3.3.13 \(\int x^2 \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx\) [213]

Optimal. Leaf size=139 \[ \frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{6} x^3 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x^2 \text {PolyLog}\left (2,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}+\frac {x \text {PolyLog}\left (3,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {\text {PolyLog}\left (4,-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3} \]

[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 = 139, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6375, 2215, 2221, 2611, 6744, 2320, 6724} \begin {gather*} -\frac {\text {Li}_4\left (-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^3}+\frac {x \text {Li}_3\left (-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b^2}-\frac {x^2 \text {Li}_2\left (-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{6} x^3 \log \left ((1-d) 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 = 0.07, size = 119, normalized size = 0.86 \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 = 1.70, size = 1745, normalized size = 12.55

method	result	size
risch	\(\text {Expression too large to display}\)	\(1745\)

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]

1/2/b^3*a^3/(d-1)*ln(1-exp(b*x+a)*(d-1)^(1/2))+1/2/b^3*a^3/(d-1)*ln(1+exp(b*x+a)*(d-1)^(1/2))-1/6*d/(d-1)*ln(1
-(d-1)*exp(2*b*x+2*a))*x^3-1/8/b^3*d/(d-1)*polylog(4,(d-1)*exp(2*b*x+2*a))+1/2/b^3*a^2/(d-1)*dilog(1-exp(b*x+a
)*(d-1)^(1/2))+1/2/b^3*a^2/(d-1)*dilog(1+exp(b*x+a)*(d-1)^(1/2))-1/3/b^3/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*a^3+
1/4/b/(d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*x^2-1/4/b^3/(d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*a^2-1/4/b^2/(d-1
)*polylog(3,(d-1)*exp(2*b*x+2*a))*x-1/3*x^3*ln(exp(b*x+a))+1/12*b*x^4+1/12*I*x^3*Pi*csgn(I*exp(2*b*x+2*a)/(exp
(2*b*x+2*a)+1))^3-1/12*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/(d-1)*ln(1-exp(b*x
+a)*(d-1)^(1/2))*x-1/2/b^2*d*a^2/(d-1)*ln(1+exp(b*x+a)*(d-1)^(1/2))*x-1/6*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))^2+1/6/b^3*d*a^3/(d-1)*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+1/6/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*x^3+1/
8/b^3/(d-1)*polylog(4,(d-1)*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)+1)
*(exp(2*b*x+2*a)*d-exp(2*b*x+2*a)-1))^2+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+1/12*I*x^3*Pi*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))*c
sgn(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))-1/12*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))-1/2/b^3*d*a^2/(d-1)*dil
og(1-exp(b*x+a)*(d-1)^(1/2))-1/2/b^3*d*a^2/(d-1)*dilog(1+exp(b*x+a)*(d-1)^(1/2))+1/3/b^3*d/(d-1)*ln(1-(d-1)*ex
p(2*b*x+2*a))*a^3-1/4/b*d/(d-1)*polylog(2,(d-1)*exp(2*b*x+2*a))*x^2+1/4/b^3*d/(d-1)*polylog(2,(d-1)*exp(2*b*x+
2*a))*a^2+1/4/b^2*d/(d-1)*polylog(3,(d-1)*exp(2*b*x+2*a))*x-1/2/b^2/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*x*a^2+1/2
/b^2*a^2/(d-1)*ln(1-exp(b*x+a)*(d-1)^(1/2))*x+1/6*I*x^3*Pi*csgn(I*d*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/6/b
^3*a^3/(d-1)*ln(exp(2*b*x+2*a)*d-exp(2*b*x+2*a)-1)+1/2/b^2*d/(d-1)*ln(1-(d-1)*exp(2*b*x+2*a))*x*a^2+1/2/b^2*a^
2/(d-1)*ln(1+exp(b*x+a)*(d-1)^(1/2))*x-1/2/b^3*d*a^3/(d-1)*ln(1-exp(b*x+a)*(d-1)^(1/2))-1/2/b^3*d*a^3/(d-1)*ln
(1+exp(b*x+a)*(d-1)^(1/2))+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)+1))*csgn(I*exp(2*b*x+2*a)/(exp(2*b*x+2*a)+1))^2-1/6*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))^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-exp(2*b*x+2*a)-1)*x^3

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Maxima [A]
time = 0.71, size = 123, normalized size = 0.88 \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. 359 vs. \(2 (112) = 224\).
time = 0.36, size = 359, normalized size = 2.58 \begin {gather*} \frac {b^{4} x^{4} - 2 \, b^{3} x^{3} \log \left (\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (d - 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (-\sqrt {d - 1} {\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 ) + 2 \, \sqrt {d - 1}\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 ) - 2 \, \sqrt {d - 1}\right ) + 12 \, b x {\rm polylog}\left (3, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 12 \, b x {\rm polylog}\left (3, -\sqrt {d - 1} {\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 (\sqrt {d - 1} {\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 (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 12 \, {\rm polylog}\left (4, -\sqrt {d - 1} {\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*cosh(b*x + a) + d*sinh(b*x + a))/((d - 2)*cosh(b*x + a) + d*sinh(b*x + a))) -
6*b^2*x^2*dilog(sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))) - 6*b^2*x^2*dilog(-sqrt(d - 1)*(cosh(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) + 2*sqrt(d - 1)) + 2*a^3*log(2*(
d - 1)*cosh(b*x + a) + 2*(d - 1)*sinh(b*x + a) - 2*sqrt(d - 1)) + 12*b*x*polylog(3, sqrt(d - 1)*(cosh(b*x + a)
+ sinh(b*x + a))) + 12*b*x*polylog(3, -sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))) - 2*(b^3*x^3 + a^3)*log(s
qrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a)) + 1) - 2*(b^3*x^3 + a^3)*log(-sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x
+ a)) + 1) - 12*polylog(4, sqrt(d - 1)*(cosh(b*x + a) + sinh(b*x + a))) - 12*polylog(4, -sqrt(d - 1)*(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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