3.3.88 \(\int \frac {x^2}{(1-a^2 x^2)^2 \tanh ^{-1}(a x)^2} \, dx\) [288]

Optimal. Leaf size=38 \[ -\frac {x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^3} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6153, 6181, 5556, 12, 3379} \begin {gather*} \frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^3}-\frac {x^2}{a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 6153

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

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

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Mathematica [A]
time = 0.06, size = 36, normalized size = 0.95 \begin {gather*} \frac {x^2}{a \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 6.84, size = 36, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\frac {1}{2 \arctanh \left (a x \right )}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{2 \arctanh \left (a x \right )}+\hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{a^{3}}\)	\(36\)
default	\(\frac {\frac {1}{2 \arctanh \left (a x \right )}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{2 \arctanh \left (a x \right )}+\hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{a^{3}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (36) = 72\).
time = 0.35, size = 111, normalized size = 2.92 \begin {gather*} \frac {4 \, a^{2} x^{2} + {\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 )}{2 \, {\left (a^{5} x^{2} - a^{3}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^2}{{\mathrm {atanh}\left (a\,x\right )}^2\,{\left (a^2\,x^2-1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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