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

Optimal. Leaf size=80 \[ \frac{1}{2} c^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac{1}{2} b^2 c^2 \log \left (1-c^2 x^2\right )+b^2 c^2 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.13302, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5916, 5982, 266, 36, 29, 31, 5948} \[ \frac{1}{2} c^2 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{x}-\frac{1}{2} b^2 c^2 \log \left (1-c^2 x^2\right )+b^2 c^2 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/x^3,x]

[Out]

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5982

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcTanh[c*x])^p)/(d +
e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[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]

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rubi steps

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

Mathematica [A]  time = 0.0676631, size = 101, normalized size = 1.26 \[ -\frac{a^2+b c^2 x^2 (a+b) \log (1-c x)-b c^2 x^2 (a-b) \log (c x+1)+2 a b c x+2 b \tanh ^{-1}(c x) (a+b c x)-2 b^2 c^2 x^2 \log (x)-b^2 \left (c^2 x^2-1\right ) \tanh ^{-1}(c x)^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.019, size = 253, normalized size = 3.2 \begin{align*} -{\frac{{a}^{2}}{2\,{x}^{2}}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{2\,{x}^{2}}}-{\frac{c{b}^{2}{\it Artanh} \left ( cx \right ) }{x}}-{\frac{{c}^{2}{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{2}}+{\frac{{c}^{2}{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{2}}-{\frac{{c}^{2}{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{8}}+{\frac{{c}^{2}{b}^{2}\ln \left ( cx-1 \right ) }{4}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{c}^{2}{b}^{2}\ln \left ( cx-1 \right ) }{2}}+{c}^{2}{b}^{2}\ln \left ( cx \right ) -{\frac{{c}^{2}{b}^{2}\ln \left ( cx+1 \right ) }{2}}-{\frac{{c}^{2}{b}^{2}}{4}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{c}^{2}{b}^{2}\ln \left ( cx+1 \right ) }{4}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }-{\frac{{c}^{2}{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{8}}-{\frac{ab{\it Artanh} \left ( cx \right ) }{{x}^{2}}}-{\frac{abc}{x}}-{\frac{{c}^{2}ab\ln \left ( cx-1 \right ) }{2}}+{\frac{{c}^{2}ab\ln \left ( cx+1 \right ) }{2}} \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^2/x^2-1/2*b^2/x^2*arctanh(c*x)^2-c*b^2*arctanh(c*x)/x-1/2*c^2*b^2*arctanh(c*x)*ln(c*x-1)+1/2*c^2*b^2*ar
ctanh(c*x)*ln(c*x+1)-1/8*c^2*b^2*ln(c*x-1)^2+1/4*c^2*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-1/2*c^2*b^2*ln(c*x-1)+c^2*b
^2*ln(c*x)-1/2*c^2*b^2*ln(c*x+1)-1/4*c^2*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/4*c^2*b^2*ln(-1/2*c*x+1/2)*ln(
c*x+1)-1/8*c^2*b^2*ln(c*x+1)^2-a*b/x^2*arctanh(c*x)-a*b*c/x-1/2*c^2*a*b*ln(c*x-1)+1/2*c^2*a*b*ln(c*x+1)

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Maxima [B]  time = 1.00399, size = 204, normalized size = 2.55 \begin{align*} \frac{1}{2} \,{\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac{2}{x}\right )} c - \frac{2 \, \operatorname{artanh}\left (c x\right )}{x^{2}}\right )} a b + \frac{1}{8} \,{\left ({\left (2 \,{\left (\log \left (c x - 1\right ) - 2\right )} \log \left (c x + 1\right ) - \log \left (c x + 1\right )^{2} - \log \left (c x - 1\right )^{2} - 4 \, \log \left (c x - 1\right ) + 8 \, \log \left (x\right )\right )} c^{2} + 4 \,{\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac{2}{x}\right )} c \operatorname{artanh}\left (c x\right )\right )} b^{2} - \frac{b^{2} \operatorname{artanh}\left (c x\right )^{2}}{2 \, x^{2}} - \frac{a^{2}}{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]

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

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Fricas [A]  time = 2.36725, size = 300, normalized size = 3.75 \begin{align*} \frac{8 \, b^{2} c^{2} x^{2} \log \left (x\right ) + 4 \,{\left (a b - b^{2}\right )} c^{2} x^{2} \log \left (c x + 1\right ) - 4 \,{\left (a b + b^{2}\right )} c^{2} x^{2} \log \left (c x - 1\right ) - 8 \, a b c x +{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} - 4 \, a^{2} - 4 \,{\left (b^{2} c x + a b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{8 \, 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]

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

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Sympy [A]  time = 1.45457, size = 126, normalized size = 1.58 \begin{align*} \begin{cases} - \frac{a^{2}}{2 x^{2}} + a b c^{2} \operatorname{atanh}{\left (c x \right )} - \frac{a b c}{x} - \frac{a b \operatorname{atanh}{\left (c x \right )}}{x^{2}} + b^{2} c^{2} \log{\left (x \right )} - b^{2} c^{2} \log{\left (x - \frac{1}{c} \right )} + \frac{b^{2} c^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{2} - b^{2} c^{2} \operatorname{atanh}{\left (c x \right )} - \frac{b^{2} c \operatorname{atanh}{\left (c x \right )}}{x} - \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{2 x^{2}} & \text{for}\: c \neq 0 \\- \frac{a^{2}}{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/(2*x**2) + a*b*c**2*atanh(c*x) - a*b*c/x - a*b*atanh(c*x)/x**2 + b**2*c**2*log(x) - b**2*c**2
*log(x - 1/c) + b**2*c**2*atanh(c*x)**2/2 - b**2*c**2*atanh(c*x) - b**2*c*atanh(c*x)/x - b**2*atanh(c*x)**2/(2
*x**2), Ne(c, 0)), (-a**2/(2*x**2), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{x^{3}}\,{d x} \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]

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