3.495 \(\int \frac{1}{(1-a^2 x^2)^{7/2} \tanh ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=93 \[ -\frac{5 x}{2 \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}-\frac{1}{2 a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}+\frac{5 \text{Chi}\left (\tanh ^{-1}(a x)\right )}{16 a}+\frac{45 \text{Chi}\left (3 \tanh ^{-1}(a x)\right )}{32 a}+\frac{25 \text{Chi}\left (5 \tanh ^{-1}(a x)\right )}{32 a} \]

[Out]

-1/(2*a*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x]^2) - (5*x)/(2*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x]) + (5*CoshIntegral[Arc
Tanh[a*x]])/(16*a) + (45*CoshIntegral[3*ArcTanh[a*x]])/(32*a) + (25*CoshIntegral[5*ArcTanh[a*x]])/(32*a)

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Rubi [A]  time = 0.404697, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5966, 6032, 6034, 5448, 3301, 5968, 3312} \[ -\frac{5 x}{2 \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}-\frac{1}{2 a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}+\frac{5 \text{Chi}\left (\tanh ^{-1}(a x)\right )}{16 a}+\frac{45 \text{Chi}\left (3 \tanh ^{-1}(a x)\right )}{32 a}+\frac{25 \text{Chi}\left (5 \tanh ^{-1}(a x)\right )}{32 a} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - a^2*x^2)^(7/2)*ArcTanh[a*x]^3),x]

[Out]

-1/(2*a*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x]^2) - (5*x)/(2*(1 - a^2*x^2)^(5/2)*ArcTanh[a*x]) + (5*CoshIntegral[Arc
Tanh[a*x]])/(16*a) + (45*CoshIntegral[3*ArcTanh[a*x]])/(32*a) + (25*CoshIntegral[5*ArcTanh[a*x]])/(32*a)

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 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])

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 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 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 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])
)

Rubi steps

\begin{align*} \int \frac{1}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)^3} \, dx &=-\frac{1}{2 a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}+\frac{1}{2} (5 a) \int \frac{x}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)^2} \, dx\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}-\frac{5 x}{2 \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5}{2} \int \frac{1}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)} \, dx+\left (10 a^2\right ) \int \frac{x^2}{\left (1-a^2 x^2\right )^{7/2} \tanh ^{-1}(a x)} \, dx\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}-\frac{5 x}{2 \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh ^5(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{2 a}+\frac{10 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}-\frac{5 x}{2 \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \operatorname{Subst}\left (\int \left (\frac{5 \cosh (x)}{8 x}+\frac{5 \cosh (3 x)}{16 x}+\frac{\cosh (5 x)}{16 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{2 a}+\frac{10 \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{8 x}+\frac{\cosh (3 x)}{16 x}+\frac{\cosh (5 x)}{16 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a}\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}-\frac{5 x}{2 \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (5 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (5 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{8 a}+\frac{25 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{32 a}-\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a}+\frac{25 \operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{16 a}\\ &=-\frac{1}{2 a \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}-\frac{5 x}{2 \left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}+\frac{5 \text{Chi}\left (\tanh ^{-1}(a x)\right )}{16 a}+\frac{45 \text{Chi}\left (3 \tanh ^{-1}(a x)\right )}{32 a}+\frac{25 \text{Chi}\left (5 \tanh ^{-1}(a x)\right )}{32 a}\\ \end{align*}

Mathematica [A]  time = 0.229125, size = 79, normalized size = 0.85 \[ \frac{-\frac{80 a x}{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)}-\frac{16}{\left (1-a^2 x^2\right )^{5/2} \tanh ^{-1}(a x)^2}+10 \text{Chi}\left (\tanh ^{-1}(a x)\right )+45 \text{Chi}\left (3 \tanh ^{-1}(a x)\right )+25 \text{Chi}\left (5 \tanh ^{-1}(a x)\right )}{32 a} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - a^2*x^2)^(7/2)*ArcTanh[a*x]^3),x]

