\(\int x^2 \arctan (a+b f^{c+d x}) \, dx\) [118]

Optimal result

Integrand size = 16, antiderivative size = 302 \[ \int x^2 \arctan \left (a+b f^{c+d x}\right ) \, dx=\frac {1}{3} x^3 \arctan \left (a+b f^{c+d x}\right )-\frac {1}{6} i x^3 \log \left (1-\frac {i b f^{c+d x}}{1-i a}\right )+\frac {1}{6} i x^3 \log \left (1+\frac {i b f^{c+d x}}{1+i a}\right )-\frac {i x^2 \operatorname {PolyLog}\left (2,\frac {i b f^{c+d x}}{1-i a}\right )}{2 d \log (f)}+\frac {i x^2 \operatorname {PolyLog}\left (2,-\frac {i b f^{c+d x}}{1+i a}\right )}{2 d \log (f)}+\frac {i x \operatorname {PolyLog}\left (3,\frac {i b f^{c+d x}}{1-i a}\right )}{d^2 \log ^2(f)}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {i b f^{c+d x}}{1+i a}\right )}{d^2 \log ^2(f)}-\frac {i \operatorname {PolyLog}\left (4,\frac {i b f^{c+d x}}{1-i a}\right )}{d^3 \log ^3(f)}+\frac {i \operatorname {PolyLog}\left (4,-\frac {i b f^{c+d x}}{1+i a}\right )}{d^3 \log ^3(f)} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 299, normalized size of antiderivative = 0.99 \[ \int x^2 \arctan \left (a+b f^{c+d x}\right ) \, dx=\frac {i \left (d^3 x^3 \log ^3(f) \log \left (1-i a-i b f^{c+d x}\right )-d^3 x^3 \log ^3(f) \log \left (1+i a+i b f^{c+d x}\right )-d^3 x^3 \log ^3(f) \log \left (\frac {i+a+b f^{c+d x}}{i+a}\right )+d^3 x^3 \log ^3(f) \log \left (1+\frac {b f^{c+d x}}{-i+a}\right )+3 d^2 x^2 \log ^2(f) \operatorname {PolyLog}\left (2,\frac {b f^{c+d x}}{i-a}\right )-3 d^2 x^2 \log ^2(f) \operatorname {PolyLog}\left (2,-\frac {b f^{c+d x}}{i+a}\right )-6 d x \log (f) \operatorname {PolyLog}\left (3,\frac {b f^{c+d x}}{i-a}\right )+6 d x \log (f) \operatorname {PolyLog}\left (3,-\frac {b f^{c+d x}}{i+a}\right )+6 \operatorname {PolyLog}\left (4,\frac {b f^{c+d x}}{i-a}\right )-6 \operatorname {PolyLog}\left (4,-\frac {b f^{c+d x}}{i+a}\right )\right )}{6 d^3 \log ^3(f)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5666, 3012, 3011, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 5666

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

\(\Big \downarrow \) 3012

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2720

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]]]
 

rule 3011

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] + Simp[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 3012

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/d)*(F^(c*(a + b*x) 
))^n], x] - Simp[(f + g*x)^(m + 1)*(Log[1 + (e/d)*(F^(c*(a + b*x)))^n]/(g*( 
m + 1))), x]) /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[ 
d, 1]
 

rule 5666

Int[ArcTan[(a_.) + (b_.)*(f_)^((c_.) + (d_.)*(x_))]*(x_)^(m_.), x_Symbol] : 
> Simp[I/2   Int[x^m*Log[1 - I*a - I*b*f^(c + d*x)], x], x] - Simp[I/2   In 
t[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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 757 vs. \(2 (268 ) = 536\).

