3.2.92 \(\int x^4 (1-a^2 x^2)^2 \tanh ^{-1}(a x) \, dx\) [192]

Optimal. Leaf size=96 \[ \frac {4 x^2}{315 a^3}+\frac {2 x^4}{315 a}-\frac {11 a x^6}{378}+\frac {a^3 x^8}{72}+\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {2}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {1}{9} a^4 x^9 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{315 a^5} \]

[Out]

4/315*x^2/a^3+2/315*x^4/a-11/378*a*x^6+1/72*a^3*x^8+1/5*x^5*arctanh(a*x)-2/7*a^2*x^7*arctanh(a*x)+1/9*a^4*x^9*
arctanh(a*x)+4/315*ln(-a^2*x^2+1)/a^5

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Rubi [A]
time = 0.14, antiderivative size = 96, 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 = {6159, 6037, 272, 45} \begin {gather*} \frac {1}{9} a^4 x^9 \tanh ^{-1}(a x)+\frac {a^3 x^8}{72}+\frac {4 x^2}{315 a^3}-\frac {2}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{315 a^5}-\frac {11 a x^6}{378}+\frac {1}{5} x^5 \tanh ^{-1}(a x)+\frac {2 x^4}{315 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^2)/(315*a^3) + (2*x^4)/(315*a) - (11*a*x^6)/378 + (a^3*x^8)/72 + (x^5*ArcTanh[a*x])/5 - (2*a^2*x^7*ArcTan
h[a*x])/7 + (a^4*x^9*ArcTanh[a*x])/9 + (4*Log[1 - a^2*x^2])/(315*a^5)

Rule 45

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 272

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 6037

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

Rule 6159

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^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx &=\int \left (x^4 \tanh ^{-1}(a x)-2 a^2 x^6 \tanh ^{-1}(a x)+a^4 x^8 \tanh ^{-1}(a x)\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int x^6 \tanh ^{-1}(a x) \, dx\right )+a^4 \int x^8 \tanh ^{-1}(a x) \, dx+\int x^4 \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {2}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {1}{9} a^4 x^9 \tanh ^{-1}(a x)-\frac {1}{5} a \int \frac {x^5}{1-a^2 x^2} \, dx+\frac {1}{7} \left (2 a^3\right ) \int \frac {x^7}{1-a^2 x^2} \, dx-\frac {1}{9} a^5 \int \frac {x^9}{1-a^2 x^2} \, dx\\ &=\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {2}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {1}{9} a^4 x^9 \tanh ^{-1}(a x)-\frac {1}{10} a \text {Subst}\left (\int \frac {x^2}{1-a^2 x} \, dx,x,x^2\right )+\frac {1}{7} a^3 \text {Subst}\left (\int \frac {x^3}{1-a^2 x} \, dx,x,x^2\right )-\frac {1}{18} a^5 \text {Subst}\left (\int \frac {x^4}{1-a^2 x} \, dx,x,x^2\right )\\ &=\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {2}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {1}{9} a^4 x^9 \tanh ^{-1}(a x)-\frac {1}{10} a \text {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}{7} a^3 \text {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 {1}{18} a^5 \text {Subst}\left (\int \left (-\frac {1}{a^8}-\frac {x}{a^6}-\frac {x^2}{a^4}-\frac {x^3}{a^2}-\frac {1}{a^8 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {4 x^2}{315 a^3}+\frac {2 x^4}{315 a}-\frac {11 a x^6}{378}+\frac {a^3 x^8}{72}+\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {2}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {1}{9} a^4 x^9 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{315 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 96, normalized size = 1.00 \begin {gather*} \frac {4 x^2}{315 a^3}+\frac {2 x^4}{315 a}-\frac {11 a x^6}{378}+\frac {a^3 x^8}{72}+\frac {1}{5} x^5 \tanh ^{-1}(a x)-\frac {2}{7} a^2 x^7 \tanh ^{-1}(a x)+\frac {1}{9} a^4 x^9 \tanh ^{-1}(a x)+\frac {4 \log \left (1-a^2 x^2\right )}{315 a^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x^2)/(315*a^3) + (2*x^4)/(315*a) - (11*a*x^6)/378 + (a^3*x^8)/72 + (x^5*ArcTanh[a*x])/5 - (2*a^2*x^7*ArcTan
h[a*x])/7 + (a^4*x^9*ArcTanh[a*x])/9 + (4*Log[1 - a^2*x^2])/(315*a^5)

