3.2.19 \(\int x^2 (a+b \tanh ^{-1}(c x^3))^2 \, dx\) [119]

Optimal. Leaf size=96 \[ \frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{3 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2-\frac {2 b \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \log \left (\frac {2}{1-c x^3}\right )}{3 c}-\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right )}{3 c} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6039, 6021, 6131, 6055, 2449, 2352} \begin {gather*} \frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2+\frac {\left (a+b \tanh ^{-1}\left (c x^3\right )\right )^2}{3 c}-\frac {2 b \log \left (\frac {2}{1-c x^3}\right ) \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{3 c}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x^3}\right )}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c*x^3])^2,x]

[Out]

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

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 6021

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

Rule 6039

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[S
implify[(m + 1)/n]]

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 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}\left (c x^3\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x^2 \left (2 a-b \log \left (1-c x^3\right )\right )^2-\frac {1}{2} b x^2 \left (-2 a+b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {1}{4} b^2 x^2 \log ^2\left (1+c x^3\right )\right ) \, dx\\ &=\frac {1}{4} \int x^2 \left (2 a-b \log \left (1-c x^3\right )\right )^2 \, dx-\frac {1}{2} b \int x^2 \left (-2 a+b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right ) \, dx+\frac {1}{4} b^2 \int x^2 \log ^2\left (1+c x^3\right ) \, dx\\ &=\frac {1}{12} \text {Subst}\left (\int (2 a-b \log (1-c x))^2 \, dx,x,x^3\right )-\frac {1}{6} b \text {Subst}\left (\int (-2 a+b \log (1-c x)) \log (1+c x) \, dx,x,x^3\right )+\frac {1}{12} b^2 \text {Subst}\left (\int \log ^2(1+c x) \, dx,x,x^3\right )\\ &=\frac {1}{6} b x^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )-\frac {\text {Subst}\left (\int (2 a-b \log (x))^2 \, dx,x,1-c x^3\right )}{12 c}+\frac {b^2 \text {Subst}\left (\int \log ^2(x) \, dx,x,1+c x^3\right )}{12 c}+\frac {1}{6} (b c) \text {Subst}\left (\int \frac {x (-2 a+b \log (1-c x))}{1+c x} \, dx,x,x^3\right )-\frac {1}{6} \left (b^2 c\right ) \text {Subst}\left (\int \frac {x \log (1+c x)}{1-c x} \, dx,x,x^3\right )\\ &=-\frac {\left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{12 c}+\frac {1}{6} b x^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {b^2 \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{12 c}-\frac {b \text {Subst}\left (\int (2 a-b \log (x)) \, dx,x,1-c x^3\right )}{6 c}-\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1+c x^3\right )}{6 c}+\frac {1}{6} (b c) \text {Subst}\left (\int \left (\frac {-2 a+b \log (1-c x)}{c}-\frac {-2 a+b \log (1-c x)}{c (1+c x)}\right ) \, dx,x,x^3\right )-\frac {1}{6} \left (b^2 c\right ) \text {Subst}\left (\int \left (-\frac {\log (1+c x)}{c}-\frac {\log (1+c x)}{c (-1+c x)}\right ) \, dx,x,x^3\right )\\ &=\frac {1}{3} a b x^3+\frac {b^2 x^3}{6}-\frac {\left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{12 c}-\frac {b^2 \left (1+c x^3\right ) \log \left (1+c x^3\right )}{6 c}+\frac {1}{6} b x^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {b^2 \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{12 c}+\frac {1}{6} b \text {Subst}\left (\int (-2 a+b \log (1-c x)) \, dx,x,x^3\right )-\frac {1}{6} b \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^3\right )+\frac {1}{6} b^2 \text {Subst}\left (\int \log (1+c x) \, dx,x,x^3\right )+\frac {1}{6} b^2 \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^3\right )+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1-c x^3\right )}{6 c}\\ &=\frac {b^2 x^3}{3}+\frac {b^2 \left (1-c x^3\right ) \log \left (1-c x^3\right )}{6 c}-\frac {\left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{12 c}+\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )}{6 c}-\frac {b^2 \left (1+c x^3\right ) \log \left (1+c x^3\right )}{6 c}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{6 c}+\frac {1}{6} b x^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {b^2 \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{12 c}-\frac {1}{6} b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^3\right )+\frac {1}{6} b^2 \text {Subst}\left (\int \log (1-c x) \, dx,x,x^3\right )-\frac {1}{6} b^2 \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^3\right )+\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1+c x^3\right )}{6 c}\\ &=\frac {b^2 x^3}{6}+\frac {b^2 \left (1-c x^3\right ) \log \left (1-c x^3\right )}{6 c}-\frac {\left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{12 c}+\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )}{6 c}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{6 c}+\frac {1}{6} b x^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {b^2 \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{12 c}+\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^3\right )}{6 c}-\frac {b^2 \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^3\right )}{6 c}-\frac {b^2 \text {Subst}\left (\int \log (x) \, dx,x,1-c x^3\right )}{6 c}\\ &=-\frac {\left (1-c x^3\right ) \left (2 a-b \log \left (1-c x^3\right )\right )^2}{12 c}+\frac {b \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (\frac {1}{2} \left (1+c x^3\right )\right )}{6 c}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )}{6 c}+\frac {1}{6} b x^3 \left (2 a-b \log \left (1-c x^3\right )\right ) \log \left (1+c x^3\right )+\frac {b^2 \left (1+c x^3\right ) \log ^2\left (1+c x^3\right )}{12 c}-\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-c x^3\right )\right )}{6 c}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+c x^3\right )\right )}{6 c}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 99, normalized size = 1.03 \begin {gather*} \frac {b^2 \left (-1+c x^3\right ) \tanh ^{-1}\left (c x^3\right )^2+2 b \tanh ^{-1}\left (c x^3\right ) \left (a c x^3-b \log \left (1+e^{-2 \tanh ^{-1}\left (c x^3\right )}\right )\right )+a \left (a c x^3+b \log \left (1-c^2 x^6\right )\right )+b^2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c x^3\right )}\right )}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcTanh[c*x^3])^2,x]

