3.94 \(\int \frac {x^4}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=98 \[ \frac {4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {4 x^3}{3 b^2} \]

[Out]

4/3*x^3/b^2+2*x^2*(b*x-arctanh(tanh(b*x+a)))/b^3+4*x*(b*x-arctanh(tanh(b*x+a)))^2/b^4-x^4/b/arctanh(tanh(b*x+a
))+4*(b*x-arctanh(tanh(b*x+a)))^3*ln(arctanh(tanh(b*x+a)))/b^5

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Rubi [A]  time = 0.08, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2168, 2159, 2158, 2157, 29} \[ \frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}+\frac {4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {4 x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

Rule 29

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

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, 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 {x^4}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {4 \int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {4 x^3}{3 b^2}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {x}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}\\ &=\frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 106, normalized size = 1.08 \[ -\frac {\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}{b^5 \tanh ^{-1}(\tanh (a+b x))}-\frac {4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac {3 x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{b^4}-\frac {x^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^3}+\frac {x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

x^3/(3*b^2) - (x^2*(-(b*x) + ArcTanh[Tanh[a + b*x]]))/b^3 + (3*x*(-(b*x) + ArcTanh[Tanh[a + b*x]])^2)/b^4 - (-
(b*x) + ArcTanh[Tanh[a + b*x]])^4/(b^5*ArcTanh[Tanh[a + b*x]]) - (4*(-(b*x) + ArcTanh[Tanh[a + b*x]])^3*Log[Ar
cTanh[Tanh[a + b*x]]])/b^5

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fricas [A]  time = 0.45, size = 73, normalized size = 0.74 \[ \frac {b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4} - 12 \, {\left (a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{3 \, {\left (b^{6} x + a b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

1/3*(b^4*x^4 - 2*a*b^3*x^3 + 6*a^2*b^2*x^2 + 9*a^3*b*x - 3*a^4 - 12*(a^3*b*x + a^4)*log(b*x + a))/(b^6*x + a*b
^5)

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giac [A]  time = 0.16, size = 62, normalized size = 0.63 \[ -\frac {4 \, a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{5}} - \frac {a^{4}}{{\left (b x + a\right )} b^{5}} + \frac {b^{4} x^{3} - 3 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x}{3 \, b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

-4*a^3*log(abs(b*x + a))/b^5 - a^4/((b*x + a)*b^5) + 1/3*(b^4*x^3 - 3*a*b^3*x^2 + 9*a^2*b^2*x)/b^6

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maple [B]  time = 0.15, size = 350, normalized size = 3.57 \[ \frac {x^{3}}{3 b^{2}}-\frac {a \,x^{2}}{b^{3}}-\frac {x^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3}}+\frac {3 a^{2} x}{b^{4}}+\frac {6 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x}{b^{4}}+\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} x}{b^{4}}-\frac {4 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a^{3}}{b^{5}}-\frac {12 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{5}}-\frac {12 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{5}}-\frac {4 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{5}}-\frac {a^{4}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {4 a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {6 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {4 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arctanh(tanh(b*x+a))^2,x)

[Out]

1/3*x^3/b^2-1/b^3*a*x^2-1/b^3*x^2*(arctanh(tanh(b*x+a))-b*x-a)+3/b^4*a^2*x+6/b^4*a*(arctanh(tanh(b*x+a))-b*x-a
)*x+3/b^4*(arctanh(tanh(b*x+a))-b*x-a)^2*x-4/b^5*ln(arctanh(tanh(b*x+a)))*a^3-12/b^5*ln(arctanh(tanh(b*x+a)))*
a^2*(arctanh(tanh(b*x+a))-b*x-a)-12/b^5*ln(arctanh(tanh(b*x+a)))*a*(arctanh(tanh(b*x+a))-b*x-a)^2-4/b^5*ln(arc
tanh(tanh(b*x+a)))*(arctanh(tanh(b*x+a))-b*x-a)^3-1/b^5/arctanh(tanh(b*x+a))*a^4-4/b^5/arctanh(tanh(b*x+a))*a^
3*(arctanh(tanh(b*x+a))-b*x-a)-6/b^5/arctanh(tanh(b*x+a))*a^2*(arctanh(tanh(b*x+a))-b*x-a)^2-4/b^5/arctanh(tan
h(b*x+a))*a*(arctanh(tanh(b*x+a))-b*x-a)^3-1/b^5/arctanh(tanh(b*x+a))*(arctanh(tanh(b*x+a))-b*x-a)^4

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maxima [A]  time = 0.75, size = 70, normalized size = 0.71 \[ \frac {b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4}}{3 \, {\left (b^{6} x + a b^{5}\right )}} - \frac {4 \, a^{3} \log \left (b x + a\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

1/3*(b^4*x^4 - 2*a*b^3*x^3 + 6*a^2*b^2*x^2 + 9*a^3*b*x - 3*a^4)/(b^6*x + a*b^5) - 4*a^3*log(b*x + a)/b^5

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mupad [B]  time = 0.18, size = 669, normalized size = 6.83 \[ \frac {x^3}{3\,b^2}-\frac {{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^4+24\,a^2\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+16\,a^4-8\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-32\,a^3\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2\,b\,\left (8\,a\,b^4+8\,b^5\,x-4\,b^4\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )}+\frac {x^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2\,b^3}+\frac {3\,x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^4}+\frac {\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )}{2\,b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/atanh(tanh(a + b*x))^2,x)

[Out]

x^3/(3*b^2) - ((2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)
) + 2*b*x)^4 + 24*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*
x) + 1)) + 2*b*x)^2 + 16*a^4 - 8*a*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(
2*a)*exp(2*b*x) + 1)) + 2*b*x)^3 - 32*a^3*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(
2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(2*b*(8*a*b^4 + 8*b^5*x - 4*b^4*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp
(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))) + (x^2*(log(2/(exp(2*a)*exp(2*b*x) + 1))
- log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(2*b^3) + (3*x*(log(2/(exp(2*a)*exp(2*b*x)
+ 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2)/(4*b^4) + (log(log((2*exp(2*a)*exp(
2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) - log(2/(exp(2*a)*exp(2*b*x) + 1)))*((2*a - log((2*exp(2*a)*exp(2*b*x))/(ex
p(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^3 - 8*a^3 - 6*a*(2*a - log((2*exp(2*a)*exp
(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)^2 + 12*a^2*(2*a - log((2*exp(2
*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)))/(2*b^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/atanh(tanh(b*x+a))**2,x)

[Out]

Integral(x**4/atanh(tanh(a + b*x))**2, x)

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