3.1.9 \(\int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx\) [9]

Optimal. Leaf size=37 \[ -\frac {b c}{2 x}+\frac {1}{2} b c^2 \tanh ^{-1}(c x)-\frac {a+b \tanh ^{-1}(c x)}{2 x^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6037, 331, 212} \begin {gather*} -\frac {a+b \tanh ^{-1}(c x)}{2 x^2}+\frac {1}{2} b c^2 \tanh ^{-1}(c x)-\frac {b c}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/x^3,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 331

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 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
]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 59, normalized size = 1.59 \begin {gather*} -\frac {a}{2 x^2}-\frac {b c}{2 x}-\frac {b \tanh ^{-1}(c x)}{2 x^2}-\frac {1}{4} b c^2 \log (1-c x)+\frac {1}{4} b c^2 \log (1+c x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/x^3,x]

[Out]

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

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Maple [A]
time = 0.02, size = 55, normalized size = 1.49

method	result	size
derivativedivides	\(c^{2} \left (-\frac {a}{2 c^{2} x^{2}}-\frac {b \arctanh \left (c x \right )}{2 c^{2} x^{2}}-\frac {b \ln \left (c x -1\right )}{4}+\frac {b \ln \left (c x +1\right )}{4}-\frac {b}{2 c x}\right )\)	\(55\)
default	\(c^{2} \left (-\frac {a}{2 c^{2} x^{2}}-\frac {b \arctanh \left (c x \right )}{2 c^{2} x^{2}}-\frac {b \ln \left (c x -1\right )}{4}+\frac {b \ln \left (c x +1\right )}{4}-\frac {b}{2 c x}\right )\)	\(55\)
risch	\(-\frac {b \ln \left (c x +1\right )}{4 x^{2}}-\frac {b \,x^{2} \ln \left (-c x +1\right ) c^{2}-b \,c^{2} \ln \left (-c x -1\right ) x^{2}+2 b c x -b \ln \left (-c x +1\right )+2 a}{4 x^{2}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 45, normalized size = 1.22 \begin {gather*} \frac {1}{4} \, {\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 )} b - \frac {a}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 43, normalized size = 1.16 \begin {gather*} -\frac {2 \, b c x - {\left (b c^{2} x^{2} - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.26, size = 36, normalized size = 0.97 \begin {gather*} - \frac {a}{2 x^{2}} + \frac {b c^{2} \operatorname {atanh}{\left (c x \right )}}{2} - \frac {b c}{2 x} - \frac {b \operatorname {atanh}{\left (c x \right )}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))/x**3,x)

[Out]

-a/(2*x**2) + b*c**2*atanh(c*x)/2 - b*c/(2*x) - b*atanh(c*x)/(2*x**2)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (31) = 62\).
time = 0.41, size = 135, normalized size = 3.65 \begin {gather*} {\left (\frac {{\left (c x + 1\right )} b c \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\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 {\frac {2 \, {\left (c x + 1\right )} a c}{c x - 1} + \frac {{\left (c x + 1\right )} b c}{c x - 1} + b c}{\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))/x^3,x, algorithm="giac")

[Out]

((c*x + 1)*b*c*log(-(c*x + 1)/(c*x - 1))/((c*x - 1)*((c*x + 1)^2/(c*x - 1)^2 + 2*(c*x + 1)/(c*x - 1) + 1)) + (
2*(c*x + 1)*a*c/(c*x - 1) + (c*x + 1)*b*c/(c*x - 1) + 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 = 0.73, size = 46, normalized size = 1.24 \begin {gather*} \frac {b\,c\,\mathrm {atan}\left (\frac {c^2\,x}{\sqrt {-c^2}}\right )\,\sqrt {-c^2}}{2}-\frac {\frac {a}{2}+\frac {b\,\mathrm {atanh}\left (c\,x\right )}{2}+\frac {b\,c\,x}{2}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/x^3,x)

[Out]

(b*c*atan((c^2*x)/(-c^2)^(1/2))*(-c^2)^(1/2))/2 - (a/2 + (b*atanh(c*x))/2 + (b*c*x)/2)/x^2

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