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

Optimal. Leaf size=86 \[ \frac {1}{7} a^4 x^7 \tanh ^{-1}(a x)+\frac {a^3 x^6}{42}-\frac {2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{105 a^3}-\frac {9 a x^4}{140}+\frac {1}{3} x^3 \tanh ^{-1}(a x)+\frac {4 x^2}{105 a} \]

[Out]

4/105*x^2/a-9/140*a*x^4+1/42*a^3*x^6+1/3*x^3*arctanh(a*x)-2/5*a^2*x^5*arctanh(a*x)+1/7*a^4*x^7*arctanh(a*x)+4/
105*ln(-a^2*x^2+1)/a^3

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Rubi [A]  time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6012, 5916, 266, 43} \[ \frac {a^3 x^6}{42}+\frac {4 \log \left (1-a^2 x^2\right )}{105 a^3}+\frac {1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac {2}{5} a^2 x^5 \tanh ^{-1}(a x)-\frac {9 a x^4}{140}+\frac {4 x^2}{105 a}+\frac {1}{3} x^3 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^2)/(105*a) - (9*a*x^4)/140 + (a^3*x^6)/42 + (x^3*ArcTanh[a*x])/3 - (2*a^2*x^5*ArcTanh[a*x])/5 + (a^4*x^7*
ArcTanh[a*x])/7 + (4*Log[1 - a^2*x^2])/(105*a^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6012

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Int[E
xpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*ArcTanh[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[
c^2*d + e, 0] && IGtQ[p, 0] && IGtQ[q, 1]

Rubi steps

\begin {align*} \int x^2 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx &=\int \left (x^2 \tanh ^{-1}(a x)-2 a^2 x^4 \tanh ^{-1}(a x)+a^4 x^6 \tanh ^{-1}(a x)\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int x^4 \tanh ^{-1}(a x) \, dx\right )+a^4 \int x^6 \tanh ^{-1}(a x) \, dx+\int x^2 \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(a x)-\frac {2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac {1}{3} a \int \frac {x^3}{1-a^2 x^2} \, dx+\frac {1}{5} \left (2 a^3\right ) \int \frac {x^5}{1-a^2 x^2} \, dx-\frac {1}{7} a^5 \int \frac {x^7}{1-a^2 x^2} \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(a x)-\frac {2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac {1}{6} a \operatorname {Subst}\left (\int \frac {x}{1-a^2 x} \, dx,x,x^2\right )+\frac {1}{5} a^3 \operatorname {Subst}\left (\int \frac {x^2}{1-a^2 x} \, dx,x,x^2\right )-\frac {1}{14} a^5 \operatorname {Subst}\left (\int \frac {x^3}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac {1}{3} x^3 \tanh ^{-1}(a x)-\frac {2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {1}{7} a^4 x^7 \tanh ^{-1}(a x)-\frac {1}{6} a \operatorname {Subst}\left (\int \left (-\frac {1}{a^2}-\frac {1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )+\frac {1}{5} a^3 \operatorname {Subst}\left (\int \left (-\frac {1}{a^4}-\frac {x}{a^2}-\frac {1}{a^4 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )-\frac {1}{14} a^5 \operatorname {Subst}\left (\int \left (-\frac {1}{a^6}-\frac {x}{a^4}-\frac {x^2}{a^2}-\frac {1}{a^6 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {4 x^2}{105 a}-\frac {9 a x^4}{140}+\frac {a^3 x^6}{42}+\frac {1}{3} x^3 \tanh ^{-1}(a x)-\frac {2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {1}{7} a^4 x^7 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{105 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 86, normalized size = 1.00 \[ \frac {1}{7} a^4 x^7 \tanh ^{-1}(a x)+\frac {a^3 x^6}{42}-\frac {2}{5} a^2 x^5 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{105 a^3}-\frac {9 a x^4}{140}+\frac {1}{3} x^3 \tanh ^{-1}(a x)+\frac {4 x^2}{105 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^2)/(105*a) - (9*a*x^4)/140 + (a^3*x^6)/42 + (x^3*ArcTanh[a*x])/3 - (2*a^2*x^5*ArcTanh[a*x])/5 + (a^4*x^7*
ArcTanh[a*x])/7 + (4*Log[1 - a^2*x^2])/(105*a^3)

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fricas [A]  time = 0.78, size = 84, normalized size = 0.98 \[ \frac {10 \, a^{6} x^{6} - 27 \, a^{4} x^{4} + 16 \, a^{2} x^{2} + 2 \, {\left (15 \, a^{7} x^{7} - 42 \, a^{5} x^{5} + 35 \, a^{3} x^{3}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 16 \, \log \left (a^{2} x^{2} - 1\right )}{420 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(10*a^6*x^6 - 27*a^4*x^4 + 16*a^2*x^2 + 2*(15*a^7*x^7 - 42*a^5*x^5 + 35*a^3*x^3)*log(-(a*x + 1)/(a*x - 1
)) + 16*log(a^2*x^2 - 1))/a^3

