3.141 \(\int \frac{a+b \tanh ^{-1}(\frac{c}{x})}{x^3} \, dx\)

Optimal. Leaf size=43 \[ -\frac{a+b \tanh ^{-1}\left (\frac{c}{x}\right )}{2 x^2}+\frac{b \tanh ^{-1}\left (\frac{x}{c}\right )}{2 c^2}-\frac{b}{2 c x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0266653, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6097, 263, 325, 207} \[ -\frac{a+b \tanh ^{-1}\left (\frac{c}{x}\right )}{2 x^2}+\frac{b \tanh ^{-1}\left (\frac{x}{c}\right )}{2 c^2}-\frac{b}{2 c x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x])/x^3,x]

[Out]

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

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]

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0090805, size = 60, normalized size = 1.4 \[ -\frac{a}{2 x^2}-\frac{b \log (x-c)}{4 c^2}+\frac{b \log (c+x)}{4 c^2}-\frac{b \tanh ^{-1}\left (\frac{c}{x}\right )}{2 x^2}-\frac{b}{2 c x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c/x])/x^3,x]

[Out]

-a/(2*x^2) - b/(2*c*x) - (b*ArcTanh[c/x])/(2*x^2) - (b*Log[-c + x])/(4*c^2) + (b*Log[c + x])/(4*c^2)

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 57, normalized size = 1.3 \begin{align*} -{\frac{a}{2\,{x}^{2}}}-{\frac{b}{2\,{x}^{2}}{\it Artanh} \left ({\frac{c}{x}} \right ) }-{\frac{b}{2\,cx}}-{\frac{b}{4\,{c}^{2}}\ln \left ({\frac{c}{x}}-1 \right ) }+{\frac{b}{4\,{c}^{2}}\ln \left ( 1+{\frac{c}{x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c/x))/x^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.95463, size = 70, normalized size = 1.63 \begin{align*} \frac{1}{4} \,{\left (c{\left (\frac{\log \left (c + x\right )}{c^{3}} - \frac{\log \left (-c + x\right )}{c^{3}} - \frac{2}{c^{2} x}\right )} - \frac{2 \, \operatorname{artanh}\left (\frac{c}{x}\right )}{x^{2}}\right )} b - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.76097, size = 103, normalized size = 2.4 \begin{align*} -\frac{2 \, a c^{2} + 2 \, b c x +{\left (b c^{2} - b x^{2}\right )} \log \left (-\frac{c + x}{c - x}\right )}{4 \, c^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 2.11399, size = 44, normalized size = 1.02 \begin{align*} \begin{cases} - \frac{a}{2 x^{2}} - \frac{b \operatorname{atanh}{\left (\frac{c}{x} \right )}}{2 x^{2}} - \frac{b}{2 c x} + \frac{b \operatorname{atanh}{\left (\frac{c}{x} \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\- \frac{a}{2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a/(2*x**2) - b*atanh(c/x)/(2*x**2) - b/(2*c*x) + b*atanh(c/x)/(2*c**2), Ne(c, 0)), (-a/(2*x**2), T
rue))

________________________________________________________________________________________

Giac [A]  time = 1.16435, size = 77, normalized size = 1.79 \begin{align*} \frac{b \log \left (c + x\right )}{4 \, c^{2}} - \frac{b \log \left (c - x\right )}{4 \, c^{2}} - \frac{b \log \left (-\frac{c + x}{c - x}\right )}{4 \, x^{2}} - \frac{a c + b x}{2 \, c x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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