3.3.98 \(\int x^2 \tanh ^{-1}(c+d \coth (a+b x)) \, dx\) [298]

Optimal. Leaf size=303 \[ \frac {1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {PolyLog}\left (2,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {PolyLog}\left (2,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \text {PolyLog}\left (3,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \text {PolyLog}\left (3,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\text {PolyLog}\left (4,\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac {\text {PolyLog}\left (4,\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3} \]

[Out]

1/3*x^3*arctanh(c+d*coth(b*x+a))+1/6*x^3*ln(1-(1-c-d)*exp(2*b*x+2*a)/(1-c+d))-1/6*x^3*ln(1-(1+c+d)*exp(2*b*x+2
*a)/(1+c-d))+1/4*x^2*polylog(2,(1-c-d)*exp(2*b*x+2*a)/(1-c+d))/b-1/4*x^2*polylog(2,(1+c+d)*exp(2*b*x+2*a)/(1+c
-d))/b-1/4*x*polylog(3,(1-c-d)*exp(2*b*x+2*a)/(1-c+d))/b^2+1/4*x*polylog(3,(1+c+d)*exp(2*b*x+2*a)/(1+c-d))/b^2
+1/8*polylog(4,(1-c-d)*exp(2*b*x+2*a)/(1-c+d))/b^3-1/8*polylog(4,(1+c+d)*exp(2*b*x+2*a)/(1+c-d))/b^3

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Rubi [A]
time = 0.32, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6380, 2221, 2611, 6744, 2320, 6724} \begin {gather*} \frac {\text {Li}_4\left (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{8 b^3}-\frac {\text {Li}_4\left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{8 b^3}-\frac {x \text {Li}_3\left (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b^2}+\frac {x \text {Li}_3\left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b^2}+\frac {x^2 \text {Li}_2\left (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac {1}{6} x^3 \log \left (1-\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )+\frac {1}{3} x^3 \tanh ^{-1}(d \coth (a+b x)+c) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[c + d*Coth[a + b*x]],x]

[Out]

(x^3*ArcTanh[c + d*Coth[a + b*x]])/3 + (x^3*Log[1 - ((1 - c - d)*E^(2*a + 2*b*x))/(1 - c + d)])/6 - (x^3*Log[1
- ((1 + c + d)*E^(2*a + 2*b*x))/(1 + c - d)])/6 + (x^2*PolyLog[2, ((1 - c - d)*E^(2*a + 2*b*x))/(1 - c + d)])
/(4*b) - (x^2*PolyLog[2, ((1 + c + d)*E^(2*a + 2*b*x))/(1 + c - d)])/(4*b) - (x*PolyLog[3, ((1 - c - d)*E^(2*a
+ 2*b*x))/(1 - c + d)])/(4*b^2) + (x*PolyLog[3, ((1 + c + d)*E^(2*a + 2*b*x))/(1 + c - d)])/(4*b^2) + PolyLog
[4, ((1 - c - d)*E^(2*a + 2*b*x))/(1 - c + d)]/(8*b^3) - PolyLog[4, ((1 + c + d)*E^(2*a + 2*b*x))/(1 + c - d)]
/(8*b^3)

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 6380

Int[ArcTanh[(c_.) + Coth[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m
+ 1)*(ArcTanh[c + d*Coth[a + b*x]]/(f*(m + 1))), x] + (-Dist[b*((1 - c - d)/(f*(m + 1))), Int[(e + f*x)^(m +
1)*(E^(2*a + 2*b*x)/(1 - c + d - (1 - c - d)*E^(2*a + 2*b*x))), x], x] + Dist[b*((1 + c + d)/(f*(m + 1))), Int
[(e + f*x)^(m + 1)*(E^(2*a + 2*b*x)/(1 + c - d - (1 + c + d)*E^(2*a + 2*b*x))), x], x]) /; FreeQ[{a, b, c, d,
e, f}, x] && IGtQ[m, 0] && NeQ[(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 \tanh ^{-1}(c+d \coth (a+b x)) \, dx &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))-\frac {1}{3} (b (1-c-d)) \int \frac {e^{2 a+2 b x} x^3}{1-c+d+(-1+c+d) e^{2 a+2 b x}} \, dx+\frac {1}{3} (b (1+c+d)) \int \frac {e^{2 a+2 b x} x^3}{1+c-d+(-1-c-d) e^{2 a+2 b x}} \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {1}{2} \int x^2 \log \left (1+\frac {(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx-\frac {1}{2} \int x^2 \log \left (1+\frac {(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}+\frac {\int x \text {Li}_2\left (-\frac {(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{2 b}-\frac {\int x \text {Li}_2\left (-\frac {(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx}{2 b}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \text {Li}_3\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \text {Li}_3\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}-\frac {\int \text {Li}_3\left (-\frac {(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{4 b^2}+\frac {\int \text {Li}_3\left (-\frac {(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx}{4 b^2}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \text {Li}_3\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \text {Li}_3\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\text {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(-1+c+d) x}{-1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}-\frac {\text {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(1+c+d) x}{1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {Li}_2\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {Li}_2\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \text {Li}_3\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \text {Li}_3\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\text {Li}_4\left (\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac {\text {Li}_4\left (\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3}\\ \end {align*}

