3.1.21 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{x^3} \, dx\) [21]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6037, 6129, 272, 36, 29, 31, 6095} \begin {gather*} \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) \end {gather*}

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 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 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 272

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 6037

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

Rule 6095

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]

Rule 6129

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]

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 ) \text {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 ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^2 c^4\right ) \text {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*}

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Mathematica [A]
time = 0.04, size = 101, normalized size = 1.26 \begin {gather*} -\frac {a^2+2 a b c x+2 b (a+b c x) \tanh ^{-1}(c x)-b^2 \left (-1+c^2 x^2\right ) \tanh ^{-1}(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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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])/x^2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(233\) vs. \(2(74)=148\).
time = 0.04, size = 234, normalized size = 2.92

method	result	size
risch	\(\frac {b^{2} \left (c^{2} x^{2}-1\right ) \ln \left (c x +1\right )^{2}}{8 x^{2}}-\frac {b \left (b \,x^{2} \ln \left (-c x +1\right ) c^{2}+2 b c x -b \ln \left (-c x +1\right )+2 a \right ) \ln \left (c x +1\right )}{4 x^{2}}+\frac {b^{2} c^{2} x^{2} \ln \left (-c x +1\right )^{2}+4 b \,c^{2} \ln \left (-c x -1\right ) x^{2} a -4 b^{2} c^{2} \ln \left (-c x -1\right ) x^{2}-4 a b \,c^{2} x^{2} \ln \left (-c x +1\right )-4 b^{2} c^{2} \ln \left (-c x +1\right ) x^{2}+8 b^{2} c^{2} \ln \left (x \right ) x^{2}+4 b^{2} c x \ln \left (-c x +1\right )-8 a b c x -b^{2} \ln \left (-c x +1\right )^{2}+4 b \ln \left (-c x +1\right ) a -4 a^{2}}{8 x^{2}}\)	\(231\)
derivativedivides	\(c^{2} \left (-\frac {a^{2}}{2 c^{2} x^{2}}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{2 c^{2} x^{2}}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{2}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {b^{2} \arctanh \left (c x \right )}{c x}+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{4}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{8}-\frac {b^{2} \ln \left (c x -1\right )}{2}+b^{2} \ln \left (c x \right )-\frac {b^{2} \ln \left (c x +1\right )}{2}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{4}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{4}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{8}-\frac {a b \arctanh \left (c x \right )}{c^{2} x^{2}}-\frac {a b \ln \left (c x -1\right )}{2}+\frac {a b \ln \left (c x +1\right )}{2}-\frac {a b}{c x}\right )\)	\(234\)
default	\(c^{2} \left (-\frac {a^{2}}{2 c^{2} x^{2}}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{2 c^{2} x^{2}}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{2}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{2}-\frac {b^{2} \arctanh \left (c x \right )}{c x}+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{4}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{8}-\frac {b^{2} \ln \left (c x -1\right )}{2}+b^{2} \ln \left (c x \right )-\frac {b^{2} \ln \left (c x +1\right )}{2}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{4}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{4}-\frac {b^{2} \ln \left (c x +1\right )^{2}}{8}-\frac {a b \arctanh \left (c x \right )}{c^{2} x^{2}}-\frac {a b \ln \left (c x -1\right )}{2}+\frac {a b \ln \left (c x +1\right )}{2}-\frac {a b}{c x}\right )\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (74) = 148\).
time = 0.27, size = 151, normalized size = 1.89 \begin {gather*} \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 {gather*}

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 = 0.36, size = 135, normalized size = 1.69 \begin {gather*} \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 {gather*}

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 = 0.40, size = 126, normalized size = 1.58 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (74) = 148\).
time = 0.43, size = 278, normalized size = 3.48 \begin {gather*} \frac {1}{2} \, {\left (2 \, b^{2} c \log \left (-\frac {c x + 1}{c x - 1} - 1\right ) - 2 \, b^{2} c \log \left (-\frac {c x + 1}{c x - 1}\right ) + \frac {{\left (c x + 1\right )} b^{2} c \log \left (-\frac {c x + 1}{c x - 1}\right )^{2}}{{\left (c x - 1\right )} {\left (\frac {{\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {2 \, {\left (c x + 1\right )}}{c x - 1} + 1\right )}} + \frac {2 \, {\left (\frac {2 \, {\left (c x + 1\right )} a b c}{c x - 1} + \frac {{\left (c x + 1\right )} b^{2} c}{c x - 1} + b^{2} c\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {2 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {4 \, {\left (\frac {{\left (c x + 1\right )} a^{2} c}{c x - 1} + \frac {{\left (c x + 1\right )} a b c}{c x - 1} + a b c\right )}}{\frac {{\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {2 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.49, size = 246, normalized size = 3.08 \begin {gather*} \frac {b^2\,c^2\,{\ln \left (c\,x+1\right )}^2}{8}-\frac {a^2}{2\,x^2}+\frac {b^2\,c^2\,{\ln \left (1-c\,x\right )}^2}{8}-\frac {b^2\,{\ln \left (c\,x+1\right )}^2}{8\,x^2}-\frac {b^2\,{\ln \left (1-c\,x\right )}^2}{8\,x^2}+b^2\,c^2\,\ln \left (x\right )-\frac {b^2\,c^2\,\ln \left (c\,x-1\right )}{2}-\frac {b^2\,c^2\,\ln \left (c\,x+1\right )}{2}-\frac {a\,b\,\ln \left (c\,x+1\right )}{2\,x^2}+\frac {a\,b\,\ln \left (1-c\,x\right )}{2\,x^2}+\frac {b^2\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )}{4\,x^2}-\frac {a\,b\,c}{x}-\frac {b^2\,c\,\ln \left (c\,x+1\right )}{2\,x}+\frac {b^2\,c\,\ln \left (1-c\,x\right )}{2\,x}-\frac {a\,b\,c^2\,\ln \left (c\,x-1\right )}{2}+\frac {a\,b\,c^2\,\ln \left (c\,x+1\right )}{2}-\frac {b^2\,c^2\,\ln \left (c\,x+1\right )\,\ln \left (1-c\,x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^2/x^3,x)

[Out]

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

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