3.4.51 \(\int \frac {x^6}{(1-a^2 x^2)^4 \tanh ^{-1}(a x)} \, dx\) [351]

Optimal. Leaf size=55 \[ \frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^7}-\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{16 a^7}+\frac {\text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^7}-\frac {5 \log \left (\tanh ^{-1}(a x)\right )}{16 a^7} \]

[Out]

15/32*Chi(2*arctanh(a*x))/a^7-3/16*Chi(4*arctanh(a*x))/a^7+1/32*Chi(6*arctanh(a*x))/a^7-5/16*ln(arctanh(a*x))/
a^7

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Rubi [A]
time = 0.09, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6181, 3393, 3382} \begin {gather*} \frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^7}-\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{16 a^7}+\frac {\text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^7}-\frac {5 \log \left (\tanh ^{-1}(a x)\right )}{16 a^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/((1 - a^2*x^2)^4*ArcTanh[a*x]),x]

[Out]

(15*CoshIntegral[2*ArcTanh[a*x]])/(32*a^7) - (3*CoshIntegral[4*ArcTanh[a*x]])/(16*a^7) + CoshIntegral[6*ArcTan
h[a*x]]/(32*a^7) - (5*Log[ArcTanh[a*x]])/(16*a^7)

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 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^6}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\sinh ^6(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^7}\\ &=-\frac {\text {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^7}\\ &=-\frac {5 \log \left (\tanh ^{-1}(a x)\right )}{16 a^7}+\frac {\text {Subst}\left (\int \frac {\cosh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^7}-\frac {3 \text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a^7}+\frac {15 \text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^7}\\ &=\frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^7}-\frac {3 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{16 a^7}+\frac {\text {Chi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^7}-\frac {5 \log \left (\tanh ^{-1}(a x)\right )}{16 a^7}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 40, normalized size = 0.73 \begin {gather*} \frac {15 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )-6 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )+\text {Chi}\left (6 \tanh ^{-1}(a x)\right )-10 \log \left (\tanh ^{-1}(a x)\right )}{32 a^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/((1 - a^2*x^2)^4*ArcTanh[a*x]),x]

[Out]

(15*CoshIntegral[2*ArcTanh[a*x]] - 6*CoshIntegral[4*ArcTanh[a*x]] + CoshIntegral[6*ArcTanh[a*x]] - 10*Log[ArcT
anh[a*x]])/(32*a^7)

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Maple [A]
time = 1.40, size = 40, normalized size = 0.73

method	result	size
derivativedivides	\(\frac {-\frac {5 \ln \left (\arctanh \left (a x \right )\right )}{16}+\frac {15 \hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )}{32}-\frac {3 \hyperbolicCosineIntegral \left (4 \arctanh \left (a x \right )\right )}{16}+\frac {\hyperbolicCosineIntegral \left (6 \arctanh \left (a x \right )\right )}{32}}{a^{7}}\)	\(40\)
default	\(\frac {-\frac {5 \ln \left (\arctanh \left (a x \right )\right )}{16}+\frac {15 \hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )}{32}-\frac {3 \hyperbolicCosineIntegral \left (4 \arctanh \left (a x \right )\right )}{16}+\frac {\hyperbolicCosineIntegral \left (6 \arctanh \left (a x \right )\right )}{32}}{a^{7}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(-a^2*x^2+1)^4/arctanh(a*x),x,method=_RETURNVERBOSE)

[Out]

1/a^7*(-5/16*ln(arctanh(a*x))+15/32*Chi(2*arctanh(a*x))-3/16*Chi(4*arctanh(a*x))+1/32*Chi(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(x^6/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="maxima")

[Out]

integrate(x^6/((a^2*x^2 - 1)^4*arctanh(a*x)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (47) = 94\).
time = 0.38, size = 220, normalized size = 4.00 \begin {gather*} -\frac {20 \, \log \left (\log \left (-\frac {a x + 1}{a x - 1}\right )\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 ) - \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 ) + 6 \, \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + 6 \, \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - 15 \, \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - 15 \, \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )}{64 \, a^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="fricas")

[Out]

-1/64*(20*log(log(-(a*x + 1)/(a*x - 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)) - 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)) + 6*log_i
ntegral((a^2*x^2 + 2*a*x + 1)/(a^2*x^2 - 2*a*x + 1)) + 6*log_integral((a^2*x^2 - 2*a*x + 1)/(a^2*x^2 + 2*a*x +
1)) - 15*log_integral(-(a*x + 1)/(a*x - 1)) - 15*log_integral(-(a*x - 1)/(a*x + 1)))/a^7

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6}}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname {atanh}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(-a**2*x**2+1)**4/atanh(a*x),x)

[Out]

Integral(x**6/((a*x - 1)**4*(a*x + 1)**4*atanh(a*x)), 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^6/(-a^2*x^2+1)^4/arctanh(a*x),x, algorithm="giac")

[Out]

integrate(x^6/((a^2*x^2 - 1)^4*arctanh(a*x)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^6}{\mathrm {atanh}\left (a\,x\right )\,{\left (a^2\,x^2-1\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(atanh(a*x)*(a^2*x^2 - 1)^4),x)

[Out]

int(x^6/(atanh(a*x)*(a^2*x^2 - 1)^4), x)

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