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Mathematica [A]
time = 7.64, size = 265, normalized size = 0.87 \begin {gather*} \frac {1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac {4 b^3 x^3 \log \left (1+\frac {(-1+c+d) e^{2 (a+b x)}}{1-c+d}\right )-4 b^3 x^3 \log \left (1+\frac {(1+c+d) e^{2 (a+b x)}}{-1-c+d}\right )+6 b^2 x^2 \text {PolyLog}\left (2,\frac {(-1+c+d) e^{2 (a+b x)}}{-1+c-d}\right )-6 b^2 x^2 \text {PolyLog}\left (2,\frac {(1+c+d) e^{2 (a+b x)}}{1+c-d}\right )-6 b x \text {PolyLog}\left (3,\frac {(-1+c+d) e^{2 (a+b x)}}{-1+c-d}\right )+6 b x \text {PolyLog}\left (3,\frac {(1+c+d) e^{2 (a+b x)}}{1+c-d}\right )+3 \text {PolyLog}\left (4,\frac {(-1+c+d) e^{2 (a+b x)}}{-1+c-d}\right )-3 \text {PolyLog}\left (4,\frac {(1+c+d) e^{2 (a+b x)}}{1+c-d}\right )}{24 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTanh[c + d*Coth[a + b*x]],x]

[Out]

(x^3*ArcTanh[c + d*Coth[a + b*x]])/3 + (4*b^3*x^3*Log[1 + ((-1 + c + d)*E^(2*(a + b*x)))/(1 - c + d)] - 4*b^3*
x^3*Log[1 + ((1 + c + d)*E^(2*(a + b*x)))/(-1 - c + d)] + 6*b^2*x^2*PolyLog[2, ((-1 + c + d)*E^(2*(a + b*x)))/
(-1 + c - d)] - 6*b^2*x^2*PolyLog[2, ((1 + c + d)*E^(2*(a + b*x)))/(1 + c - d)] - 6*b*x*PolyLog[3, ((-1 + c +
d)*E^(2*(a + b*x)))/(-1 + c - d)] + 6*b*x*PolyLog[3, ((1 + c + d)*E^(2*(a + b*x)))/(1 + c - d)] + 3*PolyLog[4,
((-1 + c + d)*E^(2*(a + b*x)))/(-1 + c - d)] - 3*PolyLog[4, ((1 + c + d)*E^(2*(a + b*x)))/(1 + c - d)])/(24*b
^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 10.25, size = 5294, normalized size = 17.47

method	result	size
risch	\(\text {Expression too large to display}\)	\(5294\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(c+d*coth(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [A]
time = 0.49, size = 277, normalized size = 0.91 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + c\right ) - \frac {1}{18} \, b d {\left (\frac {4 \, b^{3} x^{3} \log \left (-\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right ) - 6 \, b x {\rm Li}_{3}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}) + 3 \, {\rm Li}_{4}(\frac {{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1})}{b^{4} d} - \frac {4 \, b^{3} x^{3} \log \left (-\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right ) - 6 \, b x {\rm Li}_{3}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}) + 3 \, {\rm Li}_{4}(\frac {{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1})}{b^{4} d}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(c+d*coth(b*x+a)),x, algorithm="maxima")

[Out]