Time = 1.11 (sec) , antiderivative size = 758, normalized size of antiderivative = 2.51

method	result	size
risch	\(\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) c^{2}}{2 d^{3} \ln \left (f \right )}+\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{2}}{2 d \ln \left (f \right )}+\frac {i \operatorname {polylog}\left (4, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right )}{d^{3} \ln \left (f \right )^{3}}-\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a +i}{a +i}\right ) x}{2 d^{2}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{3}}{6}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x \,c^{2}}{2 d^{2}}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x \,c^{2}}{2 d^{2}}+\frac {i \operatorname {polylog}\left (3, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x}{d^{2} \ln \left (f \right )^{2}}-\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{2}}{2 d^{3} \ln \left (f \right )}-\frac {i \operatorname {polylog}\left (2, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right ) x^{2}}{2 d \ln \left (f \right )}+\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x^{3}}{6}-\frac {i \operatorname {polylog}\left (3, \frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) x}{d^{2} \ln \left (f \right )^{2}}-\frac {i c^{2} \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a +i}{a +i}\right )}{2 d^{3} \ln \left (f \right )}-\frac {i \operatorname {polylog}\left (4, \frac {i b \,f^{d x} f^{c}}{-i a +1}\right )}{d^{3} \ln \left (f \right )^{3}}+\frac {i c^{2} \operatorname {dilog}\left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right )}{2 d^{3} \ln \left (f \right )}-\frac {i c^{3} \ln \left (\frac {b \,f^{d x} f^{c}+a +i}{a +i}\right )}{2 d^{3}}+\frac {i x^{3} \ln \left (1-i \left (a +b \,f^{d x +c}\right )\right )}{6}-\frac {i c^{3} \ln \left (i f^{d x} f^{c} b +i a +1\right )}{6 d^{3}}+\frac {i c^{2} \ln \left (\frac {b \,f^{d x} f^{c}+a -i}{a -i}\right ) x}{2 d^{2}}-\frac {i \ln \left (1-\frac {i b \,f^{d x} f^{c}}{-i a -1}\right ) c^{3}}{3 d^{3}}-\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 ) 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^{3} \ln \left (1-i a -i f^{d x} f^{c} b \right )}{6 d^{3}}\)	\(758\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.25 \[ \int x^2 \arctan \left (a+b f^{c+d x}\right ) \, dx=\frac {2 \, d^{3} x^{3} \arctan \left (b f^{d x + c} + a\right ) \log \left (f\right )^{3} + 3 i \, d^{2} x^{2} {\rm Li}_2\left (-\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right )^{2} - 3 i \, d^{2} x^{2} {\rm Li}_2\left (-\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1} + 1\right ) \log \left (f\right )^{2} + i \, c^{3} \log \left (b f^{d x + c} + a + i\right ) \log \left (f\right )^{3} - i \, c^{3} \log \left (b f^{d x + c} + a - i\right ) \log \left (f\right )^{3} + {\left (i \, d^{3} x^{3} + i \, c^{3}\right )} \log \left (f\right )^{3} \log \left (\frac {a^{2} + {\left (a b + i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) + {\left (-i \, d^{3} x^{3} - i \, c^{3}\right )} \log \left (f\right )^{3} \log \left (\frac {a^{2} + {\left (a b - i \, b\right )} f^{d x + c} + 1}{a^{2} + 1}\right ) - 6 i \, d x \log \left (f\right ) {\rm polylog}\left (3, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) + 6 i \, d x \log \left (f\right ) {\rm polylog}\left (3, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) + 6 i \, {\rm polylog}\left (4, -\frac {{\left (a b + i \, b\right )} f^{d x + c}}{a^{2} + 1}\right ) - 6 i \, {\rm polylog}\left (4, -\frac {{\left (a b - i \, b\right )} f^{d x + c}}{a^{2} + 1}\right )}{6 \, d^{3} \log \left (f\right )^{3}} \] Input:

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

Output:

1/6*(2*d^3*x^3*arctan(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)
 

Sympy [F]

\[ \int x^2 \arctan \left (a+b f^{c+d x}\right ) \, dx=\int x^{2} \operatorname {atan}{\left (a + b f^{c + d x} \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int x^2 \arctan \left (a+b f^{c+d x}\right ) \, dx=\int { x^{2} \arctan \left (b f^{d x + c} + a\right ) \,d x } \] Input:

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

Giac [F]

\[ \int x^2 \arctan \left (a+b f^{c+d x}\right ) \, dx=\int { x^{2} \arctan \left (b f^{d x + c} + a\right ) \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \arctan \left (a+b f^{c+d x}\right ) \, dx=\int x^2\,\mathrm {atan}\left (a+b\,f^{c+d\,x}\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x^2 \arctan \left (a+b f^{c+d x}\right ) \, dx=\int \mathit {atan} \left (f^{d x +c} b +a \right ) x^{2}d x \] Input:

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

Output:

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