3.1.76 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^4} \, dx\) [76]

Optimal. Leaf size=77 \[ 4 b^4 x-\frac {2 b^2 \tanh ^{-1}(\tanh (a+b x))^2}{x}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{3 x^2}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{3 x^3}-4 b^3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log (x) \]

[Out]

4*b^4*x-2*b^2*arctanh(tanh(b*x+a))^2/x-2/3*b*arctanh(tanh(b*x+a))^3/x^2-1/3*arctanh(tanh(b*x+a))^4/x^3-4*b^3*(
b*x-arctanh(tanh(b*x+a)))*ln(x)

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Rubi [A]
time = 0.04, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2199, 2189, 29} \begin {gather*} -4 b^3 \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )-\frac {2 b^2 \tanh ^{-1}(\tanh (a+b x))^2}{x}-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{3 x^3}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{3 x^2}+4 b^4 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^4,x]

[Out]

4*b^4*x - (2*b^2*ArcTanh[Tanh[a + b*x]]^2)/x - (2*b*ArcTanh[Tanh[a + b*x]]^3)/(3*x^2) - ArcTanh[Tanh[a + b*x]]
^4/(3*x^3) - 4*b^3*(b*x - ArcTanh[Tanh[a + b*x]])*Log[x]

Rule 29

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

Rule 2189

Int[(v_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[b*(x/a), x] - Dist[(b*u
- a*v)/a, Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rule 2199

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[u^(m + 1)*(v^
n/(a*(m + 1))), x] - Dist[b*(n/(a*(m + 1))), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m
, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] &&  !(ILtQ[m + n, -2] && (Fra
ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] &&  !IntegerQ[m]
) || (ILtQ[m, 0] &&  !IntegerQ[n]))

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 82, normalized size = 1.06 \begin {gather*} -\frac {6 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))^2+2 b x \tanh ^{-1}(\tanh (a+b x))^3+\tanh ^{-1}(\tanh (a+b x))^4+2 b^4 x^4 (5+6 \log (x))-2 b^3 x^3 \tanh ^{-1}(\tanh (a+b x)) (11+6 \log (x))}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^4,x]

[Out]

-1/3*(6*b^2*x^2*ArcTanh[Tanh[a + b*x]]^2 + 2*b*x*ArcTanh[Tanh[a + b*x]]^3 + ArcTanh[Tanh[a + b*x]]^4 + 2*b^4*x
^4*(5 + 6*Log[x]) - 2*b^3*x^3*ArcTanh[Tanh[a + b*x]]*(11 + 6*Log[x]))/x^3

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Maple [A]
time = 1.00, size = 77, normalized size = 1.00

method	result	size
default	\(-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{3 x^{3}}+\frac {4 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{2 x^{2}}+\frac {3 b \left (-\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{2}}{x}+2 b \left (\ln \left (x \right ) \arctanh \left (\tanh \left (b x +a \right )\right )-b \left (x \ln \left (x \right )-x \right )\right )\right )}{2}\right )}{3}\)	\(77\)
risch	\(\text {Expression too large to display}\)	\(15664\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*arctanh(tanh(b*x+a))^4/x^3+4/3*b*(-1/2*arctanh(tanh(b*x+a))^3/x^2+3/2*b*(-arctanh(tanh(b*x+a))^2/x+2*b*(l
n(x)*arctanh(tanh(b*x+a))-b*(x*ln(x)-x))))

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Maxima [A]
time = 0.38, size = 91, normalized size = 1.18 \begin {gather*} 2 \, {\left (2 \, {\left (b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right ) \log \left (x\right ) - {\left (b {\left (x + \frac {a}{b}\right )} \log \left (x\right ) - b {\left (x + \frac {a \log \left (x\right )}{b}\right )}\right )} b\right )} b - \frac {b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2}}{x}\right )} b - \frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{3 \, x^{2}} - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^4,x, algorithm="maxima")

[Out]

2*(2*(b*arctanh(tanh(b*x + a))*log(x) - (b*(x + a/b)*log(x) - b*(x + a*log(x)/b))*b)*b - b*arctanh(tanh(b*x +
a))^2/x)*b - 2/3*b*arctanh(tanh(b*x + a))^3/x^2 - 1/3*arctanh(tanh(b*x + a))^4/x^3

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Fricas [A]
time = 0.33, size = 48, normalized size = 0.62 \begin {gather*} \frac {3 \, b^{4} x^{4} + 12 \, a b^{3} x^{3} \log \left (x\right ) - 18 \, a^{2} b^{2} x^{2} - 6 \, a^{3} b x - a^{4}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^4,x, algorithm="fricas")

[Out]

1/3*(3*b^4*x^4 + 12*a*b^3*x^3*log(x) - 18*a^2*b^2*x^2 - 6*a^3*b*x - a^4)/x^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**4,x)

[Out]

Integral(atanh(tanh(a + b*x))**4/x**4, x)

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Giac [A]
time = 0.39, size = 42, normalized size = 0.55 \begin {gather*} b^{4} x + 4 \, a b^{3} \log \left ({\left | x \right |}\right ) - \frac {18 \, a^{2} b^{2} x^{2} + 6 \, a^{3} b x + a^{4}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^4,x, algorithm="giac")

[Out]

b^4*x + 4*a*b^3*log(abs(x)) - 1/3*(18*a^2*b^2*x^2 + 6*a^3*b*x + a^4)/x^3

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Mupad [B]
time = 1.34, size = 571, normalized size = 7.42 \begin {gather*} \frac {11\,b^3\,\ln \left (\frac {{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{3}-\frac {{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^4}{48\,x^3}-\frac {11\,b^3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{3}-\frac {10\,b^4\,x}{3}-\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^4}{48\,x^3}+\frac {b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{12\,x^2}-2\,b^3\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )+\frac {\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{12\,x^3}+\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{12\,x^3}-\frac {b\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{12\,x^2}+2\,b^3\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (x\right )-4\,b^4\,x\,\ln \left (x\right )-\frac {b^2\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{2\,x}-\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{8\,x^3}-\frac {b^2\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{2\,x}+\frac {b^2\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{x}+\frac {b\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{4\,x^2}-\frac {b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}{4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^4,x)

[Out]

(11*b^3*log(exp(2*b*x)/(exp(2*a)*exp(2*b*x) + 1)))/3 - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^4/
(48*x^3) - (11*b^3*log(1/(exp(2*a)*exp(2*b*x) + 1)))/3 - (10*b^4*x)/3 - log(1/(exp(2*a)*exp(2*b*x) + 1))^4/(48
*x^3) + (b*log(1/(exp(2*a)*exp(2*b*x) + 1))^3)/(12*x^2) - 2*b^3*log(1/(exp(2*a)*exp(2*b*x) + 1))*log(x) + (log
(1/(exp(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3)/(12*x^3) + (log(1/(exp(2
*a)*exp(2*b*x) + 1))^3*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/(12*x^3) - (b*log((exp(2*a)*exp(2
*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3)/(12*x^2) + 2*b^3*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log
(x) - 4*b^4*x*log(x) - (b^2*log(1/(exp(2*a)*exp(2*b*x) + 1))^2)/(2*x) - (log(1/(exp(2*a)*exp(2*b*x) + 1))^2*lo
g((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2)/(8*x^3) - (b^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2
*b*x) + 1))^2)/(2*x) + (b^2*log(1/(exp(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) +
1)))/x + (b*log(1/(exp(2*a)*exp(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2)/(4*x^2) -
(b*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/(4*x^2)

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