3.1.10 \(\int \frac {a+b \tanh ^{-1}(c x)}{x^4} \, dx\) [10]

Optimal. Leaf size=54 \[ -\frac {b c}{6 x^2}-\frac {a+b \tanh ^{-1}(c x)}{3 x^3}+\frac {1}{3} b c^3 \log (x)-\frac {1}{6} b c^3 \log \left (1-c^2 x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 54, 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)}{3 x^3}+\frac {1}{3} b c^3 \log (x)-\frac {1}{6} b c^3 \log \left (1-c^2 x^2\right )-\frac {b c}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/x^4,x]

[Out]

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

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

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Mathematica [A]
time = 0.01, size = 59, normalized size = 1.09 \begin {gather*} -\frac {a}{3 x^3}-\frac {b c}{6 x^2}-\frac {b \tanh ^{-1}(c x)}{3 x^3}+\frac {1}{3} b c^3 \log (x)-\frac {1}{6} b c^3 \log \left (1-c^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/x^4,x]

[Out]

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

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Maple [A]
time = 0.02, size = 62, normalized size = 1.15

method	result	size
derivativedivides	\(c^{3} \left (-\frac {a}{3 c^{3} x^{3}}-\frac {b \arctanh \left (c x \right )}{3 c^{3} x^{3}}-\frac {b \ln \left (c x -1\right )}{6}-\frac {b}{6 c^{2} x^{2}}+\frac {b \ln \left (c x \right )}{3}-\frac {b \ln \left (c x +1\right )}{6}\right )\)	\(62\)
default	\(c^{3} \left (-\frac {a}{3 c^{3} x^{3}}-\frac {b \arctanh \left (c x \right )}{3 c^{3} x^{3}}-\frac {b \ln \left (c x -1\right )}{6}-\frac {b}{6 c^{2} x^{2}}+\frac {b \ln \left (c x \right )}{3}-\frac {b \ln \left (c x +1\right )}{6}\right )\)	\(62\)
risch	\(-\frac {b \ln \left (c x +1\right )}{6 x^{3}}+\frac {2 b \,c^{3} \ln \left (x \right ) x^{3}-b \,c^{3} \ln \left (c^{2} x^{2}-1\right ) x^{3}-b c x +b \ln \left (-c x +1\right )-2 a}{6 x^{3}}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 49, normalized size = 0.91 \begin {gather*} -\frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac {1}{x^{2}}\right )} c + \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 59, normalized size = 1.09 \begin {gather*} -\frac {b c^{3} x^{3} \log \left (c^{2} x^{2} - 1\right ) - 2 \, b c^{3} x^{3} \log \left (x\right ) + b c x + b \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.41, size = 70, normalized size = 1.30 \begin {gather*} \begin {cases} - \frac {a}{3 x^{3}} + \frac {b c^{3} \log {\left (x \right )}}{3} - \frac {b c^{3} \log {\left (x - \frac {1}{c} \right )}}{3} - \frac {b c^{3} \operatorname {atanh}{\left (c x \right )}}{3} - \frac {b c}{6 x^{2}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\- \frac {a}{3 x^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x**4,x)

[Out]

Piecewise((-a/(3*x**3) + b*c**3*log(x)/3 - b*c**3*log(x - 1/c)/3 - b*c**3*atanh(c*x)/3 - b*c/(6*x**2) - b*atan
h(c*x)/(3*x**3), Ne(c, 0)), (-a/(3*x**3), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (46) = 92\).
time = 0.44, size = 251, normalized size = 4.65 \begin {gather*} \frac {1}{3} \, {\left (b c^{2} \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - b c^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (\frac {3 \, {\left (c x + 1\right )}^{2} b c^{2}}{{\left (c x - 1\right )}^{2}} + b c^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {2 \, {\left (\frac {3 \, {\left (c x + 1\right )}^{2} a c^{2}}{{\left (c x - 1\right )}^{2}} + a c^{2} + \frac {{\left (c x + 1\right )}^{2} b c^{2}}{{\left (c x - 1\right )}^{2}} + \frac {{\left (c x + 1\right )} b c^{2}}{c x - 1}\right )}}{\frac {{\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {3 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {3 \, {\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^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.73, size = 46, normalized size = 0.85 \begin {gather*} \frac {b\,c^3\,\ln \left (x\right )}{3}-\frac {b\,c^3\,\ln \left (c^2\,x^2-1\right )}{6}-\frac {\frac {a}{3}+\frac {b\,\mathrm {atanh}\left (c\,x\right )}{3}+\frac {b\,c\,x}{6}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/x^4,x)

[Out]

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

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