3.16 \(\int x^2 (a+b \tanh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=130 \[ \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^3}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{3 c^3}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^3}+\frac {b^2 x}{3 c^2} \]

[Out]

1/3*b^2*x/c^2-1/3*b^2*arctanh(c*x)/c^3+1/3*b*x^2*(a+b*arctanh(c*x))/c+1/3*(a+b*arctanh(c*x))^2/c^3+1/3*x^3*(a+
b*arctanh(c*x))^2-2/3*b*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^3-1/3*b^2*polylog(2,1-2/(-c*x+1))/c^3

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Rubi [A]  time = 0.20, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5916, 5980, 321, 206, 5984, 5918, 2402, 2315} \[ -\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{3 c^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^3}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+\frac {b^2 x}{3 c^2}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c*x])^2,x]

[Out]

(b^2*x)/(3*c^2) - (b^2*ArcTanh[c*x])/(3*c^3) + (b*x^2*(a + b*ArcTanh[c*x]))/(3*c) + (a + b*ArcTanh[c*x])^2/(3*
c^3) + (x^3*(a + b*ArcTanh[c*x])^2)/3 - (2*b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(3*c^3) - (b^2*PolyLog[2,
1 - 2/(1 - c*x)])/(3*c^3)

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 321

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 5918

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

Rule 5980

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

Rule 5984

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

Rubi steps

\begin {align*} \int x^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{3} (2 b c) \int \frac {x^3 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {(2 b) \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c}-\frac {(2 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c}\\ &=\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{3} b^2 \int \frac {x^2}{1-c^2 x^2} \, dx-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{3 c^2}\\ &=\frac {b^2 x}{3 c^2}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^3}-\frac {b^2 \int \frac {1}{1-c^2 x^2} \, dx}{3 c^2}+\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{3 c^2}\\ &=\frac {b^2 x}{3 c^2}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^3}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^3}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{3 c^3}\\ &=\frac {b^2 x}{3 c^2}-\frac {b^2 \tanh ^{-1}(c x)}{3 c^3}+\frac {b x^2 \left (a+b \tanh ^{-1}(c x)\right )}{3 c}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{3 c^3}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{3 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 122, normalized size = 0.94 \[ \frac {a^2 c^3 x^3+a b c^2 x^2+a b \log \left (c^2 x^2-1\right )+b \tanh ^{-1}(c x) \left (2 a c^3 x^3+b c^2 x^2-2 b \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-b\right )+b^2 \left (c^3 x^3-1\right ) \tanh ^{-1}(c x)^2+b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+b^2 c x}{3 c^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2*(a + b*ArcTanh[c*x])^2,x]

[Out]

(b^2*c*x + a*b*c^2*x^2 + a^2*c^3*x^3 + b^2*(-1 + c^3*x^3)*ArcTanh[c*x]^2 + b*ArcTanh[c*x]*(-b + b*c^2*x^2 + 2*
a*c^3*x^3 - 2*b*Log[1 + E^(-2*ArcTanh[c*x])]) + a*b*Log[-1 + c^2*x^2] + b^2*PolyLog[2, -E^(-2*ArcTanh[c*x])])/
(3*c^3)

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{2} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x^{2} \operatorname {artanh}\left (c x\right ) + a^{2} x^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^2*arctanh(c*x)^2 + 2*a*b*x^2*arctanh(c*x) + a^2*x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x))^2,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2*x^2, x)

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maple [B]  time = 0.01, size = 270, normalized size = 2.08 \[ \frac {x^{3} a^{2}}{3}+\frac {b^{2} x^{3} \arctanh \left (c x \right )^{2}}{3}+\frac {b^{2} \arctanh \left (c x \right ) x^{2}}{3 c}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{3 c^{3}}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{3 c^{3}}+\frac {b^{2} x}{3 c^{2}}+\frac {b^{2} \ln \left (c x -1\right )}{6 c^{3}}-\frac {b^{2} \ln \left (c x +1\right )}{6 c^{3}}+\frac {b^{2} \ln \left (c x -1\right )^{2}}{12 c^{3}}-\frac {b^{2} \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{3 c^{3}}-\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{6 c^{3}}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{12 c^{3}}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{6 c^{3}}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{6 c^{3}}+\frac {2 a b \,x^{3} \arctanh \left (c x \right )}{3}+\frac {a b \,x^{2}}{3 c}+\frac {a b \ln \left (c x -1\right )}{3 c^{3}}+\frac {a b \ln \left (c x +1\right )}{3 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctanh(c*x))^2,x)

[Out]

