3.2.6 \(\int \frac {a+b \tanh ^{-1}(c x^3)}{x^{10}} \, dx\) [106]

Optimal. Leaf size=56 \[ -\frac {b c}{18 x^6}-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{3} b c^3 \log (x)-\frac {1}{18} b c^3 \log \left (1-c^2 x^6\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 56, 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, 46} \begin {gather*} -\frac {a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{3} b c^3 \log (x)-\frac {1}{18} b c^3 \log \left (1-c^2 x^6\right )-\frac {b c}{18 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^3])/x^10,x]

[Out]

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

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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 \frac {a+b \tanh ^{-1}\left (c x^3\right )}{x^{10}} \, dx &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{3} (b c) \int \frac {1}{x^7 \left (1-c^2 x^6\right )} \, dx\\ &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{18} (b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^6\right )\\ &=-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{18} (b c) \text {Subst}\left (\int \left (\frac {1}{x^2}+\frac {c^2}{x}-\frac {c^4}{-1+c^2 x}\right ) \, dx,x,x^6\right )\\ &=-\frac {b c}{18 x^6}-\frac {a+b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{3} b c^3 \log (x)-\frac {1}{18} b c^3 \log \left (1-c^2 x^6\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 61, normalized size = 1.09 \begin {gather*} -\frac {a}{9 x^9}-\frac {b c}{18 x^6}-\frac {b \tanh ^{-1}\left (c x^3\right )}{9 x^9}+\frac {1}{3} b c^3 \log (x)-\frac {1}{18} b c^3 \log \left (1-c^2 x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^3])/x^10,x]

[Out]

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

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Maple [A]
time = 0.04, size = 63, normalized size = 1.12

method	result	size
default	\(-\frac {a}{9 x^{9}}-\frac {b \arctanh \left (c \,x^{3}\right )}{9 x^{9}}-\frac {b \,c^{3} \ln \left (c \,x^{3}+1\right )}{18}-\frac {b \,c^{3} \ln \left (c \,x^{3}-1\right )}{18}-\frac {b c}{18 x^{6}}+\frac {b \,c^{3} \ln \left (x \right )}{3}\)	\(63\)
risch	\(-\frac {b \ln \left (c \,x^{3}+1\right )}{18 x^{9}}+\frac {6 b \,c^{3} \ln \left (x \right ) x^{9}-b \,c^{3} \ln \left (c^{2} x^{6}-1\right ) x^{9}-b c \,x^{3}+b \ln \left (-c \,x^{3}+1\right )-2 a}{18 x^{9}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^3))/x^10,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 51, normalized size = 0.91 \begin {gather*} -\frac {1}{18} \, {\left ({\left (c^{2} \log \left (c^{2} x^{6} - 1\right ) - c^{2} \log \left (x^{6}\right ) + \frac {1}{x^{6}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x^{3}\right )}{x^{9}}\right )} b - \frac {a}{9 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 65, normalized size = 1.16 \begin {gather*} -\frac {b c^{3} x^{9} \log \left (c^{2} x^{6} - 1\right ) - 6 \, b c^{3} x^{9} \log \left (x\right ) + b c x^{3} + b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, a}{18 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="fricas")

[Out]

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

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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((a+b*atanh(c*x**3))/x**10,x)

[Out]

Timed out

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Giac [A]
time = 0.42, size = 65, normalized size = 1.16 \begin {gather*} -\frac {1}{18} \, b c^{3} \log \left (c^{2} x^{6} - 1\right ) + \frac {1}{3} \, b c^{3} \log \left (x\right ) - \frac {b \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right )}{18 \, x^{9}} - \frac {b c x^{3} + 2 \, a}{18 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^3))/x^10,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.90, size = 67, normalized size = 1.20 \begin {gather*} \frac {b\,c^3\,\ln \left (x\right )}{3}-\frac {b\,c^3\,\ln \left (c^2\,x^6-1\right )}{18}-\frac {a}{9\,x^9}-\frac {b\,c}{18\,x^6}-\frac {b\,\ln \left (c\,x^3+1\right )}{18\,x^9}+\frac {b\,\ln \left (1-c\,x^3\right )}{18\,x^9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^3))/x^10,x)

[Out]

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

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