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

Optimal. Leaf size=66 \[ -\frac {1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {15 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a}+\frac {3 \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]

[Out]

-1/a/(-a^2*x^2+1)^3/arctanh(a*x)+15/16*Shi(2*arctanh(a*x))/a+3/4*Shi(4*arctanh(a*x))/a+3/16*Shi(6*arctanh(a*x)
)/a

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5966, 6034, 5448, 3298} \[ -\frac {1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {15 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a}+\frac {3 \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(a*(1 - a^2*x^2)^3*ArcTanh[a*x])) + (15*SinhIntegral[2*ArcTanh[a*x]])/(16*a) + (3*SinhIntegral[4*ArcTanh[a
*x]])/(4*a) + (3*SinhIntegral[6*ArcTanh[a*x]])/(16*a)

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 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 {1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx &=-\frac {1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+(6 a) \int \frac {x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {6 \operatorname {Subst}\left (\int \frac {\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {6 \operatorname {Subst}\left (\int \left (\frac {5 \sinh (2 x)}{32 x}+\frac {\sinh (4 x)}{8 x}+\frac {\sinh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a}+\frac {15 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}\\ &=-\frac {1}{a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {15 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{4 a}+\frac {3 \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.30, size = 56, normalized size = 0.85 \[ \frac {\frac {1}{\left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)}+\frac {15}{16} \text {Shi}\left (2 \tanh ^{-1}(a x)\right )+\frac {3}{4} \text {Shi}\left (4 \tanh ^{-1}(a x)\right )+\frac {3}{16} \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1/((-1 + a^2*x^2)^3*ArcTanh[a*x]) + (15*SinhIntegral[2*ArcTanh[a*x]])/16 + (3*SinhIntegral[4*ArcTanh[a*x]])/4
+ (3*SinhIntegral[6*ArcTanh[a*x]])/16)/a

________________________________________________________________________________________

fricas [B]  time = 0.49, size = 413, normalized size = 6.26 \[ \frac {3 \, {\left ({\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 4 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 4 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 64}{32 \, {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(3*((a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)/(a^3*x^3 - 3*a^
2*x^2 + 3*a*x - 1)) - (a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)/(a
^3*x^3 + 3*a^2*x^2 + 3*a*x + 1)) + 4*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral((a^2*x^2 + 2*a*x + 1)/
(a^2*x^2 - 2*a*x + 1)) - 4*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 +
2*a*x + 1)) + 5*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a*x + 1)/(a*x - 1)) - 5*(a^6*x^6 - 3*a^4
*x^4 + 3*a^2*x^2 - 1)*log_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x - 1)) + 64)/((a^7*x^6 - 3*a^5*x^
4 + 3*a^3*x^2 - a)*log(-(a*x + 1)/(a*x - 1)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.28, size = 86, normalized size = 1.30 \[ \frac {-\frac {5}{16 \arctanh \left (a x \right )}-\frac {15 \cosh \left (2 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {15 \Shi \left (2 \arctanh \left (a x \right )\right )}{16}-\frac {3 \cosh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )}+\frac {3 \Shi \left (4 \arctanh \left (a x \right )\right )}{4}-\frac {\cosh \left (6 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {3 \Shi \left (6 \arctanh \left (a x \right )\right )}{16}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-5/16/arctanh(a*x)-15/32/arctanh(a*x)*cosh(2*arctanh(a*x))+15/16*Shi(2*arctanh(a*x))-3/16/arctanh(a*x)*co
sh(4*arctanh(a*x))+3/4*Shi(4*arctanh(a*x))-1/32/arctanh(a*x)*cosh(6*arctanh(a*x))+3/16*Shi(6*arctanh(a*x)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -12 \, a \int -\frac {x}{{\left (a^{8} x^{8} - 4 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) - {\left (a^{8} x^{8} - 4 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )}\,{d x} + \frac {2}{{\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) - {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12*a*integrate(-x/((a^8*x^8 - 4*a^6*x^6 + 6*a^4*x^4 - 4*a^2*x^2 + 1)*log(a*x + 1) - (a^8*x^8 - 4*a^6*x^6 + 6*
a^4*x^4 - 4*a^2*x^2 + 1)*log(-a*x + 1)), x) + 2/((a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)*log(a*x + 1) - (a^7*x^6
- 3*a^5*x^4 + 3*a^3*x^2 - a)*log(-a*x + 1))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________