3.1.12 \(\int \frac {a+b \tanh ^{-1}(c x)}{x^6} \, dx\) [12]

Optimal. Leaf size=65 \[ -\frac {b c}{20 x^4}-\frac {b c^3}{10 x^2}-\frac {a+b \tanh ^{-1}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)-\frac {1}{10} b c^5 \log \left (1-c^2 x^2\right ) \]

[Out]

-1/20*b*c/x^4-1/10*b*c^3/x^2+1/5*(-a-b*arctanh(c*x))/x^5+1/5*b*c^5*ln(x)-1/10*b*c^5*ln(-c^2*x^2+1)

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Rubi [A]
time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6037, 272, 46} \begin {gather*} -\frac {a+b \tanh ^{-1}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)-\frac {b c^3}{10 x^2}-\frac {1}{10} b c^5 \log \left (1-c^2 x^2\right )-\frac {b c}{20 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/x^6,x]

[Out]

-1/20*(b*c)/x^4 - (b*c^3)/(10*x^2) - (a + b*ArcTanh[c*x])/(5*x^5) + (b*c^5*Log[x])/5 - (b*c^5*Log[1 - c^2*x^2]
)/10

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}(c x)}{x^6} \, dx &=-\frac {a+b \tanh ^{-1}(c x)}{5 x^5}+\frac {1}{5} (b c) \int \frac {1}{x^5 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {a+b \tanh ^{-1}(c x)}{5 x^5}+\frac {1}{10} (b c) \text {Subst}\left (\int \frac {1}{x^3 \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {a+b \tanh ^{-1}(c x)}{5 x^5}+\frac {1}{10} (b c) \text {Subst}\left (\int \left (\frac {1}{x^3}+\frac {c^2}{x^2}+\frac {c^4}{x}-\frac {c^6}{-1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac {b c}{20 x^4}-\frac {b c^3}{10 x^2}-\frac {a+b \tanh ^{-1}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)-\frac {1}{10} b c^5 \log \left (1-c^2 x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 70, normalized size = 1.08 \begin {gather*} -\frac {a}{5 x^5}-\frac {b c}{20 x^4}-\frac {b c^3}{10 x^2}-\frac {b \tanh ^{-1}(c x)}{5 x^5}+\frac {1}{5} b c^5 \log (x)-\frac {1}{10} b c^5 \log \left (1-c^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/x^6,x]

[Out]

-1/5*a/x^5 - (b*c)/(20*x^4) - (b*c^3)/(10*x^2) - (b*ArcTanh[c*x])/(5*x^5) + (b*c^5*Log[x])/5 - (b*c^5*Log[1 -
c^2*x^2])/10

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Maple [A]
time = 0.02, size = 71, normalized size = 1.09

method	result	size
derivativedivides	\(c^{5} \left (-\frac {a}{5 c^{5} x^{5}}-\frac {b \arctanh \left (c x \right )}{5 c^{5} x^{5}}-\frac {b \ln \left (c x -1\right )}{10}-\frac {b}{20 c^{4} x^{4}}-\frac {b}{10 c^{2} x^{2}}+\frac {b \ln \left (c x \right )}{5}-\frac {b \ln \left (c x +1\right )}{10}\right )\)	\(71\)
default	\(c^{5} \left (-\frac {a}{5 c^{5} x^{5}}-\frac {b \arctanh \left (c x \right )}{5 c^{5} x^{5}}-\frac {b \ln \left (c x -1\right )}{10}-\frac {b}{20 c^{4} x^{4}}-\frac {b}{10 c^{2} x^{2}}+\frac {b \ln \left (c x \right )}{5}-\frac {b \ln \left (c x +1\right )}{10}\right )\)	\(71\)
risch	\(-\frac {b \ln \left (c x +1\right )}{10 x^{5}}+\frac {4 b \,c^{5} \ln \left (x \right ) x^{5}-2 b \,c^{5} \ln \left (c^{2} x^{2}-1\right ) x^{5}-2 b \,c^{3} x^{3}-b c x +2 b \ln \left (-c x +1\right )-4 a}{20 x^{5}}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x^6,x,method=_RETURNVERBOSE)

[Out]

c^5*(-1/5*a/c^5/x^5-1/5*b/c^5/x^5*arctanh(c*x)-1/10*b*ln(c*x-1)-1/20*b/c^4/x^4-1/10*b/c^2/x^2+1/5*b*ln(c*x)-1/
10*b*ln(c*x+1))

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Maxima [A]
time = 0.26, size = 61, normalized size = 0.94 \begin {gather*} -\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} - 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) + \frac {2 \, c^{2} x^{2} + 1}{x^{4}}\right )} c + \frac {4 \, \operatorname {artanh}\left (c x\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^6,x, algorithm="maxima")

[Out]

-1/20*((2*c^4*log(c^2*x^2 - 1) - 2*c^4*log(x^2) + (2*c^2*x^2 + 1)/x^4)*c + 4*arctanh(c*x)/x^5)*b - 1/5*a/x^5

