3.354 \(\int x^2 \tanh ^{-1}(a+b f^{c+d x}) \, dx\)

Optimal. Leaf size=264 \[ -\frac{x \text{PolyLog}\left (3,\frac{b f^{c+d x}}{1-a}\right )}{d^2 \log ^2(f)}+\frac{x \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+1}\right )}{d^2 \log ^2(f)}+\frac{\text{PolyLog}\left (4,\frac{b f^{c+d x}}{1-a}\right )}{d^3 \log ^3(f)}-\frac{\text{PolyLog}\left (4,-\frac{b f^{c+d x}}{a+1}\right )}{d^3 \log ^3(f)}+\frac{x^2 \text{PolyLog}\left (2,\frac{b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac{x^2 \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+1}\right )}{2 d \log (f)}-\frac{1}{6} x^3 \log \left (-a-b f^{c+d x}+1\right )+\frac{1}{6} x^3 \log \left (a+b f^{c+d x}+1\right )+\frac{1}{6} x^3 \log \left (1-\frac{b f^{c+d x}}{1-a}\right )-\frac{1}{6} x^3 \log \left (\frac{b f^{c+d x}}{a+1}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.196633, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6213, 2532, 2531, 6609, 2282, 6589} \[ -\frac{x \text{PolyLog}\left (3,\frac{b f^{c+d x}}{1-a}\right )}{d^2 \log ^2(f)}+\frac{x \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+1}\right )}{d^2 \log ^2(f)}+\frac{\text{PolyLog}\left (4,\frac{b f^{c+d x}}{1-a}\right )}{d^3 \log ^3(f)}-\frac{\text{PolyLog}\left (4,-\frac{b f^{c+d x}}{a+1}\right )}{d^3 \log ^3(f)}+\frac{x^2 \text{PolyLog}\left (2,\frac{b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac{x^2 \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+1}\right )}{2 d \log (f)}-\frac{1}{6} x^3 \log \left (-a-b f^{c+d x}+1\right )+\frac{1}{6} x^3 \log \left (a+b f^{c+d x}+1\right )+\frac{1}{6} x^3 \log \left (1-\frac{b f^{c+d x}}{1-a}\right )-\frac{1}{6} x^3 \log \left (\frac{b f^{c+d x}}{a+1}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[a + b*f^(c + d*x)],x]

[Out]

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

Rule 6213

Int[ArcTanh[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[1/2, Int[x^m*Log[1 + a + b*
f^(c + d*x)], x], x] - Dist[1/2, Int[x^m*Log[1 - a - b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x] && IG
tQ[m, 0]

Rule 2532

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

Rule 2531

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 6609

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]

Rule 2282

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 6589

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]

Rubi steps

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

Mathematica [A]  time = 0.0879626, size = 235, normalized size = 0.89 \[ \frac{3 d^2 x^2 \log ^2(f) \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a-1}\right )-3 d^2 x^2 \log ^2(f) \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+1}\right )+6 \text{PolyLog}\left (4,-\frac{b f^{c+d x}}{a-1}\right )-6 \text{PolyLog}\left (4,-\frac{b f^{c+d x}}{a+1}\right )-6 d x \log (f) \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a-1}\right )+6 d x \log (f) \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+1}\right )+d^3 x^3 \log ^3(f) \log \left (\frac{b f^{c+d x}}{a-1}+1\right )-d^3 x^3 \log ^3(f) \log \left (\frac{b f^{c+d x}}{a+1}+1\right )+2 d^3 x^3 \log ^3(f) \tanh ^{-1}\left (a+b f^{c+d x}\right )}{6 d^3 \log ^3(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTanh[a + b*f^(c + d*x)],x]

[Out]

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

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Maple [B]  time = 0.126, size = 672, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(a+b*f^(d*x+c)),x)

[Out]

