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

Optimal. Leaf size=302 \[ \frac{i x \text{PolyLog}\left (3,\frac{i b f^{c+d x}}{1-i a}\right )}{d^2 \log ^2(f)}-\frac{i x \text{PolyLog}\left (3,-\frac{i b f^{c+d x}}{1+i a}\right )}{d^2 \log ^2(f)}-\frac{i \text{PolyLog}\left (4,\frac{i b f^{c+d x}}{1-i a}\right )}{d^3 \log ^3(f)}+\frac{i \text{PolyLog}\left (4,-\frac{i b f^{c+d x}}{1+i a}\right )}{d^3 \log ^3(f)}-\frac{i x^2 \text{PolyLog}\left (2,\frac{i b f^{c+d x}}{1-i a}\right )}{2 d \log (f)}+\frac{i x^2 \text{PolyLog}\left (2,-\frac{i b f^{c+d x}}{1+i a}\right )}{2 d \log (f)}-\frac{1}{6} i x^3 \log \left (1-\frac{i b f^{c+d x}}{1-i a}\right )+\frac{1}{6} i x^3 \log \left (1+\frac{i b f^{c+d x}}{1+i a}\right )+\frac{1}{3} x^3 \tan ^{-1}\left (a+b f^{c+d x}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.201092, antiderivative size = 313, normalized size of antiderivative = 1.04, 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 = {5143, 2532, 2531, 6609, 2282, 6589} \[ -\frac{i x \text{PolyLog}\left (3,\frac{b f^{c+d x}}{-a+i}\right )}{d^2 \log ^2(f)}+\frac{i x \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+i}\right )}{d^2 \log ^2(f)}+\frac{i \text{PolyLog}\left (4,\frac{b f^{c+d x}}{-a+i}\right )}{d^3 \log ^3(f)}-\frac{i \text{PolyLog}\left (4,-\frac{b f^{c+d x}}{a+i}\right )}{d^3 \log ^3(f)}+\frac{i x^2 \text{PolyLog}\left (2,\frac{b f^{c+d x}}{-a+i}\right )}{2 d \log (f)}-\frac{i x^2 \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+i}\right )}{2 d \log (f)}+\frac{1}{6} i x^3 \log \left (-i a-i b f^{c+d x}+1\right )-\frac{1}{6} i x^3 \log \left (i a+i b f^{c+d x}+1\right )+\frac{1}{6} i x^3 \log \left (1-\frac{b f^{c+d x}}{-a+i}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{b f^{c+d x}}{a+i}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[x^2*ArcTan[a + b*f^(c + d*x)],x]

[Out]

