3.104 \(\int \frac {x^3}{\cosh ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=172 \[ -\frac {2 \sqrt {\pi } \text {erf}\left (2 \sqrt {\cosh ^{-1}(a x)}\right )}{3 a^4}-\frac {\sqrt {2 \pi } \text {erf}\left (\sqrt {2} \sqrt {\cosh ^{-1}(a x)}\right )}{3 a^4}+\frac {2 \sqrt {\pi } \text {erfi}\left (2 \sqrt {\cosh ^{-1}(a x)}\right )}{3 a^4}+\frac {\sqrt {2 \pi } \text {erfi}\left (\sqrt {2} \sqrt {\cosh ^{-1}(a x)}\right )}{3 a^4}+\frac {4 x^2}{a^2 \sqrt {\cosh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\cosh ^{-1}(a x)}}-\frac {2 x^3 \sqrt {a x-1} \sqrt {a x+1}}{3 a \cosh ^{-1}(a x)^{3/2}} \]

[Out]

-2/3*erf(2*arccosh(a*x)^(1/2))*Pi^(1/2)/a^4+2/3*erfi(2*arccosh(a*x)^(1/2))*Pi^(1/2)/a^4-1/3*erf(2^(1/2)*arccos
h(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4+1/3*erfi(2^(1/2)*arccosh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a^4-2/3*x^3*(a*x-1)^(
1/2)*(a*x+1)^(1/2)/a/arccosh(a*x)^(3/2)+4*x^2/a^2/arccosh(a*x)^(1/2)-16/3*x^4/arccosh(a*x)^(1/2)

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Rubi [A]  time = 0.76, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {5668, 5775, 5670, 5448, 3308, 2180, 2204, 2205, 12} \[ -\frac {2 \sqrt {\pi } \text {Erf}\left (2 \sqrt {\cosh ^{-1}(a x)}\right )}{3 a^4}-\frac {\sqrt {2 \pi } \text {Erf}\left (\sqrt {2} \sqrt {\cosh ^{-1}(a x)}\right )}{3 a^4}+\frac {2 \sqrt {\pi } \text {Erfi}\left (2 \sqrt {\cosh ^{-1}(a x)}\right )}{3 a^4}+\frac {\sqrt {2 \pi } \text {Erfi}\left (\sqrt {2} \sqrt {\cosh ^{-1}(a x)}\right )}{3 a^4}+\frac {4 x^2}{a^2 \sqrt {\cosh ^{-1}(a x)}}-\frac {16 x^4}{3 \sqrt {\cosh ^{-1}(a x)}}-\frac {2 x^3 \sqrt {a x-1} \sqrt {a x+1}}{3 a \cosh ^{-1}(a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcCosh[a*x]^(5/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3308

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 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 5668

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

Rule 5670

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Cosh[x]^m*Sinh[x], x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5775

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_
.)*(x_)]), x_Symbol] :> Simp[((f*x)^m*(a + b*ArcCosh[c*x])^(n + 1))/(b*c*Sqrt[-(d1*d2)]*(n + 1)), x] - Dist[(f
*m)/(b*c*Sqrt[-(d1*d2)]*(n + 1)), Int[(f*x)^(m - 1)*(a + b*ArcCosh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d1
, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && LtQ[n, -1] && GtQ[d1, 0] && LtQ[d2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.80, size = 175, normalized size = 1.02 \[ \frac {-\sinh \left (4 \cosh ^{-1}(a x)\right )-4 \cosh ^{-1}(a x) \left (e^{-4 \cosh ^{-1}(a x)}+e^{4 \cosh ^{-1}(a x)}-2 \sqrt {-\cosh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-4 \cosh ^{-1}(a x)\right )-2 \sqrt {\cosh ^{-1}(a x)} \Gamma \left (\frac {1}{2},4 \cosh ^{-1}(a x)\right )\right )-2 \left (\sinh \left (2 \cosh ^{-1}(a x)\right )+2 \cosh ^{-1}(a x) \left (e^{-2 \cosh ^{-1}(a x)}+e^{2 \cosh ^{-1}(a x)}-\sqrt {2} \sqrt {-\cosh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 \cosh ^{-1}(a x)\right )-\sqrt {2} \sqrt {\cosh ^{-1}(a x)} \Gamma \left (\frac {1}{2},2 \cosh ^{-1}(a x)\right )\right )\right )}{12 a^4 \cosh ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3/ArcCosh[a*x]^(5/2),x]

[Out]

(-4*ArcCosh[a*x]*(E^(-4*ArcCosh[a*x]) + E^(4*ArcCosh[a*x]) - 2*Sqrt[-ArcCosh[a*x]]*Gamma[1/2, -4*ArcCosh[a*x]]
- 2*Sqrt[ArcCosh[a*x]]*Gamma[1/2, 4*ArcCosh[a*x]]) - 2*(2*ArcCosh[a*x]*(E^(-2*ArcCosh[a*x]) + E^(2*ArcCosh[a*
x]) - Sqrt[2]*Sqrt[-ArcCosh[a*x]]*Gamma[1/2, -2*ArcCosh[a*x]] - Sqrt[2]*Sqrt[ArcCosh[a*x]]*Gamma[1/2, 2*ArcCos
h[a*x]]) + Sinh[2*ArcCosh[a*x]]) - Sinh[4*ArcCosh[a*x]])/(12*a^4*ArcCosh[a*x]^(3/2))

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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/arccosh(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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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/arccosh(a*x)^(5/2),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 \[ \int \frac {x^{3}}{\mathrm {arccosh}\left (a x \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arccosh(a*x)^(5/2),x)

[Out]

int(x^3/arccosh(a*x)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {arcosh}\left (a x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arccosh(a*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3/arccosh(a*x)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3}{{\mathrm {acosh}\left (a\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/acosh(a*x)^(5/2),x)

[Out]

int(x^3/acosh(a*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {acosh}^{\frac {5}{2}}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/acosh(a*x)**(5/2),x)

[Out]

Integral(x**3/acosh(a*x)**(5/2), x)

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