3.320 \(\int \frac {\sqrt {\tanh ^{-1}(a x)}}{(1-a^2 x^2)^3} \, dx\)

Optimal. Leaf size=168 \[ \frac {\sqrt {\pi } \text {erf}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}+\frac {\sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}-\frac {\sqrt {\pi } \text {erfi}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}-\frac {\sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}+\frac {\tanh ^{-1}(a x)^{3/2}}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{32 a} \]

[Out]

1/4*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+1/256*erf(2*arctanh(a*x)^(1/2))*Pi^(1/2)/a-1/256*erfi(2*arctanh(a*x)^(1/2))*Pi^(1/
2)/a+1/4*sinh(2*arctanh(a*x))*arctanh(a*x)^(1/2)/a+1/32*sinh(4*arctanh(a*x))*arctanh(a*x)^(1/2)/a

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Rubi [A]  time = 0.20, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5968, 3312, 3296, 3308, 2180, 2204, 2205} \[ \frac {\sqrt {\pi } \text {Erf}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}+\frac {\sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}-\frac {\sqrt {\pi } \text {Erfi}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}-\frac {\sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}+\frac {\tanh ^{-1}(a x)^{3/2}}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{32 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[a*x]]/(1 - a^2*x^2)^3,x]

[Out]

ArcTanh[a*x]^(3/2)/(4*a) + (Sqrt[Pi]*Erf[2*Sqrt[ArcTanh[a*x]]])/(256*a) + (Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[ArcTanh
[a*x]]])/(16*a) - (Sqrt[Pi]*Erfi[2*Sqrt[ArcTanh[a*x]]])/(256*a) - (Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcTanh[a*x]]]
)/(16*a) + (Sqrt[ArcTanh[a*x]]*Sinh[2*ArcTanh[a*x]])/(4*a) + (Sqrt[ArcTanh[a*x]]*Sinh[4*ArcTanh[a*x]])/(32*a)

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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 3312

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 5968

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c, Subst[Int[(
a + b*x)^p/Cosh[x]^(2*(q + 1)), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0]
&& ILtQ[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 )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \sqrt {x} \cosh ^4(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {3 \sqrt {x}}{8}+\frac {1}{2} \sqrt {x} \cosh (2 x)+\frac {1}{8} \sqrt {x} \cosh (4 x)\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {\tanh ^{-1}(a x)^{3/2}}{4 a}+\frac {\operatorname {Subst}\left (\int \sqrt {x} \cosh (4 x) \, dx,x,\tanh ^{-1}(a x)\right )}{8 a}+\frac {\operatorname {Subst}\left (\int \sqrt {x} \cosh (2 x) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a}\\ &=\frac {\tanh ^{-1}(a x)^{3/2}}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{32 a}-\frac {\operatorname {Subst}\left (\int \frac {\sinh (4 x)}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{64 a}-\frac {\operatorname {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a}\\ &=\frac {\tanh ^{-1}(a x)^{3/2}}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{32 a}+\frac {\operatorname {Subst}\left (\int \frac {e^{-4 x}}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{128 a}-\frac {\operatorname {Subst}\left (\int \frac {e^{4 x}}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{128 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 {\tanh ^{-1}(a x)^{3/2}}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{32 a}+\frac {\operatorname {Subst}\left (\int e^{-4 x^2} \, dx,x,\sqrt {\tanh ^{-1}(a x)}\right )}{64 a}-\frac {\operatorname {Subst}\left (\int e^{4 x^2} \, dx,x,\sqrt {\tanh ^{-1}(a x)}\right )}{64 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 {\tanh ^{-1}(a x)^{3/2}}{4 a}+\frac {\sqrt {\pi } \text {erf}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}+\frac {\sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}-\frac {\sqrt {\pi } \text {erfi}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}-\frac {\sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{16 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{4 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{32 a}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 152, normalized size = 0.90 \[ \frac {\frac {32 \sqrt {\tanh ^{-1}(a x)} \left (-3 a^3 x^3+2 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)+5 a x\right )}{\left (a^2 x^2-1\right )^2}+\frac {\sqrt {\tanh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-4 \tanh ^{-1}(a x)\right )}{\sqrt {-\tanh ^{-1}(a x)}}+\frac {8 \sqrt {2} \sqrt {\tanh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 \tanh ^{-1}(a x)\right )}{\sqrt {-\tanh ^{-1}(a x)}}-8 \sqrt {2} \Gamma \left (\frac {1}{2},2 \tanh ^{-1}(a x)\right )-\Gamma \left (\frac {1}{2},4 \tanh ^{-1}(a x)\right )}{256 a} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[ArcTanh[a*x]]/(1 - a^2*x^2)^3,x]

[Out]

((32*Sqrt[ArcTanh[a*x]]*(5*a*x - 3*a^3*x^3 + 2*(-1 + a^2*x^2)^2*ArcTanh[a*x]))/(-1 + a^2*x^2)^2 + (Sqrt[ArcTan
h[a*x]]*Gamma[1/2, -4*ArcTanh[a*x]])/Sqrt[-ArcTanh[a*x]] + (8*Sqrt[2]*Sqrt[ArcTanh[a*x]]*Gamma[1/2, -2*ArcTanh
[a*x]])/Sqrt[-ArcTanh[a*x]] - 8*Sqrt[2]*Gamma[1/2, 2*ArcTanh[a*x]] - Gamma[1/2, 4*ArcTanh[a*x]])/(256*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)^3,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 \[ \int -\frac {\sqrt {\operatorname {artanh}\left (a x\right )}}{{\left (a^{2} x^{2} - 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^3,x, algorithm="giac")

[Out]

integrate(-sqrt(arctanh(a*x))/(a^2*x^2 - 1)^3, x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\arctanh \left (a x \right )}}{\left (-a^{2} x^{2}+1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^3,x)

[Out]

int(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {\sqrt {\operatorname {artanh}\left (a x\right )}}{{\left (a^{2} x^{2} - 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^3,x, algorithm="maxima")

[Out]

-integrate(sqrt(arctanh(a*x))/(a^2*x^2 - 1)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)^(1/2)/(a^2*x^2 - 1)^3,x)

[Out]

int(-atanh(a*x)^(1/2)/(a^2*x^2 - 1)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {\sqrt {\operatorname {atanh}{\left (a x \right )}}}{a^{6} x^{6} - 3 a^{4} x^{4} + 3 a^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**(1/2)/(-a**2*x**2+1)**3,x)

[Out]

-Integral(sqrt(atanh(a*x))/(a**6*x**6 - 3*a**4*x**4 + 3*a**2*x**2 - 1), x)

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