^{3/2}} \]

[Out]

-4/5*x^2/a^2/arcsinh(a*x)^(3/2)-16/15*x^4/arcsinh(a*x)^(3/2)+16/15*erf(2*arcsinh(a*x)^(1/2))*Pi^(1/2)/a^4+16/1
5*erfi(2*arcsinh(a*x)^(1/2))*Pi^(1/2)/a^4-4/15*erf(2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4-4/15*erfi(
2^(1/2)*arcsinh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4-2/5*x^3*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^(5/2)-16/5*x*(a^2*x^
2+1)^(1/2)/a^3/arcsinh(a*x)^(1/2)-128/15*x^3*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5667, 5774, 5665, 3307, 2180, 2204, 2205} \[ \frac {16 \sqrt {\pi } \text {Erf}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {4 \sqrt {2 \pi } \text {Erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}+\frac {16 \sqrt {\pi } \text {Erfi}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {4 \sqrt {2 \pi } \text {Erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {128 x^3 \sqrt {a^2 x^2+1}}{15 a \sqrt {\sinh ^{-1}(a x)}}-\frac {2 x^3 \sqrt {a^2 x^2+1}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x^2}{5 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x \sqrt {a^2 x^2+1}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {16 x^4}{15 \sinh ^{-1}(a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcSinh[a*x]^(7/2),x]

[Out]

(-2*x^3*Sqrt[1 + a^2*x^2])/(5*a*ArcSinh[a*x]^(5/2)) - (4*x^2)/(5*a^2*ArcSinh[a*x]^(3/2)) - (16*x^4)/(15*ArcSin
h[a*x]^(3/2)) - (16*x*Sqrt[1 + a^2*x^2])/(5*a^3*Sqrt[ArcSinh[a*x]]) - (128*x^3*Sqrt[1 + a^2*x^2])/(15*a*Sqrt[A
rcSinh[a*x]]) + (16*Sqrt[Pi]*Erf[2*Sqrt[ArcSinh[a*x]]])/(15*a^4) - (4*Sqrt[2*Pi]*Erf[Sqrt[2]*Sqrt[ArcSinh[a*x]
]])/(15*a^4) + (16*Sqrt[Pi]*Erfi[2*Sqrt[ArcSinh[a*x]]])/(15*a^4) - (4*Sqrt[2*Pi]*Erfi[Sqrt[2]*Sqrt[ArcSinh[a*x
]]])/(15*a^4)

Rule 2180

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] &&  !$UseGamma === True

Rule 2204

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 2205

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 3307

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 5665

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

Rule 5667

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSinh[c*x])^(n +
1))/Sqrt[1 + c^2*x^2], x], x] - Dist[m/(b*c*(n + 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] && LtQ[n, -2]