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Fricas [A]
time = 0.38, size = 70, normalized size = 1.08 \begin {gather*} -\frac {2 \, b c^{5} x^{5} \log \left (c^{2} x^{2} - 1\right ) - 4 \, b c^{5} x^{5} \log \left (x\right ) + 2 \, b c^{3} x^{3} + b c x + 2 \, b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 4 \, a}{20 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^6,x, algorithm="fricas")

[Out]

-1/20*(2*b*c^5*x^5*log(c^2*x^2 - 1) - 4*b*c^5*x^5*log(x) + 2*b*c^3*x^3 + b*c*x + 2*b*log(-(c*x + 1)/(c*x - 1))
+ 4*a)/x^5

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Sympy [A]
time = 0.57, size = 80, normalized size = 1.23 \begin {gather*} \begin {cases} - \frac {a}{5 x^{5}} + \frac {b c^{5} \log {\left (x \right )}}{5} - \frac {b c^{5} \log {\left (x - \frac {1}{c} \right )}}{5} - \frac {b c^{5} \operatorname {atanh}{\left (c x \right )}}{5} - \frac {b c^{3}}{10 x^{2}} - \frac {b c}{20 x^{4}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{5 x^{5}} & \text {for}\: c \neq 0 \\- \frac {a}{5 x^{5}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x**6,x)

[Out]

Piecewise((-a/(5*x**5) + b*c**5*log(x)/5 - b*c**5*log(x - 1/c)/5 - b*c**5*atanh(c*x)/5 - b*c**3/(10*x**2) - b*
c/(20*x**4) - b*atanh(c*x)/(5*x**5), Ne(c, 0)), (-a/(5*x**5), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (55) = 110\).
time = 0.41, size = 397, normalized size = 6.11 \begin {gather*} \frac {1}{5} \, {\left (b c^{4} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - b c^{4} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {5 \, {\left (c x + 1\right )}^{4} b c^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{2} b c^{4}}{{\left (c x - 1\right )}^{2}} + b c^{4}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{5}}{{\left (c x - 1\right )}^{5}} + \frac {5 \, {\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {2 \, {\left (\frac {5 \, {\left (c x + 1\right )}^{4} a c^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{2} a c^{4}}{{\left (c x - 1\right )}^{2}} + a c^{4} + \frac {2 \, {\left (c x + 1\right )}^{4} b c^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3} b c^{4}}{{\left (c x - 1\right )}^{3}} + \frac {4 \, {\left (c x + 1\right )}^{2} b c^{4}}{{\left (c x - 1\right )}^{2}} + \frac {2 \, {\left (c x + 1\right )} b c^{4}}{c x - 1}\right )}}{\frac {{\left (c x + 1\right )}^{5}}{{\left (c x - 1\right )}^{5}} + \frac {5 \, {\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {10 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {10 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^6,x, algorithm="giac")

[Out]

1/5*(b*c^4*log(-(c*x + 1)/(c*x - 1) - 1) - b*c^4*log(-(c*x + 1)/(c*x - 1)) + (5*(c*x + 1)^4*b*c^4/(c*x - 1)^4
+ 10*(c*x + 1)^2*b*c^4/(c*x - 1)^2 + b*c^4)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^5/(c*x - 1)^5 + 5*(c*x + 1)^4
/(c*x - 1)^4 + 10*(c*x + 1)^3/(c*x - 1)^3 + 10*(c*x + 1)^2/(c*x - 1)^2 + 5*(c*x + 1)/(c*x - 1) + 1) + 2*(5*(c*
x + 1)^4*a*c^4/(c*x - 1)^4 + 10*(c*x + 1)^2*a*c^4/(c*x - 1)^2 + a*c^4 + 2*(c*x + 1)^4*b*c^4/(c*x - 1)^4 + 4*(c
*x + 1)^3*b*c^4/(c*x - 1)^3 + 4*(c*x + 1)^2*b*c^4/(c*x - 1)^2 + 2*(c*x + 1)*b*c^4/(c*x - 1))/((c*x + 1)^5/(c*x
- 1)^5 + 5*(c*x + 1)^4/(c*x - 1)^4 + 10*(c*x + 1)^3/(c*x - 1)^3 + 10*(c*x + 1)^2/(c*x - 1)^2 + 5*(c*x + 1)/(c
*x - 1) + 1))*c

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Mupad [B]
time = 0.91, size = 71, normalized size = 1.09 \begin {gather*} \frac {b\,c^5\,\ln \left (x\right )}{5}-\frac {b\,c^5\,\ln \left (c^2\,x^2-1\right )}{10}-\frac {\frac {b\,c^3\,x^3}{2}+\frac {b\,c\,x}{4}+a}{5\,x^5}-\frac {b\,\ln \left (c\,x+1\right )}{10\,x^5}+\frac {b\,\ln \left (1-c\,x\right )}{10\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/x^6,x)

[Out]

(b*c^5*log(x))/5 - (b*c^5*log(c^2*x^2 - 1))/10 - (a + (b*c^3*x^3)/2 + (b*c*x)/4)/(5*x^5) - (b*log(c*x + 1))/(1
0*x^5) + (b*log(1 - c*x))/(10*x^5)

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