1/6*x^3*ln(1+a+b*f^(d*x+c))-1/6*x^3*ln(1-a-b*f^(d*x+c))+1/6*ln(1-b*f^(d*x)*f^c/(1-a))*x^3-1/2/d^2*ln(1-b*f^(d*
x)*f^c/(1-a))*x*c^2-1/3/d^3*ln(1-b*f^(d*x)*f^c/(1-a))*c^3+1/2/ln(f)/d*polylog(2,b*f^(d*x)*f^c/(1-a))*x^2-1/2/l
n(f)/d^3*polylog(2,b*f^(d*x)*f^c/(1-a))*c^2-1/ln(f)^2/d^2*polylog(3,b*f^(d*x)*f^c/(1-a))*x+1/ln(f)^3/d^3*polyl
og(4,b*f^(d*x)*f^c/(1-a))-1/6/d^3*c^3*ln(1-a-b*f^(d*x)*f^c)+1/2/ln(f)/d^3*c^2*dilog((b*f^(d*x)*f^c+a-1)/(a-1))
+1/2/d^2*c^2*ln((b*f^(d*x)*f^c+a-1)/(a-1))*x+1/2/d^3*c^3*ln((b*f^(d*x)*f^c+a-1)/(a-1))-1/6*ln(1-b*f^(d*x)*f^c/
(-1-a))*x^3+1/2/d^2*ln(1-b*f^(d*x)*f^c/(-1-a))*x*c^2+1/3/d^3*ln(1-b*f^(d*x)*f^c/(-1-a))*c^3-1/2/ln(f)/d*polylo
g(2,b*f^(d*x)*f^c/(-1-a))*x^2+1/2/ln(f)/d^3*polylog(2,b*f^(d*x)*f^c/(-1-a))*c^2+1/ln(f)^2/d^2*polylog(3,b*f^(d
*x)*f^c/(-1-a))*x-1/ln(f)^3/d^3*polylog(4,b*f^(d*x)*f^c/(-1-a))+1/6/d^3*c^3*ln(1+a+b*f^(d*x)*f^c)-1/2/ln(f)/d^
3*c^2*dilog((1+a+b*f^(d*x)*f^c)/(1+a))-1/2/d^2*c^2*ln((1+a+b*f^(d*x)*f^c)/(1+a))*x-1/2/d^3*c^3*ln((1+a+b*f^(d*
x)*f^c)/(1+a))

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Maxima [A]  time = 1.06354, size = 338, normalized size = 1.28 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{artanh}\left (b f^{d x + c} + a\right ) - \frac{1}{6} \, b d{\left (\frac{\log \left (\frac{b f^{d x} f^{c}}{a + 1} + 1\right ) \log \left (f^{d x}\right )^{3} + 3 \,{\rm Li}_2\left (-\frac{b f^{d x} f^{c}}{a + 1}\right ) \log \left (f^{d x}\right )^{2} - 6 \, \log \left (f^{d x}\right ){\rm Li}_{3}(-\frac{b f^{d x} f^{c}}{a + 1}) + 6 \,{\rm Li}_{4}(-\frac{b f^{d x} f^{c}}{a + 1})}{b d^{4} \log \left (f\right )^{4}} - \frac{\log \left (\frac{b f^{d x} f^{c}}{a - 1} + 1\right ) \log \left (f^{d x}\right )^{3} + 3 \,{\rm Li}_2\left (-\frac{b f^{d x} f^{c}}{a - 1}\right ) \log \left (f^{d x}\right )^{2} - 6 \, \log \left (f^{d x}\right ){\rm Li}_{3}(-\frac{b f^{d x} f^{c}}{a - 1}) + 6 \,{\rm Li}_{4}(-\frac{b f^{d x} f^{c}}{a - 1})}{b d^{4} \log \left (f\right )^{4}}\right )} \log \left (f\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.60706, size = 1454, normalized size = 5.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(d^3*x^3*log(f)^3*log(-(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a + 1)/(b*cosh((d*x + c)*log
(f)) + b*sinh((d*x + c)*log(f)) + a - 1)) - 3*d^2*x^2*dilog(-(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(
f)) + a + 1)/(a + 1) + 1)*log(f)^2 + 3*d^2*x^2*dilog(-(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a
- 1)/(a - 1) + 1)*log(f)^2 + c^3*log(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a + 1)*log(f)^3 -
c^3*log(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a - 1)*log(f)^3 - (d^3*x^3 + c^3)*log(f)^3*log((
b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a + 1)/(a + 1)) + (d^3*x^3 + c^3)*log(f)^3*log((b*cosh((
d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)) + a - 1)/(a - 1)) + 6*d*x*log(f)*polylog(3, -(b*cosh((d*x + c)*log
(f)) + b*sinh((d*x + c)*log(f)))/(a + 1)) - 6*d*x*log(f)*polylog(3, -(b*cosh((d*x + c)*log(f)) + b*sinh((d*x +
c)*log(f)))/(a - 1)) - 6*polylog(4, -(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)))/(a + 1)) + 6*polyl
og(4, -(b*cosh((d*x + c)*log(f)) + b*sinh((d*x + c)*log(f)))/(a - 1)))/(d^3*log(f)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{artanh}\left (b f^{d x + c} + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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