3.1.16 \(\int x^2 (a+b \tanh ^{-1}(c x))^2 \, dx\) [16]

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

[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 = 0.15, 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 = {6037, 6127, 327, 212, 6131, 6055, 2449, 2352} \begin {gather*} \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} \end {gather*}

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 212

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

Rule 327

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 2352

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

Rule 2449

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 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 6055

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 6127

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 6131

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 ) \text {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 = 0.18, size = 122, normalized size = 0.94 \begin {gather*} \frac {b^2 c x+a b c^2 x^2+a^2 c^3 x^3+b^2 \left (-1+c^3 x^3\right ) \tanh ^{-1}(c x)^2+b \tanh ^{-1}(c x) \left (-b+b c^2 x^2+2 a c^3 x^3-2 b \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )+a b \log \left (-1+c^2 x^2\right )+b^2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )}{3 c^3} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(244\) vs. \(2(116)=232\).
time = 0.06, size = 245, normalized size = 1.88

method	result	size
derivativedivides	\(\frac {\frac {c^{3} x^{3} a^{2}}{3}+\frac {b^{2} c^{3} x^{3} \arctanh \left (c x \right )^{2}}{3}+\frac {b^{2} \arctanh \left (c x \right ) c^{2} x^{2}}{3}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{3}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{3}+\frac {b^{2} c x}{3}+\frac {b^{2} \ln \left (c x -1\right )}{6}-\frac {b^{2} \ln \left (c x +1\right )}{6}-\frac {b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{3}-\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} \ln \left (c x -1\right )^{2}}{12}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{6}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{12}+\frac {2 a b \,c^{3} x^{3} \arctanh \left (c x \right )}{3}+\frac {a b \,c^{2} x^{2}}{3}+\frac {a b \ln \left (c x -1\right )}{3}+\frac {a b \ln \left (c x +1\right )}{3}}{c^{3}}\)	\(245\)
default	\(\frac {\frac {c^{3} x^{3} a^{2}}{3}+\frac {b^{2} c^{3} x^{3} \arctanh \left (c x \right )^{2}}{3}+\frac {b^{2} \arctanh \left (c x \right ) c^{2} x^{2}}{3}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{3}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{3}+\frac {b^{2} c x}{3}+\frac {b^{2} \ln \left (c x -1\right )}{6}-\frac {b^{2} \ln \left (c x +1\right )}{6}-\frac {b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{3}-\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} \ln \left (c x -1\right )^{2}}{12}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{6}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{12}+\frac {2 a b \,c^{3} x^{3} \arctanh \left (c x \right )}{3}+\frac {a b \,c^{2} x^{2}}{3}+\frac {a b \ln \left (c x -1\right )}{3}+\frac {a b \ln \left (c x +1\right )}{3}}{c^{3}}\)	\(245\)
risch	\(\frac {b^{2} x}{3 c^{2}}+\frac {a^{2} x^{3}}{3}+\frac {b^{2} \ln \left (c x +1\right )^{2} x^{3}}{12}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{12 c^{3}}-\frac {b^{2} \ln \left (c x +1\right )}{6 c^{3}}-\frac {11 a b}{9 c^{3}}+\frac {b^{2} \ln \left (-c x +1\right )^{2} x^{3}}{12}-\frac {b^{2} \ln \left (-c x +1\right )^{2}}{12 c^{3}}+\frac {11 b^{2} \ln \left (-c x +1\right )}{18 c^{3}}+\frac {b^{2} \ln \left (c x +1\right ) x^{2}}{6 c}+\frac {a b \ln \left (-c x +1\right )}{3 c^{3}}-\frac {a b \ln \left (-c x +1\right ) x^{3}}{3}-\frac {b^{2} \ln \left (-c x +1\right ) x^{2}}{6 c}+\frac {a b \,x^{2}}{3 c}-\frac {a^{2}}{3 c^{3}}-\frac {17 b^{2}}{54 c^{3}}+\frac {b a \ln \left (c x +1\right ) x^{3}}{3}+\frac {b a \ln \left (c x +1\right )}{3 c^{3}}-\frac {b^{2} \ln \left (-c x +1\right ) \ln \left (c x +1\right ) x^{3}}{6}-\frac {b^{2} \ln \left (-c x +1\right ) \ln \left (c x +1\right )}{6 c^{3}}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{3 c^{3}}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{3 c^{3}}-\frac {b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{3 c^{3}}-\frac {4 b^{2} \ln \left (c x -1\right )}{9 c^{3}}\)	\(350\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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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