3.3.26 \(\int x^2 \cot ^{-1}(a+b f^{c+d x}) \, dx\) [226]

3.3.26.1 Optimal result

Integrand size = 16, antiderivative size = 313 \[ \int x^2 \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i x \operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{i+a}\right )}{d^2 \log ^2(f)}-\frac {i \operatorname {PolyLog}\left (4,\frac {b f^{c+d x}}{i-a}\right )}{d^3 \log ^3(f)}+\frac {i \operatorname {PolyLog}\left (4,-\frac {b f^{c+d x}}{i+a}\right )}{d^3 \log ^3(f)} \]

output

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

3.3.26.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.00 \[ \int x^2 \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=-\frac {1}{6} i x^3 \log \left (1-\frac {b f^{c+d x}}{i-a}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {b f^{c+d x}}{i+a}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )-\frac {1}{6} i x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{i-a}\right )}{2 d \log (f)}+\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{i+a}\right )}{2 d \log (f)}+\frac {i x \operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{i-a}\right )}{d^2 \log ^2(f)}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{i+a}\right )}{d^2 \log ^2(f)}-\frac {i \operatorname {PolyLog}\left (4,\frac {b f^{c+d x}}{i-a}\right )}{d^3 \log ^3(f)}+\frac {i \operatorname {PolyLog}\left (4,-\frac {b f^{c+d x}}{i+a}\right )}{d^3 \log ^3(f)} \]

input

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

output

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

3.3.26.3 Rubi [A] (verified)

Time = 4.23 (sec) , antiderivative size = 621, normalized size of antiderivative = 1.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, 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.

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

\(\Big \downarrow \) 5667

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

\(\Big \downarrow \) 3031

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} i \left (\frac {1}{3} i b d \log (f) \left (\frac {6 i \operatorname {PolyLog}\left (4,\frac {b f^{c+d x}}{i-a}\right )}{b d^4 \log ^4(f)}-\frac {6 i \operatorname {PolyLog}\left (4,-\frac {b f^{c+d x}}{a}\right )}{b d^4 \log ^4(f)}-\frac {6 i x \operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{i-a}\right )}{b d^3 \log ^3(f)}+\frac {6 i x \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{a}\right )}{b d^3 \log ^3(f)}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{i-a}\right )}{b d^2 \log ^2(f)}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a}\right )}{b d^2 \log ^2(f)}+\frac {i x^3 \log \left (1-\frac {b f^{c+d x}}{-a+i}\right )}{b d \log (f)}-\frac {i x^3 \log \left (\frac {b f^{c+d x}}{a}+1\right )}{b d \log (f)}\right )+\frac {1}{3} x^3 \log \left (1-\frac {i}{a+b f^{c+d x}}\right )\right )-\frac {1}{2} i \left (\frac {1}{3} i b d \log (f) \left (-\frac {6 i \operatorname {PolyLog}\left (4,-\frac {b f^{c+d x}}{a}\right )}{b d^4 \log ^4(f)}+\frac {6 i \operatorname {PolyLog}\left (4,-\frac {b f^{c+d x}}{a+i}\right )}{b d^4 \log ^4(f)}+\frac {6 i x \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{a}\right )}{b d^3 \log ^3(f)}-\frac {6 i x \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{a+i}\right )}{b d^3 \log ^3(f)}-\frac {3 i x^2 \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a}\right )}{b d^2 \log ^2(f)}+\frac {3 i x^2 \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{a+i}\right )}{b d^2 \log ^2(f)}-\frac {i x^3 \log \left (\frac {b f^{c+d x}}{a}+1\right )}{b d \log (f)}+\frac {i x^3 \log \left (1+\frac {b f^{c+d x}}{a+i}\right )}{b d \log (f)}\right )+\frac {1}{3} x^3 \log \left (1+\frac {i}{a+b f^{c+d x}}\right )\right )\)

input

Int[x^2*ArcCot[a + b*f^(c + d*x)],x]
 

output

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

3.3.26.3.1 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] 
]
 

3.3.26.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 763 vs. \(2 (281 ) = 562\).