[Out]

(b^2*(-1 + c*x^3)*ArcTanh[c*x^3]^2 + 2*b*ArcTanh[c*x^3]*(a*c*x^3 - b*Log[1 + E^(-2*ArcTanh[c*x^3])]) + a*(a*c*
x^3 + b*Log[1 - c^2*x^6]) + b^2*PolyLog[2, -E^(-2*ArcTanh[c*x^3])])/(3*c)

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Maple [A]
time = 0.19, size = 137, normalized size = 1.43

method	result	size
derivativedivides	\(\frac {c \,x^{3} a^{2}+\arctanh \left (c \,x^{3}\right )^{2} b^{2} c \,x^{3}+b^{2} \arctanh \left (c \,x^{3}\right )^{2}-2 \arctanh \left (c \,x^{3}\right ) \ln \left (1+\frac {\left (c \,x^{3}+1\right )^{2}}{-c^{2} x^{6}+1}\right ) b^{2}-\polylog \left (2, -\frac {\left (c \,x^{3}+1\right )^{2}}{-c^{2} x^{6}+1}\right ) b^{2}+2 a b c \,x^{3} \arctanh \left (c \,x^{3}\right )+a b \ln \left (-c^{2} x^{6}+1\right )}{3 c}\)	\(137\)
default	\(\frac {c \,x^{3} a^{2}+\arctanh \left (c \,x^{3}\right )^{2} b^{2} c \,x^{3}+b^{2} \arctanh \left (c \,x^{3}\right )^{2}-2 \arctanh \left (c \,x^{3}\right ) \ln \left (1+\frac {\left (c \,x^{3}+1\right )^{2}}{-c^{2} x^{6}+1}\right ) b^{2}-\polylog \left (2, -\frac {\left (c \,x^{3}+1\right )^{2}}{-c^{2} x^{6}+1}\right ) b^{2}+2 a b c \,x^{3} \arctanh \left (c \,x^{3}\right )+a b \ln \left (-c^{2} x^{6}+1\right )}{3 c}\)	\(137\)
risch	\(\frac {a^{2} x^{3}}{3}-\frac {a^{2}}{3 c}-\frac {b^{2}}{3 c}-\frac {2 a b}{3 c}+\frac {\ln \left (-c \,x^{3}+1\right )^{2} x^{3} b^{2}}{12}-\frac {\ln \left (-c \,x^{3}+1\right )^{2} b^{2}}{12 c}+\frac {\ln \left (-c \,x^{3}+1\right ) b^{2}}{3 c}+\frac {b^{2} \ln \left (c \,x^{3}+1\right )^{2} x^{3}}{12}+\frac {b^{2} \ln \left (c \,x^{3}+1\right )^{2}}{12 c}-\frac {\ln \left (-c \,x^{3}+1\right ) x^{3} a b}{3}+\frac {\ln \left (-c \,x^{3}+1\right ) a b}{3 c}-\frac {b^{2} \dilog \left (\frac {c \,x^{3}}{2}+\frac {1}{2}\right )}{3 c}-\frac {b^{2} \ln \left (c \,x^{3}-1\right )}{3 c}+\frac {b a \ln \left (c \,x^{3}+1\right ) x^{3}}{3}+\frac {b a \ln \left (c \,x^{3}+1\right )}{3 c}-\frac {b^{2} \ln \left (-c \,x^{3}+1\right ) \ln \left (c \,x^{3}+1\right ) x^{3}}{6}-\frac {b^{2} \ln \left (-c \,x^{3}+1\right ) \ln \left (c \,x^{3}+1\right )}{6 c}+\frac {b^{2} \ln \left (\frac {1}{2}-\frac {c \,x^{3}}{2}\right ) \ln \left (c \,x^{3}+1\right )}{3 c}-\frac {b^{2} \ln \left (\frac {1}{2}-\frac {c \,x^{3}}{2}\right ) \ln \left (\frac {c \,x^{3}}{2}+\frac {1}{2}\right )}{3 c}\)	\(320\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctanh(c*x^3))^2,x,method=_RETURNVERBOSE)

[Out]

1/3/c*(c*x^3*a^2+arctanh(c*x^3)^2*b^2*c*x^3+b^2*arctanh(c*x^3)^2-2*arctanh(c*x^3)*ln(1+(c*x^3+1)^2/(-c^2*x^6+1
))*b^2-polylog(2,-(c*x^3+1)^2/(-c^2*x^6+1))*b^2+2*a*b*c*x^3*arctanh(c*x^3)+a*b*ln(-c^2*x^6+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^3))^2,x, algorithm="maxima")

[Out]

1/3*a^2*x^3 + 1/12*(x^3*log(-c*x^3 + 1)^2 - c^2*(2*x^3/c^2 - log(c*x^3 + 1)/c^3 + log(c*x^3 - 1)/c^3) - 2*(x^3
/c + log(c*x^3 - 1)/c^2)*c*log(-c*x^3 + 1) + 18*c*integrate(x^5*log(c*x^3 + 1)/(c^2*x^6 - 1), x) + (c*x^3*log(
c*x^3 + 1)^2 + 2*(c*x^3 - (c*x^3 + 1)*log(c*x^3 + 1))*log(-c*x^3 + 1))/c + (2*c*x^3 + log(c*x^3 - 1)^2 + 2*log
(c*x^3 - 1))/c - log(c^2*x^6 - 1)/c + 6*integrate(x^2*log(c*x^3 + 1)/(c^2*x^6 - 1), x))*b^2 + 1/3*(2*c*x^3*arc
tanh(c*x^3) + log(-c^2*x^6 + 1))*a*b/c

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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^3))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^2*arctanh(c*x^3)^2 + 2*a*b*x^2*arctanh(c*x^3) + a^2*x^2, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*atanh(c*x**3))**2,x)

[Out]

Timed out

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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^3))^2,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^3) + 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^3\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^3))^2,x)

[Out]

int(x^2*(a + b*atanh(c*x^3))^2, x)

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