3.3.84 \(\int x^2 \tanh ^{-1}(c+d \tanh (a+b x)) \, dx\) [284]

Optimal. Leaf size=307 \[ \frac {1}{3} x^3 \tanh ^{-1}(c+d \tanh (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac {x^2 \text {PolyLog}\left (2,-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac {x^2 \text {PolyLog}\left (2,-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac {x \text {PolyLog}\left (3,-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac {x \text {PolyLog}\left (3,-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac {\text {PolyLog}\left (4,-\frac {(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac {\text {PolyLog}\left (4,-\frac {(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3} \]

[Out]

1/3*x^3*arctanh(c+d*tanh(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 = 0.33, antiderivative size = 307, 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 = {6378, 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 (\frac {(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac {1}{6} x^3 \log \left (\frac {(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+\frac {1}{3} x^3 \tanh ^{-1}(d \tanh (a+b x)+c) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*ArcTanh[c + d*Tanh[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 6378

Int[ArcTanh[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m
+ 1)*(ArcTanh[c + d*Tanh[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 \tanh (a+b x)) \, dx &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tanh (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 \tanh (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 \tanh (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 \tanh (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 \tanh (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 \tanh (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 = 8.59, size = 271, normalized size = 0.88 \begin {gather*} \frac {1}{3} x^3 \tanh ^{-1}(c+d \tanh (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*Tanh[a + b*x]],x]

[Out]

(x^3*ArcTanh[c + d*Tanh[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 = 9.97, size = 5366, normalized size = 17.48

method	result	size
risch	\(\text {Expression too large to display}\)	\(5366\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [A]
time = 0.48, size = 281, normalized size = 0.92 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {artanh}\left (d \tanh \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*tanh(b*x+a)),x, algorithm="maxima")

[Out]

1/3*x^3*arctanh(d*tanh(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. 900 vs. \(2 (263) = 526\).
time = 0.39, size = 900, normalized size = 2.93 \begin {gather*} \frac {b^{3} x^{3} \log \left (-\frac {{\left (c + 1\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (c - 1\right )} \cosh \left (b x + a\right ) + d \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*tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/6*(b^3*x^3*log(-((c + 1)*cosh(b*x + a) + d*sinh(b*x + a))/((c - 1)*cosh(b*x + a) + d*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*l
og(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*po
lylog(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*poly
log(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 \tanh {\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*tanh(b*x+a)),x)

[Out]

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

[Out]

integrate(x^2*arctanh(d*tanh(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 {tanh}\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*tanh(a + b*x)),x)

[Out]

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

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