Time = 1.54 (sec) , antiderivative size = 764, normalized size of antiderivative = 2.44

method	result	size
risch	\(\frac {i x^{3} \ln \left (1+i \left (a +b \,f^{d x +c}\right )\right )}{6}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{3}}{6}-\frac {i x^{3} \ln \left (1-i \left (a +b \,f^{d x +c}\right )\right )}{6}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{3}}{6}+\frac {\pi \,x^{3}}{6}-\frac {i \operatorname {polylog}\left (4, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right )}{\ln \left (f \right )^{3} d^{3}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{3}}{3 d^{3}}+\frac {i c^{3} \ln \left (i f^{d x} f^{c} b +i a +1\right )}{6 d^{3}}+\frac {i \operatorname {polylog}\left (4, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right )}{\ln \left (f \right )^{3} d^{3}}+\frac {i c^{3} \ln \left (\frac {b \,f^{d x} f^{c}+a +i}{i+a}\right )}{2 d^{3}}-\frac {i c^{3} \ln \left (1-i a -i f^{d x} f^{c} b \right )}{6 d^{3}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c^{3}}{3 d^{3}}-\frac {i c^{3} \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right )}{2 d^{3}}+\frac {i c^{2} \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a +i}{i+a}\right )}{2 \ln \left (f \right ) d^{3}}-\frac {i \operatorname {polylog}\left (3, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x}{\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^{2} x}{2 d^{2}}-\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c^{2}}{2 \ln \left (f \right ) d^{3}}+\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a +i}{i+a}\right ) x}{2 d^{2}}+\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{2}}{2 \ln \left (f \right ) d}+\frac {i \operatorname {polylog}\left (3, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x}{\ln \left (f \right )^{2} d^{2}}-\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) x}{2 d^{2}}-\frac {i c^{2} \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right )}{2 \ln \left (f \right ) d^{3}}+\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{2}}{2 \ln \left (f \right ) d^{3}}-\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{2}}{2 \ln \left (f \right ) d}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{2} x}{2 d^{2}}\)	\(764\)

input

int(x^2*arccot(a+b*f^(d*x+c)),x,method=_RETURNVERBOSE)
 

output

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

3.3.26.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.21 \[ \int x^2 \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\frac {2 \, d^{3} x^{3} \operatorname {arccot}\left (b f^{d x + c} + a\right ) \log \left (f\right )^{3} - 3 i \, d^{2} x^{2} {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right )^{2} + 3 i \, d^{2} x^{2} {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right )^{2} - i \, c^{3} \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right )^{3} + i \, c^{3} \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right )^{3} + {\left (-i \, d^{3} x^{3} - i \, c^{3}\right )} \log \left (f\right )^{3} \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (i \, d^{3} x^{3} + i \, c^{3}\right )} \log \left (f\right )^{3} \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + 6 i \, d x \log \left (f\right ) {\rm polylog}\left (3, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 6 i \, d x \log \left (f\right ) {\rm polylog}\left (3, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 6 i \, {\rm polylog}\left (4, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) + 6 i \, {\rm polylog}\left (4, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{6 \, d^{3} \log \left (f\right )^{3}} \]

input

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

output

1/6*(2*d^3*x^3*arccot(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*dil 
og(-(a^2 + (a*b - I*b)*f^(d*x + c) + 1)/(a^2 + 1) + 1)*log(f)^2 - I*c^3*lo 
g(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)*polylog(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)
 

3.3.26.6 Sympy [F]

\[ \int x^2 \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int x^{2} \operatorname {acot}{\left (a + b f^{c + d x} \right )}\, dx \]

input

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

output

Integral(x**2*acot(a + b*f**(c + d*x)), x)
 

3.3.26.7 Maxima [F]

\[ \int x^2 \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int { x^{2} \operatorname {arccot}\left (b f^{d x + c} + a\right ) \,d x } \]

input

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

output

b*d*f^c*integrate(1/3*f^(d*x)*x^3/(b^2*f^(2*d*x)*f^(2*c) + 2*a*b*f^(d*x)*f 
^c + a^2 + 1), x)*log(f) + 1/3*x^3*arctan(1/(b*f^(d*x)*f^c + a))
 

3.3.26.8 Giac [F]

\[ \int x^2 \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int { x^{2} \operatorname {arccot}\left (b f^{d x + c} + a\right ) \,d x } \]

input

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

output

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

3.3.26.9 Mupad [F(-1)]

Timed out. \[ \int x^2 \cot ^{-1}\left (a+b f^{c+d x}\right ) \, dx=\int x^2\,\mathrm {acot}\left (a+b\,f^{c+d\,x}\right ) \,d x \]

input

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

output

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