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Maple [A]
time = 0.56, size = 90, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {\frac {\arctanh \left (a x \right ) a^{9} x^{9}}{9}-\frac {2 \arctanh \left (a x \right ) a^{7} x^{7}}{7}+\frac {\arctanh \left (a x \right ) a^{5} x^{5}}{5}+\frac {a^{8} x^{8}}{72}-\frac {11 a^{6} x^{6}}{378}+\frac {2 a^{4} x^{4}}{315}+\frac {4 a^{2} x^{2}}{315}+\frac {4 \ln \left (a x -1\right )}{315}+\frac {4 \ln \left (a x +1\right )}{315}}{a^{5}}\)	\(90\)
default	\(\frac {\frac {\arctanh \left (a x \right ) a^{9} x^{9}}{9}-\frac {2 \arctanh \left (a x \right ) a^{7} x^{7}}{7}+\frac {\arctanh \left (a x \right ) a^{5} x^{5}}{5}+\frac {a^{8} x^{8}}{72}-\frac {11 a^{6} x^{6}}{378}+\frac {2 a^{4} x^{4}}{315}+\frac {4 a^{2} x^{2}}{315}+\frac {4 \ln \left (a x -1\right )}{315}+\frac {4 \ln \left (a x +1\right )}{315}}{a^{5}}\)	\(90\)
risch	\(\left (\frac {1}{18} a^{4} x^{9}-\frac {1}{7} a^{2} x^{7}+\frac {1}{10} x^{5}\right ) \ln \left (a x +1\right )-\frac {a^{4} x^{9} \ln \left (-a x +1\right )}{18}+\frac {a^{3} x^{8}}{72}+\frac {a^{2} x^{7} \ln \left (-a x +1\right )}{7}-\frac {11 x^{6} a}{378}-\frac {\ln \left (-a x +1\right ) x^{5}}{10}+\frac {2 x^{4}}{315 a}+\frac {4 x^{2}}{315 a^{3}}+\frac {4 \ln \left (a^{2} x^{2}-1\right )}{315 a^{5}}\)	\(118\)
meijerg	\(-\frac {-\frac {x^{2} a^{2} \left (15 a^{6} x^{6}+20 a^{4} x^{4}+30 a^{2} x^{2}+60\right )}{270}+\frac {2 x^{10} a^{10} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{9 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{9}}{4 a^{5}}-\frac {\frac {a^{2} x^{2} \left (4 a^{4} x^{4}+6 a^{2} x^{2}+12\right )}{42}-\frac {2 a^{8} x^{8} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{7 \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{7}}{2 a^{5}}-\frac {-\frac {a^{2} x^{2} \left (3 a^{2} x^{2}+6\right )}{15}+\frac {2 a^{6} x^{6} \left (\ln \left (1-\sqrt {a^{2} x^{2}}\right )-\ln \left (1+\sqrt {a^{2} x^{2}}\right )\right )}{5 \sqrt {a^{2} x^{2}}}-\frac {2 \ln \left (-a^{2} x^{2}+1\right )}{5}}{4 a^{5}}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^5*(1/9*arctanh(a*x)*a^9*x^9-2/7*arctanh(a*x)*a^7*x^7+1/5*arctanh(a*x)*a^5*x^5+1/72*a^8*x^8-11/378*a^6*x^6+
2/315*a^4*x^4+4/315*a^2*x^2+4/315*ln(a*x-1)+4/315*ln(a*x+1))

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Maxima [A]
time = 0.25, size = 89, normalized size = 0.93 \begin {gather*} \frac {1}{7560} \, a {\left (\frac {105 \, a^{6} x^{8} - 220 \, a^{4} x^{6} + 48 \, a^{2} x^{4} + 96 \, x^{2}}{a^{4}} + \frac {96 \, \log \left (a x + 1\right )}{a^{6}} + \frac {96 \, \log \left (a x - 1\right )}{a^{6}}\right )} + \frac {1}{315} \, {\left (35 \, a^{4} x^{9} - 90 \, a^{2} x^{7} + 63 \, x^{5}\right )} \operatorname {artanh}\left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7560*a*((105*a^6*x^8 - 220*a^4*x^6 + 48*a^2*x^4 + 96*x^2)/a^4 + 96*log(a*x + 1)/a^6 + 96*log(a*x - 1)/a^6) +
1/315*(35*a^4*x^9 - 90*a^2*x^7 + 63*x^5)*arctanh(a*x)

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Fricas [A]
time = 0.46, size = 92, normalized size = 0.96 \begin {gather*} \frac {105 \, a^{8} x^{8} - 220 \, a^{6} x^{6} + 48 \, a^{4} x^{4} + 96 \, a^{2} x^{2} + 12 \, {\left (35 \, a^{9} x^{9} - 90 \, a^{7} x^{7} + 63 \, a^{5} x^{5}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) + 96 \, \log \left (a^{2} x^{2} - 1\right )}{7560 \, a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7560*(105*a^8*x^8 - 220*a^6*x^6 + 48*a^4*x^4 + 96*a^2*x^2 + 12*(35*a^9*x^9 - 90*a^7*x^7 + 63*a^5*x^5)*log(-(
a*x + 1)/(a*x - 1)) + 96*log(a^2*x^2 - 1))/a^5

