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

Optimal. Leaf size=100 \[ \frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]

[Out]

-1/2*x/a/(-a^2*x^2+1)^2/arctanh(a*x)^2-2/a^2/(-a^2*x^2+1)^2/arctanh(a*x)+3/2/a^2/(-a^2*x^2+1)/arctanh(a*x)+1/2
*Shi(2*arctanh(a*x))/a^2+Shi(4*arctanh(a*x))/a^2

________________________________________________________________________________________

Rubi [A]  time = 0.46, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6032, 6028, 5966, 6034, 5448, 12, 3298} \[ \frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(2*a*(1 - a^2*x^2)^2*ArcTanh[a*x]^2) - 2/(a^2*(1 - a^2*x^2)^2*ArcTanh[a*x]) + 3/(2*a^2*(1 - a^2*x^2)*ArcTan
h[a*x]) + SinhIntegral[2*ArcTanh[a*x]]/(2*a^2) + SinhIntegral[4*ArcTanh[a*x]]/a^2

Rule 12

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

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

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 5966

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

Rule 6028

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int
[x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*A
rcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegersQ[p, 2*q] && LtQ[q, -1] &&
IGtQ[m, 1] && NeQ[p, -1]

Rule 6032

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(x^m*(d
+ e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^(p + 1))/(b*c*d*(p + 1)), x] + (Dist[(c*(m + 2*q + 2))/(b*(p + 1)), Int
[x^(m + 1)*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x] - Dist[m/(b*c*(p + 1)), Int[x^(m - 1)*(d + e*x^2
)^q*(a + b*ArcTanh[c*x])^(p + 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] &&
LtQ[q, -1] && LtQ[p, -1] && NeQ[m + 2*q + 2, 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 {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx &=-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {\int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a}+\frac {1}{2} (3 a) \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {1}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+2 \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+\frac {3 \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a}-\frac {3 \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx}{2 a}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-3 \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+6 \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+\frac {2 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}-\frac {3 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac {6 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^2}+\frac {\operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^2}-\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac {6 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a^2}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^2}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2}{a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {3}{2 a^2 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{2 a^2}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 96, normalized size = 0.96 \[ -\frac {-\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )-2 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )+3 a^2 x^2 \tanh ^{-1}(a x)+a x+\tanh ^{-1}(a x)}{2 a^2 \left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*x + ArcTanh[a*x] + 3*a^2*x^2*ArcTanh[a*x] - (-1 + a^2*x^2)^2*ArcTanh[a*x]^2*SinhIntegral[2*ArcTanh[a*x
]] - 2*(-1 + a^2*x^2)^2*ArcTanh[a*x]^2*SinhIntegral[4*ArcTanh[a*x]])/(a^2*(-1 + a^2*x^2)^2*ArcTanh[a*x]^2)

________________________________________________________________________________________

fricas [B]  time = 0.56, size = 256, normalized size = 2.56 \[ \frac {{\left (2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 2 \, {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 8 \, a x - 4 \, {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{4 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*(a^4*x^4 - 2*a^2*x^2 + 1)*log_integral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) - 2*(a^4*x^4 - 2*a
^2*x^2 + 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x + 1)) + (a^4*x^4 - 2*a^2*x^2 + 1)*log_integral
(-(a*x + 1)/(a*x - 1)) - (a^4*x^4 - 2*a^2*x^2 + 1)*log_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x - 1
))^2 - 8*a*x - 4*(3*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))/((a^6*x^4 - 2*a^4*x^2 + a^2)*log(-(a*x + 1)/(a*x -
1))^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.18, size = 82, normalized size = 0.82 \[ \frac {-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\frac {\Shi \left (2 \arctanh \left (a x \right )\right )}{2}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\Shi \left (4 \arctanh \left (a x \right )\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(-1/8*sinh(2*arctanh(a*x))/arctanh(a*x)^2-1/4/arctanh(a*x)*cosh(2*arctanh(a*x))+1/2*Shi(2*arctanh(a*x))-
1/16/arctanh(a*x)^2*sinh(4*arctanh(a*x))-1/4/arctanh(a*x)*cosh(4*arctanh(a*x))+Shi(4*arctanh(a*x)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, a x + {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) - {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )}{{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) + {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (-a x + 1\right )^{2}} + \int -\frac {2 \, {\left (3 \, a^{2} x^{3} + 5 \, x\right )}}{{\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (a x + 1\right ) - {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*x + (3*a^2*x^2 + 1)*log(a*x + 1) - (3*a^2*x^2 + 1)*log(-a*x + 1))/((a^6*x^4 - 2*a^4*x^2 + a^2)*log(a*x +
1)^2 - 2*(a^6*x^4 - 2*a^4*x^2 + a^2)*log(a*x + 1)*log(-a*x + 1) + (a^6*x^4 - 2*a^4*x^2 + a^2)*log(-a*x + 1)^2
) + integrate(-2*(3*a^2*x^3 + 5*x)/((a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log(a*x + 1) - (a^6*x^6 - 3*a^4*x^4
+ 3*a^2*x^2 - 1)*log(-a*x + 1)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________