3.5.47 \(\int x^4 (1-a^2 x^2)^{3/2} \tanh ^{-1}(a x) \, dx\) [447]

Optimal. Leaf size=292 \[ \frac {3 \sqrt {1-a^2 x^2}}{128 a^5}+\frac {\left (1-a^2 x^2\right )^{3/2}}{192 a^5}-\frac {3 \left (1-a^2 x^2\right )^{5/2}}{80 a^5}+\frac {\left (1-a^2 x^2\right )^{7/2}}{56 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{128 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{64 a^2}+\frac {3}{16} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} a^2 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {3 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{64 a^5}-\frac {3 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{128 a^5}+\frac {3 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{128 a^5} \]

[Out]

1/192*(-a^2*x^2+1)^(3/2)/a^5-3/80*(-a^2*x^2+1)^(5/2)/a^5+1/56*(-a^2*x^2+1)^(7/2)/a^5-3/64*arctan((-a*x+1)^(1/2
)/(a*x+1)^(1/2))*arctanh(a*x)/a^5-3/128*I*polylog(2,-I*(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^5+3/128*I*polylog(2,I*(
-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^5+3/128*(-a^2*x^2+1)^(1/2)/a^5-3/128*x*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^4-1/64
*x^3*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a^2+3/16*x^5*arctanh(a*x)*(-a^2*x^2+1)^(1/2)-1/8*a^2*x^7*arctanh(a*x)*(-a
^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.56, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6161, 6157, 6163, 272, 45, 267, 6097} \begin {gather*} -\frac {3 \text {ArcTan}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{64 a^5}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{128 a^5}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{128 a^5}-\frac {1}{8} a^2 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {3}{16} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{64 a^2}+\frac {\left (1-a^2 x^2\right )^{7/2}}{56 a^5}-\frac {3 \left (1-a^2 x^2\right )^{5/2}}{80 a^5}+\frac {\left (1-a^2 x^2\right )^{3/2}}{192 a^5}+\frac {3 \sqrt {1-a^2 x^2}}{128 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{128 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(1 - a^2*x^2)^(3/2)*ArcTanh[a*x],x]

[Out]

(3*Sqrt[1 - a^2*x^2])/(128*a^5) + (1 - a^2*x^2)^(3/2)/(192*a^5) - (3*(1 - a^2*x^2)^(5/2))/(80*a^5) + (1 - a^2*
x^2)^(7/2)/(56*a^5) - (3*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(128*a^4) - (x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(6
4*a^2) + (3*x^5*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/16 - (a^2*x^7*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/8 - (3*ArcTan[Sq
rt[1 - a*x]/Sqrt[1 + a*x]]*ArcTanh[a*x])/(64*a^5) - (((3*I)/128)*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]
])/a^5 + (((3*I)/128)*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/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 267

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

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 6097

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

Rule 6157

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

Rule 6161

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]))

Rule 6163

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

Rubi steps

\begin {align*} \int x^4 \left (1-a^2 x^2\right )^{3/2} \tanh ^{-1}(a x) \, dx &=-\left (a^2 \int x^6 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\right )+\int x^4 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{6} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} a^2 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {1}{6} \int \frac {x^4 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{6} a \int \frac {x^5}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{8} a^2 \int \frac {x^6 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx+\frac {1}{8} a^3 \int \frac {x^7}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{24 a^2}+\frac {3}{16} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} a^2 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {5}{48} \int \frac {x^4 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx+\frac {\int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{8 a^2}+\frac {\int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx}{24 a}-\frac {1}{48} a \int \frac {x^5}{\sqrt {1-a^2 x^2}} \, dx-\frac {1}{12} a \text {Subst}\left (\int \frac {x^2}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )+\frac {1}{16} a^3 \text {Subst}\left (\int \frac {x^3}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{64 a^2}+\frac {3}{16} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} a^2 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+\frac {\int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{16 a^4}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{16 a^3}-\frac {5 \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{64 a^2}+\frac {\text {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )}{48 a}-\frac {5 \int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx}{192 a}-\frac {1}{96} a \text {Subst}\left (\int \frac {x^2}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )-\frac {1}{12} a \text {Subst}\left (\int \left (\frac {1}{a^4 \sqrt {1-a^2 x}}-\frac {2 \sqrt {1-a^2 x}}{a^4}+\frac {\left (1-a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )+\frac {1}{16} a^3 \text {Subst}\left (\int \left (\frac {1}{a^6 \sqrt {1-a^2 x}}-\frac {3 \sqrt {1-a^2 x}}{a^6}+\frac {3 \left (1-a^2 x\right )^{3/2}}{a^6}-\frac {\left (1-a^2 x\right )^{5/2}}{a^6}\right ) \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-a^2 x^2}}{48 a^5}+\frac {\left (1-a^2 x^2\right )^{3/2}}{72 a^5}-\frac {\left (1-a^2 x^2\right )^{5/2}}{24 a^5}+\frac {\left (1-a^2 x^2\right )^{7/2}}{56 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{128 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{64 a^2}+\frac {3}{16} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} a^2 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {\tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{8 a^5}-\frac {i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^5}+\frac {i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{16 a^5}-\frac {5 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{128 a^4}-\frac {5 \int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{128 a^3}-\frac {5 \text {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )}{384 a}+\frac {\text {Subst}\left (\int \left (\frac {1}{a^2 \sqrt {1-a^2 x}}-\frac {\sqrt {1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )}{48 a}-\frac {1}{96} a \text {Subst}\left (\int \left (\frac {1}{a^4 \sqrt {1-a^2 x}}-\frac {2 \sqrt {1-a^2 x}}{a^4}+\frac {\left (1-a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-a^2 x^2}}{384 a^5}+\frac {\left (1-a^2 x^2\right )^{3/2}}{72 a^5}-\frac {3 \left (1-a^2 x^2\right )^{5/2}}{80 a^5}+\frac {\left (1-a^2 x^2\right )^{7/2}}{56 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{128 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{64 a^2}+\frac {3}{16} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} a^2 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{64 a^5}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{128 a^5}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{128 a^5}-\frac {5 \text {Subst}\left (\int \left (\frac {1}{a^2 \sqrt {1-a^2 x}}-\frac {\sqrt {1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )}{384 a}\\ &=\frac {3 \sqrt {1-a^2 x^2}}{128 a^5}+\frac {\left (1-a^2 x^2\right )^{3/2}}{192 a^5}-\frac {3 \left (1-a^2 x^2\right )^{5/2}}{80 a^5}+\frac {\left (1-a^2 x^2\right )^{7/2}}{56 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{128 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{64 a^2}+\frac {3}{16} x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {1}{8} a^2 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{64 a^5}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{128 a^5}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{128 a^5}\\ \end {align*}

