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

Optimal. Leaf size=92 \[ \frac {6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac {6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 x^2}{b^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 92, 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} \[ -\frac {2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac {6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac {x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 x^2}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^4/ArcTanh[Tanh[a + b*x]]^3,x]

[Out]

(3*x^2)/b^3 + (6*x*(b*x - ArcTanh[Tanh[a + b*x]]))/b^4 - x^4/(2*b*ArcTanh[Tanh[a + b*x]]^2) - (2*x^3)/(b^2*Arc
Tanh[Tanh[a + b*x]]) + (6*(b*x - ArcTanh[Tanh[a + b*x]])^2*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))^3} \, dx &=-\frac {x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {2 \int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx}{b}\\ &=-\frac {x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {6 \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {3 x^2}{b^3}-\frac {x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (6 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {x}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac {3 x^2}{b^3}+\frac {6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (6 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac {3 x^2}{b^3}+\frac {6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (6 \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^5}\\ &=\frac {3 x^2}{b^3}+\frac {6 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^4}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {6 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 114, normalized size = 1.24 \[ -\frac {\left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^4}{2 b^5 \tanh ^{-1}(\tanh (a+b x))^2}+\frac {4 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^3}{b^5 \tanh ^{-1}(\tanh (a+b x))}+\frac {6 \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )^2 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}-\frac {3 x \left (\tanh ^{-1}(\tanh (a+b x))-b x\right )}{b^4}+\frac {x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.92, size = 95, normalized size = 1.03 \[ \frac {b^{4} x^{4} - 4 \, a b^{3} x^{3} - 11 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x + 7 \, a^{4} + 12 \, {\left (a^{2} b^{2} x^{2} + 2 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^4*x^4 - 4*a*b^3*x^3 - 11*a^2*b^2*x^2 + 2*a^3*b*x + 7*a^4 + 12*(a^2*b^2*x^2 + 2*a^3*b*x + a^4)*log(b*x +
 a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

________________________________________________________________________________________

giac [A]  time = 0.20, size = 61, normalized size = 0.66 \[ \frac {6 \, a^{2} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac {b^{3} x^{2} - 6 \, a b^{2} x}{2 \, b^{6}} + \frac {8 \, a^{3} b x + 7 \, a^{4}}{2 \, {\left (b x + a\right )}^{2} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*a^2*log(abs(b*x + a))/b^5 + 1/2*(b^3*x^2 - 6*a*b^2*x)/b^6 + 1/2*(8*a^3*b*x + 7*a^4)/((b*x + a)^2*b^5)

________________________________________________________________________________________

maple [B]  time = 0.15, size = 371, normalized size = 4.03 \[ \frac {x^{2}}{2 b^{3}}-\frac {3 a x}{b^{4}}-\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x}{b^{4}}+\frac {6 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a^{2}}{b^{5}}+\frac {12 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{5}}+\frac {6 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{5}}-\frac {a^{4}}{2 b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {2 a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {2 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{2 b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {4 a^{3}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {12 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {12 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}+\frac {4 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.38, size = 81, normalized size = 0.88 \[ \frac {b^{4} x^{4} - 4 \, a b^{3} x^{3} - 11 \, a^{2} b^{2} x^{2} + 2 \, a^{3} b x + 7 \, a^{4}}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} + \frac {6 \, a^{2} \log \left (b x + a\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^4*x^4 - 4*a*b^3*x^3 - 11*a^2*b^2*x^2 + 2*a^3*b*x + 7*a^4)/(b^7*x^2 + 2*a*b^6*x + a^2*b^5) + 6*a^2*log(b
*x + a)/b^5

________________________________________________________________________________________

mupad [B]  time = 1.34, size = 867, normalized size = 9.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/atanh(tanh(a + b*x))^3,x)

[Out]

((7*((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)))/(4*b) - x*(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 - 32*a^3 - 24*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 + 48*a^2*(2*a - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*e
xp(2*b*x) + 1)) + log(2/(exp(2*a)*exp(2*b*x) + 1)) + 2*b*x)))/(2*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)^2 + x*(16*a*b^5 - 8*b^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)) + 8*a^2*b^4 + 8*b^6*x^2
- 8*a*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/(2*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)))*(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))/(2*b^5) + (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*b^4)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________