3.1.20 \(\int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx\) [20]

3.1.20.1 Optimal result

Integrand size = 23, antiderivative size = 287 \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=-\frac {b^2 (a+b \arctan (c+d x))}{d e^4 (c+d x)}-\frac {b (a+b \arctan (c+d x))^2}{2 d e^4}-\frac {b (a+b \arctan (c+d x))^2}{2 d e^4 (c+d x)^2}+\frac {i (a+b \arctan (c+d x))^3}{3 d e^4}-\frac {(a+b \arctan (c+d x))^3}{3 d e^4 (c+d x)^3}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left (1+(c+d x)^2\right )}{2 d e^4}-\frac {b (a+b \arctan (c+d x))^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4} \]

output

-b^2*(a+b*arctan(d*x+c))/d/e^4/(d*x+c)-1/2*b*(a+b*arctan(d*x+c))^2/d/e^4-1 
/2*b*(a+b*arctan(d*x+c))^2/d/e^4/(d*x+c)^2+1/3*I*(a+b*arctan(d*x+c))^3/d/e 
^4-1/3*(a+b*arctan(d*x+c))^3/d/e^4/(d*x+c)^3+b^3*ln(d*x+c)/d/e^4-1/2*b^3*l 
n(1+(d*x+c)^2)/d/e^4-b*(a+b*arctan(d*x+c))^2*ln(2-2/(1-I*(d*x+c)))/d/e^4+I 
*b^2*(a+b*arctan(d*x+c))*polylog(2,-1+2/(1-I*(d*x+c)))/d/e^4-1/2*b^3*polyl 
og(3,-1+2/(1-I*(d*x+c)))/d/e^4
 

3.1.20.2 Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\frac {-\frac {2 a^3}{(c+d x)^3}-\frac {3 a^2 b}{(c+d x)^2}-\frac {6 a^2 b \arctan (c+d x)}{(c+d x)^3}-6 a^2 b \log (c+d x)+3 a^2 b \log \left (1+c^2+2 c d x+d^2 x^2\right )+6 a b^2 \left (-\frac {(c+d x)^2+\arctan (c+d x)^2}{(c+d x)^3}+\arctan (c+d x) \left (-1-\frac {1}{(c+d x)^2}+i \arctan (c+d x)-2 \log \left (1-e^{2 i \arctan (c+d x)}\right )\right )+i \operatorname {PolyLog}\left (2,e^{2 i \arctan (c+d x)}\right )\right )+6 b^3 \left (\frac {i \pi ^3}{24}-\frac {\arctan (c+d x)}{c+d x}-\frac {1}{2} \arctan (c+d x)^2-\frac {\arctan (c+d x)^2}{2 (c+d x)^2}-\frac {1}{3} i \arctan (c+d x)^3-\frac {\arctan (c+d x)^3}{3 (c+d x)^3}-\arctan (c+d x)^2 \log \left (1-e^{-2 i \arctan (c+d x)}\right )+\log (c+d x)+\log \left (\frac {1}{\sqrt {1+(c+d x)^2}}\right )-i \arctan (c+d x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c+d x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c+d x)}\right )\right )}{6 d e^4} \]

input

Integrate[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^4,x]
 

output

((-2*a^3)/(c + d*x)^3 - (3*a^2*b)/(c + d*x)^2 - (6*a^2*b*ArcTan[c + d*x])/ 
(c + d*x)^3 - 6*a^2*b*Log[c + d*x] + 3*a^2*b*Log[1 + c^2 + 2*c*d*x + d^2*x 
^2] + 6*a*b^2*(-(((c + d*x)^2 + ArcTan[c + d*x]^2)/(c + d*x)^3) + ArcTan[c 
 + d*x]*(-1 - (c + d*x)^(-2) + I*ArcTan[c + d*x] - 2*Log[1 - E^((2*I)*ArcT 
an[c + d*x])]) + I*PolyLog[2, E^((2*I)*ArcTan[c + d*x])]) + 6*b^3*((I/24)* 
Pi^3 - ArcTan[c + d*x]/(c + d*x) - ArcTan[c + d*x]^2/2 - ArcTan[c + d*x]^2 
/(2*(c + d*x)^2) - (I/3)*ArcTan[c + d*x]^3 - ArcTan[c + d*x]^3/(3*(c + d*x 
)^3) - ArcTan[c + d*x]^2*Log[1 - E^((-2*I)*ArcTan[c + d*x])] + Log[c + d*x 
] + Log[1/Sqrt[1 + (c + d*x)^2]] - I*ArcTan[c + d*x]*PolyLog[2, E^((-2*I)* 
ArcTan[c + d*x])] - PolyLog[3, E^((-2*I)*ArcTan[c + d*x])]/2))/(6*d*e^4)
 

3.1.20.3 Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.87, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {5566, 27, 5361, 5453, 5361, 5453, 5361, 243, 47, 14, 16, 5419, 5459, 5403, 5527, 7164}

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.

