3.1.100 \(\int x^8 (a+b \tanh ^{-1}(c x^3)) \, dx\) [100]

Optimal. Leaf size=48 \[ \frac {b x^6}{18 c}+\frac {1}{9} x^9 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-c^2 x^6\right )}{18 c^3} \]

[Out]

1/18*b*x^6/c+1/9*x^9*(a+b*arctanh(c*x^3))+1/18*b*ln(-c^2*x^6+1)/c^3

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Rubi [A]
time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6037, 272, 45} \begin {gather*} \frac {1}{9} x^9 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-c^2 x^6\right )}{18 c^3}+\frac {b x^6}{18 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8*(a + b*ArcTanh[c*x^3]),x]

[Out]

(b*x^6)/(18*c) + (x^9*(a + b*ArcTanh[c*x^3]))/9 + (b*Log[1 - c^2*x^6])/(18*c^3)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

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^8 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\frac {1}{9} x^9 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {1}{3} (b c) \int \frac {x^{11}}{1-c^2 x^6} \, dx\\ &=\frac {1}{9} x^9 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {1}{18} (b c) \text {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^6\right )\\ &=\frac {1}{9} x^9 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )-\frac {1}{18} (b c) \text {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^6\right )\\ &=\frac {b x^6}{18 c}+\frac {1}{9} x^9 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )+\frac {b \log \left (1-c^2 x^6\right )}{18 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 53, normalized size = 1.10 \begin {gather*} \frac {b x^6}{18 c}+\frac {a x^9}{9}+\frac {1}{9} b x^9 \tanh ^{-1}\left (c x^3\right )+\frac {b \log \left (1-c^2 x^6\right )}{18 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8*(a + b*ArcTanh[c*x^3]),x]

[Out]

(b*x^6)/(18*c) + (a*x^9)/9 + (b*x^9*ArcTanh[c*x^3])/9 + (b*Log[1 - c^2*x^6])/(18*c^3)

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Maple [A]
time = 0.05, size = 45, normalized size = 0.94

method	result	size
default	\(\frac {x^{9} a}{9}+\frac {b \,x^{9} \arctanh \left (c \,x^{3}\right )}{9}+\frac {b \,x^{6}}{18 c}+\frac {b \ln \left (c^{2} x^{6}-1\right )}{18 c^{3}}\)	\(45\)
risch	\(\frac {x^{9} b \ln \left (c \,x^{3}+1\right )}{18}-\frac {x^{9} b \ln \left (-c \,x^{3}+1\right )}{18}+\frac {x^{9} a}{9}+\frac {b \,x^{6}}{18 c}+\frac {b \ln \left (c^{2} x^{6}-1\right )}{18 c^{3}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a+b*arctanh(c*x^3)),x,method=_RETURNVERBOSE)

[Out]

1/9*x^9*a+1/9*b*x^9*arctanh(c*x^3)+1/18*b*x^6/c+1/18*b/c^3*ln(c^2*x^6-1)

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Maxima [A]
time = 0.27, size = 46, normalized size = 0.96 \begin {gather*} \frac {1}{9} \, a x^{9} + \frac {1}{18} \, {\left (2 \, x^{9} \operatorname {artanh}\left (c x^{3}\right ) + {\left (\frac {x^{6}}{c^{2}} + \frac {\log \left (c^{2} x^{6} - 1\right )}{c^{4}}\right )} c\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(a+b*arctanh(c*x^3)),x, algorithm="maxima")

[Out]

1/9*a*x^9 + 1/18*(2*x^9*arctanh(c*x^3) + (x^6/c^2 + log(c^2*x^6 - 1)/c^4)*c)*b

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Fricas [A]
time = 0.34, size = 62, normalized size = 1.29 \begin {gather*} \frac {b c^{3} x^{9} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, a c^{3} x^{9} + b c^{2} x^{6} + b \log \left (c^{2} x^{6} - 1\right )}{18 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(a+b*arctanh(c*x^3)),x, algorithm="fricas")

[Out]

1/18*(b*c^3*x^9*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 2*a*c^3*x^9 + b*c^2*x^6 + b*log(c^2*x^6 - 1))/c^3

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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**8*(a+b*atanh(c*x**3)),x)

[Out]

Timed out

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Giac [A]
time = 0.44, size = 57, normalized size = 1.19 \begin {gather*} \frac {1}{18} \, b x^{9} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + \frac {1}{9} \, a x^{9} + \frac {b x^{6}}{18 \, c} + \frac {b \log \left (c^{2} x^{6} - 1\right )}{18 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(a+b*arctanh(c*x^3)),x, algorithm="giac")

[Out]

1/18*b*x^9*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 1/9*a*x^9 + 1/18*b*x^6/c + 1/18*b*log(c^2*x^6 - 1)/c^3

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Mupad [B]
time = 0.82, size = 61, normalized size = 1.27 \begin {gather*} \frac {a\,x^9}{9}+\frac {b\,\ln \left (c^2\,x^6-1\right )}{18\,c^3}+\frac {b\,x^6}{18\,c}+\frac {b\,x^9\,\ln \left (c\,x^3+1\right )}{18}-\frac {b\,x^9\,\ln \left (1-c\,x^3\right )}{18} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(a + b*atanh(c*x^3)),x)

[Out]

(a*x^9)/9 + (b*log(c^2*x^6 - 1))/(18*c^3) + (b*x^6)/(18*c) + (b*x^9*log(c*x^3 + 1))/18 - (b*x^9*log(1 - c*x^3)
)/18

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