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Sympy [A]
time = 0.63, size = 100, normalized size = 1.04 \begin {gather*} \begin {cases} \frac {a^{4} x^{9} \operatorname {atanh}{\left (a x \right )}}{9} + \frac {a^{3} x^{8}}{72} - \frac {2 a^{2} x^{7} \operatorname {atanh}{\left (a x \right )}}{7} - \frac {11 a x^{6}}{378} + \frac {x^{5} \operatorname {atanh}{\left (a x \right )}}{5} + \frac {2 x^{4}}{315 a} + \frac {4 x^{2}}{315 a^{3}} + \frac {8 \log {\left (x - \frac {1}{a} \right )}}{315 a^{5}} + \frac {8 \operatorname {atanh}{\left (a x \right )}}{315 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x**9*atanh(a*x)/9 + a**3*x**8/72 - 2*a**2*x**7*atanh(a*x)/7 - 11*a*x**6/378 + x**5*atanh(a*x)/
5 + 2*x**4/(315*a) + 4*x**2/(315*a**3) + 8*log(x - 1/a)/(315*a**5) + 8*atanh(a*x)/(315*a**5), Ne(a, 0)), (0, T
rue))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 383 vs. \(2 (80) = 160\).
time = 0.40, size = 383, normalized size = 3.99 \begin {gather*} \frac {4}{945} \, a {\left (\frac {6 \, \log \left (\frac {{\left | -a x - 1 \right |}}{{\left | a x - 1 \right |}}\right )}{a^{6}} - \frac {6 \, \log \left ({\left | -\frac {a x + 1}{a x - 1} + 1 \right |}\right )}{a^{6}} - \frac {\frac {6 \, {\left (a x + 1\right )}^{7}}{{\left (a x - 1\right )}^{7}} - \frac {45 \, {\left (a x + 1\right )}^{6}}{{\left (a x - 1\right )}^{6}} - \frac {274 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} - \frac {214 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} - \frac {274 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} - \frac {45 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} + \frac {6 \, {\left (a x + 1\right )}}{a x - 1}}{a^{6} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{8}} + \frac {6 \, {\left (\frac {210 \, {\left (a x + 1\right )}^{6}}{{\left (a x - 1\right )}^{6}} + \frac {315 \, {\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} + \frac {441 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {126 \, {\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 {9 \, {\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 )}^{9}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/945*a*(6*log(abs(-a*x - 1)/abs(a*x - 1))/a^6 - 6*log(abs(-(a*x + 1)/(a*x - 1) + 1))/a^6 - (6*(a*x + 1)^7/(a*
x - 1)^7 - 45*(a*x + 1)^6/(a*x - 1)^6 - 274*(a*x + 1)^5/(a*x - 1)^5 - 214*(a*x + 1)^4/(a*x - 1)^4 - 274*(a*x +
1)^3/(a*x - 1)^3 - 45*(a*x + 1)^2/(a*x - 1)^2 + 6*(a*x + 1)/(a*x - 1))/(a^6*((a*x + 1)/(a*x - 1) - 1)^8) + 6*
(210*(a*x + 1)^6/(a*x - 1)^6 + 315*(a*x + 1)^5/(a*x - 1)^5 + 441*(a*x + 1)^4/(a*x - 1)^4 + 126*(a*x + 1)^3/(a*
x - 1)^3 + 36*(a*x + 1)^2/(a*x - 1)^2 - 9*(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)^9))

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Mupad [B]
time = 1.01, size = 106, normalized size = 1.10 \begin {gather*} \frac {4\,\ln \left (a^2\,x^2-1\right )}{315\,a^5}-\frac {11\,a\,x^6}{378}+\ln \left (a\,x+1\right )\,\left (\frac {a^4\,x^9}{18}-\frac {a^2\,x^7}{7}+\frac {x^5}{10}\right )-\ln \left (1-a\,x\right )\,\left (\frac {a^4\,x^9}{18}-\frac {a^2\,x^7}{7}+\frac {x^5}{10}\right )+\frac {2\,x^4}{315\,a}+\frac {4\,x^2}{315\,a^3}+\frac {a^3\,x^8}{72} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*atanh(a*x)*(a^2*x^2 - 1)^2,x)

[Out]

(4*log(a^2*x^2 - 1))/(315*a^5) - (11*a*x^6)/378 + log(a*x + 1)*(x^5/10 - (a^2*x^7)/7 + (a^4*x^9)/18) - log(1 -
a*x)*(x^5/10 - (a^2*x^7)/7 + (a^4*x^9)/18) + (2*x^4)/(315*a) + (4*x^2)/(315*a^3) + (a^3*x^8)/72

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