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

Optimal. Leaf size=196 \[ \frac{i \text{PolyLog}\left (2,1-\frac{2}{1-i \left (a+b f^{c+d x}\right )}\right )}{2 d \log (f)}-\frac{i \text{PolyLog}\left (2,1-\frac{2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{2 d \log (f)}-\frac{\log \left (\frac{2}{1-i \left (a+b f^{c+d x}\right )}\right ) \tan ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}+\frac{\log \left (\frac{2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right ) \tan ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.14815, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2282, 5047, 4856, 2402, 2315, 2447} \[ \frac{i \text{PolyLog}\left (2,1-\frac{2}{1-i \left (a+b f^{c+d x}\right )}\right )}{2 d \log (f)}-\frac{i \text{PolyLog}\left (2,1-\frac{2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{2 d \log (f)}-\frac{\log \left (\frac{2}{1-i \left (a+b f^{c+d x}\right )}\right ) \tan ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)}+\frac{\log \left (\frac{2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right ) \tan ^{-1}\left (a+b f^{c+d x}\right )}{d \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*f^(c + d*x)],x]

[Out]

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

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 5047

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x]
&& IGtQ[p, 0]

Rule 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

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

Mathematica [A]  time = 0.180523, size = 167, normalized size = 0.85 \[ x \tan ^{-1}\left (a+b f^{c+d x}\right )-\frac{b \left (\text{PolyLog}\left (2,-\frac{b^2 f^{c+d x}}{a b-\sqrt{-b^2}}\right )-\text{PolyLog}\left (2,-\frac{b^2 f^{c+d x}}{a b+\sqrt{-b^2}}\right )+d x \log (f) \left (\log \left (\frac{b^2 f^{c+d x}}{a b-\sqrt{-b^2}}+1\right )-\log \left (\frac{b^2 f^{c+d x}}{a b+\sqrt{-b^2}}+1\right )\right )\right )}{2 \sqrt{-b^2} d \log (f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 186, normalized size = 1. \begin{align*}{\frac{\ln \left ( b{f}^{dx+c} \right ) \arctan \left ( a+b{f}^{dx+c} \right ) }{d\ln \left ( f \right ) }}+{\frac{{\frac{i}{2}}\ln \left ( b{f}^{dx+c} \right ) }{d\ln \left ( f \right ) }\ln \left ({\frac{-b{f}^{dx+c}-a+i}{i-a}} \right ) }-{\frac{{\frac{i}{2}}\ln \left ( b{f}^{dx+c} \right ) }{d\ln \left ( f \right ) }\ln \left ({\frac{b{f}^{dx+c}+a+i}{i+a}} \right ) }+{\frac{{\frac{i}{2}}}{d\ln \left ( f \right ) }{\it dilog} \left ({\frac{-b{f}^{dx+c}-a+i}{i-a}} \right ) }-{\frac{{\frac{i}{2}}}{d\ln \left ( f \right ) }{\it dilog} \left ({\frac{b{f}^{dx+c}+a+i}{i+a}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/ln(f)*ln(b*f^(d*x+c))*arctan(a+b*f^(d*x+c))+1/2*I/d/ln(f)*ln(b*f^(d*x+c))*ln((-b*f^(d*x+c)-a+I)/(I-a))-1/2
*I/d/ln(f)*ln(b*f^(d*x+c))*ln((b*f^(d*x+c)+a+I)/(I+a))+1/2*I/d/ln(f)*dilog((-b*f^(d*x+c)-a+I)/(I-a))-1/2*I/d/l
n(f)*dilog((b*f^(d*x+c)+a+I)/(I+a))

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Maxima [A]  time = 1.69977, size = 302, normalized size = 1.54 \begin{align*} \frac{\arctan \left (b f^{d x + c} + a\right ) \log \left (f^{d x + c}\right )}{d \log \left (f\right )} - \frac{\arctan \left (\frac{b f^{d x + c}}{a^{2} + 1}, -\frac{a b f^{d x + c}}{a^{2} + 1}\right ) \log \left (b^{2} f^{2 \, d x + 2 \, c} + 2 \, a b f^{d x + c} + a^{2} + 1\right ) - \arctan \left (b f^{d x + c} + a\right ) \log \left (\frac{b^{2} f^{2 \, d x + 2 \, c}}{a^{2} + 1}\right ) + 2 \, \arctan \left (\frac{b^{2} f^{d x + c} + a b}{b}\right ) \log \left (f^{d x + c}\right ) + i \,{\rm Li}_2\left (\frac{i \, b f^{d x + c} + i \, a + 1}{i \, a + 1}\right ) - i \,{\rm Li}_2\left (\frac{i \, b f^{d x + c} + i \, a - 1}{i \, a - 1}\right )}{2 \, d \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(b*f^(d*x + c) + a)*log(f^(d*x + c))/(d*log(f)) - 1/2*(arctan2(b*f^(d*x + c)/(a^2 + 1), -a*b*f^(d*x + c)
/(a^2 + 1))*log(b^2*f^(2*d*x + 2*c) + 2*a*b*f^(d*x + c) + a^2 + 1) - arctan(b*f^(d*x + c) + a)*log(b^2*f^(2*d*
x + 2*c)/(a^2 + 1)) + 2*arctan((b^2*f^(d*x + c) + a*b)/b)*log(f^(d*x + c)) + I*dilog((I*b*f^(d*x + c) + I*a +
1)/(I*a + 1)) - I*dilog((I*b*f^(d*x + c) + I*a - 1)/(I*a - 1)))/(d*log(f))

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Fricas [A]  time = 2.41286, size = 554, normalized size = 2.83 \begin{align*} \frac{2 \, d x \arctan \left (b f^{d x + c} + a\right ) \log \left (f\right ) + i \, c \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right ) - i \, c \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right ) +{\left (i \, d x + i \, c\right )} \log \left (f\right ) \log \left (\frac{a^{2} +{\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) +{\left (-i \, d x - i \, c\right )} \log \left (f\right ) \log \left (\frac{a^{2} +{\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + i \,{\rm Li}_2\left (-\frac{a^{2} +{\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) - i \,{\rm Li}_2\left (-\frac{a^{2} +{\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right )}{2 \, d \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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