3.348 \(\int \frac {\sqrt {\tanh ^{-1}(a x)}}{(1-a^2 x^2)^4} \, dx\)

Optimal. Leaf size=252 \[ \frac {3 \sqrt {\pi } \text {erf}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{512 a}+\frac {15 \sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}+\frac {\sqrt {\frac {\pi }{6}} \text {erf}\left (\sqrt {6} \sqrt {\tanh ^{-1}(a x)}\right )}{768 a}-\frac {3 \sqrt {\pi } \text {erfi}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{512 a}-\frac {15 \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}-\frac {\sqrt {\frac {\pi }{6}} \text {erfi}\left (\sqrt {6} \sqrt {\tanh ^{-1}(a x)}\right )}{768 a}+\frac {5 \tanh ^{-1}(a x)^{3/2}}{24 a}+\frac {15 \sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{64 a}+\frac {3 \sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (6 \tanh ^{-1}(a x)\right )}{192 a} \]

[Out]

5/24*arctanh(a*x)^(3/2)/a+1/4608*erf(6^(1/2)*arctanh(a*x)^(1/2))*6^(1/2)*Pi^(1/2)/a-1/4608*erfi(6^(1/2)*arctan
h(a*x)^(1/2))*6^(1/2)*Pi^(1/2)/a+15/512*erf(2^(1/2)*arctanh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a-15/512*erfi(2^(1/2)
*arctanh(a*x)^(1/2))*2^(1/2)*Pi^(1/2)/a+3/512*erf(2*arctanh(a*x)^(1/2))*Pi^(1/2)/a-3/512*erfi(2*arctanh(a*x)^(
1/2))*Pi^(1/2)/a+15/64*sinh(2*arctanh(a*x))*arctanh(a*x)^(1/2)/a+3/64*sinh(4*arctanh(a*x))*arctanh(a*x)^(1/2)/
a+1/192*sinh(6*arctanh(a*x))*arctanh(a*x)^(1/2)/a

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 21, 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} \[ \frac {3 \sqrt {\pi } \text {Erf}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{512 a}+\frac {15 \sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}+\frac {\sqrt {\frac {\pi }{6}} \text {Erf}\left (\sqrt {6} \sqrt {\tanh ^{-1}(a x)}\right )}{768 a}-\frac {3 \sqrt {\pi } \text {Erfi}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{512 a}-\frac {15 \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}-\frac {\sqrt {\frac {\pi }{6}} \text {Erfi}\left (\sqrt {6} \sqrt {\tanh ^{-1}(a x)}\right )}{768 a}+\frac {5 \tanh ^{-1}(a x)^{3/2}}{24 a}+\frac {15 \sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{64 a}+\frac {3 \sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (6 \tanh ^{-1}(a x)\right )}{192 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[ArcTanh[a*x]]/(1 - a^2*x^2)^4,x]

[Out]

