3.280 \(\int \frac {\sqrt {\tanh ^{-1}(a x)}}{(1-a^2 x^2)^2} \, dx\)

Optimal. Leaf size=103 \[ \frac {x \sqrt {\tanh ^{-1}(a x)}}{2 \left (1-a^2 x^2\right )}+\frac {\sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}-\frac {\sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}+\frac {\tanh ^{-1}(a x)^{3/2}}{3 a} \]

[Out]

1/3*arctanh(a*x)^(3/2)/a+1/32*erf(2^(1/2)*arctanh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a-1/32*erfi(2^(1/2)*arctanh(a*x
)^(1/2))*2^(1/2)*Pi^(1/2)/a+1/2*x*arctanh(a*x)^(1/2)/(-a^2*x^2+1)

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Rubi [A]  time = 0.14, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {5956, 6034, 5448, 12, 3308, 2180, 2204, 2205} \[ \frac {x \sqrt {\tanh ^{-1}(a x)}}{2 \left (1-a^2 x^2\right )}+\frac {\sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}-\frac {\sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}+\frac {\tanh ^{-1}(a x)^{3/2}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[a*x]]/(1 - a^2*x^2)^2,x]

[Out]

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

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 5956

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*(a + b*ArcTanh[c*x
])^p)/(2*d*(d + e*x^2)), x] + (-Dist[(b*c*p)/2, Int[(x*(a + b*ArcTanh[c*x])^(p - 1))/(d + e*x^2)^2, x], x] + S
imp[(a + b*ArcTanh[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] &&
GtQ[p, 0]

Rule 6034

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c^(
m + 1), Subst[Int[((a + b*x)^p*Sinh[x]^m)/Cosh[x]^(m + 2*(q + 1)), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c,
d, e, p}, x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] && (IntegerQ[q] || GtQ[d, 0])

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 87, normalized size = 0.84 \[ \sqrt {\tanh ^{-1}(a x)} \left (\frac {\tanh ^{-1}(a x)}{3 a}-\frac {x}{2 \left (a^2 x^2-1\right )}\right )-\frac {\sqrt {\frac {\pi }{2}} \left (\text {erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )-\text {erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )\right )}{16 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[ArcTanh[a*x]]/(1 - a^2*x^2)^2,x]

[Out]

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

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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(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {artanh}\left (a x\right )}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^2,x, algorithm="giac")

[Out]

integrate(sqrt(arctanh(a*x))/(a^2*x^2 - 1)^2, x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\arctanh \left (a x \right )}}{\left (-a^{2} x^{2}+1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^2,x)

[Out]

int(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {artanh}\left (a x\right )}}{{\left (a^{2} x^{2} - 1\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^2,x, algorithm="maxima")

[Out]

integrate(sqrt(arctanh(a*x))/(a^2*x^2 - 1)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^(1/2)/(a^2*x^2 - 1)^2,x)

[Out]

int(atanh(a*x)^(1/2)/(a^2*x^2 - 1)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {atanh}{\left (a x \right )}}}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**(1/2)/(-a**2*x**2+1)**2,x)

[Out]

Integral(sqrt(atanh(a*x))/((a*x - 1)**2*(a*x + 1)**2), x)

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