3.2.76 \(\int \frac {x^4}{\coth ^{-1}(\tanh (a+b x))^3} \, dx\) [176]

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

[Out]

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

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Rubi [A]
time = 0.05, 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 = {2199, 2190, 2189, 2188, 29} \begin {gather*} \frac {6 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac {6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {3 x^2}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/ArcCoth[Tanh[a + b*x]]^3,x]

[Out]

(3*x^2)/b^3 + (6*x*(b*x - ArcCoth[Tanh[a + b*x]]))/b^4 - x^4/(2*b*ArcCoth[Tanh[a + b*x]]^2) - (2*x^3)/(b^2*Arc
Coth[Tanh[a + b*x]]) + (6*(b*x - ArcCoth[Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[a + b*x]]])/b^5

Rule 29

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

Rule 2188

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 2189

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 2190

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 2199

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}{\coth ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}+\frac {2 \int \frac {x^3}{\coth ^{-1}(\tanh (a+b x))^2} \, dx}{b}\\ &=-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {6 \int \frac {x^2}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {3 x^2}{b^3}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}-\frac {\left (6 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {x}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac {3 x^2}{b^3}+\frac {6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {\left (6 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac {3 x^2}{b^3}+\frac {6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {\left (6 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{b^5}\\ &=\frac {3 x^2}{b^3}+\frac {6 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^4}{2 b \coth ^{-1}(\tanh (a+b x))^2}-\frac {2 x^3}{b^2 \coth ^{-1}(\tanh (a+b x))}+\frac {6 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 114, normalized size = 1.24 \begin {gather*} \frac {x^2}{2 b^3}-\frac {3 x \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )}{b^4}+\frac {4 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^3}{b^5 \coth ^{-1}(\tanh (a+b x))}-\frac {\left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^4}{2 b^5 \coth ^{-1}(\tanh (a+b x))^2}+\frac {6 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^2/(2*b^3) - (3*x*(-(b*x) + ArcCoth[Tanh[a + b*x]]))/b^4 + (4*(-(b*x) + ArcCoth[Tanh[a + b*x]])^3)/(b^5*ArcCo
th[Tanh[a + b*x]]) - (-(b*x) + ArcCoth[Tanh[a + b*x]])^4/(2*b^5*ArcCoth[Tanh[a + b*x]]^2) + (6*(-(b*x) + ArcCo
th[Tanh[a + b*x]])^2*Log[ArcCoth[Tanh[a + b*x]]])/b^5

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.45, size = 29456, normalized size = 320.17

method	result	size
risch	\(\text {Expression too large to display}\)	\(29456\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/arccoth(tanh(b*x+a))^3,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [C] Result contains complex when optimal does not.
time = 0.88, size = 198, normalized size = 2.15 \begin {gather*} \frac {16 \, b^{4} x^{4} + 7 \, \pi ^{4} + 56 i \, \pi ^{3} a - 168 \, \pi ^{2} a^{2} - 224 i \, \pi a^{3} + 112 \, a^{4} - 32 \, {\left (-i \, \pi b^{3} + 2 \, a b^{3}\right )} x^{3} + 44 \, {\left (\pi ^{2} b^{2} + 4 i \, \pi a b^{2} - 4 \, a^{2} b^{2}\right )} x^{2} - 4 \, {\left (-i \, \pi ^{3} b + 6 \, \pi ^{2} a b + 12 i \, \pi a^{2} b - 8 \, a^{3} b\right )} x}{8 \, {\left (4 \, b^{7} x^{2} - \pi ^{2} b^{5} - 4 i \, \pi a b^{5} + 4 \, a^{2} b^{5} - 4 \, {\left (i \, \pi b^{6} - 2 \, a b^{6}\right )} x\right )}} - \frac {3 \, {\left (\pi ^{2} + 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{2 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(16*b^4*x^4 + 7*pi^4 + 56*I*pi^3*a - 168*pi^2*a^2 - 224*I*pi*a^3 + 112*a^4 - 32*(-I*pi*b^3 + 2*a*b^3)*x^3
+ 44*(pi^2*b^2 + 4*I*pi*a*b^2 - 4*a^2*b^2)*x^2 - 4*(-I*pi^3*b + 6*pi^2*a*b + 12*I*pi*a^2*b - 8*a^3*b)*x)/(4*b^
7*x^2 - pi^2*b^5 - 4*I*pi*a*b^5 + 4*a^2*b^5 - 4*(I*pi*b^6 - 2*a*b^6)*x) - 3/2*(pi^2 + 4*I*pi*a - 4*a^2)*log(-I
*pi + 2*b*x + 2*a)/b^5

