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

Optimal. Leaf size=157 \[ -\frac{8}{15 a c^3 \sqrt{c-a^2 c x^2}}-\frac{4}{45 a c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{8 x \tanh ^{-1}(a x)}{15 c^3 \sqrt{c-a^2 c x^2}}+\frac{4 x \tanh ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac{1}{25 a c \left (c-a^2 c x^2\right )^{5/2}}+\frac{x \tanh ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-1/(25*a*c*(c - a^2*c*x^2)^(5/2)) - 4/(45*a*c^2*(c - a^2*c*x^2)^(3/2)) - 8/(15*a*c^3*Sqrt[c - a^2*c*x^2]) + (x
*ArcTanh[a*x])/(5*c*(c - a^2*c*x^2)^(5/2)) + (4*x*ArcTanh[a*x])/(15*c^2*(c - a^2*c*x^2)^(3/2)) + (8*x*ArcTanh[
a*x])/(15*c^3*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.101478, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {5960, 5958} \[ -\frac{8}{15 a c^3 \sqrt{c-a^2 c x^2}}-\frac{4}{45 a c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{8 x \tanh ^{-1}(a x)}{15 c^3 \sqrt{c-a^2 c x^2}}+\frac{4 x \tanh ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac{1}{25 a c \left (c-a^2 c x^2\right )^{5/2}}+\frac{x \tanh ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(25*a*c*(c - a^2*c*x^2)^(5/2)) - 4/(45*a*c^2*(c - a^2*c*x^2)^(3/2)) - 8/(15*a*c^3*Sqrt[c - a^2*c*x^2]) + (x
*ArcTanh[a*x])/(5*c*(c - a^2*c*x^2)^(5/2)) + (4*x*ArcTanh[a*x])/(15*c^2*(c - a^2*c*x^2)^(3/2)) + (8*x*ArcTanh[
a*x])/(15*c^3*Sqrt[c - a^2*c*x^2])

Rule 5960

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(b*(d + e*x^2)^(q + 1))
/(4*c*d*(q + 1)^2), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x] -
Simp[(x*(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]))/(2*d*(q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*
d + e, 0] && LtQ[q, -1] && NeQ[q, -3/2]

Rule 5958

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> -Simp[b/(c*d*Sqrt[d + e*x^2]
), x] + Simp[(x*(a + b*ArcTanh[c*x]))/(d*Sqrt[d + e*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0
]

Rubi steps

\begin{align*} \int \frac{\tanh ^{-1}(a x)}{\left (c-a^2 c x^2\right )^{7/2}} \, dx &=-\frac{1}{25 a c \left (c-a^2 c x^2\right )^{5/2}}+\frac{x \tanh ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}}+\frac{4 \int \frac{\tanh ^{-1}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx}{5 c}\\ &=-\frac{1}{25 a c \left (c-a^2 c x^2\right )^{5/2}}-\frac{4}{45 a c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{x \tanh ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}}+\frac{4 x \tanh ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{8 \int \frac{\tanh ^{-1}(a x)}{\left (c-a^2 c x^2\right )^{3/2}} \, dx}{15 c^2}\\ &=-\frac{1}{25 a c \left (c-a^2 c x^2\right )^{5/2}}-\frac{4}{45 a c^2 \left (c-a^2 c x^2\right )^{3/2}}-\frac{8}{15 a c^3 \sqrt{c-a^2 c x^2}}+\frac{x \tanh ^{-1}(a x)}{5 c \left (c-a^2 c x^2\right )^{5/2}}+\frac{4 x \tanh ^{-1}(a x)}{15 c^2 \left (c-a^2 c x^2\right )^{3/2}}+\frac{8 x \tanh ^{-1}(a x)}{15 c^3 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0820581, size = 80, normalized size = 0.51 \[ \frac{\sqrt{c-a^2 c x^2} \left (120 a^4 x^4-260 a^2 x^2-15 a x \left (8 a^4 x^4-20 a^2 x^2+15\right ) \tanh ^{-1}(a x)+149\right )}{225 a c^4 \left (a^2 x^2-1\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - a^2*c*x^2]*(149 - 260*a^2*x^2 + 120*a^4*x^4 - 15*a*x*(15 - 20*a^2*x^2 + 8*a^4*x^4)*ArcTanh[a*x]))/(2
25*a*c^4*(-1 + a^2*x^2)^3)