[Out]

(-16/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]^2) - (80*a*x)/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]) + 10*CoshIntegral[ArcT
anh[a*x]] + 45*CoshIntegral[3*ArcTanh[a*x]] + 25*CoshIntegral[5*ArcTanh[a*x]])/(32*a)

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Maple [B]  time = 0.174, size = 272, normalized size = 2.9 \begin{align*}{\frac{1}{32\,a \left ({\it Artanh} \left ( ax \right ) \right ) ^{2} \left ({a}^{2}{x}^{2}-1 \right ) } \left ( 10\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{\it Chi} \left ({\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+45\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{\it Chi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+25\, \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}{\it Chi} \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-5\,{\it Artanh} \left ( ax \right ) \sinh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-15\,{\it Artanh} \left ( ax \right ) \sinh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-5\,\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}-\cosh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){x}^{2}{a}^{2}+10\,\sqrt{-{a}^{2}{x}^{2}+1}ax{\it Artanh} \left ( ax \right ) -10\,{\it Chi} \left ({\it Artanh} \left ( ax \right ) \right ) \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}-45\,{\it Chi} \left ( 3\,{\it Artanh} \left ( ax \right ) \right ) \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}-25\,{\it Chi} \left ( 5\,{\it Artanh} \left ( ax \right ) \right ) \left ({\it Artanh} \left ( ax \right ) \right ) ^{2}+5\,\sinh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) +15\,\sinh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ){\it Artanh} \left ( ax \right ) +10\,\sqrt{-{a}^{2}{x}^{2}+1}+5\,\cosh \left ( 3\,{\it Artanh} \left ( ax \right ) \right ) +\cosh \left ( 5\,{\it Artanh} \left ( ax \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^3,x)

[Out]

1/32/a*(10*arctanh(a*x)^2*Chi(arctanh(a*x))*x^2*a^2+45*arctanh(a*x)^2*Chi(3*arctanh(a*x))*x^2*a^2+25*arctanh(a
*x)^2*Chi(5*arctanh(a*x))*x^2*a^2-5*arctanh(a*x)*sinh(5*arctanh(a*x))*x^2*a^2-15*arctanh(a*x)*sinh(3*arctanh(a
*x))*x^2*a^2-5*cosh(3*arctanh(a*x))*x^2*a^2-cosh(5*arctanh(a*x))*x^2*a^2+10*(-a^2*x^2+1)^(1/2)*a*x*arctanh(a*x
)-10*Chi(arctanh(a*x))*arctanh(a*x)^2-45*Chi(3*arctanh(a*x))*arctanh(a*x)^2-25*Chi(5*arctanh(a*x))*arctanh(a*x
)^2+5*sinh(5*arctanh(a*x))*arctanh(a*x)+15*sinh(3*arctanh(a*x))*arctanh(a*x)+10*(-a^2*x^2+1)^(1/2)+5*cosh(3*ar
ctanh(a*x))+cosh(5*arctanh(a*x)))/arctanh(a*x)^2/(a^2*x^2-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{7}{2}} \operatorname{artanh}\left (a x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^3,x, algorithm="maxima")

[Out]

integrate(1/((-a^2*x^2 + 1)^(7/2)*arctanh(a*x)^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} x^{2} + 1}}{{\left (a^{8} x^{8} - 4 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 1\right )} \operatorname{artanh}\left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^3,x, algorithm="fricas")

[Out]

integral(sqrt(-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)*arctanh(a*x)^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a**2*x**2+1)**(7/2)/atanh(a*x)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{7}{2}} \operatorname{artanh}\left (a x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-a^2*x^2+1)^(7/2)/arctanh(a*x)^3,x, algorithm="giac")

[Out]

integrate(1/((-a^2*x^2 + 1)^(7/2)*arctanh(a*x)^3), x)