3.142 \(\int \frac {a+b \tanh ^{-1}(\frac {c}{x})}{x^4} \, dx\)

Optimal. Leaf size=57 \[ -\frac {a+b \tanh ^{-1}\left (\frac {c}{x}\right )}{3 x^3}+\frac {b \log (x)}{3 c^3}-\frac {b \log \left (c^2-x^2\right )}{6 c^3}-\frac {b}{6 c x^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6097, 263, 266, 44} \[ -\frac {a+b \tanh ^{-1}\left (\frac {c}{x}\right )}{3 x^3}-\frac {b \log \left (c^2-x^2\right )}{6 c^3}+\frac {b \log (x)}{3 c^3}-\frac {b}{6 c x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x])/x^4,x]

[Out]

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

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 263

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

Rule 266

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 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 62, normalized size = 1.09 \[ -\frac {a}{3 x^3}+\frac {b \log (x)}{3 c^3}-\frac {b \log \left (x^2-c^2\right )}{6 c^3}-\frac {b \tanh ^{-1}\left (\frac {c}{x}\right )}{3 x^3}-\frac {b}{6 c x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c/x])/x^4,x]

[Out]

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

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fricas [A]  time = 1.12, size = 62, normalized size = 1.09 \[ -\frac {b x^{3} \log \left (-c^{2} + x^{2}\right ) - 2 \, b x^{3} \log \relax (x) + b c^{3} \log \left (-\frac {c + x}{c - x}\right ) + 2 \, a c^{3} + b c^{2} x}{6 \, c^{3} x^{3}} \]

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

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giac [B]  time = 0.31, size = 234, normalized size = 4.11 \[ -\frac {\frac {{\left (b + \frac {3 \, b {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}}\right )} \log \left (-\frac {c + x}{c - x}\right )}{\frac {{\left (c + x\right )}^{3} c^{2}}{{\left (c - x\right )}^{3}} - \frac {3 \, {\left (c + x\right )}^{2} c^{2}}{{\left (c - x\right )}^{2}} + \frac {3 \, {\left (c + x\right )} c^{2}}{c - x} - c^{2}} + \frac {2 \, {\left (a + \frac {3 \, a {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} + \frac {b {\left (c + x\right )}^{2}}{{\left (c - x\right )}^{2}} - \frac {b {\left (c + x\right )}}{c - x}\right )}}{\frac {{\left (c + x\right )}^{3} c^{2}}{{\left (c - x\right )}^{3}} - \frac {3 \, {\left (c + x\right )}^{2} c^{2}}{{\left (c - x\right )}^{2}} + \frac {3 \, {\left (c + x\right )} c^{2}}{c - x} - c^{2}} - \frac {b \log \left (-\frac {c + x}{c - x} + 1\right )}{c^{2}} + \frac {b \log \left (-\frac {c + x}{c - x}\right )}{c^{2}}}{3 \, c} \]

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

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maple [A]  time = 0.03, size = 57, normalized size = 1.00 \[ -\frac {a}{3 x^{3}}-\frac {b \arctanh \left (\frac {c}{x}\right )}{3 x^{3}}-\frac {b}{6 c \,x^{2}}-\frac {b \ln \left (\frac {c}{x}-1\right )}{6 c^{3}}-\frac {b \ln \left (1+\frac {c}{x}\right )}{6 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 55, normalized size = 0.96 \[ -\frac {1}{6} \, {\left (c {\left (\frac {\log \left (-c^{2} + x^{2}\right )}{c^{4}} - \frac {\log \left (x^{2}\right )}{c^{4}} + \frac {1}{c^{2} x^{2}}\right )} + \frac {2 \, \operatorname {artanh}\left (\frac {c}{x}\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \]

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

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mupad [B]  time = 0.75, size = 59, normalized size = 1.04 \[ -\frac {\frac {a}{3}+\frac {b\,\mathrm {atanh}\left (\frac {c}{x}\right )}{3}}{x^3}-\frac {\frac {b\,x^3\,\ln \left (x^2-c^2\right )}{6}-\frac {b\,x^3\,\ln \relax (x)}{3}+\frac {b\,c^2\,x}{6}}{c^3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c/x))/x^4,x)

[Out]

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

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sympy [A]  time = 1.29, size = 68, normalized size = 1.19 \[ \begin {cases} - \frac {a}{3 x^{3}} - \frac {b \operatorname {atanh}{\left (\frac {c}{x} \right )}}{3 x^{3}} - \frac {b}{6 c x^{2}} + \frac {b \log {\relax (x )}}{3 c^{3}} - \frac {b \log {\left (- c + x \right )}}{3 c^{3}} - \frac {b \operatorname {atanh}{\left (\frac {c}{x} \right )}}{3 c^{3}} & \text {for}\: c \neq 0 \\- \frac {a}{3 x^{3}} & \text {otherwise} \end {cases} \]

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

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