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

Optimal. Leaf size=144 \[ \frac {\left (1-a^2 x^2\right )^3}{42 a}+\frac {3 \left (1-a^2 x^2\right )^2}{70 a}+\frac {4 \left (1-a^2 x^2\right )}{35 a}+\frac {8 \log \left (1-a^2 x^2\right )}{35 a}+\frac {1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac {6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {16}{35} x \tanh ^{-1}(a x) \]

[Out]

4/35*(-a^2*x^2+1)/a+3/70*(-a^2*x^2+1)^2/a+1/42*(-a^2*x^2+1)^3/a+16/35*x*arctanh(a*x)+8/35*x*(-a^2*x^2+1)*arcta
nh(a*x)+6/35*x*(-a^2*x^2+1)^2*arctanh(a*x)+1/7*x*(-a^2*x^2+1)^3*arctanh(a*x)+8/35*ln(-a^2*x^2+1)/a

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Rubi [A]  time = 0.07, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {5942, 5910, 260} \[ \frac {\left (1-a^2 x^2\right )^3}{42 a}+\frac {3 \left (1-a^2 x^2\right )^2}{70 a}+\frac {4 \left (1-a^2 x^2\right )}{35 a}+\frac {8 \log \left (1-a^2 x^2\right )}{35 a}+\frac {1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac {6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {16}{35} x \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^3*ArcTanh[a*x],x]

[Out]

(4*(1 - a^2*x^2))/(35*a) + (3*(1 - a^2*x^2)^2)/(70*a) + (1 - a^2*x^2)^3/(42*a) + (16*x*ArcTanh[a*x])/35 + (8*x
*(1 - a^2*x^2)*ArcTanh[a*x])/35 + (6*x*(1 - a^2*x^2)^2*ArcTanh[a*x])/35 + (x*(1 - a^2*x^2)^3*ArcTanh[a*x])/7 +
(8*Log[1 - a^2*x^2])/(35*a)

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5942

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

Rubi steps

\begin {align*} \int \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x) \, dx &=\frac {\left (1-a^2 x^2\right )^3}{42 a}+\frac {1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac {6}{7} \int \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx\\ &=\frac {3 \left (1-a^2 x^2\right )^2}{70 a}+\frac {\left (1-a^2 x^2\right )^3}{42 a}+\frac {6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac {24}{35} \int \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx\\ &=\frac {4 \left (1-a^2 x^2\right )}{35 a}+\frac {3 \left (1-a^2 x^2\right )^2}{70 a}+\frac {\left (1-a^2 x^2\right )^3}{42 a}+\frac {8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac {16}{35} \int \tanh ^{-1}(a x) \, dx\\ &=\frac {4 \left (1-a^2 x^2\right )}{35 a}+\frac {3 \left (1-a^2 x^2\right )^2}{70 a}+\frac {\left (1-a^2 x^2\right )^3}{42 a}+\frac {16}{35} x \tanh ^{-1}(a x)+\frac {8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)-\frac {1}{35} (16 a) \int \frac {x}{1-a^2 x^2} \, dx\\ &=\frac {4 \left (1-a^2 x^2\right )}{35 a}+\frac {3 \left (1-a^2 x^2\right )^2}{70 a}+\frac {\left (1-a^2 x^2\right )^3}{42 a}+\frac {16}{35} x \tanh ^{-1}(a x)+\frac {8}{35} x \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)+\frac {6}{35} x \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)+\frac {1}{7} x \left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)+\frac {8 \log \left (1-a^2 x^2\right )}{35 a}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 79, normalized size = 0.55 \[ \frac {-5 a^6 x^6+24 a^4 x^4-57 a^2 x^2+48 \log \left (1-a^2 x^2\right )-6 a x \left (5 a^6 x^6-21 a^4 x^4+35 a^2 x^2-35\right ) \tanh ^{-1}(a x)}{210 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-57*a^2*x^2 + 24*a^4*x^4 - 5*a^6*x^6 - 6*a*x*(-35 + 35*a^2*x^2 - 21*a^4*x^4 + 5*a^6*x^6)*ArcTanh[a*x] + 48*Lo
g[1 - a^2*x^2])/(210*a)

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fricas [A]  time = 0.52, size = 88, normalized size = 0.61 \[ -\frac {5 \, a^{6} x^{6} - 24 \, a^{4} x^{4} + 57 \, a^{2} x^{2} + 3 \, {\left (5 \, a^{7} x^{7} - 21 \, a^{5} x^{5} + 35 \, a^{3} x^{3} - 35 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 48 \, \log \left (a^{2} x^{2} - 1\right )}{210 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/210*(5*a^6*x^6 - 24*a^4*x^4 + 57*a^2*x^2 + 3*(5*a^7*x^7 - 21*a^5*x^5 + 35*a^3*x^3 - 35*a*x)*log(-(a*x + 1)/
(a*x - 1)) - 48*log(a^2*x^2 - 1))/a

