3.3.98 \(\int \frac {1}{(1-a^2 x^2)^2 \tanh ^{-1}(a x)^5} \, dx\) [298]

Optimal. Leaf size=120 \[ -\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{3 a} \]

[Out]

-1/4/a/(-a^2*x^2+1)/arctanh(a*x)^4-1/6*x/(-a^2*x^2+1)/arctanh(a*x)^3+1/12*(-a^2*x^2-1)/a/(-a^2*x^2+1)/arctanh(
a*x)^2-1/3*x/(-a^2*x^2+1)/arctanh(a*x)+1/3*Chi(2*arctanh(a*x))/a

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Rubi [A]
time = 0.21, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6113, 6143, 6179, 6181, 3393, 3382, 6115} \begin {gather*} -\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {a^2 x^2+1}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}+\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{3 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^2*ArcTanh[a*x]^5),x]

[Out]

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

Rule 3382

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 3393

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 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 6115

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 6143

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 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 )^2 \tanh ^{-1}(a x)^5} \, dx &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}+\frac {1}{2} a \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^4} \, dx\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}+\frac {1}{3} a \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {1}{3} \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx+\frac {1}{3} a^2 \int \frac {x^2}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {\cosh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}+\frac {\text {Subst}\left (\int \frac {\sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {\text {Subst}\left (\int \left (\frac {1}{2 x}-\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}+\frac {\text {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{3 a}\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+2 \frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{6 a}\\ &=-\frac {1}{4 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^4}-\frac {x}{6 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^3}-\frac {1+a^2 x^2}{12 a \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)^2}-\frac {x}{3 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{3 a}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 84, normalized size = 0.70 \begin {gather*} \frac {3+2 a x \tanh ^{-1}(a x)+\left (1+a^2 x^2\right ) \tanh ^{-1}(a x)^2+4 a x \tanh ^{-1}(a x)^3+4 \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^4 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{12 a \left (-1+a^2 x^2\right ) \tanh ^{-1}(a x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^2*ArcTanh[a*x]^5),x]

[Out]

(3 + 2*a*x*ArcTanh[a*x] + (1 + a^2*x^2)*ArcTanh[a*x]^2 + 4*a*x*ArcTanh[a*x]^3 + 4*(-1 + a^2*x^2)*ArcTanh[a*x]^
4*CoshIntegral[2*ArcTanh[a*x]])/(12*a*(-1 + a^2*x^2)*ArcTanh[a*x]^4)

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Maple [A]
time = 6.87, size = 83, normalized size = 0.69

method	result	size
derivativedivides	\(\frac {-\frac {1}{8 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )^{4}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{2}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )}{3}}{a}\)	\(83\)
default	\(\frac {-\frac {1}{8 \arctanh \left (a x \right )^{4}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{8 \arctanh \left (a x \right )^{4}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{3}}-\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{12 \arctanh \left (a x \right )^{2}}-\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{6 \arctanh \left (a x \right )}+\frac {\hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )}{3}}{a}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)^2/arctanh(a*x)^5,x,method=_RETURNVERBOSE)

[Out]

1/a*(-1/8/arctanh(a*x)^4-1/8/arctanh(a*x)^4*cosh(2*arctanh(a*x))-1/12*sinh(2*arctanh(a*x))/arctanh(a*x)^3-1/12
/arctanh(a*x)^2*cosh(2*arctanh(a*x))-1/6*sinh(2*arctanh(a*x))/arctanh(a*x)+1/3*Chi(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(1/(-a^2*x^2+1)^2/arctanh(a*x)^5,x, algorithm="maxima")

[Out]

1/3*(2*a*x*log(a*x + 1)^3 - 2*a*x*log(-a*x + 1)^3 + 4*a*x*log(a*x + 1) + (a^2*x^2 + 1)*log(a*x + 1)^2 + (a^2*x
^2 + 6*a*x*log(a*x + 1) + 1)*log(-a*x + 1)^2 - 2*(3*a*x*log(a*x + 1)^2 + 2*a*x + (a^2*x^2 + 1)*log(a*x + 1))*l
og(-a*x + 1) + 12)/((a^3*x^2 - a)*log(a*x + 1)^4 - 4*(a^3*x^2 - a)*log(a*x + 1)^3*log(-a*x + 1) + 6*(a^3*x^2 -
a)*log(a*x + 1)^2*log(-a*x + 1)^2 - 4*(a^3*x^2 - a)*log(a*x + 1)*log(-a*x + 1)^3 + (a^3*x^2 - a)*log(-a*x + 1
)^4) - integrate(-2/3*(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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Fricas [A]
time = 0.36, size = 171, normalized size = 1.42 \begin {gather*} \frac {4 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right )^{3} + {\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 )^{4} + 8 \, a x \log \left (-\frac {a x + 1}{a x - 1}\right ) + 2 \, {\left (a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 24}{6 \, {\left (a^{3} x^{2} - a\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^2/arctanh(a*x)^5,x, algorithm="fricas")

[Out]

1/6*(4*a*x*log(-(a*x + 1)/(a*x - 1))^3 + ((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))^4 + 8*a*x*log(-(a*x + 1)/(a*x - 1)) + 2*(a^2*x^2 +
1)*log(-(a*x + 1)/(a*x - 1))^2 + 24)/((a^3*x^2 - a)*log(-(a*x + 1)/(a*x - 1))^4)

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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 )^{2} \left (a x + 1\right )^{2} \operatorname {atanh}^{5}{\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)**2/atanh(a*x)**5,x)

[Out]

Integral(1/((a*x - 1)**2*(a*x + 1)**2*atanh(a*x)**5), 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)^2/arctanh(a*x)^5,x, algorithm="giac")

[Out]

integrate(1/((a^2*x^2 - 1)^2*arctanh(a*x)^5), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {atanh}\left (a\,x\right )}^5\,{\left (a^2\,x^2-1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(atanh(a*x)^5*(a^2*x^2 - 1)^2),x)

[Out]

int(1/(atanh(a*x)^5*(a^2*x^2 - 1)^2), x)

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