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

Optimal. Leaf size=81 \[ \frac {\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac {x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}+\frac {x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{2 b^2}+\frac {x^3}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 81, 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 = {2159, 2158, 2157, 29} \[ \frac {x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{2 b^2}+\frac {x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^3}+\frac {\left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac {x^3}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3/ArcTanh[Tanh[a + b*x]],x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 79, normalized size = 0.98 \[ -\frac {\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac {x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2}{b^3}-\frac {x^2 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{2 b^2}+\frac {x^3}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 41, normalized size = 0.51 \[ \frac {2 \, b^{3} x^{3} - 3 \, a b^{2} x^{2} + 6 \, a^{2} b x - 6 \, a^{3} \log \left (b x + a\right )}{6 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 43, normalized size = 0.53 \[ -\frac {a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac {2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{6 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.14, size = 202, normalized size = 2.49 \[ \frac {x^{3}}{3 b}-\frac {a \,x^{2}}{2 b^{2}}-\frac {x^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{2 b^{2}}+\frac {a^{2} x}{b^{3}}+\frac {2 x a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3}}+\frac {x \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{3}}-\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a^{3}}{b^{4}}-\frac {3 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4}}-\frac {3 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4}}-\frac {\ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 42, normalized size = 0.52 \[ -\frac {a^{3} \log \left (b x + a\right )}{b^{4}} + \frac {2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{6 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 354, normalized size = 4.37 \[ \frac {x^3}{3\,b}+\frac {x^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{4\,b^2}+\frac {x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^3}+\frac {\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )}{8\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/atanh(tanh(a + b*x)),x)

[Out]

x^3/(3*b) + (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))/(4*b^2) + (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^3) + (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))/(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)))/(8*b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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