(5*ArcTanh[a*x]^(3/2))/(24*a) + (3*Sqrt[Pi]*Erf[2*Sqrt[ArcTanh[a*x]]])/(512*a) + (15*Sqrt[Pi/2]*Erf[Sqrt[2]*Sq
rt[ArcTanh[a*x]]])/(256*a) + (Sqrt[Pi/6]*Erf[Sqrt[6]*Sqrt[ArcTanh[a*x]]])/(768*a) - (3*Sqrt[Pi]*Erfi[2*Sqrt[Ar
cTanh[a*x]]])/(512*a) - (15*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[ArcTanh[a*x]]])/(256*a) - (Sqrt[Pi/6]*Erfi[Sqrt[6]*Sq
rt[ArcTanh[a*x]]])/(768*a) + (15*Sqrt[ArcTanh[a*x]]*Sinh[2*ArcTanh[a*x]])/(64*a) + (3*Sqrt[ArcTanh[a*x]]*Sinh[
4*ArcTanh[a*x]])/(64*a) + (Sqrt[ArcTanh[a*x]]*Sinh[6*ArcTanh[a*x]])/(192*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 )^4} \, dx &=\frac {\operatorname {Subst}\left (\int \sqrt {x} \cosh ^6(x) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {5 \sqrt {x}}{16}+\frac {15}{32} \sqrt {x} \cosh (2 x)+\frac {3}{16} \sqrt {x} \cosh (4 x)+\frac {1}{32} \sqrt {x} \cosh (6 x)\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=\frac {5 \tanh ^{-1}(a x)^{3/2}}{24 a}+\frac {\operatorname {Subst}\left (\int \sqrt {x} \cosh (6 x) \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}+\frac {3 \operatorname {Subst}\left (\int \sqrt {x} \cosh (4 x) \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {15 \operatorname {Subst}\left (\int \sqrt {x} \cosh (2 x) \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}\\ &=\frac {5 \tanh ^{-1}(a x)^{3/2}}{24 a}+\frac {15 \sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{64 a}+\frac {3 \sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (6 \tanh ^{-1}(a x)\right )}{192 a}-\frac {\operatorname {Subst}\left (\int \frac {\sinh (6 x)}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{384 a}-\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{128 a}-\frac {15 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{128 a}\\ &=\frac {5 \tanh ^{-1}(a x)^{3/2}}{24 a}+\frac {15 \sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{64 a}+\frac {3 \sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (6 \tanh ^{-1}(a x)\right )}{192 a}+\frac {\operatorname {Subst}\left (\int \frac {e^{-6 x}}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{768 a}-\frac {\operatorname {Subst}\left (\int \frac {e^{6 x}}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{768 a}+\frac {3 \operatorname {Subst}\left (\int \frac {e^{-4 x}}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{256 a}-\frac {3 \operatorname {Subst}\left (\int \frac {e^{4 x}}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{256 a}+\frac {15 \operatorname {Subst}\left (\int \frac {e^{-2 x}}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{256 a}-\frac {15 \operatorname {Subst}\left (\int \frac {e^{2 x}}{\sqrt {x}} \, dx,x,\tanh ^{-1}(a x)\right )}{256 a}\\ &=\frac {5 \tanh ^{-1}(a x)^{3/2}}{24 a}+\frac {15 \sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{64 a}+\frac {3 \sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (6 \tanh ^{-1}(a x)\right )}{192 a}+\frac {\operatorname {Subst}\left (\int e^{-6 x^2} \, dx,x,\sqrt {\tanh ^{-1}(a x)}\right )}{384 a}-\frac {\operatorname {Subst}\left (\int e^{6 x^2} \, dx,x,\sqrt {\tanh ^{-1}(a x)}\right )}{384 a}+\frac {3 \operatorname {Subst}\left (\int e^{-4 x^2} \, dx,x,\sqrt {\tanh ^{-1}(a x)}\right )}{128 a}-\frac {3 \operatorname {Subst}\left (\int e^{4 x^2} \, dx,x,\sqrt {\tanh ^{-1}(a x)}\right )}{128 a}+\frac {15 \operatorname {Subst}\left (\int e^{-2 x^2} \, dx,x,\sqrt {\tanh ^{-1}(a x)}\right )}{128 a}-\frac {15 \operatorname {Subst}\left (\int e^{2 x^2} \, dx,x,\sqrt {\tanh ^{-1}(a x)}\right )}{128 a}\\ &=\frac {5 \tanh ^{-1}(a x)^{3/2}}{24 a}+\frac {3 \sqrt {\pi } \text {erf}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{512 a}+\frac {15 \sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}+\frac {\sqrt {\frac {\pi }{6}} \text {erf}\left (\sqrt {6} \sqrt {\tanh ^{-1}(a x)}\right )}{768 a}-\frac {3 \sqrt {\pi } \text {erfi}\left (2 \sqrt {\tanh ^{-1}(a x)}\right )}{512 a}-\frac {15 \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {\tanh ^{-1}(a x)}\right )}{256 a}-\frac {\sqrt {\frac {\pi }{6}} \text {erfi}\left (\sqrt {6} \sqrt {\tanh ^{-1}(a x)}\right )}{768 a}+\frac {15 \sqrt {\tanh ^{-1}(a x)} \sinh \left (2 \tanh ^{-1}(a x)\right )}{64 a}+\frac {3 \sqrt {\tanh ^{-1}(a x)} \sinh \left (4 \tanh ^{-1}(a x)\right )}{64 a}+\frac {\sqrt {\tanh ^{-1}(a x)} \sinh \left (6 \tanh ^{-1}(a x)\right )}{192 a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.75, size = 257, normalized size = 1.02 \[ \frac {-\frac {3168 x \sqrt {\tanh ^{-1}(a x)}}{\left (a^2 x^2-1\right )^3}+\frac {3840 a^2 x^3 \sqrt {\tanh ^{-1}(a x)}}{\left (a^2 x^2-1\right )^3}-\frac {1440 a^4 x^5 \sqrt {\tanh ^{-1}(a x)}}{\left (a^2 x^2-1\right )^3}+\frac {960 \tanh ^{-1}(a x)^{3/2}}{a}+\frac {\sqrt {6} \sqrt {\tanh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-6 \tanh ^{-1}(a x)\right )}{a \sqrt {-\tanh ^{-1}(a x)}}+\frac {27 \sqrt {\tanh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-4 \tanh ^{-1}(a x)\right )}{a \sqrt {-\tanh ^{-1}(a x)}}+\frac {135 \sqrt {2} \sqrt {\tanh ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 \tanh ^{-1}(a x)\right )}{a \sqrt {-\tanh ^{-1}(a x)}}-\frac {135 \sqrt {2} \Gamma \left (\frac {1}{2},2 \tanh ^{-1}(a x)\right )}{a}-\frac {27 \Gamma \left (\frac {1}{2},4 \tanh ^{-1}(a x)\right )}{a}-\frac {\sqrt {6} \Gamma \left (\frac {1}{2},6 \tanh ^{-1}(a x)\right )}{a}}{4608} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[ArcTanh[a*x]]/(1 - a^2*x^2)^4,x]

