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

Optimal. Leaf size=87 \[ \frac {1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac {\tanh ^{-1}(a x)}{24 a^4}+\frac {a^3 x^7}{56}+\frac {x}{24 a^3}-\frac {1}{3} a^2 x^6 \tanh ^{-1}(a x)-\frac {a x^5}{24}+\frac {1}{4} x^4 \tanh ^{-1}(a x)+\frac {x^3}{72 a} \]

[Out]

1/24*x/a^3+1/72*x^3/a-1/24*a*x^5+1/56*a^3*x^7-1/24*arctanh(a*x)/a^4+1/4*x^4*arctanh(a*x)-1/3*a^2*x^6*arctanh(a
*x)+1/8*a^4*x^8*arctanh(a*x)

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Rubi [A]  time = 0.14, antiderivative size = 87, 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, 302, 206} \[ \frac {a^3 x^7}{56}+\frac {1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac {1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac {x}{24 a^3}-\frac {\tanh ^{-1}(a x)}{24 a^4}-\frac {a x^5}{24}+\frac {x^3}{72 a}+\frac {1}{4} x^4 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(24*a^3) + x^3/(72*a) - (a*x^5)/24 + (a^3*x^7)/56 - ArcTanh[a*x]/(24*a^4) + (x^4*ArcTanh[a*x])/4 - (a^2*x^6*
ArcTanh[a*x])/3 + (a^4*x^8*ArcTanh[a*x])/8

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^3 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x) \, dx &=\int \left (x^3 \tanh ^{-1}(a x)-2 a^2 x^5 \tanh ^{-1}(a x)+a^4 x^7 \tanh ^{-1}(a x)\right ) \, dx\\ &=-\left (\left (2 a^2\right ) \int x^5 \tanh ^{-1}(a x) \, dx\right )+a^4 \int x^7 \tanh ^{-1}(a x) \, dx+\int x^3 \tanh ^{-1}(a x) \, dx\\ &=\frac {1}{4} x^4 \tanh ^{-1}(a x)-\frac {1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac {1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac {1}{4} a \int \frac {x^4}{1-a^2 x^2} \, dx+\frac {1}{3} a^3 \int \frac {x^6}{1-a^2 x^2} \, dx-\frac {1}{8} a^5 \int \frac {x^8}{1-a^2 x^2} \, dx\\ &=\frac {1}{4} x^4 \tanh ^{-1}(a x)-\frac {1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac {1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac {1}{4} a \int \left (-\frac {1}{a^4}-\frac {x^2}{a^2}+\frac {1}{a^4 \left (1-a^2 x^2\right )}\right ) \, dx+\frac {1}{3} a^3 \int \left (-\frac {1}{a^6}-\frac {x^2}{a^4}-\frac {x^4}{a^2}+\frac {1}{a^6 \left (1-a^2 x^2\right )}\right ) \, dx-\frac {1}{8} a^5 \int \left (-\frac {1}{a^8}-\frac {x^2}{a^6}-\frac {x^4}{a^4}-\frac {x^6}{a^2}+\frac {1}{a^8 \left (1-a^2 x^2\right )}\right ) \, dx\\ &=\frac {x}{24 a^3}+\frac {x^3}{72 a}-\frac {a x^5}{24}+\frac {a^3 x^7}{56}+\frac {1}{4} x^4 \tanh ^{-1}(a x)-\frac {1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac {1}{8} a^4 x^8 \tanh ^{-1}(a x)-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{8 a^3}-\frac {\int \frac {1}{1-a^2 x^2} \, dx}{4 a^3}+\frac {\int \frac {1}{1-a^2 x^2} \, dx}{3 a^3}\\ &=\frac {x}{24 a^3}+\frac {x^3}{72 a}-\frac {a x^5}{24}+\frac {a^3 x^7}{56}-\frac {\tanh ^{-1}(a x)}{24 a^4}+\frac {1}{4} x^4 \tanh ^{-1}(a x)-\frac {1}{3} a^2 x^6 \tanh ^{-1}(a x)+\frac {1}{8} a^4 x^8 \tanh ^{-1}(a x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 103, normalized size = 1.18 \[ \frac {1}{8} a^4 x^8 \tanh ^{-1}(a x)+\frac {\log (1-a x)}{48 a^4}-\frac {\log (a x+1)}{48 a^4}+\frac {a^3 x^7}{56}+\frac {x}{24 a^3}-\frac {1}{3} a^2 x^6 \tanh ^{-1}(a x)-\frac {a x^5}{24}+\frac {1}{4} x^4 \tanh ^{-1}(a x)+\frac {x^3}{72 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(24*a^3) + x^3/(72*a) - (a*x^5)/24 + (a^3*x^7)/56 + (x^4*ArcTanh[a*x])/4 - (a^2*x^6*ArcTanh[a*x])/3 + (a^4*x
^8*ArcTanh[a*x])/8 + Log[1 - a*x]/(48*a^4) - Log[1 + a*x]/(48*a^4)

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fricas [A]  time = 0.51, size = 77, normalized size = 0.89 \[ \frac {18 \, a^{7} x^{7} - 42 \, a^{5} x^{5} + 14 \, a^{3} x^{3} + 42 \, a x + 21 \, {\left (3 \, a^{8} x^{8} - 8 \, a^{6} x^{6} + 6 \, a^{4} x^{4} - 1\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )}{1008 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1008*(18*a^7*x^7 - 42*a^5*x^5 + 14*a^3*x^3 + 42*a*x + 21*(3*a^8*x^8 - 8*a^6*x^6 + 6*a^4*x^4 - 1)*log(-(a*x +
1)/(a*x - 1)))/a^4

