3.464 \(\int \frac {\tanh ^{-1}(a x)}{(1-a^2 x^2)^{9/2}} \, dx\)

Optimal. Leaf size=177 \[ -\frac {16}{35 a \sqrt {1-a^2 x^2}}-\frac {8}{105 a \left (1-a^2 x^2\right )^{3/2}}-\frac {6}{175 a \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{49 a \left (1-a^2 x^2\right )^{7/2}}+\frac {16 x \tanh ^{-1}(a x)}{35 \sqrt {1-a^2 x^2}}+\frac {8 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{5/2}}+\frac {x \tanh ^{-1}(a x)}{7 \left (1-a^2 x^2\right )^{7/2}} \]

[Out]

-1/49/a/(-a^2*x^2+1)^(7/2)-6/175/a/(-a^2*x^2+1)^(5/2)-8/105/a/(-a^2*x^2+1)^(3/2)+1/7*x*arctanh(a*x)/(-a^2*x^2+
1)^(7/2)+6/35*x*arctanh(a*x)/(-a^2*x^2+1)^(5/2)+8/35*x*arctanh(a*x)/(-a^2*x^2+1)^(3/2)-16/35/a/(-a^2*x^2+1)^(1
/2)+16/35*x*arctanh(a*x)/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5960, 5958} \[ -\frac {16}{35 a \sqrt {1-a^2 x^2}}-\frac {8}{105 a \left (1-a^2 x^2\right )^{3/2}}-\frac {6}{175 a \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{49 a \left (1-a^2 x^2\right )^{7/2}}+\frac {16 x \tanh ^{-1}(a x)}{35 \sqrt {1-a^2 x^2}}+\frac {8 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{3/2}}+\frac {6 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{5/2}}+\frac {x \tanh ^{-1}(a x)}{7 \left (1-a^2 x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(1 - a^2*x^2)^(9/2),x]

[Out]

-1/(49*a*(1 - a^2*x^2)^(7/2)) - 6/(175*a*(1 - a^2*x^2)^(5/2)) - 8/(105*a*(1 - a^2*x^2)^(3/2)) - 16/(35*a*Sqrt[
1 - a^2*x^2]) + (x*ArcTanh[a*x])/(7*(1 - a^2*x^2)^(7/2)) + (6*x*ArcTanh[a*x])/(35*(1 - a^2*x^2)^(5/2)) + (8*x*
ArcTanh[a*x])/(35*(1 - a^2*x^2)^(3/2)) + (16*x*ArcTanh[a*x])/(35*Sqrt[1 - a^2*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 (1-a^2 x^2\right )^{9/2}} \, dx &=-\frac {1}{49 a \left (1-a^2 x^2\right )^{7/2}}+\frac {x \tanh ^{-1}(a x)}{7 \left (1-a^2 x^2\right )^{7/2}}+\frac {6}{7} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {1}{49 a \left (1-a^2 x^2\right )^{7/2}}-\frac {6}{175 a \left (1-a^2 x^2\right )^{5/2}}+\frac {x \tanh ^{-1}(a x)}{7 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{5/2}}+\frac {24}{35} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx\\ &=-\frac {1}{49 a \left (1-a^2 x^2\right )^{7/2}}-\frac {6}{175 a \left (1-a^2 x^2\right )^{5/2}}-\frac {8}{105 a \left (1-a^2 x^2\right )^{3/2}}+\frac {x \tanh ^{-1}(a x)}{7 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{3/2}}+\frac {16}{35} \int \frac {\tanh ^{-1}(a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac {1}{49 a \left (1-a^2 x^2\right )^{7/2}}-\frac {6}{175 a \left (1-a^2 x^2\right )^{5/2}}-\frac {8}{105 a \left (1-a^2 x^2\right )^{3/2}}-\frac {16}{35 a \sqrt {1-a^2 x^2}}+\frac {x \tanh ^{-1}(a x)}{7 \left (1-a^2 x^2\right )^{7/2}}+\frac {6 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{5/2}}+\frac {8 x \tanh ^{-1}(a x)}{35 \left (1-a^2 x^2\right )^{3/2}}+\frac {16 x \tanh ^{-1}(a x)}{35 \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 81, normalized size = 0.46 \[ \frac {1680 a^6 x^6-5320 a^4 x^4+5726 a^2 x^2-105 a x \left (16 a^6 x^6-56 a^4 x^4+70 a^2 x^2-35\right ) \tanh ^{-1}(a x)-2161}{3675 a \left (1-a^2 x^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]/(1 - a^2*x^2)^(9/2),x]

[Out]

(-2161 + 5726*a^2*x^2 - 5320*a^4*x^4 + 1680*a^6*x^6 - 105*a*x*(-35 + 70*a^2*x^2 - 56*a^4*x^4 + 16*a^6*x^6)*Arc
Tanh[a*x])/(3675*a*(1 - a^2*x^2)^(7/2))

