3.3.8 \(\int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx\) [208]

Optimal. Leaf size=128 \[ \frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^2 \text {PolyLog}\left (2,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {x \text {PolyLog}\left (3,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {\text {PolyLog}\left (4,-\left ((1+d) e^{2 a+2 b x}\right )\right )}{8 b^3} \]

[Out]

1/12*b*x^4+1/3*x^3*arccoth(1+d+d*tanh(b*x+a))-1/6*x^3*ln(1+(1+d)*exp(2*b*x+2*a))-1/4*x^2*polylog(2,-(1+d)*exp(
2*b*x+2*a))/b+1/4*x*polylog(3,-(1+d)*exp(2*b*x+2*a))/b^2-1/8*polylog(4,-(1+d)*exp(2*b*x+2*a))/b^3

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Rubi [A]
time = 0.18, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6375, 2215, 2221, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {\text {Li}_4\left (-\left ((d+1) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {x \text {Li}_3\left (-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {x^2 \text {Li}_2\left (-\left ((d+1) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{6} x^3 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac {1}{3} x^3 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac {b x^4}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCoth[1 + d + d*Tanh[a + b*x]],x]

[Out]

(b*x^4)/12 + (x^3*ArcCoth[1 + d + d*Tanh[a + b*x]])/3 - (x^3*Log[1 + (1 + d)*E^(2*a + 2*b*x)])/6 - (x^2*PolyLo
g[2, -((1 + d)*E^(2*a + 2*b*x))])/(4*b) + (x*PolyLog[3, -((1 + d)*E^(2*a + 2*b*x))])/(4*b^2) - PolyLog[4, -((1
+ d)*E^(2*a + 2*b*x))]/(8*b^3)

Rule 2215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n))
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6375

Int[ArcCoth[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m
+ 1)*(ArcCoth[c + d*Tanh[a + b*x]]/(f*(m + 1))), x] + Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c - d + c*E^
(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - d)^2, 1]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps

\begin {align*} \int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx &=\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))+\frac {1}{3} b \int \frac {x^3}{1+(1+d) e^{2 a+2 b x}} \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{3} (b (1+d)) \int \frac {e^{2 a+2 b x} x^3}{1+(1+d) e^{2 a+2 b x}} \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )+\frac {1}{2} \int x^2 \log \left (1+(1+d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^2 \text {Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac {\int x \text {Li}_2\left (-(1+d) e^{2 a+2 b x}\right ) \, dx}{2 b}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^2 \text {Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac {x \text {Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {\int \text {Li}_3\left ((-1-d) e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^2 \text {Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac {x \text {Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {\text {Subst}\left (\int \frac {\text {Li}_3((-1-d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac {x^2 \text {Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac {x \text {Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {\text {Li}_4\left (-(1+d) e^{2 a+2 b x}\right )}{8 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 118, normalized size = 0.92 \begin {gather*} \frac {1}{24} \left (8 x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-4 x^3 \log \left (1+\frac {e^{-2 (a+b x)}}{1+d}\right )+\frac {6 x^2 \text {PolyLog}\left (2,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{b}+\frac {6 x \text {PolyLog}\left (3,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{b^2}+\frac {3 \text {PolyLog}\left (4,-\frac {e^{-2 (a+b x)}}{1+d}\right )}{b^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCoth[1 + d + d*Tanh[a + b*x]],x]

[Out]

(8*x^3*ArcCoth[1 + d + d*Tanh[a + b*x]] - 4*x^3*Log[1 + 1/((1 + d)*E^(2*(a + b*x)))] + (6*x^2*PolyLog[2, -(1/(
(1 + d)*E^(2*(a + b*x))))])/b + (6*x*PolyLog[3, -(1/((1 + d)*E^(2*(a + b*x))))])/b^2 + (3*PolyLog[4, -(1/((1 +
d)*E^(2*(a + b*x))))])/b^3)/24

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.52, size = 1667, normalized size = 13.02

method	result	size
risch	\(\text {Expression too large to display}\)	\(1667\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccoth(1+d+d*tanh(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