1/3*x^3*a^2+1/3*b^2*x^3*arctanh(c*x)^2+1/3/c*b^2*arctanh(c*x)*x^2+1/3/c^3*b^2*arctanh(c*x)*ln(c*x-1)+1/3/c^3*b
^2*arctanh(c*x)*ln(c*x+1)+1/3*b^2*x/c^2+1/6/c^3*b^2*ln(c*x-1)-1/6/c^3*b^2*ln(c*x+1)+1/12/c^3*b^2*ln(c*x-1)^2-1
/3/c^3*b^2*dilog(1/2+1/2*c*x)-1/6/c^3*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-1/12/c^3*b^2*ln(c*x+1)^2-1/6/c^3*b^2*ln(-1
/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/6/c^3*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+2/3*a*b*x^3*arctanh(c*x)+1/3*a*b*x^2/c+1/3/
c^3*a*b*ln(c*x-1)+1/3/c^3*a*b*ln(c*x+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a^{2} x^{3} + \frac {1}{3} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {x^{2}}{c^{2}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4}}\right )}\right )} a b - \frac {1}{216} \, {\left (2 \, c^{4} {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{6}} - \frac {3 \, \log \left (c x + 1\right )}{c^{7}} + \frac {3 \, \log \left (c x - 1\right )}{c^{7}}\right )} - 3 \, c^{3} {\left (\frac {x^{2}}{c^{4}} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{6}}\right )} - 648 \, c^{3} \int \frac {x^{3} \log \left (c x + 1\right )}{9 \, {\left (c^{4} x^{2} - c^{2}\right )}}\,{d x} + 9 \, c^{2} {\left (\frac {2 \, x}{c^{4}} - \frac {\log \left (c x + 1\right )}{c^{5}} + \frac {\log \left (c x - 1\right )}{c^{5}}\right )} - 324 \, c \int \frac {x \log \left (c x + 1\right )}{9 \, {\left (c^{4} x^{2} - c^{2}\right )}}\,{d x} - \frac {6 \, {\left (3 \, c^{3} x^{3} \log \left (c x + 1\right )^{2} + {\left (2 \, c^{3} x^{3} - 3 \, c^{2} x^{2} + 6 \, c x - 6 \, {\left (c^{3} x^{3} + 1\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )\right )}}{c^{3}} - \frac {2 \, {\left (c x - 1\right )}^{3} {\left (9 \, \log \left (-c x + 1\right )^{2} - 6 \, \log \left (-c x + 1\right ) + 2\right )} + 27 \, {\left (c x - 1\right )}^{2} {\left (2 \, \log \left (-c x + 1\right )^{2} - 2 \, \log \left (-c x + 1\right ) + 1\right )} + 54 \, {\left (c x - 1\right )} {\left (\log \left (-c x + 1\right )^{2} - 2 \, \log \left (-c x + 1\right ) + 2\right )}}{c^{3}} + \frac {18 \, \log \left (9 \, c^{4} x^{2} - 9 \, c^{2}\right )}{c^{3}} - 324 \, \int \frac {\log \left (c x + 1\right )}{9 \, {\left (c^{4} x^{2} - c^{2}\right )}}\,{d x}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x))^2,x, algorithm="maxima")

[Out]

1/3*a^2*x^3 + 1/3*(2*x^3*arctanh(c*x) + c*(x^2/c^2 + log(c^2*x^2 - 1)/c^4))*a*b - 1/216*(2*c^4*(2*(c^2*x^3 + 3
*x)/c^6 - 3*log(c*x + 1)/c^7 + 3*log(c*x - 1)/c^7) - 3*c^3*(x^2/c^4 + log(c^2*x^2 - 1)/c^6) - 648*c^3*integrat
e(1/9*x^3*log(c*x + 1)/(c^4*x^2 - c^2), x) + 9*c^2*(2*x/c^4 - log(c*x + 1)/c^5 + log(c*x - 1)/c^5) - 324*c*int
egrate(1/9*x*log(c*x + 1)/(c^4*x^2 - c^2), x) - 6*(3*c^3*x^3*log(c*x + 1)^2 + (2*c^3*x^3 - 3*c^2*x^2 + 6*c*x -
6*(c^3*x^3 + 1)*log(c*x + 1))*log(-c*x + 1))/c^3 - (2*(c*x - 1)^3*(9*log(-c*x + 1)^2 - 6*log(-c*x + 1) + 2) +
27*(c*x - 1)^2*(2*log(-c*x + 1)^2 - 2*log(-c*x + 1) + 1) + 54*(c*x - 1)*(log(-c*x + 1)^2 - 2*log(-c*x + 1) +
2))/c^3 + 18*log(9*c^4*x^2 - 9*c^2)/c^3 - 324*integrate(1/9*log(c*x + 1)/(c^4*x^2 - c^2), x))*b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atanh(c*x))^2,x)

[Out]

int(x^2*(a + b*atanh(c*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*atanh(c*x))**2,x)

[Out]

Integral(x**2*(a + b*atanh(c*x))**2, x)

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