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (90) = 180\).
time = 0.37, size = 493, normalized size = 5.36 \begin {gather*} \frac {64 \, b^{6} x^{6} - 128 \, a b^{5} x^{5} - 7 \, \pi ^{6} - 28 \, \pi ^{4} a^{2} + 112 \, \pi ^{2} a^{4} + 448 \, a^{6} + 32 \, {\left (\pi ^{2} b^{4} - 36 \, a^{2} b^{4}\right )} x^{4} - 512 \, {\left (\pi ^{2} a b^{3} + 3 \, a^{3} b^{3}\right )} x^{3} - 32 \, {\left (\pi ^{4} b^{2} + 32 \, \pi ^{2} a^{2} b^{2}\right )} x^{2} - 32 \, {\left (5 \, \pi ^{4} a b + 12 \, \pi ^{2} a^{3} b - 32 \, a^{5} b\right )} x - 96 \, {\left (16 \, \pi a b^{4} x^{4} + 64 \, \pi a^{2} b^{3} x^{3} + \pi ^{5} a + 8 \, \pi ^{3} a^{3} + 16 \, \pi a^{5} + 8 \, {\left (\pi ^{3} a b^{2} + 12 \, \pi a^{3} b^{2}\right )} x^{2} + 16 \, {\left (\pi ^{3} a^{2} b + 4 \, \pi a^{4} b\right )} x\right )} \arctan \left (-\frac {2 \, b x + 2 \, a - \sqrt {4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}}}{\pi }\right ) - 6 \, {\left (\pi ^{6} + 4 \, \pi ^{4} a^{2} - 16 \, \pi ^{2} a^{4} - 64 \, a^{6} + 16 \, {\left (\pi ^{2} b^{4} - 4 \, a^{2} b^{4}\right )} x^{4} + 64 \, {\left (\pi ^{2} a b^{3} - 4 \, a^{3} b^{3}\right )} x^{3} + 8 \, {\left (\pi ^{4} b^{2} + 8 \, \pi ^{2} a^{2} b^{2} - 48 \, a^{4} b^{2}\right )} x^{2} + 16 \, {\left (\pi ^{4} a b - 16 \, a^{5} b\right )} x\right )} \log \left (4 \, b^{2} x^{2} + 8 \, a b x + \pi ^{2} + 4 \, a^{2}\right )}{8 \, {\left (16 \, b^{9} x^{4} + 64 \, a b^{8} x^{3} + \pi ^{4} b^{5} + 8 \, \pi ^{2} a^{2} b^{5} + 16 \, a^{4} b^{5} + 8 \, {\left (\pi ^{2} b^{7} + 12 \, a^{2} b^{7}\right )} x^{2} + 16 \, {\left (\pi ^{2} a b^{6} + 4 \, a^{3} b^{6}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(64*b^6*x^6 - 128*a*b^5*x^5 - 7*pi^6 - 28*pi^4*a^2 + 112*pi^2*a^4 + 448*a^6 + 32*(pi^2*b^4 - 36*a^2*b^4)*x
^4 - 512*(pi^2*a*b^3 + 3*a^3*b^3)*x^3 - 32*(pi^4*b^2 + 32*pi^2*a^2*b^2)*x^2 - 32*(5*pi^4*a*b + 12*pi^2*a^3*b -
32*a^5*b)*x - 96*(16*pi*a*b^4*x^4 + 64*pi*a^2*b^3*x^3 + pi^5*a + 8*pi^3*a^3 + 16*pi*a^5 + 8*(pi^3*a*b^2 + 12*
pi*a^3*b^2)*x^2 + 16*(pi^3*a^2*b + 4*pi*a^4*b)*x)*arctan(-(2*b*x + 2*a - sqrt(4*b^2*x^2 + 8*a*b*x + pi^2 + 4*a
^2))/pi) - 6*(pi^6 + 4*pi^4*a^2 - 16*pi^2*a^4 - 64*a^6 + 16*(pi^2*b^4 - 4*a^2*b^4)*x^4 + 64*(pi^2*a*b^3 - 4*a^
3*b^3)*x^3 + 8*(pi^4*b^2 + 8*pi^2*a^2*b^2 - 48*a^4*b^2)*x^2 + 16*(pi^4*a*b - 16*a^5*b)*x)*log(4*b^2*x^2 + 8*a*
b*x + pi^2 + 4*a^2))/(16*b^9*x^4 + 64*a*b^8*x^3 + pi^4*b^5 + 8*pi^2*a^2*b^5 + 16*a^4*b^5 + 8*(pi^2*b^7 + 12*a^
2*b^7)*x^2 + 16*(pi^2*a*b^6 + 4*a^3*b^6)*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\operatorname {acoth}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 0.42, size = 163, normalized size = 1.77 \begin {gather*} -\frac {16 \, \pi ^{3} b x - 96 i \, \pi ^{2} a b x - 192 \, \pi a^{2} b x + 128 i \, a^{3} b x + 7 i \, \pi ^{4} + 56 \, \pi ^{3} a - 168 i \, \pi ^{2} a^{2} - 224 \, \pi a^{3} + 112 i \, a^{4}}{-32 i \, b^{7} x^{2} + 32 \, \pi b^{6} x - 64 i \, a b^{6} x + 8 i \, \pi ^{2} b^{5} + 32 \, \pi a b^{5} - 32 i \, a^{2} b^{5}} + \frac {x^{2}}{2 \, b^{3}} - \frac {3 \, {\left (i \, \pi + 2 \, a\right )} x}{2 \, b^{4}} - \frac {3 \, {\left (\pi ^{2} - 4 i \, \pi a - 4 \, a^{2}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{2 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(16*pi^3*b*x - 96*I*pi^2*a*b*x - 192*pi*a^2*b*x + 128*I*a^3*b*x + 7*I*pi^4 + 56*pi^3*a - 168*I*pi^2*a^2 - 224
*pi*a^3 + 112*I*a^4)/(-32*I*b^7*x^2 + 32*pi*b^6*x - 64*I*a*b^6*x + 8*I*pi^2*b^5 + 32*pi*a*b^5 - 32*I*a^2*b^5)
+ 1/2*x^2/b^3 - 3/2*(I*pi + 2*a)*x/b^4 - 3/2*(pi^2 - 4*I*pi*a - 4*a^2)*log(I*pi + 2*b*x + 2*a)/b^5

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Mupad [B]
time = 1.38, size = 867, normalized size = 9.42 \begin {gather*} \frac {\frac {7\,\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 )}^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 )\right )}{4\,b}-x\,\left (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 )}^3-32\,a^3-24\,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+48\,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^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 )}^2+x\,\left (16\,a\,b^5-8\,b^5\,\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\,a^2\,b^4+8\,b^6\,x^2-8\,a\,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 )}+\frac {x^2}{2\,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 (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 )}{2\,b^5}+\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\,b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/acoth(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)*ex
p(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)*exp(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)

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