3.75 \(\int \frac {\tanh ^{-1}(\tanh (a+b x))^4}{x^3} \, dx\)

Optimal. Leaf size=87 \[ -6 b^3 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+3 b^2 \tanh ^{-1}(\tanh (a+b x))^2+6 b^2 \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{2 x^2}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{x} \]

[Out]

-6*b^3*x*(b*x-arctanh(tanh(b*x+a)))+3*b^2*arctanh(tanh(b*x+a))^2-2*b*arctanh(tanh(b*x+a))^3/x-1/2*arctanh(tanh
(b*x+a))^4/x^2+6*b^2*(b*x-arctanh(tanh(b*x+a)))^2*ln(x)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2168, 2159, 2158, 29} \[ 3 b^2 \tanh ^{-1}(\tanh (a+b x))^2-6 b^3 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )+6 b^2 \log (x) \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{2 x^2}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[Tanh[a + b*x]]^4/x^3,x]

[Out]

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

Rule 29

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

Rule 2158

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 2159

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^n/(a*n), x] - Dis
t[(b*u - a*v)/a, Int[v^(n - 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && GtQ[n, 0] && Ne
Q[n, 1]

Rule 2168

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 81, normalized size = 0.93 \[ -6 b^3 x (2 \log (x)+1) \tanh ^{-1}(\tanh (a+b x))+3 b^2 (2 \log (x)+3) \tanh ^{-1}(\tanh (a+b x))^2-\frac {\tanh ^{-1}(\tanh (a+b x))^4}{2 x^2}-\frac {2 b \tanh ^{-1}(\tanh (a+b x))^3}{x}+6 b^4 x^2 \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[Tanh[a + b*x]]^4/x^3,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 47, normalized size = 0.54 \[ \frac {b^{4} x^{4} + 8 \, a b^{3} x^{3} + 12 \, a^{2} b^{2} x^{2} \log \relax (x) - 8 \, a^{3} b x - a^{4}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.23, size = 43, normalized size = 0.49 \[ \frac {1}{2} \, b^{4} x^{2} + 4 \, a b^{3} x + 6 \, a^{2} b^{2} \log \left ({\left | x \right |}\right ) - \frac {8 \, a^{3} b x + a^{4}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^3,x, algorithm="giac")

[Out]

1/2*b^4*x^2 + 4*a*b^3*x + 6*a^2*b^2*log(abs(x)) - 1/2*(8*a^3*b*x + a^4)/x^2

________________________________________________________________________________________

maple [A]  time = 0.18, size = 93, normalized size = 1.07 \[ -\frac {\arctanh \left (\tanh \left (b x +a \right )\right )^{4}}{2 x^{2}}-\frac {2 b \arctanh \left (\tanh \left (b x +a \right )\right )^{3}}{x}+6 b^{2} \ln \relax (x ) \arctanh \left (\tanh \left (b x +a \right )\right )^{2}+6 b^{4} x^{2} \ln \relax (x )-9 x^{2} b^{4}-12 b^{3} \arctanh \left (\tanh \left (b x +a \right )\right ) \ln \relax (x ) x +12 b^{3} \arctanh \left (\tanh \left (b x +a \right )\right ) x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(tanh(b*x+a))^4/x^3,x)

[Out]

-1/2*arctanh(tanh(b*x+a))^4/x^2-2*b*arctanh(tanh(b*x+a))^3/x+6*b^2*ln(x)*arctanh(tanh(b*x+a))^2+6*b^4*x^2*ln(x
)-9*x^2*b^4-12*b^3*arctanh(tanh(b*x+a))*ln(x)*x+12*b^3*arctanh(tanh(b*x+a))*x

________________________________________________________________________________________

maxima [A]  time = 0.85, size = 83, normalized size = 0.95 \[ -\frac {2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{3}}{x} + 3 \, {\left (2 \, b \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \relax (x) + {\left (b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} \log \relax (x) - 2 \, \operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{2} \log \relax (x)\right )} b\right )} b - \frac {\operatorname {artanh}\left (\tanh \left (b x + a\right )\right )^{4}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(tanh(b*x+a))^4/x^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.54, size = 672, normalized size = 7.72 \[ \frac {9\,b^2\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2}{4}-\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}{32\,x^2}-\frac {{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^4}{32\,x^2}+\frac {9\,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}{4}-3\,b^3\,x\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\frac {b\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^3}{4\,x}+\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}{8\,x^2}+\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 )}{8\,x^2}-\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}{4\,x}+\frac {3\,b^2\,{\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )}^2\,\ln \relax (x)}{2}-\frac {3\,{\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}{16\,x^2}+\frac {3\,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\,\ln \relax (x)}{2}+6\,b^4\,x^2\,\ln \relax (x)-\frac {9\,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 )}{2}+3\,b^3\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\frac {3\,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}-\frac {3\,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}-3\,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 )\,\ln \relax (x)+6\,b^3\,x\,\ln \left (\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \relax (x)-6\,b^3\,x\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tanh(a + b*x))^4/x^3,x)

[Out]

(9*b^2*log(1/(exp(2*a)*exp(2*b*x) + 1))^2)/4 - log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^4/(32*x^2)
- log(1/(exp(2*a)*exp(2*b*x) + 1))^4/(32*x^2) + (9*b^2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2
)/4 - 3*b^3*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + (b*log(1/(exp(2*a)*exp(2*b*x) + 1))^3)/(4
*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)/(8*x^2) + (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)))/(8*x^2) - (b*log((exp(2*
a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^3)/(4*x) + (3*b^2*log(1/(exp(2*a)*exp(2*b*x) + 1))^2*log(x))/2 - (3*
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))^2)/(16*x^2) + (3*b^2*l
og((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))^2*log(x))/2 + 6*b^4*x^2*log(x) - (9*b^2*log(1/(exp(2*a)*ex
p(2*b*x) + 1))*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)))/2 + 3*b^3*x*log(1/(exp(2*a)*exp(2*b*x) +
1)) + (3*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) - (3
*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) - 3*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))*log(x) + 6*b^3*x*log(1/(exp(
2*a)*exp(2*b*x) + 1))*log(x) - 6*b^3*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1))*log(x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}^{4}{\left (\tanh {\left (a + b x \right )} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(tanh(b*x+a))**4/x**3,x)

[Out]

Integral(atanh(tanh(a + b*x))**4/x**3, x)

________________________________________________________________________________________