3.55 \(\int \frac{a+b \tanh ^{-1}(c x^2)}{x^3} \, dx\)

Optimal. Leaf size=40 \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{2 x^2}-\frac{1}{4} b c \log \left (1-c^2 x^4\right )+b c \log (x) \]

[Out]

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

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Rubi [A]  time = 0.0255877, antiderivative size = 40, 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, 266, 36, 29, 31} \[ -\frac{a+b \tanh ^{-1}\left (c x^2\right )}{2 x^2}-\frac{1}{4} b c \log \left (1-c^2 x^4\right )+b c \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])/x^3,x]

[Out]

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

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 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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0110482, size = 45, normalized size = 1.12 \[ -\frac{a}{2 x^2}-\frac{1}{4} b c \log \left (1-c^2 x^4\right )-\frac{b \tanh ^{-1}\left (c x^2\right )}{2 x^2}+b c \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])/x^3,x]

[Out]

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

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Maple [A]  time = 0.013, size = 49, normalized size = 1.2 \begin{align*} -{\frac{a}{2\,{x}^{2}}}-{\frac{b{\it Artanh} \left ( c{x}^{2} \right ) }{2\,{x}^{2}}}-{\frac{bc\ln \left ( c{x}^{2}+1 \right ) }{4}}+bc\ln \left ( x \right ) -{\frac{bc\ln \left ( c{x}^{2}-1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))/x^3,x)

[Out]

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

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Maxima [A]  time = 0.980565, size = 55, normalized size = 1.38 \begin{align*} -\frac{1}{4} \,{\left (c{\left (\log \left (c^{2} x^{4} - 1\right ) - \log \left (x^{4}\right )\right )} + \frac{2 \, \operatorname{artanh}\left (c x^{2}\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^2))/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.00197, size = 130, normalized size = 3.25 \begin{align*} -\frac{b c x^{2} \log \left (c^{2} x^{4} - 1\right ) - 4 \, b c x^{2} \log \left (x\right ) + b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 17.2699, size = 80, normalized size = 2. \begin{align*} \begin{cases} - \frac{a}{2 x^{2}} + b c \log{\left (x \right )} - \frac{b c \log{\left (x - i \sqrt{\frac{1}{c}} \right )}}{2} - \frac{b c \log{\left (x + i \sqrt{\frac{1}{c}} \right )}}{2} + \frac{b c \operatorname{atanh}{\left (c x^{2} \right )}}{2} - \frac{b \operatorname{atanh}{\left (c x^{2} \right )}}{2 x^{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**2))/x**3,x)

[Out]

Piecewise((-a/(2*x**2) + b*c*log(x) - b*c*log(x - I*sqrt(1/c))/2 - b*c*log(x + I*sqrt(1/c))/2 + b*c*atanh(c*x*
*2)/2 - b*atanh(c*x**2)/(2*x**2), Ne(c, 0)), (-a/(2*x**2), True))

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Giac [A]  time = 1.1719, size = 69, normalized size = 1.72 \begin{align*} -\frac{1}{4} \, b c \log \left (c^{2} x^{4} - 1\right ) + b c \log \left (x\right ) - \frac{b \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{4 \, x^{2}} - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^3,x, algorithm="giac")

[Out]

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