3.3.25 \(\int x \cot ^{-1}(a+b f^{c+d x}) \, dx\) [225]

3.3.25.1 Optimal result

Integrand size = 14, antiderivative size = 250 \[ \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i \operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{i-a}\right )}{2 d^2 \log ^2(f)}-\frac {i \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{i+a}\right )}{2 d^2 \log ^2(f)} \]

output

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

3.3.25.2 Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00 \[ \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {1}{4} i x^2 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{4} i x^2 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{4} i x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{4} i x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i \operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{i-a}\right )}{2 d^2 \log ^2(f)}-\frac {i \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{i+a}\right )}{2 d^2 \log ^2(f)} \]

input

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

output

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

3.3.25.3 Rubi [A] (verified)

Time = 4.42 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.94, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5667, 3031, 26, 27, 7292, 7293, 2009}

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.

\(\displaystyle \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx\)

\(\Big \downarrow \) 5667

\(\displaystyle \frac {1}{2} i \int x \log \left (1-\frac {i}{b f^{c+d x}+a}\right )dx-\frac {1}{2} i \int x \log \left (1+\frac {i}{b f^{c+d x}+a}\right )dx\)

\(\Big \downarrow \) 3031

\(\displaystyle \frac {1}{2} i \left (\frac {1}{2} x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{2} \int -\frac {i b d f^{c+d x} x^2 \log (f)}{\left (-b f^{c+d x}-a+i\right ) \left (b f^{c+d x}+a\right )}dx\right )-\frac {1}{2} i \left (\frac {1}{2} x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{2} \int -\frac {i b d f^{c+d x} x^2 \log (f)}{\left (b f^{c+d x}+a\right ) \left (b f^{c+d x}+a+i\right )}dx\right )\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i \int \frac {b d f^{c+d x} x^2 \log (f)}{\left (-b f^{c+d x}-a+i\right ) \left (b f^{c+d x}+a\right )}dx+\frac {1}{2} x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )\right )-\frac {1}{2} i \left (\frac {1}{2} i \int \frac {b d f^{c+d x} x^2 \log (f)}{\left (b f^{c+d x}+a\right ) \left (b f^{c+d x}+a+i\right )}dx+\frac {1}{2} x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b d \log (f) \int \frac {f^{c+d x} x^2}{\left (-b f^{c+d x}-a+i\right ) \left (b f^{c+d x}+a\right )}dx+\frac {1}{2} x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )\right )-\frac {1}{2} i \left (\frac {1}{2} i b d \log (f) \int \frac {f^{c+d x} x^2}{\left (b f^{c+d x}+a\right ) \left (b f^{c+d x}+a+i\right )}dx+\frac {1}{2} x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )\right )\)

\(\Big \downarrow \) 7292

\(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b d \log (f) \int \frac {f^{c+d x} x^2}{\left (i (i a+1)-b f^{c+d x}\right ) \left (b f^{c+d x}+a\right )}dx+\frac {1}{2} x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )\right )-\frac {1}{2} i \left (\frac {1}{2} i b d \log (f) \int \frac {f^{c+d x} x^2}{\left (b f^{c+d x}+i (1-i a)\right ) \left (b f^{c+d x}+a\right )}dx+\frac {1}{2} x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )\right )\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b d \log (f) \int \left (\frac {i f^{c+d x} x^2}{b f^{c+d x}+a-i}-\frac {i f^{c+d x} x^2}{b f^{c+d x}+a}\right )dx+\frac {1}{2} x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )\right )-\frac {1}{2} i \left (\frac {1}{2} i b d \log (f) \int \left (\frac {i f^{c+d x} x^2}{b f^{c+d x}+a+i}-\frac {i f^{c+d x} x^2}{b f^{c+d x}+a}\right )dx+\frac {1}{2} x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} i \left (\frac {1}{2} i b d \log (f) \left (-\frac {2 i \operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{i-a}\right )}{b d^3 \log ^3(f)}+\frac {2 i \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{a}\right )}{b d^3 \log ^3(f)}+\frac {2 i x \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{i-a}\right )}{b d^2 \log ^2(f)}-\frac {2 i x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a}\right )}{b d^2 \log ^2(f)}+\frac {i x^2 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )}{b d \log (f)}-\frac {i x^2 \log \left (\frac {b f^{c+d x}}{a}+1\right )}{b d \log (f)}\right )+\frac {1}{2} x^2 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )\right )-\frac {1}{2} i \left (\frac {1}{2} i b d \log (f) \left (\frac {2 i \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{a}\right )}{b d^3 \log ^3(f)}-\frac {2 i \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{a+i}\right )}{b d^3 \log ^3(f)}-\frac {2 i x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a}\right )}{b d^2 \log ^2(f)}+\frac {2 i x \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+i}\right )}{b d^2 \log ^2(f)}-\frac {i x^2 \log \left (\frac {b f^{c+d x}}{a}+1\right )}{b d \log (f)}+\frac {i x^2 \log \left (1+\frac {b f^{c+d x}}{a+i}\right )}{b d \log (f)}\right )+\frac {1}{2} x^2 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )\right )\)

