3.2.2 \(\int x^2 (a+b \tanh ^{-1}(c x^3)) \, dx\) [102]

Optimal. Leaf size=37 \[ \frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-c^2 x^6\right )}{6 c} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6037, 266} \begin {gather*} \frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-c^2 x^6\right )}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c*x^3]),x]

[Out]

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

Rule 266

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

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
]

Rubi steps

\begin {align*} \int x^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-(b c) \int \frac {x^5}{1-c^2 x^6} \, dx\\ &=\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-c^2 x^6\right )}{6 c}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 42, normalized size = 1.14 \begin {gather*} \frac {a x^3}{3}+\frac {1}{3} b x^3 \tanh ^{-1}\left (c x^3\right )+\frac {b \log \left (1-c^2 x^6\right )}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.02, size = 39, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {a c \,x^{3}+b c \,x^{3} \arctanh \left (c \,x^{3}\right )+\frac {b \ln \left (-c^{2} x^{6}+1\right )}{2}}{3 c}\)	\(39\)
default	\(\frac {a c \,x^{3}+b c \,x^{3} \arctanh \left (c \,x^{3}\right )+\frac {b \ln \left (-c^{2} x^{6}+1\right )}{2}}{3 c}\)	\(39\)
risch	\(\frac {x^{3} b \ln \left (c \,x^{3}+1\right )}{6}-\frac {b \,x^{3} \ln \left (-c \,x^{3}+1\right )}{6}+\frac {x^{3} a}{3}+\frac {b \ln \left (c^{2} x^{6}-1\right )}{6 c}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 37, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, a x^{3} + \frac {{\left (2 \, c x^{3} \operatorname {artanh}\left (c x^{3}\right ) + \log \left (-c^{2} x^{6} + 1\right )\right )} b}{6 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 50, normalized size = 1.35 \begin {gather*} \frac {b c x^{3} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, a c x^{3} + b \log \left (c^{2} x^{6} - 1\right )}{6 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*c*x^3*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 2*a*c*x^3 + b*log(c^2*x^6 - 1))/c

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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)),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (33) = 66\).
time = 0.44, size = 188, normalized size = 5.08 \begin {gather*} \frac {1}{3} \, a x^{3} + \frac {1}{3} \, b c {\left (\frac {\log \left (\frac {{\left | -c x^{3} - 1 \right |}}{{\left | c x^{3} - 1 \right |}}\right )}{c^{2}} - \frac {\log \left ({\left | -\frac {c x^{3} + 1}{c x^{3} - 1} + 1 \right |}\right )}{c^{2}} + \frac {\log \left (-\frac {\frac {c {\left (\frac {c x^{3} + 1}{c x^{3} - 1} + 1\right )}}{\frac {{\left (c x^{3} + 1\right )} c}{c x^{3} - 1} - c} + 1}{\frac {c {\left (\frac {c x^{3} + 1}{c x^{3} - 1} + 1\right )}}{\frac {{\left (c x^{3} + 1\right )} c}{c x^{3} - 1} - c} - 1}\right )}{c^{2} {\left (\frac {c x^{3} + 1}{c x^{3} - 1} - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 + 1/3*b*c*(log(abs(-c*x^3 - 1)/abs(c*x^3 - 1))/c^2 - log(abs(-(c*x^3 + 1)/(c*x^3 - 1) + 1))/c^2 + lo
g(-(c*((c*x^3 + 1)/(c*x^3 - 1) + 1)/((c*x^3 + 1)*c/(c*x^3 - 1) - c) + 1)/(c*((c*x^3 + 1)/(c*x^3 - 1) + 1)/((c*
x^3 + 1)*c/(c*x^3 - 1) - c) - 1))/(c^2*((c*x^3 + 1)/(c*x^3 - 1) - 1)))

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Mupad [B]
time = 0.78, size = 52, normalized size = 1.41 \begin {gather*} \frac {a\,x^3}{3}+\frac {b\,\ln \left (c^2\,x^6-1\right )}{6\,c}+\frac {b\,x^3\,\ln \left (c\,x^3+1\right )}{6}-\frac {b\,x^3\,\ln \left (1-c\,x^3\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atanh(c*x^3)),x)

[Out]

(a*x^3)/3 + (b*log(c^2*x^6 - 1))/(6*c) + (b*x^3*log(c*x^3 + 1))/6 - (b*x^3*log(1 - c*x^3))/6

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