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fricas [A]  time = 0.46, size = 121, normalized size = 0.68 \[ \frac {{\left (3360 \, a^{6} x^{6} - 10640 \, a^{4} x^{4} + 11452 \, a^{2} x^{2} - 105 \, {\left (16 \, a^{7} x^{7} - 56 \, a^{5} x^{5} + 70 \, a^{3} x^{3} - 35 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 4322\right )} \sqrt {-a^{2} x^{2} + 1}}{7350 \, {\left (a^{9} x^{8} - 4 \, a^{7} x^{6} + 6 \, a^{5} x^{4} - 4 \, a^{3} x^{2} + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(9/2),x, algorithm="fricas")

[Out]

1/7350*(3360*a^6*x^6 - 10640*a^4*x^4 + 11452*a^2*x^2 - 105*(16*a^7*x^7 - 56*a^5*x^5 + 70*a^3*x^3 - 35*a*x)*log
(-(a*x + 1)/(a*x - 1)) - 4322)*sqrt(-a^2*x^2 + 1)/(a^9*x^8 - 4*a^7*x^6 + 6*a^5*x^4 - 4*a^3*x^2 + a)

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giac [A]  time = 0.23, size = 138, normalized size = 0.78 \[ -\frac {\sqrt {-a^{2} x^{2} + 1} {\left (2 \, {\left (4 \, {\left (2 \, a^{6} x^{2} - 7 \, a^{4}\right )} x^{2} + 35 \, a^{2}\right )} x^{2} - 35\right )} x \log \left (-\frac {a x + 1}{a x - 1}\right )}{70 \, {\left (a^{2} x^{2} - 1\right )}^{4}} - \frac {126 \, a^{2} x^{2} + 1680 \, {\left (a^{2} x^{2} - 1\right )}^{3} - 280 \, {\left (a^{2} x^{2} - 1\right )}^{2} - 201}{3675 \, {\left (a^{2} x^{2} - 1\right )}^{3} \sqrt {-a^{2} x^{2} + 1} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(9/2),x, algorithm="giac")

[Out]

-1/70*sqrt(-a^2*x^2 + 1)*(2*(4*(2*a^6*x^2 - 7*a^4)*x^2 + 35*a^2)*x^2 - 35)*x*log(-(a*x + 1)/(a*x - 1))/(a^2*x^
2 - 1)^4 - 1/3675*(126*a^2*x^2 + 1680*(a^2*x^2 - 1)^3 - 280*(a^2*x^2 - 1)^2 - 201)/((a^2*x^2 - 1)^3*sqrt(-a^2*
x^2 + 1)*a)

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maple [A]  time = 0.44, size = 99, normalized size = 0.56 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \left (1680 \arctanh \left (a x \right ) x^{7} a^{7}-1680 x^{6} a^{6}-5880 \arctanh \left (a x \right ) x^{5} a^{5}+5320 x^{4} a^{4}+7350 a^{3} x^{3} \arctanh \left (a x \right )-5726 a^{2} x^{2}-3675 a x \arctanh \left (a x \right )+2161\right )}{3675 a \left (a^{2} x^{2}-1\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)/(-a^2*x^2+1)^(9/2),x)

[Out]

-1/3675/a*(-a^2*x^2+1)^(1/2)*(1680*arctanh(a*x)*x^7*a^7-1680*x^6*a^6-5880*arctanh(a*x)*x^5*a^5+5320*x^4*a^4+73
50*a^3*x^3*arctanh(a*x)-5726*a^2*x^2-3675*a*x*arctanh(a*x)+2161)/(a^2*x^2-1)^4

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maxima [A]  time = 0.33, size = 140, normalized size = 0.79 \[ -\frac {1}{3675} \, a {\left (\frac {1680}{\sqrt {-a^{2} x^{2} + 1} a^{2}} + \frac {280}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{2}} + \frac {126}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} a^{2}} + \frac {75}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}} a^{2}}\right )} + \frac {1}{35} \, {\left (\frac {16 \, x}{\sqrt {-a^{2} x^{2} + 1}} + \frac {8 \, x}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {6 \, x}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}}} + \frac {5 \, x}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {7}{2}}}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(-a^2*x^2+1)^(9/2),x, algorithm="maxima")

[Out]

-1/3675*a*(1680/(sqrt(-a^2*x^2 + 1)*a^2) + 280/((-a^2*x^2 + 1)^(3/2)*a^2) + 126/((-a^2*x^2 + 1)^(5/2)*a^2) + 7
5/((-a^2*x^2 + 1)^(7/2)*a^2)) + 1/35*(16*x/sqrt(-a^2*x^2 + 1) + 8*x/(-a^2*x^2 + 1)^(3/2) + 6*x/(-a^2*x^2 + 1)^
(5/2) + 5*x/(-a^2*x^2 + 1)^(7/2))*arctanh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )}{{\left (1-a^2\,x^2\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)/(1 - a^2*x^2)^(9/2),x)

[Out]

int(atanh(a*x)/(1 - a^2*x^2)^(9/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (a x \right )}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {9}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/(-a**2*x**2+1)**(9/2),x)

[Out]

Integral(atanh(a*x)/(-(a*x - 1)*(a*x + 1))**(9/2), x)

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