3.3.97 \(\int \frac {1}{(1-a^2 x^2)^2 \tanh ^{-1}(a x)^4} \, dx\) [297]

Optimal. Leaf size=97 \[ -\frac {1}{3 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1+a^2 x^2}{3 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6113, 6143, 6181, 5556, 12, 3379} \begin {gather*} -\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {a^2 x^2+1}{3 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {1}{3 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*1/(a*(1 - a^2*x^2)*ArcTanh[a*x]^3) - x/(3*(1 - a^2*x^2)*ArcTanh[a*x]^2) - (1 + a^2*x^2)/(3*a*(1 - a^2*x^2
)*ArcTanh[a*x]) + (2*SinhIntegral[2*ArcTanh[a*x]])/(3*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 3379

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 5556

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 6113

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 6143

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[x*((a + b*ArcTa
nh[c*x])^(p + 1)/(b*c*d*(p + 1)*(d + e*x^2))), x] + (Dist[4/(b^2*(p + 1)*(p + 2)), Int[x*((a + b*ArcTanh[c*x])
^(p + 2)/(d + e*x^2)^2), x], x] + Simp[(1 + c^2*x^2)*((a + b*ArcTanh[c*x])^(p + 2)/(b^2*e*(p + 1)*(p + 2)*(d +
e*x^2))), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[p, -1] && NeQ[p, -2]

Rule 6181

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

________________________________________________________________________________________

Mathematica [A]
time = 0.10, size = 73, normalized size = 0.75 \begin {gather*} \frac {1+a x \tanh ^{-1}(a x)+\left (1+a^2 x^2\right ) \tanh ^{-1}(a x)^2+2 \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^3 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 6.91, size = 68, normalized size = 0.70

method	result	size
derivativedivides	\(\frac {-\frac {1}{6 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{3 \arctanh \left (a x \right )}+\frac {2 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{3}}{a}\)	\(68\)
default	\(\frac {-\frac {1}{6 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{3 \arctanh \left (a x \right )}+\frac {2 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{3}}{a}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/6/arctanh(a*x)^3-1/6/arctanh(a*x)^3*cosh(2*arctanh(a*x))-1/6*sinh(2*arctanh(a*x))/arctanh(a*x)^2-1/3/a
rctanh(a*x)*cosh(2*arctanh(a*x))+2/3*Shi(2*arctanh(a*x)))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.41, size = 151, normalized size = 1.56 \begin {gather*} \frac {{\left ({\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (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 )^{3} + 4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 8}{3 \, {\left (a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^4\,{\left (a^2\,x^2-1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________