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

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

[Out]

-1/25/a/c/(-a^2*c*x^2+c)^(5/2)-4/45/a/c^2/(-a^2*c*x^2+c)^(3/2)+1/5*x*arctanh(a*x)/c/(-a^2*c*x^2+c)^(5/2)+4/15*
x*arctanh(a*x)/c^2/(-a^2*c*x^2+c)^(3/2)-8/15/a/c^3/(-a^2*c*x^2+c)^(1/2)+8/15*x*arctanh(a*x)/c^3/(-a^2*c*x^2+c)
^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 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
]

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]

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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.50, size = 112, normalized size = 0.71 \[ \frac {{\left (240 \, a^{4} x^{4} - 520 \, a^{2} x^{2} - 15 \, {\left (8 \, a^{5} x^{5} - 20 \, a^{3} x^{3} + 15 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 298\right )} \sqrt {-a^{2} c x^{2} + c}}{450 \, {\left (a^{7} c^{4} x^{6} - 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} - a c^{4}\right )}} \]

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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giac [A]  time = 0.31, size = 149, normalized size = 0.95 \[ -\frac {\sqrt {-a^{2} c x^{2} + c} {\left (4 \, {\left (\frac {2 \, a^{4} x^{2}}{c} - \frac {5 \, a^{2}}{c}\right )} x^{2} + \frac {15}{c}\right )} x \log \left (-\frac {a x + 1}{a x - 1}\right )}{30 \, {\left (a^{2} c x^{2} - c\right )}^{3}} - \frac {120 \, {\left (a^{2} c x^{2} - c\right )}^{2} - 20 \, {\left (a^{2} c x^{2} - c\right )} c + 9 \, c^{2}}{225 \, {\left (a^{2} c x^{2} - c\right )}^{2} \sqrt {-a^{2} c x^{2} + c} a c^{3}} \]

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
)

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maple [A]  time = 0.49, size = 250, normalized size = 1.59 \[ -\frac {\left (a x +1\right )^{2} \left (-1+5 \arctanh \left (a x \right )\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}}{800 a \left (a x -1\right )^{3} c^{4}}+\frac {5 \left (a x +1\right ) \left (-1+3 \arctanh \left (a x \right )\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}}{288 a \left (a x -1\right )^{2} c^{4}}-\frac {5 \left (\arctanh \left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}}{16 a \left (a x -1\right ) c^{4}}-\frac {5 \left (\arctanh \left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}}{16 a \left (a x +1\right ) c^{4}}+\frac {5 \left (a x -1\right ) \left (1+3 \arctanh \left (a x \right )\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}}{288 a \left (a x +1\right )^{2} c^{4}}-\frac {\left (a x -1\right )^{2} \left (1+5 \arctanh \left (a x \right )\right ) \sqrt {-\left (a x -1\right ) \left (a x +1\right ) c}}{800 \left (a x +1\right )^{3} a \,c^{4}} \]

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 = 0.35, size = 132, normalized size = 0.84 \[ -\frac {1}{225} \, a {\left (\frac {120}{\sqrt {-a^{2} c x^{2} + c} a^{2} c^{3}} + \frac {20}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c^{2}} + \frac {9}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} a^{2} c}\right )} + \frac {1}{15} \, {\left (\frac {8 \, x}{\sqrt {-a^{2} c x^{2} + c} c^{3}} + \frac {4 \, x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {3 \, x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} c}\right )} \operatorname {artanh}\left (a x\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )}{{\left (c-a^2\,c\,x^2\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (a x \right )}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}}}\, dx \]

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]

Integral(atanh(a*x)/(-c*(a*x - 1)*(a*x + 1))**(7/2), x)

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