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Mathematica [A]
time = 0.95, size = 272, normalized size = 0.93 \begin {gather*} \frac {121 \sqrt {1-a^2 x^2}+218 a^2 x^2 \sqrt {1-a^2 x^2}+216 a^4 x^4 \sqrt {1-a^2 x^2}-240 a^6 x^6 \sqrt {1-a^2 x^2}-315 a x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-210 a^3 x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)+2520 a^5 x^5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-1680 a^7 x^7 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)-315 i \tanh ^{-1}(a x) \log \left (1-i e^{-\tanh ^{-1}(a x)}\right )+315 i \tanh ^{-1}(a x) \log \left (1+i e^{-\tanh ^{-1}(a x)}\right )-315 i \text {PolyLog}\left (2,-i e^{-\tanh ^{-1}(a x)}\right )+315 i \text {PolyLog}\left (2,i e^{-\tanh ^{-1}(a x)}\right )}{13440 a^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(1 - a^2*x^2)^(3/2)*ArcTanh[a*x],x]

[Out]

(121*Sqrt[1 - a^2*x^2] + 218*a^2*x^2*Sqrt[1 - a^2*x^2] + 216*a^4*x^4*Sqrt[1 - a^2*x^2] - 240*a^6*x^6*Sqrt[1 -
a^2*x^2] - 315*a*x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 210*a^3*x^3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] + 2520*a^5*x^5*
Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - 1680*a^7*x^7*Sqrt[1 - a^2*x^2]*ArcTanh[a*x] - (315*I)*ArcTanh[a*x]*Log[1 - I/
E^ArcTanh[a*x]] + (315*I)*ArcTanh[a*x]*Log[1 + I/E^ArcTanh[a*x]] - (315*I)*PolyLog[2, (-I)/E^ArcTanh[a*x]] + (
315*I)*PolyLog[2, I/E^ArcTanh[a*x]])/(13440*a^5)

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Maple [A]
time = 1.64, size = 215, normalized size = 0.74

method	result	size
default	\(-\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (1680 \arctanh \left (a x \right ) a^{7} x^{7}+240 a^{6} x^{6}-2520 \arctanh \left (a x \right ) a^{5} x^{5}-216 a^{4} x^{4}+210 a^{3} x^{3} \arctanh \left (a x \right )-218 a^{2} x^{2}+315 a x \arctanh \left (a x \right )-121\right )}{13440 a^{5}}-\frac {3 i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{128 a^{5}}+\frac {3 i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{128 a^{5}}-\frac {3 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{128 a^{5}}+\frac {3 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{128 a^{5}}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(-a^2*x^2+1)^(3/2)*arctanh(a*x),x,method=_RETURNVERBOSE)

[Out]

-1/13440/a^5*(-(a*x-1)*(a*x+1))^(1/2)*(1680*arctanh(a*x)*a^7*x^7+240*a^6*x^6-2520*arctanh(a*x)*a^5*x^5-216*a^4
*x^4+210*a^3*x^3*arctanh(a*x)-218*a^2*x^2+315*a*x*arctanh(a*x)-121)-3/128*I*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))
*arctanh(a*x)/a^5+3/128*I*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))*arctanh(a*x)/a^5-3/128*I*dilog(1+I*(a*x+1)/(-a^2*
x^2+1)^(1/2))/a^5+3/128*I*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))/a^5

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-a^2*x^2+1)^(3/2)*arctanh(a*x),x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*x^4*arctanh(a*x), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-a^2*x^2+1)^(3/2)*arctanh(a*x),x, algorithm="fricas")

[Out]

integral(-(a^2*x^6 - x^4)*sqrt(-a^2*x^2 + 1)*arctanh(a*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \operatorname {atanh}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(-a**2*x**2+1)**(3/2)*atanh(a*x),x)

[Out]

Integral(x**4*(-(a*x - 1)*(a*x + 1))**(3/2)*atanh(a*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(-a^2*x^2+1)^(3/2)*arctanh(a*x),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*x^4*arctanh(a*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^4\,\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*atanh(a*x)*(1 - a^2*x^2)^(3/2),x)

[Out]

int(x^4*atanh(a*x)*(1 - a^2*x^2)^(3/2), x)

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