Rule 5774

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), 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] && GtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {x^3}{\sinh ^{-1}(a x)^{7/2}} \, dx &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}+\frac {6 \int \frac {x^2}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{5/2}} \, dx}{5 a}+\frac {1}{5} (8 a) \int \frac {x^4}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^{5/2}} \, dx\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x^2}{5 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x^4}{15 \sinh ^{-1}(a x)^{3/2}}+\frac {64}{15} \int \frac {x^3}{\sinh ^{-1}(a x)^{3/2}} \, dx+\frac {8 \int \frac {x}{\sinh ^{-1}(a x)^{3/2}} \, dx}{5 a^2}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x^2}{5 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x^4}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {128 x^3 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}+\frac {16 \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^4}+\frac {128 \operatorname {Subst}\left (\int \left (-\frac {\cosh (2 x)}{2 \sqrt {x}}+\frac {\cosh (4 x)}{2 \sqrt {x}}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{15 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x^2}{5 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x^4}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {128 x^3 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}+\frac {8 \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^4}+\frac {8 \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{5 a^4}-\frac {64 \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a^4}+\frac {64 \operatorname {Subst}\left (\int \frac {\cosh (4 x)}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x^2}{5 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x^4}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {128 x^3 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}+\frac {32 \operatorname {Subst}\left (\int \frac {e^{-4 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a^4}-\frac {32 \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a^4}-\frac {32 \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a^4}+\frac {32 \operatorname {Subst}\left (\int \frac {e^{4 x}}{\sqrt {x}} \, dx,x,\sinh ^{-1}(a x)\right )}{15 a^4}+\frac {16 \operatorname {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{5 a^4}+\frac {16 \operatorname {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{5 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x^2}{5 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x^4}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {128 x^3 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}+\frac {4 \sqrt {2 \pi } \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{5 a^4}+\frac {4 \sqrt {2 \pi } \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{5 a^4}+\frac {64 \operatorname {Subst}\left (\int e^{-4 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {64 \operatorname {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {64 \operatorname {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}+\frac {64 \operatorname {Subst}\left (\int e^{4 x^2} \, dx,x,\sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}\\ &=-\frac {2 x^3 \sqrt {1+a^2 x^2}}{5 a \sinh ^{-1}(a x)^{5/2}}-\frac {4 x^2}{5 a^2 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x^4}{15 \sinh ^{-1}(a x)^{3/2}}-\frac {16 x \sqrt {1+a^2 x^2}}{5 a^3 \sqrt {\sinh ^{-1}(a x)}}-\frac {128 x^3 \sqrt {1+a^2 x^2}}{15 a \sqrt {\sinh ^{-1}(a x)}}+\frac {16 \sqrt {\pi } \text {erf}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {4 \sqrt {2 \pi } \text {erf}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}+\frac {16 \sqrt {\pi } \text {erfi}\left (2 \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}-\frac {4 \sqrt {2 \pi } \text {erfi}\left (\sqrt {2} \sqrt {\sinh ^{-1}(a x)}\right )}{15 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.73, size = 210, normalized size = 0.92 \[ \frac {6 \sinh \left (2 \sinh ^{-1}(a x)\right )-3 \sinh \left (4 \sinh ^{-1}(a x)\right )+4 \sinh ^{-1}(a x) \left (e^{-2 \sinh ^{-1}(a x)} \left (1-4 \sinh ^{-1}(a x)\right )+e^{2 \sinh ^{-1}(a x)} \left (4 \sinh ^{-1}(a x)+1\right )+4 \sqrt {2} \left (-\sinh ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-2 \sinh ^{-1}(a x)\right )+4 \sqrt {2} \sinh ^{-1}(a x)^{3/2} \Gamma \left (\frac {1}{2},2 \sinh ^{-1}(a x)\right )\right )-4 \sinh ^{-1}(a x) \left (e^{-4 \sinh ^{-1}(a x)} \left (1-8 \sinh ^{-1}(a x)\right )+e^{4 \sinh ^{-1}(a x)} \left (8 \sinh ^{-1}(a x)+1\right )+16 \left (-\sinh ^{-1}(a x)\right )^{3/2} \Gamma \left (\frac {1}{2},-4 \sinh ^{-1}(a x)\right )+16 \sinh ^{-1}(a x)^{3/2} \Gamma \left (\frac {1}{2},4 \sinh ^{-1}(a x)\right )\right )}{60 a^4 \sinh ^{-1}(a x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3/ArcSinh[a*x]^(7/2),x]

[Out]

(4*ArcSinh[a*x]*((1 - 4*ArcSinh[a*x])/E^(2*ArcSinh[a*x]) + E^(2*ArcSinh[a*x])*(1 + 4*ArcSinh[a*x]) + 4*Sqrt[2]
*(-ArcSinh[a*x])^(3/2)*Gamma[1/2, -2*ArcSinh[a*x]] + 4*Sqrt[2]*ArcSinh[a*x]^(3/2)*Gamma[1/2, 2*ArcSinh[a*x]])
- 4*ArcSinh[a*x]*((1 - 8*ArcSinh[a*x])/E^(4*ArcSinh[a*x]) + E^(4*ArcSinh[a*x])*(1 + 8*ArcSinh[a*x]) + 16*(-Arc
Sinh[a*x])^(3/2)*Gamma[1/2, -4*ArcSinh[a*x]] + 16*ArcSinh[a*x]^(3/2)*Gamma[1/2, 4*ArcSinh[a*x]]) + 6*Sinh[2*Ar
cSinh[a*x]] - 3*Sinh[4*ArcSinh[a*x]])/(60*a^4*ArcSinh[a*x]^(5/2))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)^(7/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\arcsinh \left (a x \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arcsinh(a*x)^(7/2),x)

[Out]

int(x^3/arcsinh(a*x)^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {arsinh}\left (a x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)^(7/2),x, algorithm="maxima")

[Out]

integrate(x^3/arcsinh(a*x)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3}{{\mathrm {asinh}\left (a\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/asinh(a*x)^(7/2),x)

[Out]

int(x^3/asinh(a*x)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {asinh}^{\frac {7}{2}}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/asinh(a*x)**(7/2),x)

[Out]

Integral(x**3/asinh(a*x)**(7/2), x)

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