1/3*x^3*arctanh(d*coth(b*x + a) + c) - 1/18*b*d*((4*b^3*x^3*log(-(c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1) + 1)
+ 6*b^2*x^2*dilog((c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1)) - 6*b*x*polylog(3, (c + d + 1)*e^(2*b*x + 2*a)/(c -
d + 1)) + 3*polylog(4, (c + d + 1)*e^(2*b*x + 2*a)/(c - d + 1)))/(b^4*d) - (4*b^3*x^3*log(-(c + d - 1)*e^(2*b
*x + 2*a)/(c - d - 1) + 1) + 6*b^2*x^2*dilog((c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1)) - 6*b*x*polylog(3, (c +
d - 1)*e^(2*b*x + 2*a)/(c - d - 1)) + 3*polylog(4, (c + d - 1)*e^(2*b*x + 2*a)/(c - d - 1)))/(b^4*d))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 880 vs. \(2 (259) = 518\).
time = 0.51, size = 880, normalized size = 2.90 \begin {gather*} \frac {b^{3} x^{3} \log \left (-\frac {d \cosh \left (b x + a\right ) + {\left (c + 1\right )} \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + {\left (c - 1\right )} \sinh \left (b x + a\right )}\right ) - 3 \, b^{2} x^{2} {\rm Li}_2\left (\sqrt {\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 3 \, b^{2} x^{2} {\rm Li}_2\left (-\sqrt {\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (\sqrt {\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-\sqrt {\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + a^{3} \log \left (2 \, {\left (c + d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (c + d + 1\right )} \sinh \left (b x + a\right ) + 2 \, {\left (c - d + 1\right )} \sqrt {\frac {c + d + 1}{c - d + 1}}\right ) + a^{3} \log \left (2 \, {\left (c + d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (c + d + 1\right )} \sinh \left (b x + a\right ) - 2 \, {\left (c - d + 1\right )} \sqrt {\frac {c + d + 1}{c - d + 1}}\right ) - a^{3} \log \left (2 \, {\left (c + d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (c + d - 1\right )} \sinh \left (b x + a\right ) + 2 \, {\left (c - d - 1\right )} \sqrt {\frac {c + d - 1}{c - d - 1}}\right ) - a^{3} \log \left (2 \, {\left (c + d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (c + d - 1\right )} \sinh \left (b x + a\right ) - 2 \, {\left (c - d - 1\right )} \sqrt {\frac {c + d - 1}{c - d - 1}}\right ) + 6 \, b x {\rm polylog}\left (3, \sqrt {\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, b x {\rm polylog}\left (3, -\sqrt {\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b x {\rm polylog}\left (3, \sqrt {\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b x {\rm polylog}\left (3, -\sqrt {\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\sqrt {\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\sqrt {\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\sqrt {\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\sqrt {\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 6 \, {\rm polylog}\left (4, \sqrt {\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, {\rm polylog}\left (4, -\sqrt {\frac {c + d + 1}{c - d + 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, {\rm polylog}\left (4, \sqrt {\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, {\rm polylog}\left (4, -\sqrt {\frac {c + d - 1}{c - d - 1}} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{6 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(c+d*coth(b*x+a)),x, algorithm="fricas")

[Out]

1/6*(b^3*x^3*log(-(d*cosh(b*x + a) + (c + 1)*sinh(b*x + a))/(d*cosh(b*x + a) + (c - 1)*sinh(b*x + a))) - 3*b^2
*x^2*dilog(sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) - 3*b^2*x^2*dilog(-sqrt((c + d + 1)/
(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) + 3*b^2*x^2*dilog(sqrt((c + d - 1)/(c - d - 1))*(cosh(b*x + a) +
sinh(b*x + a))) + 3*b^2*x^2*dilog(-sqrt((c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))) + a^3*log(2
*(c + d + 1)*cosh(b*x + a) + 2*(c + d + 1)*sinh(b*x + a) + 2*(c - d + 1)*sqrt((c + d + 1)/(c - d + 1))) + a^3*
log(2*(c + d + 1)*cosh(b*x + a) + 2*(c + d + 1)*sinh(b*x + a) - 2*(c - d + 1)*sqrt((c + d + 1)/(c - d + 1))) -
a^3*log(2*(c + d - 1)*cosh(b*x + a) + 2*(c + d - 1)*sinh(b*x + a) + 2*(c - d - 1)*sqrt((c + d - 1)/(c - d - 1
))) - a^3*log(2*(c + d - 1)*cosh(b*x + a) + 2*(c + d - 1)*sinh(b*x + a) - 2*(c - d - 1)*sqrt((c + d - 1)/(c -
d - 1))) + 6*b*x*polylog(3, sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) + 6*b*x*polylog(3,
-sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) - 6*b*x*polylog(3, sqrt((c + d - 1)/(c - d - 1
))*(cosh(b*x + a) + sinh(b*x + a))) - 6*b*x*polylog(3, -sqrt((c + d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*
x + a))) - (b^3*x^3 + a^3)*log(sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) - (b^3*x^3 +
a^3)*log(-sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (b^3*x^3 + a^3)*log(sqrt((c +
d - 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (b^3*x^3 + a^3)*log(-sqrt((c + d - 1)/(c - d - 1))*
(cosh(b*x + a) + sinh(b*x + a)) + 1) - 6*polylog(4, sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x +
a))) - 6*polylog(4, -sqrt((c + d + 1)/(c - d + 1))*(cosh(b*x + a) + sinh(b*x + a))) + 6*polylog(4, sqrt((c + d
- 1)/(c - d - 1))*(cosh(b*x + a) + sinh(b*x + a))) + 6*polylog(4, -sqrt((c + d - 1)/(c - 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 {atanh}{\left (c + d \coth {\left (a + b x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(c+d*coth(b*x+a)),x)

[Out]

Integral(x**2*atanh(c + d*coth(a + b*x)), 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*arctanh(c+d*coth(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^2*arctanh(d*coth(b*x + a) + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(c + d*coth(a + b*x)),x)

[Out]

int(x^2*atanh(c + d*coth(a + b*x)), x)

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