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

Optimal. Leaf size=117 \[ \frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}-\frac{b^2 c^2}{12 x^2}-\frac{1}{3} b^2 c^4 \log \left (1-c^2 x^2\right )+\frac{2}{3} b^2 c^4 \log (x) \]

[Out]

-(b^2*c^2)/(12*x^2) - (b*c*(a + b*ArcTanh[c*x]))/(6*x^3) - (b*c^3*(a + b*ArcTanh[c*x]))/(2*x) + (c^4*(a + b*Ar
cTanh[c*x])^2)/4 - (a + b*ArcTanh[c*x])^2/(4*x^4) + (2*b^2*c^4*Log[x])/3 - (b^2*c^4*Log[1 - c^2*x^2])/3

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Rubi [A]  time = 0.228329, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {5916, 5982, 266, 44, 36, 29, 31, 5948} \[ \frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}-\frac{b^2 c^2}{12 x^2}-\frac{1}{3} b^2 c^4 \log \left (1-c^2 x^2\right )+\frac{2}{3} b^2 c^4 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/x^5,x]

[Out]

-(b^2*c^2)/(12*x^2) - (b*c*(a + b*ArcTanh[c*x]))/(6*x^3) - (b*c^3*(a + b*ArcTanh[c*x]))/(2*x) + (c^4*(a + b*Ar
cTanh[c*x])^2)/4 - (a + b*ArcTanh[c*x])^2/(4*x^4) + (2*b^2*c^4*Log[x])/3 - (b^2*c^4*Log[1 - c^2*x^2])/3

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 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 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^5} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{2} (b c) \int \frac{a+b \tanh ^{-1}(c x)}{x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{2} (b c) \int \frac{a+b \tanh ^{-1}(c x)}{x^4} \, dx+\frac{1}{2} \left (b c^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{6} \left (b^2 c^2\right ) \int \frac{1}{x^3 \left (1-c^2 x^2\right )} \, dx+\frac{1}{2} \left (b c^3\right ) \int \frac{a+b \tanh ^{-1}(c x)}{x^2} \, dx+\frac{1}{2} \left (b c^5\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 )}{6 x^3}-\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{12} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x\right )} \, dx,x,x^2\right )+\frac{1}{2} \left (b^2 c^4\right ) \int \frac{1}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{12} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}+\frac{c^2}{x}-\frac{c^4}{-1+c^2 x}\right ) \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac{b^2 c^2}{12 x^2}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{6} b^2 c^4 \log (x)-\frac{1}{12} b^2 c^4 \log \left (1-c^2 x^2\right )+\frac{1}{4} \left (b^2 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c^6\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b^2 c^2}{12 x^2}-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )}{6 x^3}-\frac{b c^3 \left (a+b \tanh ^{-1}(c x)\right )}{2 x}+\frac{1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^4}+\frac{2}{3} b^2 c^4 \log (x)-\frac{1}{3} b^2 c^4 \log \left (1-c^2 x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0691125, size = 164, normalized size = 1.4 \[ -\frac{3 a^2+6 a b c^3 x^3+3 a b c^4 x^4 \log (1-c x)-3 a b c^4 x^4 \log (c x+1)+2 b \tanh ^{-1}(c x) \left (3 a+3 b c^3 x^3+b c x\right )+2 a b c x+b^2 c^2 x^2-8 b^2 c^4 x^4 \log (x)+4 b^2 c^4 x^4 \log (1-c x)+4 b^2 c^4 x^4 \log (c x+1)-3 b^2 \left (c^4 x^4-1\right ) \tanh ^{-1}(c x)^2}{12 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(3*a^2 + 2*a*b*c*x + b^2*c^2*x^2 + 6*a*b*c^3*x^3 + 2*b*(3*a + b*c*x + 3*b*c^3*x^3)*ArcTanh[c*x] - 3*b^2*(-1 +
c^4*x^4)*ArcTanh[c*x]^2 - 8*b^2*c^4*x^4*Log[x] + 3*a*b*c^4*x^4*Log[1 - c*x] + 4*b^2*c^4*x^4*Log[1 - c*x] - 3*
a*b*c^4*x^4*Log[1 + c*x] + 4*b^2*c^4*x^4*Log[1 + c*x])/(12*x^4)

