Integrand size = 12, antiderivative size = 196 \[ \int \arctan \left (a+b f^{c+d x}\right ) \, dx=-\frac {\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{d \log (f)}+\frac {\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{d \log (f)}+\frac {i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )}{2 d \log (f)}-\frac {i \operatorname {PolyLog}\left (2,1-\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{2 d \log (f)} \]
-arctan(a+b*f^(d*x+c))*ln(2/(1-I*(a+b*f^(d*x+c))))/d/ln(f)+arctan(a+b*f^(d *x+c))*ln(2*b*f^(d*x+c)/(I-a)/(1-I*(a+b*f^(d*x+c))))/d/ln(f)+1/2*I*polylog (2,1-2/(1-I*(a+b*f^(d*x+c))))/d/ln(f)-1/2*I*polylog(2,1-2*b*f^(d*x+c)/(I-a )/(1-I*(a+b*f^(d*x+c))))/d/ln(f)
Time = 0.31 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.85 \[ \int \arctan \left (a+b f^{c+d x}\right ) \, dx=x \arctan \left (a+b f^{c+d x}\right )-\frac {b \left (d x \log (f) \left (\log \left (1+\frac {b^2 f^{c+d x}}{a b-\sqrt {-b^2}}\right )-\log \left (1+\frac {b^2 f^{c+d x}}{a b+\sqrt {-b^2}}\right )\right )+\operatorname {PolyLog}\left (2,-\frac {b^2 f^{c+d x}}{a b-\sqrt {-b^2}}\right )-\operatorname {PolyLog}\left (2,-\frac {b^2 f^{c+d x}}{a b+\sqrt {-b^2}}\right )\right )}{2 \sqrt {-b^2} d \log (f)} \]
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])]) + PolyL og[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])
Time = 0.54 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2720, 5570, 25, 27, 5381, 2849, 2752, 2897}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \arctan \left (a+b f^{c+d x}\right ) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int f^{-c-d x} \arctan \left (b f^{c+d x}+a\right )df^{c+d x}}{d \log (f)}\) |
\(\Big \downarrow \) 5570 |
\(\displaystyle \frac {\int f^{-c-d x} \arctan \left (b f^{c+d x}+a\right )d\left (b f^{c+d x}+a\right )}{b d \log (f)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int -f^{-c-d x} \arctan \left (b f^{c+d x}+a\right )d\left (b f^{c+d x}+a\right )}{b d \log (f)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int -\frac {f^{-c-d x} \arctan \left (b f^{c+d x}+a\right )}{b}d\left (b f^{c+d x}+a\right )}{d \log (f)}\) |
\(\Big \downarrow \) 5381 |
\(\displaystyle -\frac {-\int \frac {\log \left (\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )}{f^{2 c+2 d x}+1}d\left (b f^{c+d x}+a\right )+\int \frac {\log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )}{f^{2 c+2 d x}+1}d\left (b f^{c+d x}+a\right )+\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )-\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{d \log (f)}\) |
\(\Big \downarrow \) 2849 |
\(\displaystyle -\frac {-i \int \frac {\log \left (\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )}{1-\frac {2}{1-i \left (b f^{c+d x}+a\right )}}d\frac {1}{1-i \left (b f^{c+d x}+a\right )}+\int \frac {\log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )}{f^{2 c+2 d x}+1}d\left (b f^{c+d x}+a\right )+\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )-\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )}{d \log (f)}\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle -\frac {\int \frac {\log \left (\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )}{f^{2 c+2 d x}+1}d\left (b f^{c+d x}+a\right )+\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )-\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )}{d \log (f)}\) |
\(\Big \downarrow \) 2897 |
\(\displaystyle -\frac {\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2}{1-i \left (a+b f^{c+d x}\right )}\right )-\arctan \left (a+b f^{c+d x}\right ) \log \left (\frac {2 b f^{c+d x}}{(-a+i) \left (1-i \left (a+b f^{c+d x}\right )\right )}\right )-\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2}{1-i \left (b f^{c+d x}+a\right )}\right )+\frac {1}{2} i \operatorname {PolyLog}\left (2,1-\frac {2 b f^{c+d x}}{(i-a) \left (1-i \left (b f^{c+d x}+a\right )\right )}\right )}{d \log (f)}\) |
-((ArcTan[a + b*f^(c + d*x)]*Log[2/(1 - I*(a + b*f^(c + d*x)))] - ArcTan[a + b*f^(c + d*x)]*Log[(2*b*f^(c + d*x))/((I - a)*(1 - I*(a + b*f^(c + d*x) )))] - (I/2)*PolyLog[2, 1 - 2/(1 - I*(a + b*f^(c + d*x)))] + (I/2)*PolyLog [2, 1 - (2*b*f^(c + d*x))/((I - a)*(1 - I*(a + b*f^(c + d*x))))])/(d*Log[f ]))
3.2.16.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp [-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]
