3.363 \(\int \frac {1}{(1-a^2 x^2)^4 \tanh ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=89 \[ -\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac {9 \text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]

[Out]

-1/2/a/(-a^2*x^2+1)^3/arctanh(a*x)^2-3*x/(-a^2*x^2+1)^3/arctanh(a*x)+15/16*Chi(2*arctanh(a*x))/a+3/2*Chi(4*arc
tanh(a*x))/a+9/16*Chi(6*arctanh(a*x))/a

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Rubi [A]  time = 0.39, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5966, 6032, 6034, 5448, 3301, 5968, 3312} \[ -\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac {9 \text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*a*(1 - a^2*x^2)^3*ArcTanh[a*x]^2) - (3*x)/((1 - a^2*x^2)^3*ArcTanh[a*x]) + (15*CoshIntegral[2*ArcTanh[a*
x]])/(16*a) + (3*CoshIntegral[4*ArcTanh[a*x]])/(2*a) + (9*CoshIntegral[6*ArcTanh[a*x]])/(16*a)

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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 5968

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c, Subst[Int[(
a + b*x)^p/Cosh[x]^(2*(q + 1)), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0]
&& ILtQ[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 6032

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 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)^3} \, dx &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+(3 a) \int \frac {x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+3 \int \frac {1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\left (15 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {3 \operatorname {Subst}\left (\int \frac {\cosh ^6(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {15 \operatorname {Subst}\left (\int \frac {\cosh ^4(x) \sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {3 \operatorname {Subst}\left (\int \left (\frac {5}{16 x}+\frac {15 \cosh (2 x)}{32 x}+\frac {3 \cosh (4 x)}{16 x}+\frac {\cosh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}+\frac {15 \operatorname {Subst}\left (\int \left (-\frac {1}{16 x}-\frac {\cosh (2 x)}{32 x}+\frac {\cosh (4 x)}{16 x}+\frac {\cosh (6 x)}{32 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {3 \operatorname {Subst}\left (\int \frac {\cosh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}-\frac {15 \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}+\frac {15 \operatorname {Subst}\left (\int \frac {\cosh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}+\frac {9 \operatorname {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {15 \operatorname {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}+\frac {45 \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}\\ &=-\frac {1}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3 x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{16 a}+\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a}+\frac {9 \text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{16 a}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 83, normalized size = 0.93 \[ \frac {1}{16} \left (\frac {48 x}{\left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)}+\frac {8}{a \left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2}+\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a}+\frac {24 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{a}+\frac {9 \text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{a}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8/(a*(-1 + a^2*x^2)^3*ArcTanh[a*x]^2) + (48*x)/((-1 + a^2*x^2)^3*ArcTanh[a*x]) + (15*CoshIntegral[2*ArcTanh[a
*x]])/a + (24*CoshIntegral[4*ArcTanh[a*x]])/a + (9*CoshIntegral[6*ArcTanh[a*x]])/a)/16

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fricas [B]  time = 0.71, size = 435, normalized size = 4.89 \[ \frac {192 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 3 \, {\left (3 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + 3 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 8 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 8 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 64}{32 \, {\left (a^{7} x^{6} - 3 \, a^{5} x^{4} + 3 \, a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^4/arctanh(a*x)^3,x, algorithm="fricas")

[Out]

1/32*(192*a*x*log(-(a*x + 1)/(a*x - 1)) + 3*(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)) + 3*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_integr
al(-(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)) + 8*(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)) + 8*(a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*log_i
ntegral((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))^2 + 64)/((a^7*x^6 - 3*a^5*x^4 + 3*a^3*x^2 - a)*log(-(a*x + 1)/(a*x - 1))^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a^{2} x^{2} - 1\right )}^{4} \operatorname {artanh}\left (a x\right )^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((a^2*x^2 - 1)^4*arctanh(a*x)^3), x)

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maple [A]  time = 0.28, size = 131, normalized size = 1.47 \[ \frac {-\frac {5}{32 \arctanh \left (a x \right )^{2}}-\frac {15 \cosh \left (2 \arctanh \left (a x \right )\right )}{64 \arctanh \left (a x \right )^{2}}-\frac {15 \sinh \left (2 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {15 \Chi \left (2 \arctanh \left (a x \right )\right )}{16}-\frac {3 \cosh \left (4 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )^{2}}-\frac {3 \sinh \left (4 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )}+\frac {3 \Chi \left (4 \arctanh \left (a x \right )\right )}{2}-\frac {\cosh \left (6 \arctanh \left (a x \right )\right )}{64 \arctanh \left (a x \right )^{2}}-\frac {3 \sinh \left (6 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {9 \Chi \left (6 \arctanh \left (a x \right )\right )}{16}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-5/32/arctanh(a*x)^2-15/64/arctanh(a*x)^2*cosh(2*arctanh(a*x))-15/32*sinh(2*arctanh(a*x))/arctanh(a*x)+15
/16*Chi(2*arctanh(a*x))-3/32/arctanh(a*x)^2*cosh(4*arctanh(a*x))-3/8/arctanh(a*x)*sinh(4*arctanh(a*x))+3/2*Chi
(4*arctanh(a*x))-1/64/arctanh(a*x)^2*cosh(6*arctanh(a*x))-3/32/arctanh(a*x)*sinh(6*arctanh(a*x))+9/16*Chi(6*ar
ctanh(a*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^4/arctanh(a*x)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^3\,{\left (a^2\,x^2-1\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname {atanh}^{3}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a**2*x**2+1)**4/atanh(a*x)**3,x)

[Out]

Integral(1/((a*x - 1)**4*(a*x + 1)**4*atanh(a*x)**3), x)

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