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

Optimal. Leaf size=232 \[ \frac{i \text{PolyLog}\left (3,\frac{i b f^{c+d x}}{1-i a}\right )}{2 d^2 \log ^2(f)}-\frac{i \text{PolyLog}\left (3,-\frac{i b f^{c+d x}}{1+i a}\right )}{2 d^2 \log ^2(f)}-\frac{i x \text{PolyLog}\left (2,\frac{i b f^{c+d x}}{1-i a}\right )}{2 d \log (f)}+\frac{i x \text{PolyLog}\left (2,-\frac{i b f^{c+d x}}{1+i a}\right )}{2 d \log (f)}-\frac{1}{4} i x^2 \log \left (1-\frac{i b f^{c+d x}}{1-i a}\right )+\frac{1}{4} i x^2 \log \left (1+\frac{i b f^{c+d x}}{1+i a}\right )+\frac{1}{2} x^2 \tan ^{-1}\left (a+b f^{c+d x}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.150929, antiderivative size = 250, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5143, 2532, 2531, 2282, 6589} \[ -\frac{i \text{PolyLog}\left (3,\frac{b f^{c+d x}}{-a+i}\right )}{2 d^2 \log ^2(f)}+\frac{i \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+i}\right )}{2 d^2 \log ^2(f)}+\frac{i x \text{PolyLog}\left (2,\frac{b f^{c+d x}}{-a+i}\right )}{2 d \log (f)}-\frac{i x \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+i}\right )}{2 d \log (f)}+\frac{1}{4} i x^2 \log \left (-i a-i b f^{c+d x}+1\right )-\frac{1}{4} i x^2 \log \left (i a+i b f^{c+d x}+1\right )+\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{-a+i}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{a+i}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Int[x*ArcTan[a + b*f^(c + d*x)],x]

[Out]

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

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 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 \tan ^{-1}\left (a+b f^{c+d x}\right ) \, dx &=\frac{1}{2} i \int x \log \left (1-i a-i b f^{c+d x}\right ) \, dx-\frac{1}{2} i \int x \log \left (1+i a+i b f^{c+d x}\right ) \, dx\\ &=\frac{1}{4} i x^2 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{4} i x^2 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{1}{2} i \int x \log \left (1-\frac{i b f^{c+d x}}{1-i a}\right ) \, dx-\frac{1}{2} i \int x \log \left (1+\frac{i b f^{c+d x}}{1+i a}\right ) \, dx\\ &=\frac{1}{4} i x^2 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{4} i x^2 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{i x \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}-\frac{i x \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac{i \int \text{Li}_2\left (\frac{i b f^{c+d x}}{1-i a}\right ) \, dx}{2 d \log (f)}-\frac{i \int \text{Li}_2\left (-\frac{i b f^{c+d x}}{1+i a}\right ) \, dx}{2 d \log (f)}\\ &=\frac{1}{4} i x^2 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{4} i x^2 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{i x \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}-\frac{i x \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}-\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{b x}{i-a}\right )}{x} \, dx,x,f^{c+d x}\right )}{2 d^2 \log ^2(f)}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{b x}{i+a}\right )}{x} \, dx,x,f^{c+d x}\right )}{2 d^2 \log ^2(f)}\\ &=\frac{1}{4} i x^2 \log \left (1-i a-i b f^{c+d x}\right )-\frac{1}{4} i x^2 \log \left (1+i a+i b f^{c+d x}\right )+\frac{1}{4} i x^2 \log \left (1-\frac{b f^{c+d x}}{i-a}\right )-\frac{1}{4} i x^2 \log \left (1+\frac{b f^{c+d x}}{i+a}\right )+\frac{i x \text{Li}_2\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d \log (f)}-\frac{i x \text{Li}_2\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d \log (f)}-\frac{i \text{Li}_3\left (\frac{b f^{c+d x}}{i-a}\right )}{2 d^2 \log ^2(f)}+\frac{i \text{Li}_3\left (-\frac{b f^{c+d x}}{i+a}\right )}{2 d^2 \log ^2(f)}\\ \end{align*}

Mathematica [A]  time = 0.0872073, size = 236, normalized size = 1.02 \[ \frac{i \left (-2 \text{PolyLog}\left (3,\frac{b f^{c+d x}}{-a+i}\right )+2 \text{PolyLog}\left (3,-\frac{b f^{c+d x}}{a+i}\right )+2 d x \log (f) \text{PolyLog}\left (2,\frac{b f^{c+d x}}{-a+i}\right )-2 d x \log (f) \text{PolyLog}\left (2,-\frac{b f^{c+d x}}{a+i}\right )+d^2 x^2 \log ^2(f) \log \left (-i a-i b f^{c+d x}+1\right )-d^2 x^2 \log ^2(f) \log \left (i a+i b f^{c+d x}+1\right )-d^2 x^2 \log ^2(f) \log \left (\frac{a+b f^{c+d x}+i}{a+i}\right )+d^2 x^2 \log ^2(f) \log \left (1+\frac{b f^{c+d x}}{a-i}\right )\right )}{4 d^2 \log ^2(f)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.242, size = 672, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [C]  time = 2.47553, size = 778, normalized size = 3.35 \begin{align*} \frac{2 \, d^{2} x^{2} \arctan \left (b f^{d x + c} + a\right ) \log \left (f\right )^{2} - i \, c^{2} \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right )^{2} + i \, c^{2} \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right )^{2} + 2 i \, d x{\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 \, d x{\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 ) +{\left (i \, d^{2} x^{2} - i \, c^{2}\right )} \log \left (f\right )^{2} \log \left (\frac{a^{2} +{\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) +{\left (-i \, d^{2} x^{2} + i \, c^{2}\right )} \log \left (f\right )^{2} \log \left (\frac{a^{2} +{\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) - 2 i \,{\rm polylog}\left (3, -\frac{{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) + 2 i \,{\rm polylog}\left (3, -\frac{{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{4 \, d^{2} \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*atan(a+b*f**(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \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*arctan(a+b*f^(d*x+c)),x, algorithm="giac")

[Out]

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