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

Optimal. Leaf size=71 \[ \frac {3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 71, 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, 2158, 2157, 29} \[ -\frac {3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 86, normalized size = 1.21 \[ -\frac {3 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-b x \tanh ^{-1}(\tanh (a+b x))^2 \left (6 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+11\right )+\tanh ^{-1}(\tanh (a+b x))^3 \left (6 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+5\right )+b^3 x^3}{2 b^4 \tanh ^{-1}(\tanh (a+b x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.56, size = 83, normalized size = 1.17 \[ \frac {2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 44, normalized size = 0.62 \[ \frac {x}{b^{3}} - \frac {3 \, a \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac {6 \, a^{2} b x + 5 \, a^{3}}{2 \, {\left (b x + a\right )}^{2} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.15, size = 239, normalized size = 3.37 \[ \frac {x}{b^{3}}+\frac {a^{3}}{2 b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{2 b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{2 b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{2 b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {3 a^{2}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {6 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {3 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a}{b^{4}}-\frac {3 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.41, size = 69, normalized size = 0.97 \[ \frac {2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3}}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} - \frac {3 \, a \log \left (b x + a\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.47, size = 620, normalized size = 8.73 \[ \frac {x}{b^3}-\frac {x\,\left (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 )}^2-12\,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 )+12\,a^2\right )-\frac {5\,\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 )}{4\,b}}{b^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+x\,\left (8\,a\,b^4-4\,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 )+4\,a^2\,b^3+4\,b^5\,x^2-4\,a\,b^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 )}+\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 (3\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-3\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+6\,b\,x\right )}{2\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/atanh(tanh(a + b*x))^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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