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Maple [A]  time = 0.268, size = 250, normalized size = 1.6 \begin{align*} -{\frac{ \left ( ax+1 \right ) ^{2} \left ( -1+5\,{\it Artanh} \left ( ax \right ) \right ) }{800\,a \left ( ax-1 \right ) ^{3}{c}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) c}}+{\frac{ \left ( 5\,ax+5 \right ) \left ( -1+3\,{\it Artanh} \left ( ax \right ) \right ) }{288\,a \left ( ax-1 \right ) ^{2}{c}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) c}}-{\frac{5\,{\it Artanh} \left ( ax \right ) -5}{16\,a \left ( ax-1 \right ){c}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) c}}-{\frac{5\,{\it Artanh} \left ( ax \right ) +5}{16\,a \left ( ax+1 \right ){c}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) c}}+{\frac{ \left ( 5\,ax-5 \right ) \left ( 1+3\,{\it Artanh} \left ( ax \right ) \right ) }{288\,a \left ( ax+1 \right ) ^{2}{c}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) c}}-{\frac{ \left ( ax-1 \right ) ^{2} \left ( 1+5\,{\it Artanh} \left ( ax \right ) \right ) }{800\, \left ( ax+1 \right ) ^{3}a{c}^{4}}\sqrt{- \left ( ax-1 \right ) \left ( ax+1 \right ) c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/800*(a*x+1)^2*(-1+5*arctanh(a*x))*(-(a*x-1)*(a*x+1)*c)^(1/2)/a/(a*x-1)^3/c^4+5/288*(a*x+1)*(-1+3*arctanh(a*
x))*(-(a*x-1)*(a*x+1)*c)^(1/2)/a/(a*x-1)^2/c^4-5/16*(arctanh(a*x)-1)*(-(a*x-1)*(a*x+1)*c)^(1/2)/a/(a*x-1)/c^4-
5/16*(arctanh(a*x)+1)*(-(a*x-1)*(a*x+1)*c)^(1/2)/a/(a*x+1)/c^4+5/288*(a*x-1)*(1+3*arctanh(a*x))*(-(a*x-1)*(a*x
+1)*c)^(1/2)/a/(a*x+1)^2/c^4-1/800*(a*x-1)^2*(1+5*arctanh(a*x))*(-(a*x-1)*(a*x+1)*c)^(1/2)/(a*x+1)^3/a/c^4

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Maxima [A]  time = 1.02541, size = 178, normalized size = 1.13 \begin{align*} -\frac{1}{225} \, a{\left (\frac{120}{\sqrt{-a^{2} c x^{2} + c} a^{2} c^{3}} + \frac{20}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} a^{2} c^{2}} + \frac{9}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} a^{2} c}\right )} + \frac{1}{15} \,{\left (\frac{8 \, x}{\sqrt{-a^{2} c x^{2} + c} c^{3}} + \frac{4 \, x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} c^{2}} + \frac{3 \, x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{2}} c}\right )} \operatorname{artanh}\left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/225*a*(120/(sqrt(-a^2*c*x^2 + c)*a^2*c^3) + 20/((-a^2*c*x^2 + c)^(3/2)*a^2*c^2) + 9/((-a^2*c*x^2 + c)^(5/2)
*a^2*c)) + 1/15*(8*x/(sqrt(-a^2*c*x^2 + c)*c^3) + 4*x/((-a^2*c*x^2 + c)^(3/2)*c^2) + 3*x/((-a^2*c*x^2 + c)^(5/
2)*c))*arctanh(a*x)

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Fricas [A]  time = 1.59179, size = 244, normalized size = 1.55 \begin{align*} \frac{{\left (240 \, a^{4} x^{4} - 520 \, a^{2} x^{2} - 15 \,{\left (8 \, a^{5} x^{5} - 20 \, a^{3} x^{3} + 15 \, a x\right )} \log \left (-\frac{a x + 1}{a x - 1}\right ) + 298\right )} \sqrt{-a^{2} c x^{2} + c}}{450 \,{\left (a^{7} c^{4} x^{6} - 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} - a c^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/450*(240*a^4*x^4 - 520*a^2*x^2 - 15*(8*a^5*x^5 - 20*a^3*x^3 + 15*a*x)*log(-(a*x + 1)/(a*x - 1)) + 298)*sqrt(
-a^2*c*x^2 + c)/(a^7*c^4*x^6 - 3*a^5*c^4*x^4 + 3*a^3*c^4*x^2 - a*c^4)

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

[Out]

Timed out

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Giac [A]  time = 1.31219, size = 201, normalized size = 1.28 \begin{align*} -\frac{\sqrt{-a^{2} c x^{2} + c}{\left (4 \,{\left (\frac{2 \, a^{4} x^{2}}{c} - \frac{5 \, a^{2}}{c}\right )} x^{2} + \frac{15}{c}\right )} x \log \left (-\frac{a x + 1}{a x - 1}\right )}{30 \,{\left (a^{2} c x^{2} - c\right )}^{3}} - \frac{120 \,{\left (a^{2} c x^{2} - c\right )}^{2} - 20 \,{\left (a^{2} c x^{2} - c\right )} c + 9 \, c^{2}}{225 \,{\left (a^{2} c x^{2} - c\right )}^{2} \sqrt{-a^{2} c x^{2} + c} a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*sqrt(-a^2*c*x^2 + c)*(4*(2*a^4*x^2/c - 5*a^2/c)*x^2 + 15/c)*x*log(-(a*x + 1)/(a*x - 1))/(a^2*c*x^2 - c)^
3 - 1/225*(120*(a^2*c*x^2 - c)^2 - 20*(a^2*c*x^2 - c)*c + 9*c^2)/((a^2*c*x^2 - c)^2*sqrt(-a^2*c*x^2 + c)*a*c^3
)