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

Rule 5143

Int[ArcTan[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] :> Dist[I/2, Int[x^m*Log[1 - I*a - I
*b*f^(c + d*x)], x], x] - Dist[I/2, Int[x^m*Log[1 + I*a + I*b*f^(c + d*x)], x], x] /; FreeQ[{a, b, c, d, f}, x
] && IntegerQ[m] && 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 \tan ^{-1}\left (a+b f^{c+d x}\right ) \, dx &=\frac{1}{2} i \int x^2 \log \left (1-i a-i b f^{c+d x}\right ) \, dx-\frac{1}{2} i \int x^2 \log \left (1+i a+i b f^{c+d x}\right ) \, dx\\ &=\frac{1}{6} i x^3 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{6} i x^3 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{6} i x^3 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{1}{2} i \int x^2 \log \left (1-\frac{i b f^{c+d x}}{1-i a}\right ) \, dx-\frac{1}{2} i \int x^2 \log \left (1+\frac{i b f^{c+d x}}{1+i a}\right ) \, dx\\ &=\frac{1}{6} i x^3 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{6} i x^3 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{6} i x^3 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{i x^2 \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}-\frac{i x^2 \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac{i \int x \text{Li}_2\left (\frac{i b f^{c+d x}}{1-i a}\right ) \, dx}{d \log (f)}-\frac{i \int x \text{Li}_2\left (-\frac{i b f^{c+d x}}{1+i a}\right ) \, dx}{d \log (f)}\\ &=\frac{1}{6} i x^3 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{6} i x^3 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{6} i x^3 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{i x^2 \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}-\frac{i x^2 \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}-\frac{i x \text{Li}_3\left (\frac{b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}+\frac{i x \text{Li}_3\left (-\frac{b f^{c+d x}}{i+a}\right )}{d^2 \log ^2(f)}-\frac{i \int \text{Li}_3\left (\frac{i b f^{c+d x}}{1-i a}\right ) \, dx}{d^2 \log ^2(f)}+\frac{i \int \text{Li}_3\left (-\frac{i b f^{c+d x}}{1+i a}\right ) \, dx}{d^2 \log ^2(f)}\\ &=\frac{1}{6} i x^3 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{6} i x^3 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{6} i x^3 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{i x^2 \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}-\frac{i x^2 \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}-\frac{i x \text{Li}_3\left (\frac{b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}+\frac{i x \text{Li}_3\left (-\frac{b f^{c+d x}}{i+a}\right )}{d^2 \log ^2(f)}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{b x}{i-a}\right )}{x} \, dx,x,f^{c+d x}\right )}{d^3 \log ^3(f)}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{b x}{i+a}\right )}{x} \, dx,x,f^{c+d x}\right )}{d^3 \log ^3(f)}\\ &=\frac{1}{6} i x^3 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{6} i x^3 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{6} i x^3 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{6} i x^3 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{i x^2 \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}-\frac{i x^2 \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}-\frac{i x \text{Li}_3\left (\frac{b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}+\frac{i x \text{Li}_3\left (-\frac{b f^{c+d x}}{i+a}\right )}{d^2 \log ^2(f)}+\frac{i \text{Li}_4\left (\frac{b f^{c+d x}}{i-a}\right )}{d^3 \log ^3(f)}-\frac{i \text{Li}_4\left (-\frac{b f^{c+d x}}{i+a}\right )}{d^3 \log ^3(f)}\\ \end{align*}