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giac [B]  time = 0.20, size = 240, normalized size = 2.76 \[ \frac {4}{63} \, a {\left (\frac {\frac {28 \, {\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} - \frac {7 \, {\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}} + \frac {21 \, {\left (a x + 1\right )}^{2}}{{\left (a x - 1\right )}^{2}} - \frac {7 \, {\left (a x + 1\right )}}{a x - 1} + 1}{a^{5} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{7}} + \frac {84 \, {\left (\frac {{\left (a x + 1\right )}^{5}}{{\left (a x - 1\right )}^{5}} + \frac {{\left (a x + 1\right )}^{4}}{{\left (a x - 1\right )}^{4}} + \frac {{\left (a x + 1\right )}^{3}}{{\left (a x - 1\right )}^{3}}\right )} \log \left (-\frac {\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} + 1}{\frac {a {\left (\frac {a x + 1}{a x - 1} + 1\right )}}{\frac {{\left (a x + 1\right )} a}{a x - 1} - a} - 1}\right )}{a^{5} {\left (\frac {a x + 1}{a x - 1} - 1\right )}^{8}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/63*a*((28*(a*x + 1)^4/(a*x - 1)^4 - 7*(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)/(a^5*((a*x + 1)/(a*x - 1) - 1)^7) + 84*((a*x + 1)^5/(a*x - 1)^5 + (a*x + 1)^4/(a*x - 1)^4 + (a*x +
1)^3/(a*x - 1)^3)*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^5*((a*x + 1)/(a*x - 1) - 1)^8))

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maple [A]  time = 0.03, size = 85, normalized size = 0.98 \[ \frac {a^{4} x^{8} \arctanh \left (a x \right )}{8}-\frac {a^{2} x^{6} \arctanh \left (a x \right )}{3}+\frac {x^{4} \arctanh \left (a x \right )}{4}+\frac {a^{3} x^{7}}{56}-\frac {a \,x^{5}}{24}+\frac {x^{3}}{72 a}+\frac {x}{24 a^{3}}+\frac {\ln \left (a x -1\right )}{48 a^{4}}-\frac {\ln \left (a x +1\right )}{48 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^4*x^8*arctanh(a*x)-1/3*a^2*x^6*arctanh(a*x)+1/4*x^4*arctanh(a*x)+1/56*a^3*x^7-1/24*a*x^5+1/72*x^3/a+1/24
*x/a^3+1/48/a^4*ln(a*x-1)-1/48/a^4*ln(a*x+1)

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maxima [A]  time = 0.30, size = 88, normalized size = 1.01 \[ \frac {1}{1008} \, a {\left (\frac {2 \, {\left (9 \, a^{6} x^{7} - 21 \, a^{4} x^{5} + 7 \, a^{2} x^{3} + 21 \, x\right )}}{a^{4}} - \frac {21 \, \log \left (a x + 1\right )}{a^{5}} + \frac {21 \, \log \left (a x - 1\right )}{a^{5}}\right )} + \frac {1}{24} \, {\left (3 \, a^{4} x^{8} - 8 \, a^{2} x^{6} + 6 \, x^{4}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1008*a*(2*(9*a^6*x^7 - 21*a^4*x^5 + 7*a^2*x^3 + 21*x)/a^4 - 21*log(a*x + 1)/a^5 + 21*log(a*x - 1)/a^5) + 1/2
4*(3*a^4*x^8 - 8*a^2*x^6 + 6*x^4)*arctanh(a*x)

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mupad [B]  time = 1.21, size = 101, normalized size = 1.16 \[ \frac {x}{24\,a^3}-\frac {a\,x^5}{24}+\ln \left (a\,x+1\right )\,\left (\frac {a^4\,x^8}{16}-\frac {a^2\,x^6}{6}+\frac {x^4}{8}\right )-\ln \left (1-a\,x\right )\,\left (\frac {a^4\,x^8}{16}-\frac {a^2\,x^6}{6}+\frac {x^4}{8}\right )+\frac {x^3}{72\,a}+\frac {a^3\,x^7}{56}+\frac {\mathrm {atan}\left (a\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{24\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atanh(a*x)*(a^2*x^2 - 1)^2,x)

[Out]

x/(24*a^3) - (a*x^5)/24 + (atan(a*x*1i)*1i)/(24*a^4) + log(a*x + 1)*(x^4/8 - (a^2*x^6)/6 + (a^4*x^8)/16) - log
(1 - a*x)*(x^4/8 - (a^2*x^6)/6 + (a^4*x^8)/16) + x^3/(72*a) + (a^3*x^7)/56

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sympy [A]  time = 2.53, size = 76, normalized size = 0.87 \[ \begin {cases} \frac {a^{4} x^{8} \operatorname {atanh}{\left (a x \right )}}{8} + \frac {a^{3} x^{7}}{56} - \frac {a^{2} x^{6} \operatorname {atanh}{\left (a x \right )}}{3} - \frac {a x^{5}}{24} + \frac {x^{4} \operatorname {atanh}{\left (a x \right )}}{4} + \frac {x^{3}}{72 a} + \frac {x}{24 a^{3}} - \frac {\operatorname {atanh}{\left (a x \right )}}{24 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x**8*atanh(a*x)/8 + a**3*x**7/56 - a**2*x**6*atanh(a*x)/3 - a*x**5/24 + x**4*atanh(a*x)/4 + x*
*3/(72*a) + x/(24*a**3) - atanh(a*x)/(24*a**4), Ne(a, 0)), (0, True))

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