input

Int[x*ArcCot[a + b*f^(c + d*x)],x]
 

output

(I/2)*((x^2*Log[1 - I/(a + b*f^(c + d*x))])/2 + (I/2)*b*d*Log[f]*((I*x^2*L 
og[1 - (b*f^(c + d*x))/(I - a)])/(b*d*Log[f]) - (I*x^2*Log[1 + (b*f^(c + d 
*x))/a])/(b*d*Log[f]) + ((2*I)*x*PolyLog[2, (b*f^(c + d*x))/(I - a)])/(b*d 
^2*Log[f]^2) - ((2*I)*x*PolyLog[2, -((b*f^(c + d*x))/a)])/(b*d^2*Log[f]^2) 
 - ((2*I)*PolyLog[3, (b*f^(c + d*x))/(I - a)])/(b*d^3*Log[f]^3) + ((2*I)*P 
olyLog[3, -((b*f^(c + d*x))/a)])/(b*d^3*Log[f]^3))) - (I/2)*((x^2*Log[1 + 
I/(a + b*f^(c + d*x))])/2 + (I/2)*b*d*Log[f]*(((-I)*x^2*Log[1 + (b*f^(c + 
d*x))/a])/(b*d*Log[f]) + (I*x^2*Log[1 + (b*f^(c + d*x))/(I + a)])/(b*d*Log 
[f]) - ((2*I)*x*PolyLog[2, -((b*f^(c + d*x))/a)])/(b*d^2*Log[f]^2) + ((2*I 
)*x*PolyLog[2, -((b*f^(c + d*x))/(I + a))])/(b*d^2*Log[f]^2) + ((2*I)*Poly 
Log[3, -((b*f^(c + d*x))/a)])/(b*d^3*Log[f]^3) - ((2*I)*PolyLog[3, -((b*f^ 
(c + d*x))/(I + a))])/(b*d^3*Log[f]^3)))
 

3.3.25.3.1 Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3031

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[(a + b*x)^(m + 1) 
*(Log[u]/(b*(m + 1))), x] - Simp[1/(b*(m + 1))   Int[SimplifyIntegrand[(a + 
 b*x)^(m + 1)*(D[u, x]/u), x], x], x] /; FreeQ[{a, b, m}, x] && InverseFunc 
tionFreeQ[u, x] && NeQ[m, -1]
 

rule 5667

Int[ArcCot[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] : 
> Simp[I/2   Int[x^m*Log[1 - I/(a + b*f^(c + d*x))], x], x] - Simp[I/2   In 
t[x^m*Log[1 + I/(a + b*f^(c + d*x))], x], x] /; FreeQ[{a, b, c, d, f}, x] & 
& IntegerQ[m] && m > 0
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.3.25.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 677 vs. \(2 (218 ) = 436\).