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giac [B]  time = 0.23, size = 319, normalized size = 3.71 \[ \frac {4}{105} \, a {\left (\frac {2 \, \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{4}} - \frac {2 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{4}} - \frac {\frac {2 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} - \frac {11 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} - \frac {22 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {11 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {2 \, {\left (a x + 1\right )}}{a x - 1}}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{6}} + \frac {2 \, {\left (\frac {70 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {35 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {21 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - \frac {7 \, {\left (a x + 1\right )}}{a x - 1} + 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{4} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{7}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/105*a*(2*log(abs(-a*x - 1)/abs(a*x - 1))/a^4 - 2*log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^4 - (2*(a*x + 1)^5/(a*
x - 1)^5 - 11*(a*x + 1)^4/(a*x - 1)^4 - 22*(a*x + 1)^3/(a*x - 1)^3 - 11*(a*x + 1)^2/(a*x - 1)^2 + 2*(a*x + 1)/
(a*x - 1))/(a^4*((a*x + 1)/(a*x - 1) - 1)^6) + 2*(70*(a*x + 1)^4/(a*x - 1)^4 + 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^4*((a*x + 1)/(a*x - 1) - 1)^7))

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maple [A]  time = 0.03, size = 79, normalized size = 0.92 \[ \frac {a^{4} x^{7} \arctanh \left (a x \right )}{7}-\frac {2 a^{2} x^{5} \arctanh \left (a x \right )}{5}+\frac {x^{3} \arctanh \left (a x \right )}{3}+\frac {x^{6} a^{3}}{42}-\frac {9 x^{4} a}{140}+\frac {4 x^{2}}{105 a}+\frac {4 \ln \left (a x -1\right )}{105 a^{3}}+\frac {4 \ln \left (a x +1\right )}{105 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a^4*x^7*arctanh(a*x)-2/5*a^2*x^5*arctanh(a*x)+1/3*x^3*arctanh(a*x)+1/42*x^6*a^3-9/140*x^4*a+4/105*x^2/a+4/
105/a^3*ln(a*x-1)+4/105/a^3*ln(a*x+1)

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maxima [A]  time = 0.30, size = 81, normalized size = 0.94 \[ \frac {1}{420} \, a {\left (\frac {10 \, a^{4} x^{6} - 27 \, a^{2} x^{4} + 16 \, x^{2}}{a^{2}} + \frac {16 \, \log \left (a x + 1\right )}{a^{4}} + \frac {16 \, \log \left (a x - 1\right )}{a^{4}}\right )} + \frac {1}{105} \, {\left (15 \, a^{4} x^{7} - 42 \, a^{2} x^{5} + 35 \, x^{3}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*a*((10*a^4*x^6 - 27*a^2*x^4 + 16*x^2)/a^2 + 16*log(a*x + 1)/a^4 + 16*log(a*x - 1)/a^4) + 1/105*(15*a^4*x
^7 - 42*a^2*x^5 + 35*x^3)*arctanh(a*x)

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mupad [B]  time = 0.95, size = 71, normalized size = 0.83 \[ \frac {x^3\,\mathrm {atanh}\left (a\,x\right )}{3}-\frac {9\,a\,x^4}{140}+\frac {4\,\ln \left (a^2\,x^2-1\right )}{105\,a^3}+\frac {4\,x^2}{105\,a}+\frac {a^3\,x^6}{42}-\frac {2\,a^2\,x^5\,\mathrm {atanh}\left (a\,x\right )}{5}+\frac {a^4\,x^7\,\mathrm {atanh}\left (a\,x\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(a*x)*(a^2*x^2 - 1)^2,x)

[Out]

(x^3*atanh(a*x))/3 - (9*a*x^4)/140 + (4*log(a^2*x^2 - 1))/(105*a^3) + (4*x^2)/(105*a) + (a^3*x^6)/42 - (2*a^2*
x^5*atanh(a*x))/5 + (a^4*x^7*atanh(a*x))/7

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sympy [A]  time = 2.01, size = 90, normalized size = 1.05 \[ \begin {cases} \frac {a^{4} x^{7} \operatorname {atanh}{\left (a x \right )}}{7} + \frac {a^{3} x^{6}}{42} - \frac {2 a^{2} x^{5} \operatorname {atanh}{\left (a x \right )}}{5} - \frac {9 a x^{4}}{140} + \frac {x^{3} \operatorname {atanh}{\left (a x \right )}}{3} + \frac {4 x^{2}}{105 a} + \frac {8 \log {\left (x - \frac {1}{a} \right )}}{105 a^{3}} + \frac {8 \operatorname {atanh}{\left (a x \right )}}{105 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x**7*atanh(a*x)/7 + a**3*x**6/42 - 2*a**2*x**5*atanh(a*x)/5 - 9*a*x**4/140 + x**3*atanh(a*x)/3
+ 4*x**2/(105*a) + 8*log(x - 1/a)/(105*a**3) + 8*atanh(a*x)/(105*a**3), Ne(a, 0)), (0, True))

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