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

Optimal. Leaf size=114 \[ \frac {5 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac {9 \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac {5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \]

[Out]

-1/2*x/a/(-a^2*x^2+1)^3/arctanh(a*x)^2-3/a^2/(-a^2*x^2+1)^3/arctanh(a*x)+5/2/a^2/(-a^2*x^2+1)^2/arctanh(a*x)+5
/16*Shi(2*arctanh(a*x))/a^2+Shi(4*arctanh(a*x))/a^2+9/16*Shi(6*arctanh(a*x))/a^2

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Rubi [A]  time = 0.59, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6032, 6028, 5966, 6034, 5448, 3298} \[ \frac {5 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac {9 \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac {5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-\frac {3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(2*a*(1 - a^2*x^2)^3*ArcTanh[a*x]^2) - 3/(a^2*(1 - a^2*x^2)^3*ArcTanh[a*x]) + 5/(2*a^2*(1 - a^2*x^2)^2*ArcT
anh[a*x]) + (5*SinhIntegral[2*ArcTanh[a*x]])/(16*a^2) + SinhIntegral[4*ArcTanh[a*x]]/a^2 + (9*SinhIntegral[6*A
rcTanh[a*x]])/(16*a^2)

Rule 3298

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 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 6028

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int
[x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*A
rcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegersQ[p, 2*q] && LtQ[q, -1] &&
IGtQ[m, 1] && NeQ[p, -1]

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 {x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^3} \, dx &=-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}+\frac {\int \frac {1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx}{2 a}+\frac {1}{2} (5 a) \int \frac {x^2}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {1}{2 a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+3 \int \frac {x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\frac {5 \int \frac {1}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)^2} \, dx}{2 a}-\frac {5 \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{2 a}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}-10 \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx+15 \int \frac {x}{\left (1-a^2 x^2\right )^4 \tanh ^{-1}(a x)} \, dx+\frac {3 \operatorname {Subst}\left (\int \frac {\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {3 \operatorname {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^2}-\frac {10 \operatorname {Subst}\left (\int \frac {\cosh ^3(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac {15 \operatorname {Subst}\left (\int \frac {\cosh ^5(x) \sinh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac {15 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac {10 \operatorname {Subst}\left (\int \left (\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^2}+\frac {15 \operatorname {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^2}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {15 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac {3 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{8 a^2}+\frac {3 \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{32 a^2}+\frac {15 \operatorname {Subst}\left (\int \frac {\sinh (6 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac {5 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^2}+\frac {15 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a^2}+\frac {75 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a^2}-\frac {5 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a^2}\\ &=-\frac {x}{2 a \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2}-\frac {3}{a^2 \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)}+\frac {5}{2 a^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {5 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )}{16 a^2}+\frac {\text {Shi}\left (4 \tanh ^{-1}(a x)\right )}{a^2}+\frac {9 \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 73, normalized size = 0.64 \[ \frac {\frac {8 \left (\left (5 a^2 x^2+1\right ) \tanh ^{-1}(a x)+a x\right )}{\left (a^2 x^2-1\right )^3 \tanh ^{-1}(a x)^2}+5 \text {Shi}\left (2 \tanh ^{-1}(a x)\right )+16 \text {Shi}\left (4 \tanh ^{-1}(a x)\right )+9 \text {Shi}\left (6 \tanh ^{-1}(a x)\right )}{16 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*(a*x + (1 + 5*a^2*x^2)*ArcTanh[a*x]))/((-1 + a^2*x^2)^3*ArcTanh[a*x]^2) + 5*SinhIntegral[2*ArcTanh[a*x]] +
16*SinhIntegral[4*ArcTanh[a*x]] + 9*SinhIntegral[6*ArcTanh[a*x]])/(16*a^2)

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fricas [B]  time = 0.51, size = 447, normalized size = 3.92 \[ \frac {{\left (9 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) - 9 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}{a^{3} x^{3} + 3 \, a^{2} x^{2} + 3 \, a x + 1}\right ) + 16 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 16 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) + 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - 5 \, {\left (a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2} + 64 \, a x + 32 \, {\left (5 \, a^{2} x^{2} + 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{32 \, {\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*((9*(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)) - 9*(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)) + 16*(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)) - 16*(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))^2 + 64*a*x + 32*(5*a^2
*x^2 + 1)*log(-(a*x + 1)/(a*x - 1)))/((a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)*log(-(a*x + 1)/(a*x - 1))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 121, normalized size = 1.06 \[ \frac {-\frac {5 \sinh \left (2 \arctanh \left (a x \right )\right )}{64 \arctanh \left (a x \right )^{2}}-\frac {5 \cosh \left (2 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {5 \Shi \left (2 \arctanh \left (a x \right )\right )}{16}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\Shi \left (4 \arctanh \left (a x \right )\right )-\frac {\sinh \left (6 \arctanh \left (a x \right )\right )}{64 \arctanh \left (a x \right )^{2}}-\frac {3 \cosh \left (6 \arctanh \left (a x \right )\right )}{32 \arctanh \left (a x \right )}+\frac {9 \Shi \left (6 \arctanh \left (a x \right )\right )}{16}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(-5/64*sinh(2*arctanh(a*x))/arctanh(a*x)^2-5/32/arctanh(a*x)*cosh(2*arctanh(a*x))+5/16*Shi(2*arctanh(a*x
))-1/16/arctanh(a*x)^2*sinh(4*arctanh(a*x))-1/4/arctanh(a*x)*cosh(4*arctanh(a*x))+Shi(4*arctanh(a*x))-1/64/arc
tanh(a*x)^2*sinh(6*arctanh(a*x))-3/32/arctanh(a*x)*cosh(6*arctanh(a*x))+9/16*Shi(6*arctanh(a*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*x + (5*a^2*x^2 + 1)*log(a*x + 1) - (5*a^2*x^2 + 1)*log(-a*x + 1))/((a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2
)*log(a*x + 1)^2 - 2*(a^8*x^6 - 3*a^6*x^4 + 3*a^4*x^2 - a^2)*log(a*x + 1)*log(-a*x + 1) + (a^8*x^6 - 3*a^6*x^4
+ 3*a^4*x^2 - a^2)*log(-a*x + 1)^2) - integrate(-4*(5*a^2*x^3 + 4*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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/(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 \[ \int \frac {x}{\left (a x - 1\right )^{4} \left (a x + 1\right )^{4} \operatorname {atanh}^{3}{\left (a x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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