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[RationalFunctionExponents[u, x][[2]], Expon[Pq, x]]
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Si mp[(-(a + b*ArcTan[c*x]))*(Log[2/(1 - I*c*x)]/e), x] + (Simp[(a + b*ArcTan[ c*x])*(Log[2*c*((d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/e), x] + Simp[b*(c/e) Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Simp[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]) /; FreeQ[{a , b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]
Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && I GtQ[p, 0]
Time = 0.74 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {\ln \left (-b \,f^{d x +c}\right ) \arctan \left (a +b \,f^{d x +c}\right )-\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}+\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}}{d \ln \left (f \right )}\) | \(162\) |
default | \(\frac {\ln \left (-b \,f^{d x +c}\right ) \arctan \left (a +b \,f^{d x +c}\right )-\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}+\frac {i \ln \left (-b \,f^{d x +c}\right ) \ln \left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}-\frac {i \operatorname {dilog}\left (\frac {i+b \,f^{d x +c}+a}{i+a}\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {i-b \,f^{d x +c}-a}{i-a}\right )}{2}}{d \ln \left (f \right )}\) | \(162\) |
risch | \(-\frac {i x \ln \left (1+i \left (a +b \,f^{d x +c}\right )\right )}{2}+\frac {i \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right )}{2 \ln \left (f \right ) d}+\frac {i \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) x}{2}+\frac {i \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) c}{2 d}-\frac {i c \ln \left (i f^{d x} f^{c} b +i a +1\right )}{2 d}+\frac {i \ln \left (-i b \,f^{d x +c}-i a +1\right ) \ln \left (-\frac {i f^{d x +c} b}{i a -1}\right )}{2 d \ln \left (f \right )}+\frac {i \operatorname {dilog}\left (-\frac {i f^{d x +c} b}{i a -1}\right )}{2 d \ln \left (f \right )}\) | \(214\) |
1/d/ln(f)*(ln(-b*f^(d*x+c))*arctan(a+b*f^(d*x+c))-1/2*I*ln(-b*f^(d*x+c))*l n((I+b*f^(d*x+c)+a)/(I+a))+1/2*I*ln(-b*f^(d*x+c))*ln((I-b*f^(d*x+c)-a)/(I- a))-1/2*I*dilog((I+b*f^(d*x+c)+a)/(I+a))+1/2*I*dilog((I-b*f^(d*x+c)-a)/(I- a)))
Time = 0.28 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.08 \[ \int \arctan \left (a+b f^{c+d x}\right ) \, dx=\frac {2 \, d x \arctan \left (b f^{d x + c} + a\right ) \log \left (f\right ) + i \, c \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right ) - i \, c \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right ) + {\left (i \, d x + i \, c\right )} \log \left (f\right ) \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (-i \, d x - i \, c\right )} \log \left (f\right ) \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + i \, {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) - i \, {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right )}{2 \, d \log \left (f\right )} \]
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)*l og((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))
\[ \int \arctan \left (a+b f^{c+d x}\right ) \, dx=\int \operatorname {atan}{\left (a + b f^{c + d x} \right )}\, dx \]
Time = 0.31 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.96 \[ \int \arctan \left (a+b f^{c+d x}\right ) \, dx=\frac {{\left (d x + c\right )} \arctan \left (b f^{d x + c} + a\right )}{d} - \frac {2 \, {\left (d x + c\right )} \arctan \left (\frac {b^{2} f^{d x + c} + a b}{b}\right ) \log \left (f\right ) + {\left (\pi - \arctan \left (\frac {1}{a}\right )\right )} \log \left (b^{2} f^{2 \, d x + 2 \, c} + 2 \, a b f^{d x + c} + a^{2} + 1\right ) - \arctan \left (b f^{d x + c} + a\right ) \log \left (\frac {b^{2} f^{2 \, d x + 2 \, c}}{a^{2} + 1}\right ) + i \, {\rm Li}_2\left (\frac {i \, b f^{d x + c} + i \, a + 1}{i \, a + 1}\right ) - i \, {\rm Li}_2\left (\frac {i \, b f^{d x + c} + i \, a - 1}{i \, a - 1}\right )}{2 \, d \log \left (f\right )} \]
(d*x + c)*arctan(b*f^(d*x + c) + a)/d - 1/2*(2*(d*x + c)*arctan((b^2*f^(d* x + c) + a*b)/b)*log(f) + (pi - arctan(1/a))*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)) + 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))
\[ \int \arctan \left (a+b f^{c+d x}\right ) \, dx=\int { \arctan \left (b f^{d x + c} + a\right ) \,d x } \]
Timed out. \[ \int \arctan \left (a+b f^{c+d x}\right ) \, dx=\int \mathrm {atan}\left (a+b\,f^{c+d\,x}\right ) \,d x \]