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giac [B]  time = 0.19, size = 303, normalized size = 2.10 \[ \frac {8}{105} \, a {\left (\frac {6 \, \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{2}} - \frac {6 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{2}} - \frac {\frac {6 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} - \frac {33 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {74 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {33 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {6 \, {\left (a x + 1\right )}}{a x - 1}}{a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{6}} - \frac {6 \, {\left (\frac {35 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {21 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {7 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{2} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{7}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/105*a*(6*log(abs(-a*x - 1)/abs(a*x - 1))/a^2 - 6*log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^2 - (6*(a*x + 1)^5/(a*
x - 1)^5 - 33*(a*x + 1)^4/(a*x - 1)^4 + 74*(a*x + 1)^3/(a*x - 1)^3 - 33*(a*x + 1)^2/(a*x - 1)^2 + 6*(a*x + 1)/
(a*x - 1))/(a^2*((a*x + 1)/(a*x - 1) - 1)^6) - 6*(35*(a*x + 1)^3/(a*x - 1)^3 - 21*(a*x + 1)^2/(a*x - 1)^2 + 7*
(a*x + 1)/(a*x - 1) - 1)*log(-(a*((a*x + 1)/(a*x - 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) + 1)/(a*((a*x + 1)/(a*x
- 1) + 1)/((a*x + 1)*a/(a*x - 1) - a) - 1))/(a^2*((a*x + 1)/(a*x - 1) - 1)^7))

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maple [A]  time = 0.03, size = 88, normalized size = 0.61 \[ -\frac {a^{6} \arctanh \left (a x \right ) x^{7}}{7}+\frac {3 a^{4} \arctanh \left (a x \right ) x^{5}}{5}-a^{2} \arctanh \left (a x \right ) x^{3}+x \arctanh \left (a x \right )-\frac {a^{5} x^{6}}{42}+\frac {4 x^{4} a^{3}}{35}-\frac {19 a \,x^{2}}{70}+\frac {8 \ln \left (a x -1\right )}{35 a}+\frac {8 \ln \left (a x +1\right )}{35 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*a^6*arctanh(a*x)*x^7+3/5*a^4*arctanh(a*x)*x^5-a^2*arctanh(a*x)*x^3+x*arctanh(a*x)-1/42*a^5*x^6+4/35*x^4*a
^3-19/70*a*x^2+8/35/a*ln(a*x-1)+8/35/a*ln(a*x+1)

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maxima [A]  time = 0.31, size = 82, normalized size = 0.57 \[ -\frac {1}{210} \, {\left (5 \, a^{4} x^{6} - 24 \, a^{2} x^{4} + 57 \, x^{2} - \frac {48 \, \log \left (a x + 1\right )}{a^{2}} - \frac {48 \, \log \left (a x - 1\right )}{a^{2}}\right )} a - \frac {1}{35} \, {\left (5 \, a^{6} x^{7} - 21 \, a^{4} x^{5} + 35 \, a^{2} x^{3} - 35 \, x\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/210*(5*a^4*x^6 - 24*a^2*x^4 + 57*x^2 - 48*log(a*x + 1)/a^2 - 48*log(a*x - 1)/a^2)*a - 1/35*(5*a^6*x^7 - 21*
a^4*x^5 + 35*a^2*x^3 - 35*x)*arctanh(a*x)

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mupad [B]  time = 0.94, size = 80, normalized size = 0.56 \[ x\,\mathrm {atanh}\left (a\,x\right )-\frac {19\,a\,x^2}{70}+\frac {8\,\ln \left (a^2\,x^2-1\right )}{35\,a}+\frac {4\,a^3\,x^4}{35}-\frac {a^5\,x^6}{42}-a^2\,x^3\,\mathrm {atanh}\left (a\,x\right )+\frac {3\,a^4\,x^5\,\mathrm {atanh}\left (a\,x\right )}{5}-\frac {a^6\,x^7\,\mathrm {atanh}\left (a\,x\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-atanh(a*x)*(a^2*x^2 - 1)^3,x)

[Out]

x*atanh(a*x) - (19*a*x^2)/70 + (8*log(a^2*x^2 - 1))/(35*a) + (4*a^3*x^4)/35 - (a^5*x^6)/42 - a^2*x^3*atanh(a*x
) + (3*a^4*x^5*atanh(a*x))/5 - (a^6*x^7*atanh(a*x))/7

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sympy [A]  time = 2.04, size = 97, normalized size = 0.67 \[ \begin {cases} - \frac {a^{6} x^{7} \operatorname {atanh}{\left (a x \right )}}{7} - \frac {a^{5} x^{6}}{42} + \frac {3 a^{4} x^{5} \operatorname {atanh}{\left (a x \right )}}{5} + \frac {4 a^{3} x^{4}}{35} - a^{2} x^{3} \operatorname {atanh}{\left (a x \right )} - \frac {19 a x^{2}}{70} + x \operatorname {atanh}{\left (a x \right )} + \frac {16 \log {\left (x - \frac {1}{a} \right )}}{35 a} + \frac {16 \operatorname {atanh}{\left (a x \right )}}{35 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**6*x**7*atanh(a*x)/7 - a**5*x**6/42 + 3*a**4*x**5*atanh(a*x)/5 + 4*a**3*x**4/35 - a**2*x**3*atan
h(a*x) - 19*a*x**2/70 + x*atanh(a*x) + 16*log(x - 1/a)/(35*a) + 16*atanh(a*x)/(35*a), Ne(a, 0)), (0, True))

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