3.4.61 \(\int \frac {1}{(1-a^2 x^2)^4 \tanh ^{-1}(a x)^2} \, dx\) [361]

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

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Rubi [A]
time = 0.10, 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 = {6113, 6181, 5556, 3379} \begin {gather*} -\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 {gather*}

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 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 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 )^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 \text {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 \text {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 \text {Subst}\left (\int \frac {\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a}+\frac {15 \text {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*}

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Mathematica [A]
time = 0.14, size = 56, normalized size = 0.85 \begin {gather*} \frac {\frac {1}{\left (-1+a^2 x^2\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} \end {gather*}

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

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Maple [A]
time = 3.64, size = 86, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {-\frac {5}{16 \arctanh \left (a x \right )}-\frac {15 \cosh \left (2 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {15 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{16}-\frac {3 \cosh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )}+\frac {3 \hyperbolicSineIntegral \left (4 \arctanh \left (a x \right )\right )}{4}-\frac {\cosh \left (6 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {3 \hyperbolicSineIntegral \left (6 \arctanh \left (a x \right )\right )}{16}}{a}\)	\(86\)
default	\(\frac {-\frac {5}{16 \arctanh \left (a x \right )}-\frac {15 \cosh \left (2 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {15 \hyperbolicSineIntegral \left (2 \arctanh \left (a x \right )\right )}{16}-\frac {3 \cosh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )}+\frac {3 \hyperbolicSineIntegral \left (4 \arctanh \left (a x \right )\right )}{4}-\frac {\cosh \left (6 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {3 \hyperbolicSineIntegral \left (6 \arctanh \left (a x \right )\right )}{16}}{a}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)))

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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(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))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (58) = 116\).
time = 0.41, size = 413, normalized size = 6.26 \begin {gather*} \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 )} \end {gather*}

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)))

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

[Out]

Integral(1/((a*x - 1)**4*(a*x + 1)**4*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(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)

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

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)

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