1/12*I*x^3*Pi*csgn(I*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*
x+2*a)+1))^2-1/3*x^3*ln(exp(b*x+a))+1/12*b*x^4+1/3/b^3*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*a^3+1/12*I*x^3*Pi*cs
gn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3+1/2/b^2*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x*a^2-1/2/b^3*a^2/(1+d)*d
ilog(1+exp(b*x+a)*(-1-d)^(1/2))-1/2/b^3*a^2/(1+d)*dilog(1-exp(b*x+a)*(-1-d)^(1/2))+1/12*I*x^3*Pi*csgn(I*d*exp(
2*b*x+2*a)/(exp(2*b*x+2*a)+1))^3-1/2/b^2*d*a^2/(1+d)*ln(1+exp(b*x+a)*(-1-d)^(1/2))*x-1/2/b^2*d*a^2/(1+d)*ln(1-
exp(b*x+a)*(-1-d)^(1/2))*x+1/6/b^3*d*a^3/(1+d)*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)-1/12*I*x^3*Pi*csgn(I*exp(
2*b*x+2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/4/b*d/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*x^2+1/4
/b^3*d/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*a^2+1/4/b^2*d/(1+d)*polylog(3,-(1+d)*exp(2*b*x+2*a))*x-1/2/b^2*a
^2/(1+d)*ln(1+exp(b*x+a)*(-1-d)^(1/2))*x-1/2/b^2*a^2/(1+d)*ln(1-exp(b*x+a)*(-1-d)^(1/2))*x-1/2/b^3*d*a^3/(1+d)
*ln(1+exp(b*x+a)*(-1-d)^(1/2))-1/2/b^3*d*a^3/(1+d)*ln(1-exp(b*x+a)*(-1-d)^(1/2))-1/2/b^3*d*a^2/(1+d)*dilog(1+e
xp(b*x+a)*(-1-d)^(1/2))-1/2/b^3*d*a^2/(1+d)*dilog(1-exp(b*x+a)*(-1-d)^(1/2))+1/2/b^2/(1+d)*ln(1+(1+d)*exp(2*b*
x+2*a))*x*a^2-1/12*I*x^3*Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))*csgn(I/(exp
(2*b*x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))+1/12*I*x^3*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*cs
gn(I*d)*csgn(I*d*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))+1/12*I*x^3*Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+
2*a))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))-1/6*d/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^3+1/3/b^3/(1+d)*ln(1+
(1+d)*exp(2*b*x+2*a))*a^3-1/4/b/(1+d)*polylog(2,-(1+d)*exp(2*b*x+2*a))*x^2+1/4/b^3/(1+d)*polylog(2,-(1+d)*exp(
2*b*x+2*a))*a^2+1/4/b^2/(1+d)*polylog(3,-(1+d)*exp(2*b*x+2*a))*x-1/2/b^3*a^3/(1+d)*ln(1+exp(b*x+a)*(-1-d)^(1/2
))-1/2/b^3*a^3/(1+d)*ln(1-exp(b*x+a)*(-1-d)^(1/2))-1/8/b^3*d/(1+d)*polylog(4,-(1+d)*exp(2*b*x+2*a))-1/12*I*x^3
*Pi*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2+1/12*I*x^3*Pi*csgn(I/(exp(2*b*x+2*a
)+1))*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))^2-1/6/(1+d)*ln(1+(1+d)*exp(2*b*x+2*a))*x^
3-1/8/b^3/(1+d)*polylog(4,-(1+d)*exp(2*b*x+2*a))+1/6/b^3*a^3/(1+d)*ln(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1)-1/6*I
*x^3*Pi*csgn(I*exp(b*x+a))*csgn(I*exp(2*b*x+2*a))^2+1/12*I*x^3*Pi*csgn(I*exp(b*x+a))^2*csgn(I*exp(2*b*x+2*a))-
1/12*I*x^3*Pi*csgn(I/(exp(2*b*x+2*a)+1)*(exp(2*b*x+2*a)*d+exp(2*b*x+2*a)+1))^3+1/12*I*x^3*Pi*csgn(I*exp(2*b*x+
2*a))^3-1/12*I*x^3*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*csgn(I*d*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-
1/12*I*x^3*Pi*csgn(I*d)*csgn(I*d*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/6*x^3*ln(d)+1/6*ln(exp(2*b*x+2*a)*d+ex
p(2*b*x+2*a)+1)*x^3

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Maxima [A]
time = 0.69, size = 125, normalized size = 0.98 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right ) + \frac {1}{36} \, {\left (\frac {3 \, x^{4}}{d} - \frac {2 \, {\left (4 \, b^{3} x^{3} \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{4} d}\right )} b d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="maxima")

[Out]

1/3*x^3*arccoth(d*tanh(b*x + a) + d + 1) + 1/36*(3*x^4/d - 2*(4*b^3*x^3*log((d + 1)*e^(2*b*x + 2*a) + 1) + 6*b
^2*x^2*dilog(-(d + 1)*e^(2*b*x + 2*a)) - 6*b*x*polylog(3, -(d + 1)*e^(2*b*x + 2*a)) + 3*polylog(4, -(d + 1)*e^
(2*b*x + 2*a)))/(b^4*d))*b*d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 381 vs. \(2 (111) = 222\).
time = 0.36, size = 381, normalized size = 2.98 \begin {gather*} \frac {b^{4} x^{4} + 2 \, b^{3} x^{3} \log \left (\frac {{\left (d + 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) + \sqrt {-4 \, d - 4}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) - \sqrt {-4 \, d - 4}\right ) + 12 \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 12 \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 12 \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 \, d - 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/12*(b^4*x^4 + 2*b^3*x^3*log(((d + 2)*cosh(b*x + a) + d*sinh(b*x + a))/(d*cosh(b*x + a) + d*sinh(b*x + a))) -
6*b^2*x^2*dilog(1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))) - 6*b^2*x^2*dilog(-1/2*sqrt(-4*d - 4)*(co
sh(b*x + a) + sinh(b*x + a))) + 2*a^3*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a) + sqrt(-4*d - 4))
+ 2*a^3*log(2*(d + 1)*cosh(b*x + a) + 2*(d + 1)*sinh(b*x + a) - sqrt(-4*d - 4)) + 12*b*x*polylog(3, 1/2*sqrt(-
4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))) + 12*b*x*polylog(3, -1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x +
a))) - 2*(b^3*x^3 + a^3)*log(1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 2*(b^3*x^3 + a^3)*log(-
1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 12*polylog(4, 1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + si
nh(b*x + a))) - 12*polylog(4, -1/2*sqrt(-4*d - 4)*(cosh(b*x + a) + sinh(b*x + a))))/b^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {acoth}{\left (d \tanh {\left (a + b x \right )} + d + 1 \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acoth(1+d+d*tanh(b*x+a)),x)

[Out]

Integral(x**2*acoth(d*tanh(a + b*x) + d + 1), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccoth(1+d+d*tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^2*arccoth(d*tanh(b*x + a) + d + 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )+1\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acoth(d + d*tanh(a + b*x) + 1),x)

[Out]

int(x^2*acoth(d + d*tanh(a + b*x) + 1), x)

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