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

Optimal. Leaf size=264 \[ \frac {\text {Li}_4\left (\frac {b f^{c+d x}}{1-a}\right )}{d^3 \log ^3(f)}-\frac {\text {Li}_4\left (-\frac {b f^{c+d x}}{a+1}\right )}{d^3 \log ^3(f)}-\frac {x \text {Li}_3\left (\frac {b f^{c+d x}}{1-a}\right )}{d^2 \log ^2(f)}+\frac {x \text {Li}_3\left (-\frac {b f^{c+d x}}{a+1}\right )}{d^2 \log ^2(f)}+\frac {x^2 \text {Li}_2\left (\frac {b f^{c+d x}}{1-a}\right )}{2 d \log (f)}-\frac {x^2 \text {Li}_2\left (-\frac {b f^{c+d x}}{a+1}\right )}{2 d \log (f)}-\frac {1}{6} x^3 \log \left (-a-b f^{c+d x}+1\right )+\frac {1}{6} x^3 \log \left (a+b f^{c+d x}+1\right )+\frac {1}{6} x^3 \log \left (1-\frac {b f^{c+d x}}{1-a}\right )-\frac {1}{6} x^3 \log \left (\frac {b f^{c+d x}}{a+1}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 264, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.09, size = 235, normalized size = 0.89 \[ \frac {d^3 x^3 \log ^3(f) \log \left (\frac {b f^{c+d x}}{a-1}+1\right )-d^3 x^3 \log ^3(f) \log \left (\frac {b f^{c+d x}}{a+1}+1\right )+2 d^3 x^3 \log ^3(f) \tanh ^{-1}\left (a+b f^{c+d x}\right )+3 d^2 x^2 \log ^2(f) \text {Li}_2\left (-\frac {b f^{c+d x}}{a-1}\right )-3 d^2 x^2 \log ^2(f) \text {Li}_2\left (-\frac {b f^{c+d x}}{a+1}\right )+6 \text {Li}_4\left (-\frac {b f^{c+d x}}{a-1}\right )-6 \text {Li}_4\left (-\frac {b f^{c+d x}}{a+1}\right )-6 d x \log (f) \text {Li}_3\left (-\frac {b f^{c+d x}}{a-1}\right )+6 d x \log (f) \text {Li}_3\left (-\frac {b f^{c+d x}}{a+1}\right )}{6 d^3 \log ^3(f)} \]

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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fricas [C]  time = 0.53, size = 480, normalized size = 1.82 \[ \frac {d^{3} x^{3} \log \relax (f)^{3} \log \left (-\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right ) + a + 1}{b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right ) + a - 1}\right ) - 3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right ) + a + 1}{a + 1} + 1\right ) \log \relax (f)^{2} + 3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right ) + a - 1}{a - 1} + 1\right ) \log \relax (f)^{2} + c^{3} \log \left (b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right ) + a + 1\right ) \log \relax (f)^{3} - c^{3} \log \left (b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right ) + a - 1\right ) \log \relax (f)^{3} - {\left (d^{3} x^{3} + c^{3}\right )} \log \relax (f)^{3} \log \left (\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right ) + a + 1}{a + 1}\right ) + {\left (d^{3} x^{3} + c^{3}\right )} \log \relax (f)^{3} \log \left (\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right ) + a - 1}{a - 1}\right ) + 6 \, d x \log \relax (f) {\rm polylog}\left (3, -\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right )}{a + 1}\right ) - 6 \, d x \log \relax (f) {\rm polylog}\left (3, -\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right )}{a - 1}\right ) - 6 \, {\rm polylog}\left (4, -\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right )}{a + 1}\right ) + 6 \, {\rm polylog}\left (4, -\frac {b \cosh \left ({\left (d x + c\right )} \log \relax (f)\right ) + b \sinh \left ({\left (d x + c\right )} \log \relax (f)\right )}{a - 1}\right )}{6 \, d^{3} \log \relax (f)^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {artanh}\left (b f^{d x + c} + a\right )\,{d x} \]

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)