Mathematica [A]  time = 0.0137312, size = 334, normalized size = 1.11 \[ \frac{i x \text{PolyLog}\left (3,\frac{i b f^{c+d x}}{1-i a}\right )}{d^2 \log ^2(f)}-\frac{i x \text{PolyLog}\left (3,-\frac{i b f^{c+d x}}{1+i a}\right )}{d^2 \log ^2(f)}+\frac{i \text{PolyLog}\left (4,\frac{b f^{c+d x}}{-a+i}\right )}{d^3 \log ^3(f)}-\frac{i \text{PolyLog}\left (4,-\frac{b f^{c+d x}}{a+i}\right )}{d^3 \log ^3(f)}-\frac{i x^2 \text{PolyLog}\left (2,\frac{i b f^{c+d x}}{1-i a}\right )}{2 d \log (f)}+\frac{i x^2 \text{PolyLog}\left (2,-\frac{i b f^{c+d x}}{1+i a}\right )}{2 d \log (f)}+\frac{1}{6} i x^3 \log \left (-i a-i b f^{c+d x}+1\right )-\frac{1}{6} i x^3 \log \left (i a+i b f^{c+d x}+1\right )-\frac{1}{6} i x^3 \log \left (1-\frac{i b f^{c+d x}}{1-i a}\right )+\frac{1}{6} i x^3 \log \left (1+\frac{i b f^{c+d x}}{1+i a}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.217, size = 758, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/d^2*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*x*c^2-1/6*I/d^3*c^3*ln(I*f^(d*x)*f^c*b+I*a+1)-1/2*I/d/ln(f)*polylog(
2,I*b/(1-I*a)*f^(d*x)*f^c)*x^2+1/2*I/d/ln(f)*polylog(2,I*b/(-I*a-1)*f^(d*x)*f^c)*x^2-I/d^3/ln(f)^3*polylog(4,I
*b/(1-I*a)*f^(d*x)*f^c)-1/2*I/d^3/ln(f)*polylog(2,I*b/(-I*a-1)*f^(d*x)*f^c)*c^2+1/2*I/d^3/ln(f)*polylog(2,I*b/
(1-I*a)*f^(d*x)*f^c)*c^2-1/6*I*x^3*ln(1+I*(a+b*f^(d*x+c)))+I/d^2/ln(f)^2*polylog(3,I*b/(1-I*a)*f^(d*x)*f^c)*x-
1/2*I/d^2*c^2*ln((b*f^(d*x)*f^c+I+a)/(I+a))*x+1/2*I/d^3/ln(f)*c^2*dilog((b*f^(d*x)*f^c+a-I)/(-I+a))+1/2*I/d^2*
ln(1-I*b/(1-I*a)*f^(d*x)*f^c)*x*c^2+1/6*I*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*x^3+1/3*I/d^3*ln(1-I*b/(1-I*a)*f^(d*x
)*f^c)*c^3+1/2*I/d^3*c^3*ln((b*f^(d*x)*f^c+a-I)/(-I+a))-1/2*I/d^3*c^3*ln((b*f^(d*x)*f^c+I+a)/(I+a))-1/2*I/d^3/
ln(f)*c^2*dilog((b*f^(d*x)*f^c+I+a)/(I+a))+1/6*I*x^3*ln(1-I*(a+b*f^(d*x+c)))-1/6*I*ln(1-I*b/(1-I*a)*f^(d*x)*f^
c)*x^3+1/6*I/d^3*c^3*ln(1-I*a-I*f^(d*x)*f^c*b)+1/2*I/d^2*c^2*ln((b*f^(d*x)*f^c+a-I)/(-I+a))*x+I/d^3/ln(f)^3*po
lylog(4,I*b/(-I*a-1)*f^(d*x)*f^c)-1/3*I/d^3*ln(1-I*b/(-I*a-1)*f^(d*x)*f^c)*c^3-I/d^2/ln(f)^2*polylog(3,I*b/(-I
*a-1)*f^(d*x)*f^c)*x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.47284, size = 967, normalized size = 3.2 \begin{align*} \frac{2 \, d^{3} x^{3} \arctan \left (b f^{d x + c} + a\right ) \log \left (f\right )^{3} + 3 i \, d^{2} x^{2}{\rm Li}_2\left (-\frac{a^{2} +{\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right )^{2} - 3 i \, d^{2} x^{2}{\rm Li}_2\left (-\frac{a^{2} +{\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right )^{2} + i \, c^{3} \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right )^{3} - i \, c^{3} \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right )^{3} +{\left (i \, d^{3} x^{3} + i \, c^{3}\right )} \log \left (f\right )^{3} \log \left (\frac{a^{2} +{\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) +{\left (-i \, d^{3} x^{3} - i \, c^{3}\right )} \log \left (f\right )^{3} \log \left (\frac{a^{2} +{\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) - 6 i \, d x \log \left (f\right ){\rm polylog}\left (3, -\frac{{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) + 6 i \, d x \log \left (f\right ){\rm polylog}\left (3, -\frac{{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) + 6 i \,{\rm polylog}\left (4, -\frac{{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 6 i \,{\rm polylog}\left (4, -\frac{{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{6 \, d^{3} \log \left (f\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*d^3*x^3*arctan(b*f^(d*x + c) + a)*log(f)^3 + 3*I*d^2*x^2*dilog(-(a^2 + (a*b + I*b)*f^(d*x + c) + 1)/(a^
2 + 1) + 1)*log(f)^2 - 3*I*d^2*x^2*dilog(-(a^2 + (a*b - I*b)*f^(d*x + c) + 1)/(a^2 + 1) + 1)*log(f)^2 + I*c^3*
log(b*f^(d*x + c) + a + I)*log(f)^3 - I*c^3*log(b*f^(d*x + c) + a - I)*log(f)^3 + (I*d^3*x^3 + I*c^3)*log(f)^3
*log((a^2 + (a*b + I*b)*f^(d*x + c) + 1)/(a^2 + 1)) + (-I*d^3*x^3 - I*c^3)*log(f)^3*log((a^2 + (a*b - I*b)*f^(
d*x + c) + 1)/(a^2 + 1)) - 6*I*d*x*log(f)*polylog(3, -(a*b + I*b)*f^(d*x + c)/(a^2 + 1)) + 6*I*d*x*log(f)*poly
log(3, -(a*b - I*b)*f^(d*x + c)/(a^2 + 1)) + 6*I*polylog(4, -(a*b + I*b)*f^(d*x + c)/(a^2 + 1)) - 6*I*polylog(
4, -(a*b - I*b)*f^(d*x + c)/(a^2 + 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*atan(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} \arctan \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*arctan(a+b*f^(d*x+c)),x, algorithm="giac")

[Out]

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