3.300 \(\int \frac {1}{(1-a^2 x^2)^2 \tanh ^{-1}(a x)^7} \, dx\)

Optimal. Leaf size=177 \[ -\frac {2 x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {a^2 x^2+1}{90 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {a^2 x^2+1}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}+\frac {2 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{45 a} \]

[Out]

-1/6/a/(-a^2*x^2+1)/arctanh(a*x)^6-1/15*x/(-a^2*x^2+1)/arctanh(a*x)^5+1/60*(-a^2*x^2-1)/a/(-a^2*x^2+1)/arctanh
(a*x)^4-1/45*x/(-a^2*x^2+1)/arctanh(a*x)^3+1/90*(-a^2*x^2-1)/a/(-a^2*x^2+1)/arctanh(a*x)^2-2/45*x/(-a^2*x^2+1)
/arctanh(a*x)+2/45*Chi(2*arctanh(a*x))/a

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Rubi [A]  time = 0.34, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5966, 5996, 6032, 6034, 3312, 3301, 5968} \[ -\frac {2 x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {a^2 x^2+1}{90 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {a^2 x^2+1}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}+\frac {2 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{45 a} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^2*ArcTanh[a*x]^7),x]

[Out]

-1/(6*a*(1 - a^2*x^2)*ArcTanh[a*x]^6) - x/(15*(1 - a^2*x^2)*ArcTanh[a*x]^5) - (1 + a^2*x^2)/(60*a*(1 - a^2*x^2
)*ArcTanh[a*x]^4) - x/(45*(1 - a^2*x^2)*ArcTanh[a*x]^3) - (1 + a^2*x^2)/(90*a*(1 - a^2*x^2)*ArcTanh[a*x]^2) -
(2*x)/(45*(1 - a^2*x^2)*ArcTanh[a*x]) + (2*CoshIntegral[2*ArcTanh[a*x]])/(45*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 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 5996

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 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 )^2 \tanh ^{-1}(a x)^7} \, dx &=-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}+\frac {1}{3} a \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^6} \, dx\\ &=-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {1+a^2 x^2}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}+\frac {1}{15} a \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4} \, dx\\ &=-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {1+a^2 x^2}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{90 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}+\frac {1}{45} (2 a) \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {1+a^2 x^2}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{90 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2}{45} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+\frac {1}{45} \left (2 a^2\right ) \int \frac {x^2}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {1+a^2 x^2}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{90 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \operatorname {Subst}\left (\int \frac {\cosh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{45 a}+\frac {2 \operatorname {Subst}\left (\int \frac {\sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{45 a}\\ &=-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {1+a^2 x^2}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{90 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 \operatorname {Subst}\left (\int \left (\frac {1}{2 x}-\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{45 a}+\frac {2 \operatorname {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{45 a}\\ &=-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {1+a^2 x^2}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{90 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+2 \frac {\operatorname {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{45 a}\\ &=-\frac {1}{6 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^6}-\frac {x}{15 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^5}-\frac {1+a^2 x^2}{60 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{90 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {2 x}{45 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{45 a}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 112, normalized size = 0.63 \[ \frac {8 \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^6 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )+2 \left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^4+3 \left (a^2 x^2+1\right ) \tanh ^{-1}(a x)^2+8 a x \tanh ^{-1}(a x)^5+4 a x \tanh ^{-1}(a x)^3+12 a x \tanh ^{-1}(a x)+30}{180 a \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)^6} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^2*ArcTanh[a*x]^7),x]

[Out]

(30 + 12*a*x*ArcTanh[a*x] + 3*(1 + a^2*x^2)*ArcTanh[a*x]^2 + 4*a*x*ArcTanh[a*x]^3 + 2*(1 + a^2*x^2)*ArcTanh[a*
x]^4 + 8*a*x*ArcTanh[a*x]^5 + 8*(-1 + a^2*x^2)*ArcTanh[a*x]^6*CoshIntegral[2*ArcTanh[a*x]])/(180*a*(-1 + a^2*x
^2)*ArcTanh[a*x]^6)

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fricas [A]  time = 0.58, size = 220, normalized size = 1.24 \[ \frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{5} + {\left ({\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) + {\left (a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{6} + 8 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4} + 96 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 12 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 480}{45 \, {\left (a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^2/arctanh(a*x)^7,x, algorithm="fricas")

[Out]

1/45*(4*a*x*log(-(a*x + 1)/(a*x - 1))^5 + ((a^2*x^2 - 1)*log_integral(-(a*x + 1)/(a*x - 1)) + (a^2*x^2 - 1)*lo
g_integral(-(a*x - 1)/(a*x + 1)))*log(-(a*x + 1)/(a*x - 1))^6 + 8*a*x*log(-(a*x + 1)/(a*x - 1))^3 + 2*(a^2*x^2
+ 1)*log(-(a*x + 1)/(a*x - 1))^4 + 96*a*x*log(-(a*x + 1)/(a*x - 1)) + 12*(a^2*x^2 + 1)*log(-(a*x + 1)/(a*x -
1))^2 + 480)/((a^3*x^2 - a)*log(-(a*x + 1)/(a*x - 1))^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a^{2} x^{2} - 1\right )}^{2} \operatorname {artanh}\left (a x\right )^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^2/arctanh(a*x)^7,x, algorithm="giac")