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maple [B]  time = 0.13, size = 672, normalized size = 2.55 \[ \frac {x^{3} \ln \left (1+a +b \,f^{d x +c}\right )}{6}-\frac {x^{3} \ln \left (1-a -b \,f^{d x +c}\right )}{6}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) x^{3}}{6}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) x \,c^{2}}{2 d^{2}}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{1-a}\right ) c^{3}}{3 d^{3}}+\frac {\polylog \left (2, \frac {b \,f^{d x} f^{c}}{1-a}\right ) x^{2}}{2 \ln \relax (f ) d}-\frac {\polylog \left (2, \frac {b \,f^{d x} f^{c}}{1-a}\right ) c^{2}}{2 \ln \relax (f ) d^{3}}-\frac {\polylog \left (3, \frac {b \,f^{d x} f^{c}}{1-a}\right ) x}{\ln \relax (f )^{2} d^{2}}+\frac {\polylog \left (4, \frac {b \,f^{d x} f^{c}}{1-a}\right )}{\ln \relax (f )^{3} d^{3}}-\frac {c^{3} \ln \left (1-a -f^{c} f^{d x} b \right )}{6 d^{3}}+\frac {c^{2} \dilog \left (\frac {f^{c} f^{d x} b +a -1}{a -1}\right )}{2 \ln \relax (f ) d^{3}}+\frac {c^{2} \ln \left (\frac {f^{c} f^{d x} b +a -1}{a -1}\right ) x}{2 d^{2}}+\frac {c^{3} \ln \left (\frac {f^{c} f^{d x} b +a -1}{a -1}\right )}{2 d^{3}}-\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-1-a}\right ) x^{3}}{6}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-1-a}\right ) x \,c^{2}}{2 d^{2}}+\frac {\ln \left (1-\frac {b \,f^{d x} f^{c}}{-1-a}\right ) c^{3}}{3 d^{3}}-\frac {\polylog \left (2, \frac {b \,f^{d x} f^{c}}{-1-a}\right ) x^{2}}{2 \ln \relax (f ) d}+\frac {\polylog \left (2, \frac {b \,f^{d x} f^{c}}{-1-a}\right ) c^{2}}{2 \ln \relax (f ) d^{3}}+\frac {\polylog \left (3, \frac {b \,f^{d x} f^{c}}{-1-a}\right ) x}{\ln \relax (f )^{2} d^{2}}-\frac {\polylog \left (4, \frac {b \,f^{d x} f^{c}}{-1-a}\right )}{\ln \relax (f )^{3} d^{3}}+\frac {c^{3} \ln \left (1+a +f^{c} f^{d x} b \right )}{6 d^{3}}-\frac {c^{2} \dilog \left (\frac {1+a +f^{c} f^{d x} b}{1+a}\right )}{2 \ln \relax (f ) d^{3}}-\frac {c^{2} \ln \left (\frac {1+a +f^{c} f^{d x} b}{1+a}\right ) x}{2 d^{2}}-\frac {c^{3} \ln \left (\frac {1+a +f^{c} f^{d x} b}{1+a}\right )}{2 d^{3}} \]

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-f^c*f^(d*x)*b)+1/2/ln(f)/d^3*c^2*dilog((f^c*f^(d*x)*b+a-1)/(a-1))
+1/2/d^2*c^2*ln((f^c*f^(d*x)*b+a-1)/(a-1))*x+1/2/d^3*c^3*ln((f^c*f^(d*x)*b+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+f^c*f^(d*x)*b)-1/2/ln(f)/d^
3*c^2*dilog((1+a+f^c*f^(d*x)*b)/(1+a))-1/2/d^2*c^2*ln((1+a+f^c*f^(d*x)*b)/(1+a))*x-1/2/d^3*c^3*ln((1+a+f^c*f^(
d*x)*b)/(1+a))

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maxima [A]  time = 0.37, size = 254, normalized size = 0.96 \[ \frac {1}{3} \, x^{3} \operatorname {artanh}\left (b f^{d x + c} + a\right ) - \frac {1}{6} \, b d {\left (\frac {d^{3} x^{3} \log \left (\frac {b f^{d x} f^{c}}{a + 1} + 1\right ) \log \relax (f)^{3} + 3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {b f^{d x} f^{c}}{a + 1}\right ) \log \relax (f)^{2} - 6 \, d x \log \relax (f) {\rm Li}_{3}(-\frac {b f^{d x} f^{c}}{a + 1}) + 6 \, {\rm Li}_{4}(-\frac {b f^{d x} f^{c}}{a + 1})}{b d^{4} \log \relax (f)^{4}} - \frac {d^{3} x^{3} \log \left (\frac {b f^{d x} f^{c}}{a - 1} + 1\right ) \log \relax (f)^{3} + 3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {b f^{d x} f^{c}}{a - 1}\right ) \log \relax (f)^{2} - 6 \, d x \log \relax (f) {\rm Li}_{3}(-\frac {b f^{d x} f^{c}}{a - 1}) + 6 \, {\rm Li}_{4}(-\frac {b f^{d x} f^{c}}{a - 1})}{b d^{4} \log \relax (f)^{4}}\right )} \log \relax (f) \]

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*((d^3*x^3*log(b*f^(d*x)*f^c/(a + 1) + 1)*log(f)^3 + 3*d^2*x^2*dil
og(-b*f^(d*x)*f^c/(a + 1))*log(f)^2 - 6*d*x*log(f)*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) - (d^3*x^3*log(b*f^(d*x)*f^c/(a - 1) + 1)*log(f)^3 + 3*d^2*x^2*dilog(-b*f^(d*
x)*f^c/(a - 1))*log(f)^2 - 6*d*x*log(f)*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,\mathrm {atanh}\left (a+b\,f^{c+d\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh(a + b*f^(c + d*x)),x)

[Out]

int(x^2*atanh(a + b*f^(c + d*x)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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