3.163 \(\int \frac{a+b \tanh ^{-1}(\frac{c}{x^2})}{x^5} \, dx\)

Optimal. Leaf size=45 \[ -\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{4 x^4}+\frac{b \tanh ^{-1}\left (\frac{x^2}{c}\right )}{4 c^2}-\frac{b}{4 c x^2} \]

[Out]

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

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Rubi [A]  time = 0.0320788, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6097, 263, 275, 325, 207} \[ -\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{4 x^4}+\frac{b \tanh ^{-1}\left (\frac{x^2}{c}\right )}{4 c^2}-\frac{b}{4 c x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x^2])/x^5,x]

[Out]

-b/(4*c*x^2) - (a + b*ArcTanh[c/x^2])/(4*x^4) + (b*ArcTanh[x^2/c])/(4*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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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^2}\right )}{x^5} \, dx &=-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{4 x^4}-\frac{1}{2} (b c) \int \frac{1}{\left (1-\frac{c^2}{x^4}\right ) x^7} \, dx\\ &=-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{4 x^4}-\frac{1}{2} (b c) \int \frac{1}{x^3 \left (-c^2+x^4\right )} \, dx\\ &=-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{4 x^4}-\frac{1}{4} (b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (-c^2+x^2\right )} \, dx,x,x^2\right )\\ &=-\frac{b}{4 c x^2}-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{4 x^4}-\frac{b \operatorname{Subst}\left (\int \frac{1}{-c^2+x^2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac{b}{4 c x^2}-\frac{a+b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{4 x^4}+\frac{b \tanh ^{-1}\left (\frac{x^2}{c}\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0113542, size = 64, normalized size = 1.42 \[ -\frac{a}{4 x^4}-\frac{b \log \left (x^2-c\right )}{8 c^2}+\frac{b \log \left (c+x^2\right )}{8 c^2}-\frac{b}{4 c x^2}-\frac{b \tanh ^{-1}\left (\frac{c}{x^2}\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c/x^2])/x^5,x]

[Out]

-a/(4*x^4) - b/(4*c*x^2) - (b*ArcTanh[c/x^2])/(4*x^4) - (b*Log[-c + x^2])/(8*c^2) + (b*Log[c + x^2])/(8*c^2)

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Maple [A]  time = 0.009, size = 57, normalized size = 1.3 \begin{align*} -{\frac{a}{4\,{x}^{4}}}-{\frac{b}{4\,{x}^{4}}{\it Artanh} \left ({\frac{c}{{x}^{2}}} \right ) }-{\frac{b}{4\,c{x}^{2}}}-{\frac{b}{8\,{c}^{2}}\ln \left ({\frac{c}{{x}^{2}}}-1 \right ) }+{\frac{b}{8\,{c}^{2}}\ln \left ( 1+{\frac{c}{{x}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c/x^2))/x^5,x)

[Out]

-1/4*a/x^4-1/4*b/x^4*arctanh(c/x^2)-1/4*b/c/x^2-1/8*b/c^2*ln(c/x^2-1)+1/8*b/c^2*ln(1+c/x^2)

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Maxima [A]  time = 0.967116, size = 76, normalized size = 1.69 \begin{align*} \frac{1}{8} \,{\left (c{\left (\frac{\log \left (x^{2} + c\right )}{c^{3}} - \frac{\log \left (x^{2} - c\right )}{c^{3}} - \frac{2}{c^{2} x^{2}}\right )} - \frac{2 \, \operatorname{artanh}\left (\frac{c}{x^{2}}\right )}{x^{4}}\right )} b - \frac{a}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))/x^5,x, algorithm="maxima")

[Out]

1/8*(c*(log(x^2 + c)/c^3 - log(x^2 - c)/c^3 - 2/(c^2*x^2)) - 2*arctanh(c/x^2)/x^4)*b - 1/4*a/x^4

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Fricas [A]  time = 1.54632, size = 109, normalized size = 2.42 \begin{align*} -\frac{2 \, b c x^{2} + 2 \, a c^{2} -{\left (b x^{4} - b c^{2}\right )} \log \left (\frac{x^{2} + c}{x^{2} - c}\right )}{8 \, c^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))/x^5,x, algorithm="fricas")

[Out]

-1/8*(2*b*c*x^2 + 2*a*c^2 - (b*x^4 - b*c^2)*log((x^2 + c)/(x^2 - c)))/(c^2*x^4)

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Sympy [A]  time = 45.3849, size = 49, normalized size = 1.09 \begin{align*} \begin{cases} - \frac{a}{4 x^{4}} - \frac{b \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}}{4 x^{4}} - \frac{b}{4 c x^{2}} + \frac{b \operatorname{atanh}{\left (\frac{c}{x^{2}} \right )}}{4 c^{2}} & \text{for}\: c \neq 0 \\- \frac{a}{4 x^{4}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c/x**2))/x**5,x)

[Out]

Piecewise((-a/(4*x**4) - b*atanh(c/x**2)/(4*x**4) - b/(4*c*x**2) + b*atanh(c/x**2)/(4*c**2), Ne(c, 0)), (-a/(4
*x**4), True))

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Giac [A]  time = 1.2143, size = 89, normalized size = 1.98 \begin{align*} \frac{b \log \left (x^{2} + c\right )}{8 \, c^{2}} - \frac{b \log \left (-x^{2} + c\right )}{8 \, c^{2}} - \frac{b \log \left (\frac{x^{2} + c}{x^{2} - c}\right )}{8 \, x^{4}} - \frac{b x^{2} + a c}{4 \, c x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c/x^2))/x^5,x, algorithm="giac")

[Out]

1/8*b*log(x^2 + c)/c^2 - 1/8*b*log(-x^2 + c)/c^2 - 1/8*b*log((x^2 + c)/(x^2 - c))/x^4 - 1/4*(b*x^2 + a*c)/(c*x
^4)