3.4.42 \(\int \frac {1}{(1-a^2 x^2)^3 \tanh ^{-1}(a x)^4} \, dx\) [342]

Optimal. Leaf size=125 \[ -\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}-\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {4 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a} \]

[Out]

-1/3/a/(-a^2*x^2+1)^2/arctanh(a*x)^3-2/3*x/(-a^2*x^2+1)^2/arctanh(a*x)^2-8/3/a/(-a^2*x^2+1)^2/arctanh(a*x)+2/a
/(-a^2*x^2+1)/arctanh(a*x)+2/3*Shi(2*arctanh(a*x))/a+4/3*Shi(4*arctanh(a*x))/a

________________________________________________________________________________________

Rubi [A]
time = 0.35, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6113, 6179, 6175, 6181, 5556, 12, 3379} \begin {gather*} -\frac {2 x}{3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {8}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {1}{3 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^3}+\frac {2 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}+\frac {4 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^3*ArcTanh[a*x]^4),x]

[Out]

-1/3*1/(a*(1 - a^2*x^2)^2*ArcTanh[a*x]^3) - (2*x)/(3*(1 - a^2*x^2)^2*ArcTanh[a*x]^2) - 8/(3*a*(1 - a^2*x^2)^2*
ArcTanh[a*x]) + 2/(a*(1 - a^2*x^2)*ArcTanh[a*x]) + (2*SinhIntegral[2*ArcTanh[a*x]])/(3*a) + (4*SinhIntegral[4*
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 6175

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 6179

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

________________________________________________________________________________________

Mathematica [A]
time = 0.15, size = 108, normalized size = 0.86 \begin {gather*} -\frac {1+2 a x \tanh ^{-1}(a x)+2 \tanh ^{-1}(a x)^2+6 a^2 x^2 \tanh ^{-1}(a x)^2-2 \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)^3 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )-4 \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)^3 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{3 a \left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^3*ArcTanh[a*x]^4),x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 3.63, size = 122, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {-\frac {1}{8 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{3 \arctanh \left (a x \right )}+\frac {2 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{3}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{24 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{3 \arctanh \left (a x \right )}+\frac {4 \hyperbolicSineIntegral \left (4 \arctanh \left (a x \right )\right )}{3}}{a}\)	\(122\)
default	\(\frac {-\frac {1}{8 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{3 \arctanh \left (a x \right )}+\frac {2 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{3}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{24 \arctanh \left (a x \right )^{3}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{3 \arctanh \left (a x \right )}+\frac {4 \hyperbolicSineIntegral \left (4 \arctanh \left (a x \right )\right )}{3}}{a}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)^3/arctanh(a*x)^4,x,method=_RETURNVERBOSE)

[Out]

1/a*(-1/8/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))-1/24/arctanh(a*x)^3*cosh(4*arctanh(a*x))-1/12/arctanh
(a*x)^2*sinh(4*arctanh(a*x))-1/3/arctanh(a*x)*cosh(4*arctanh(a*x))+4/3*Shi(4*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)^3/arctanh(a*x)^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (111) = 222\).
time = 0.34, size = 272, normalized size = 2.18 \begin {gather*} \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 )^{3} - 8 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) - 4 \, {\left (3 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} - 8}{3 \, {\left (a^{5} x^{4} - 2 \, 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)^3/arctanh(a*x)^4,x, algorithm="fricas")

[Out]

1/3*((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
))^3 - 8*a*x*log(-(a*x + 1)/(a*x - 1)) - 4*(3*a^2*x^2 + 1)*log(-(a*x + 1)/(a*x - 1))^2 - 8)/((a^5*x^4 - 2*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}{a^{6} x^{6} \operatorname {atanh}^{4}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}^{4}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}^{4}{\left (a x \right )} - \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)**3/atanh(a*x)**4,x)

[Out]

-Integral(1/(a**6*x**6*atanh(a*x)**4 - 3*a**4*x**4*atanh(a*x)**4 + 3*a**2*x**2*atanh(a*x)**4 - 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)^3/arctanh(a*x)^4,x, algorithm="giac")

[Out]

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

[Out]

-int(1/(atanh(a*x)^4*(a^2*x^2 - 1)^3), x)

________________________________________________________________________________________