3.161 \(\int x^4 (1-a^2 x^2) \tanh ^{-1}(a x) \, dx\)

Optimal. Leaf size=72 \[ \frac {x^2}{35 a^3}-\frac {1}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{35 a^5}-\frac {a x^6}{42}+\frac {1}{5} x^5 \tanh ^{-1}(a x)+\frac {x^4}{70 a} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6014, 5916, 266, 43} \[ \frac {x^2}{35 a^3}+\frac {\log \left (1-a^2 x^2\right )}{35 a^5}-\frac {1}{7} a^2 x^7 \tanh ^{-1}(a x)-\frac {a x^6}{42}+\frac {x^4}{70 a}+\frac {1}{5} x^5 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*(1 - a^2*x^2)*ArcTanh[a*x],x]

[Out]

x^2/(35*a^3) + x^4/(70*a) - (a*x^6)/42 + (x^5*ArcTanh[a*x])/5 - (a^2*x^7*ArcTanh[a*x])/7 + Log[1 - a^2*x^2]/(3
5*a^5)

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 6014

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

Rubi steps

\begin {align*} \int x^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x) \, dx &=-\left (a^2 \int x^6 \tanh ^{-1}(a x) \, dx\right )+\int x^4 \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {1}{7} a^2 x^7 \tanh ^{-1}(a x)-\frac {1}{5} a \int \frac {x^5}{1-a^2 x^2} \, dx+\frac {1}{7} a^3 \int \frac {x^7}{1-a^2 x^2} \, dx\\ &=\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {1}{7} a^2 x^7 \tanh ^{-1}(a x)-\frac {1}{10} a \operatorname {Subst}\left (\int \frac {x^2}{1-a^2 x} \, dx,x,x^2\right )+\frac {1}{14} a^3 \operatorname {Subst}\left (\int \frac {x^3}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {1}{7} a^2 x^7 \tanh ^{-1}(a x)-\frac {1}{10} a \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^3 \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 {x^2}{35 a^3}+\frac {x^4}{70 a}-\frac {a x^6}{42}+\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {1}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{35 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 72, normalized size = 1.00 \[ \frac {x^2}{35 a^3}-\frac {1}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {\log \left (1-a^2 x^2\right )}{35 a^5}-\frac {a x^6}{42}+\frac {1}{5} x^5 \tanh ^{-1}(a x)+\frac {x^4}{70 a} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(1 - a^2*x^2)*ArcTanh[a*x],x]

[Out]

x^2/(35*a^3) + x^4/(70*a) - (a*x^6)/42 + (x^5*ArcTanh[a*x])/5 - (a^2*x^7*ArcTanh[a*x])/7 + Log[1 - a^2*x^2]/(3
5*a^5)

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fricas [A]  time = 0.56, size = 76, normalized size = 1.06 \[ -\frac {5 \, a^{6} x^{6} - 3 \, a^{4} x^{4} - 6 \, a^{2} x^{2} + 3 \, {\left (5 \, a^{7} x^{7} - 7 \, a^{5} x^{5}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 6 \, \log \left (a^{2} x^{2} - 1\right )}{210 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="fricas")

[Out]

-1/210*(5*a^6*x^6 - 3*a^4*x^4 - 6*a^2*x^2 + 3*(5*a^7*x^7 - 7*a^5*x^5)*log(-(a*x + 1)/(a*x - 1)) - 6*log(a^2*x^
2 - 1))/a^5

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giac [B]  time = 0.23, size = 335, normalized size = 4.65 \[ \frac {2}{105} \, a {\left (\frac {3 \, \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{6}} - \frac {3 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{6}} - \frac {\frac {3 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} + \frac {36 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {2 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {36 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {3 \, {\left (a x + 1\right )}}{a x - 1}}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{6}} - \frac {3 \, {\left (\frac {35 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} + \frac {35 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {70 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {14 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {7 \, {\left (a x + 1\right )}}{a x - 1} - 1\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{7}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="giac")

[Out]

2/105*a*(3*log(abs(-a*x - 1)/abs(a*x - 1))/a^6 - 3*log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^6 - (3*(a*x + 1)^5/(a*
x - 1)^5 + 36*(a*x + 1)^4/(a*x - 1)^4 + 2*(a*x + 1)^3/(a*x - 1)^3 + 36*(a*x + 1)^2/(a*x - 1)^2 + 3*(a*x + 1)/(
a*x - 1))/(a^6*((a*x + 1)/(a*x - 1) - 1)^6) - 3*(35*(a*x + 1)^5/(a*x - 1)^5 + 35*(a*x + 1)^4/(a*x - 1)^4 + 70*
(a*x + 1)^3/(a*x - 1)^3 + 14*(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^6*((a
*x + 1)/(a*x - 1) - 1)^7))

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maple [A]  time = 0.02, size = 67, normalized size = 0.93 \[ -\frac {a^{2} x^{7} \arctanh \left (a x \right )}{7}+\frac {x^{5} \arctanh \left (a x \right )}{5}-\frac {x^{6} a}{42}+\frac {x^{4}}{70 a}+\frac {x^{2}}{35 a^{3}}+\frac {\ln \left (a x -1\right )}{35 a^{5}}+\frac {\ln \left (a x +1\right )}{35 a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(-a^2*x^2+1)*arctanh(a*x),x)

[Out]

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

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maxima [A]  time = 0.31, size = 73, normalized size = 1.01 \[ -\frac {1}{210} \, a {\left (\frac {5 \, a^{4} x^{6} - 3 \, a^{2} x^{4} - 6 \, x^{2}}{a^{4}} - \frac {6 \, \log \left (a x + 1\right )}{a^{6}} - \frac {6 \, \log \left (a x - 1\right )}{a^{6}}\right )} - \frac {1}{35} \, {\left (5 \, a^{2} x^{7} - 7 \, x^{5}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-a^2*x^2+1)*arctanh(a*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.99, size = 61, normalized size = 0.85 \[ \frac {\frac {\ln \left (a^2\,x^2-1\right )}{35}+\frac {a^2\,x^2}{35}+\frac {a^4\,x^4}{70}}{a^5}-\frac {a\,x^6}{42}+\frac {x^5\,\mathrm {atanh}\left (a\,x\right )}{5}-\frac {a^2\,x^7\,\mathrm {atanh}\left (a\,x\right )}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^4*atanh(a*x)*(a^2*x^2 - 1),x)

[Out]

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

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sympy [A]  time = 1.98, size = 71, normalized size = 0.99 \[ \begin {cases} - \frac {a^{2} x^{7} \operatorname {atanh}{\left (a x \right )}}{7} - \frac {a x^{6}}{42} + \frac {x^{5} \operatorname {atanh}{\left (a x \right )}}{5} + \frac {x^{4}}{70 a} + \frac {x^{2}}{35 a^{3}} + \frac {2 \log {\left (x - \frac {1}{a} \right )}}{35 a^{5}} + \frac {2 \operatorname {atanh}{\left (a x \right )}}{35 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(-a**2*x**2+1)*atanh(a*x),x)

[Out]

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

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