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Maple [B]  time = 0.026, size = 290, normalized size = 2.5 \begin{align*} -{\frac{{a}^{2}}{4\,{x}^{4}}}-{\frac{{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{4\,{x}^{4}}}-{\frac{{c}^{4}{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx-1 \right ) }{4}}-{\frac{c{b}^{2}{\it Artanh} \left ( cx \right ) }{6\,{x}^{3}}}-{\frac{{c}^{3}{b}^{2}{\it Artanh} \left ( cx \right ) }{2\,x}}+{\frac{{c}^{4}{b}^{2}{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{4}}-{\frac{{c}^{4}{b}^{2} \left ( \ln \left ( cx-1 \right ) \right ) ^{2}}{16}}+{\frac{{c}^{4}{b}^{2}\ln \left ( cx-1 \right ) }{8}\ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{{c}^{4}{b}^{2}}{8}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{{c}^{4}{b}^{2}\ln \left ( cx+1 \right ) }{8}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }-{\frac{{c}^{4}{b}^{2} \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{16}}-{\frac{{c}^{4}{b}^{2}\ln \left ( cx-1 \right ) }{3}}-{\frac{{b}^{2}{c}^{2}}{12\,{x}^{2}}}+{\frac{2\,{c}^{4}{b}^{2}\ln \left ( cx \right ) }{3}}-{\frac{{c}^{4}{b}^{2}\ln \left ( cx+1 \right ) }{3}}-{\frac{ab{\it Artanh} \left ( cx \right ) }{2\,{x}^{4}}}-{\frac{{c}^{4}ab\ln \left ( cx-1 \right ) }{4}}-{\frac{abc}{6\,{x}^{3}}}-{\frac{{c}^{3}ab}{2\,x}}+{\frac{{c}^{4}ab\ln \left ( cx+1 \right ) }{4}} \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^2/x^4-1/4*b^2/x^4*arctanh(c*x)^2-1/4*c^4*b^2*arctanh(c*x)*ln(c*x-1)-1/6*c*b^2*arctanh(c*x)/x^3-1/2*c^3*
b^2*arctanh(c*x)/x+1/4*c^4*b^2*arctanh(c*x)*ln(c*x+1)-1/16*c^4*b^2*ln(c*x-1)^2+1/8*c^4*b^2*ln(c*x-1)*ln(1/2+1/
2*c*x)-1/8*c^4*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/8*c^4*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-1/16*c^4*b^2*ln(c*x
+1)^2-1/3*c^4*b^2*ln(c*x-1)-1/12*b^2*c^2/x^2+2/3*c^4*b^2*ln(c*x)-1/3*c^4*b^2*ln(c*x+1)-1/2*a*b/x^4*arctanh(c*x
)-1/4*c^4*a*b*ln(c*x-1)-1/6*a*b*c/x^3-1/2*c^3*a*b/x+1/4*c^4*a*b*ln(c*x+1)

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Maxima [B]  time = 0.997564, size = 302, normalized size = 2.58 \begin{align*} \frac{1}{12} \,{\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac{2 \,{\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac{6 \, \operatorname{artanh}\left (c x\right )}{x^{4}}\right )} a b + \frac{1}{48} \,{\left ({\left (32 \, c^{2} \log \left (x\right ) - \frac{3 \, c^{2} x^{2} \log \left (c x + 1\right )^{2} + 3 \, c^{2} x^{2} \log \left (c x - 1\right )^{2} + 16 \, c^{2} x^{2} \log \left (c x - 1\right ) - 2 \,{\left (3 \, c^{2} x^{2} \log \left (c x - 1\right ) - 8 \, c^{2} x^{2}\right )} \log \left (c x + 1\right ) + 4}{x^{2}}\right )} c^{2} + 4 \,{\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac{2 \,{\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c \operatorname{artanh}\left (c x\right )\right )} b^{2} - \frac{b^{2} \operatorname{artanh}\left (c x\right )^{2}}{4 \, x^{4}} - \frac{a^{2}}{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]