Time = 1.04 (sec) , antiderivative size = 678, normalized size of antiderivative = 2.71

method	result	size
risch	\(-\frac {i x^{2} \ln \left (1-i \left (a +b \,f^{d x +c}\right )\right )}{4}+\frac {\pi \,x^{2}}{4}-\frac {i c \ln \left (\frac {b \,f^{d x} f^{c}+a +i}{i+a}\right ) x}{2 d}-\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c}{2 \ln \left (f \right ) d^{2}}+\frac {i c \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) x}{2 d}+\frac {i c^{2} \ln \left (1-i a -i f^{d x} f^{c} b \right )}{4 d^{2}}-\frac {i c \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a +i}{i+a}\right )}{2 \ln \left (f \right ) d^{2}}+\frac {i x^{2} \ln \left (1+i \left (a +b \,f^{d x +c}\right )\right )}{4}+\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c}{2 \ln \left (f \right ) d^{2}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{2}}{4}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{2}}{4 d^{2}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c x}{2 d}+\frac {i \operatorname {polylog}\left (3, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right )}{2 \ln \left (f \right )^{2} d^{2}}+\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x}{2 \ln \left (f \right ) d}-\frac {i c^{2} \ln \left (i f^{d x} f^{c} b +i a +1\right )}{4 d^{2}}-\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a +i}{i+a}\right )}{2 d^{2}}+\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right )}{2 d^{2}}-\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x}{2 \ln \left (f \right ) d}-\frac {i \operatorname {polylog}\left (3, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right )}{2 \ln \left (f \right )^{2} d^{2}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c x}{2 d}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c^{2}}{4 d^{2}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{2}}{4}+\frac {i c \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right )}{2 \ln \left (f \right ) d^{2}}\)	\(678\)

input

int(x*arccot(a+b*f^(d*x+c)),x,method=_RETURNVERBOSE)
 

output

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

3.3.25.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.22 \[ \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\frac {2 \, d^{2} x^{2} \operatorname {arccot}\left (b f^{d x + c} + a\right ) \log \left (f\right )^{2} + i \, c^{2} \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right )^{2} - i \, c^{2} \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right )^{2} - 2 i \, d x {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right ) + 2 i \, d x {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right ) + {\left (-i \, d^{2} x^{2} + i \, c^{2}\right )} \log \left (f\right )^{2} \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (i \, d^{2} x^{2} - i \, c^{2}\right )} \log \left (f\right )^{2} \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + 2 i \, {\rm polylog}\left (3, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 2 i \, {\rm polylog}\left (3, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{4 \, d^{2} \log \left (f\right )^{2}} \]

input

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

output

1/4*(2*d^2*x^2*arccot(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*polylo 
g(3, -(a*b - I*b)*f^(d*x + c)/(a^2 + 1)))/(d^2*log(f)^2)
 

3.3.25.6 Sympy [F]

\[ \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int x \operatorname {acot}{\left (a + b f^{c + d x} \right )}\, dx \]

input

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

output

Integral(x*acot(a + b*f**(c + d*x)), x)
 

3.3.25.7 Maxima [F]

\[ \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int { x \operatorname {arccot}\left (b f^{d x + c} + a\right ) \,d x } \]

input

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

output

b*d*f^c*integrate(1/2*f^(d*x)*x^2/(b^2*f^(2*d*x)*f^(2*c) + 2*a*b*f^(d*x)*f 
^c + a^2 + 1), x)*log(f) + 1/2*x^2*arctan(1/(b*f^(d*x)*f^c + a))
 

3.3.25.8 Giac [F]

\[ \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int { x \operatorname {arccot}\left (b f^{d x + c} + a\right ) \,d x } \]

input

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

output

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

3.3.25.9 Mupad [F(-1)]

Timed out. \[ \int x \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int x\,\mathrm {acot}\left (a+b\,f^{c+d\,x}\right ) \,d x \]

input

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

output

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