3.95 \(\int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx\)

Optimal. Leaf size=75 \[ \frac {3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac {3 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^3}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {3 x^2}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {3 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^3}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {3 x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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Mathematica [A]  time = 0.06, size = 83, normalized size = 1.11 \[ \frac {\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}{b^4 \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {2 x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^3}+\frac {x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcTanh[Tanh[a + b*x]]^2,x]

[Out]

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

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fricas [A]  time = 0.45, size = 62, normalized size = 0.83 \[ \frac {b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3} + 6 \, {\left (a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 48, normalized size = 0.64 \[ \frac {3 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac {a^{3}}{{\left (b x + a\right )} b^{4}} + \frac {b^{2} x^{2} - 4 \, a b x}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.15, size = 223, normalized size = 2.97 \[ \frac {x^{2}}{2 b^{2}}-\frac {2 a x}{b^{3}}-\frac {2 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x}{b^{3}}+\frac {3 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a^{2}}{b^{4}}+\frac {6 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4}}+\frac {3 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4}}+\frac {a^{3}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arctanh(tanh(b*x+a))^2,x)

[Out]

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

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maxima [A]  time = 0.75, size = 59, normalized size = 0.79 \[ \frac {b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3}}{2 \, {\left (b^{5} x + a b^{4}\right )}} + \frac {3 \, a^{2} \log \left (b x + a\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arctanh(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.03, size = 490, normalized size = 6.53 \[ \frac {x^2}{2\,b^2}+\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\,{\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 )}{4\,b^4}-\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 )}^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 )}{4\,b\,\left (2\,a\,b^3+2\,b^4\,x-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 )\right )}+\frac {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 )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/atanh(tanh(a + b*x))^2,x)

[Out]

x^2/(2*b^2) + (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*(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))/(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)^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/(e
xp(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)) + l
og(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x))/(4*b*(2*a*b^3 + 2*b^4*x - b^3*(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))) + (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))/b^3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/atanh(tanh(b*x+a))**2,x)

[Out]

Integral(x**3/atanh(tanh(a + b*x))**2, x)

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