1/12*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*a*b + 1/48*((3
2*c^2*log(x) - (3*c^2*x^2*log(c*x + 1)^2 + 3*c^2*x^2*log(c*x - 1)^2 + 16*c^2*x^2*log(c*x - 1) - 2*(3*c^2*x^2*l
og(c*x - 1) - 8*c^2*x^2)*log(c*x + 1) + 4)/x^2)*c^2 + 4*(3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^
2 + 1)/x^3)*c*arctanh(c*x))*b^2 - 1/4*b^2*arctanh(c*x)^2/x^4 - 1/4*a^2/x^4

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Fricas [A]  time = 2.37999, size = 386, normalized size = 3.3 \begin{align*} \frac{32 \, b^{2} c^{4} x^{4} \log \left (x\right ) + 4 \,{\left (3 \, a b - 4 \, b^{2}\right )} c^{4} x^{4} \log \left (c x + 1\right ) - 4 \,{\left (3 \, a b + 4 \, b^{2}\right )} c^{4} x^{4} \log \left (c x - 1\right ) - 24 \, a b c^{3} x^{3} - 4 \, b^{2} c^{2} x^{2} - 8 \, a b c x + 3 \,{\left (b^{2} c^{4} x^{4} - b^{2}\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )^{2} - 12 \, a^{2} - 4 \,{\left (3 \, b^{2} c^{3} x^{3} + b^{2} c x + 3 \, a b\right )} \log \left (-\frac{c x + 1}{c x - 1}\right )}{48 \, 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]

1/48*(32*b^2*c^4*x^4*log(x) + 4*(3*a*b - 4*b^2)*c^4*x^4*log(c*x + 1) - 4*(3*a*b + 4*b^2)*c^4*x^4*log(c*x - 1)
- 24*a*b*c^3*x^3 - 4*b^2*c^2*x^2 - 8*a*b*c*x + 3*(b^2*c^4*x^4 - b^2)*log(-(c*x + 1)/(c*x - 1))^2 - 12*a^2 - 4*
(3*b^2*c^3*x^3 + b^2*c*x + 3*a*b)*log(-(c*x + 1)/(c*x - 1)))/x^4

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Sympy [A]  time = 2.60062, size = 184, normalized size = 1.57 \begin{align*} \begin{cases} - \frac{a^{2}}{4 x^{4}} + \frac{a b c^{4} \operatorname{atanh}{\left (c x \right )}}{2} - \frac{a b c^{3}}{2 x} - \frac{a b c}{6 x^{3}} - \frac{a b \operatorname{atanh}{\left (c x \right )}}{2 x^{4}} + \frac{2 b^{2} c^{4} \log{\left (x \right )}}{3} - \frac{2 b^{2} c^{4} \log{\left (x - \frac{1}{c} \right )}}{3} + \frac{b^{2} c^{4} \operatorname{atanh}^{2}{\left (c x \right )}}{4} - \frac{2 b^{2} c^{4} \operatorname{atanh}{\left (c x \right )}}{3} - \frac{b^{2} c^{3} \operatorname{atanh}{\left (c x \right )}}{2 x} - \frac{b^{2} c^{2}}{12 x^{2}} - \frac{b^{2} c \operatorname{atanh}{\left (c x \right )}}{6 x^{3}} - \frac{b^{2} \operatorname{atanh}^{2}{\left (c x \right )}}{4 x^{4}} & \text{for}\: c \neq 0 \\- \frac{a^{2}}{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**2/(4*x**4) + a*b*c**4*atanh(c*x)/2 - a*b*c**3/(2*x) - a*b*c/(6*x**3) - a*b*atanh(c*x)/(2*x**4)
+ 2*b**2*c**4*log(x)/3 - 2*b**2*c**4*log(x - 1/c)/3 + b**2*c**4*atanh(c*x)**2/4 - 2*b**2*c**4*atanh(c*x)/3 - b
**2*c**3*atanh(c*x)/(2*x) - b**2*c**2/(12*x**2) - b**2*c*atanh(c*x)/(6*x**3) - b**2*atanh(c*x)**2/(4*x**4), Ne
(c, 0)), (-a**2/(4*x**4), 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^{5}}\,{d x} \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]

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