[Out]

integrate(1/((a^2*x^2 - 1)^2*arctanh(a*x)^7), x)

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maple [A]  time = 0.18, size = 113, normalized size = 0.64 \[ \frac {-\frac {1}{12 \arctanh \left (a x \right )^{6}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{6}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{30 \arctanh \left (a x \right )^{5}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{60 \arctanh \left (a x \right )^{4}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{90 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{90 \arctanh \left (a x \right )^{2}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{45 \arctanh \left (a x \right )}+\frac {2 \Chi \left (2 \arctanh \left (a x \right )\right )}{45}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)^2/arctanh(a*x)^7,x)

[Out]

1/a*(-1/12/arctanh(a*x)^6-1/12/arctanh(a*x)^6*cosh(2*arctanh(a*x))-1/30/arctanh(a*x)^5*sinh(2*arctanh(a*x))-1/
60/arctanh(a*x)^4*cosh(2*arctanh(a*x))-1/90/arctanh(a*x)^3*sinh(2*arctanh(a*x))-1/90/arctanh(a*x)^2*cosh(2*arc
tanh(a*x))-1/45*sinh(2*arctanh(a*x))/arctanh(a*x)+2/45*Chi(2*arctanh(a*x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, {\left (2 \, a x \log \left (a x + 1\right )^{5} - 2 \, a x \log \left (-a x + 1\right )^{5} + 4 \, a x \log \left (a x + 1\right )^{3} + {\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{4} + {\left (a^{2} x^{2} + 10 \, a x \log \left (a x + 1\right ) + 1\right )} \log \left (-a x + 1\right )^{4} - 4 \, {\left (5 \, a x \log \left (a x + 1\right )^{2} + a x + {\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right )^{3} + 48 \, a x \log \left (a x + 1\right ) + 6 \, {\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 2 \, {\left (10 \, a x \log \left (a x + 1\right )^{3} + 3 \, a^{2} x^{2} + 6 \, a x \log \left (a x + 1\right ) + 3 \, {\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{2} + 3\right )} \log \left (-a x + 1\right )^{2} - 2 \, {\left (5 \, a x \log \left (a x + 1\right )^{4} + 6 \, a x \log \left (a x + 1\right )^{2} + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )^{3} + 24 \, a x + 6 \, {\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right )\right )} \log \left (-a x + 1\right ) + 240\right )}}{45 \, {\left ({\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{6} - 6 \, {\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{5} \log \left (-a x + 1\right ) + 15 \, {\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{4} \log \left (-a x + 1\right )^{2} - 20 \, {\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{3} \log \left (-a x + 1\right )^{3} + 15 \, {\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right )^{4} - 6 \, {\left (a^{3} x^{2} - a\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{5} + {\left (a^{3} x^{2} - a\right )} \log \left (-a x + 1\right )^{6}\right )}} - \int -\frac {4 \, {\left (a^{2} x^{2} + 1\right )}}{45 \, {\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \log \left (-a x + 1\right )\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^2/arctanh(a*x)^7,x, algorithm="maxima")

[Out]

2/45*(2*a*x*log(a*x + 1)^5 - 2*a*x*log(-a*x + 1)^5 + 4*a*x*log(a*x + 1)^3 + (a^2*x^2 + 1)*log(a*x + 1)^4 + (a^
2*x^2 + 10*a*x*log(a*x + 1) + 1)*log(-a*x + 1)^4 - 4*(5*a*x*log(a*x + 1)^2 + a*x + (a^2*x^2 + 1)*log(a*x + 1))
*log(-a*x + 1)^3 + 48*a*x*log(a*x + 1) + 6*(a^2*x^2 + 1)*log(a*x + 1)^2 + 2*(10*a*x*log(a*x + 1)^3 + 3*a^2*x^2
+ 6*a*x*log(a*x + 1) + 3*(a^2*x^2 + 1)*log(a*x + 1)^2 + 3)*log(-a*x + 1)^2 - 2*(5*a*x*log(a*x + 1)^4 + 6*a*x*
log(a*x + 1)^2 + 2*(a^2*x^2 + 1)*log(a*x + 1)^3 + 24*a*x + 6*(a^2*x^2 + 1)*log(a*x + 1))*log(-a*x + 1) + 240)/
((a^3*x^2 - a)*log(a*x + 1)^6 - 6*(a^3*x^2 - a)*log(a*x + 1)^5*log(-a*x + 1) + 15*(a^3*x^2 - a)*log(a*x + 1)^4
*log(-a*x + 1)^2 - 20*(a^3*x^2 - a)*log(a*x + 1)^3*log(-a*x + 1)^3 + 15*(a^3*x^2 - a)*log(a*x + 1)^2*log(-a*x
+ 1)^4 - 6*(a^3*x^2 - a)*log(a*x + 1)*log(-a*x + 1)^5 + (a^3*x^2 - a)*log(-a*x + 1)^6) - integrate(-4/45*(a^2*
x^2 + 1)/((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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^7\,{\left (a^2\,x^2-1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(atanh(a*x)^7*(a^2*x^2 - 1)^2),x)

[Out]

int(1/(atanh(a*x)^7*(a^2*x^2 - 1)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a x - 1\right )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{7}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a**2*x**2+1)**2/atanh(a*x)**7,x)

[Out]

Integral(1/((a*x - 1)**2*(a*x + 1)**2*atanh(a*x)**7), x)

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