[Out]

((-3168*x*Sqrt[ArcTanh[a*x]])/(-1 + a^2*x^2)^3 + (3840*a^2*x^3*Sqrt[ArcTanh[a*x]])/(-1 + a^2*x^2)^3 - (1440*a^
4*x^5*Sqrt[ArcTanh[a*x]])/(-1 + a^2*x^2)^3 + (960*ArcTanh[a*x]^(3/2))/a + (Sqrt[6]*Sqrt[ArcTanh[a*x]]*Gamma[1/
2, -6*ArcTanh[a*x]])/(a*Sqrt[-ArcTanh[a*x]]) + (27*Sqrt[ArcTanh[a*x]]*Gamma[1/2, -4*ArcTanh[a*x]])/(a*Sqrt[-Ar
cTanh[a*x]]) + (135*Sqrt[2]*Sqrt[ArcTanh[a*x]]*Gamma[1/2, -2*ArcTanh[a*x]])/(a*Sqrt[-ArcTanh[a*x]]) - (135*Sqr
t[2]*Gamma[1/2, 2*ArcTanh[a*x]])/a - (27*Gamma[1/2, 4*ArcTanh[a*x]])/a - (Sqrt[6]*Gamma[1/2, 6*ArcTanh[a*x]])/
a)/4608

________________________________________________________________________________________

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)^4,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {artanh}\left (a x\right )}}{{\left (a^{2} x^{2} - 1\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^4,x, algorithm="giac")

[Out]

integrate(sqrt(arctanh(a*x))/(a^2*x^2 - 1)^4, x)

________________________________________________________________________________________

maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\arctanh \left (a x \right )}}{\left (-a^{2} x^{2}+1\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^4,x)

[Out]

int(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^4,x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\operatorname {artanh}\left (a x\right )}}{{\left (a^{2} x^{2} - 1\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^(1/2)/(-a^2*x^2+1)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(arctanh(a*x))/(a^2*x^2 - 1)^4, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {\mathrm {atanh}\left (a\,x\right )}}{{\left (a^2\,x^2-1\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^(1/2)/(a^2*x^2 - 1)^4,x)

[Out]

int(atanh(a*x)^(1/2)/(a^2*x^2 - 1)^4, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**(1/2)/(-a**2*x**2+1)**4,x)

[Out]

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

________________________________________________________________________________________