\(\displaystyle \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx\)

\(\Big \downarrow \) 5566

\(\displaystyle \frac {\int \frac {(a+b \arctan (c+d x))^3}{e^4 (c+d x)^4}d(c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(a+b \arctan (c+d x))^3}{(c+d x)^4}d(c+d x)}{d e^4}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {b \int \frac {(a+b \arctan (c+d x))^2}{(c+d x)^3 \left ((c+d x)^2+1\right )}d(c+d x)-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 5453

\(\displaystyle \frac {b \left (\int \frac {(a+b \arctan (c+d x))^2}{(c+d x)^3}d(c+d x)-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {b \left (b \int \frac {a+b \arctan (c+d x)}{(c+d x)^2 \left ((c+d x)^2+1\right )}d(c+d x)-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 5453

\(\displaystyle \frac {b \left (b \left (\int \frac {a+b \arctan (c+d x)}{(c+d x)^2}d(c+d x)-\int \frac {a+b \arctan (c+d x)}{(c+d x)^2+1}d(c+d x)\right )-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {b \left (b \left (-\int \frac {a+b \arctan (c+d x)}{(c+d x)^2+1}d(c+d x)+b \int \frac {1}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {a+b \arctan (c+d x)}{c+d x}\right )-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {b \left (b \left (-\int \frac {a+b \arctan (c+d x)}{(c+d x)^2+1}d(c+d x)+\frac {1}{2} b \int \frac {1}{(c+d x)^2 \left ((c+d x)^2+1\right )}d(c+d x)^2-\frac {a+b \arctan (c+d x)}{c+d x}\right )-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 47

\(\displaystyle \frac {b \left (b \left (-\int \frac {a+b \arctan (c+d x)}{(c+d x)^2+1}d(c+d x)+\frac {1}{2} b \left (\int \frac {1}{(c+d x)^2}d(c+d x)^2-\int \frac {1}{(c+d x)^2+1}d(c+d x)^2\right )-\frac {a+b \arctan (c+d x)}{c+d x}\right )-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 14

\(\displaystyle \frac {b \left (-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)+b \left (-\int \frac {a+b \arctan (c+d x)}{(c+d x)^2+1}d(c+d x)+\frac {1}{2} b \left (\log \left ((c+d x)^2\right )-\int \frac {1}{(c+d x)^2+1}d(c+d x)^2\right )-\frac {a+b \arctan (c+d x)}{c+d x}\right )-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {b \left (-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)+b \left (-\int \frac {a+b \arctan (c+d x)}{(c+d x)^2+1}d(c+d x)-\frac {a+b \arctan (c+d x)}{c+d x}+\frac {1}{2} b \left (\log \left ((c+d x)^2\right )-\log \left ((c+d x)^2+1\right )\right )\right )-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 5419

\(\displaystyle \frac {b \left (-\int \frac {(a+b \arctan (c+d x))^2}{(c+d x) \left ((c+d x)^2+1\right )}d(c+d x)-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}+b \left (-\frac {(a+b \arctan (c+d x))^2}{2 b}-\frac {a+b \arctan (c+d x)}{c+d x}+\frac {1}{2} b \left (\log \left ((c+d x)^2\right )-\log \left ((c+d x)^2+1\right )\right )\right )\right )-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}}{d e^4}\)

\(\Big \downarrow \) 5459

\(\displaystyle \frac {-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}+b \left (-i \int \frac {(a+b \arctan (c+d x))^2}{(c+d x) (c+d x+i)}d(c+d x)+\frac {i (a+b \arctan (c+d x))^3}{3 b}-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}+b \left (-\frac {(a+b \arctan (c+d x))^2}{2 b}-\frac {a+b \arctan (c+d x)}{c+d x}+\frac {1}{2} b \left (\log \left ((c+d x)^2\right )-\log \left ((c+d x)^2+1\right )\right )\right )\right )}{d e^4}\)

\(\Big \downarrow \) 5403

\(\displaystyle \frac {-\frac {(a+b \arctan (c+d x))^3}{3 (c+d x)^3}+b \left (-i \left (2 i b \int \frac {(a+b \arctan (c+d x)) \log \left (2-\frac {2}{1-i (c+d x)}\right )}{(c+d x)^2+1}d(c+d x)-i \log \left (2-\frac {2}{1-i (c+d x)}\right ) (a+b \arctan (c+d x))^2\right )+\frac {i (a+b \arctan (c+d x))^3}{3 b}-\frac {(a+b \arctan (c+d x))^2}{2 (c+d x)^2}+b \left (-\frac {(a+b \arctan (c+d x))^2}{2 b}-\frac {a+b \arctan (c+d x)}{c+d x}+\frac {1}{2} b \left (\log \left ((c+d x)^2\right )-\log \left ((c+d x)^2+1\right )\right )\right )\right )}{d e^4}\)

\(\Big \downarrow \) 5527

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

\(\Big \downarrow \) 7164

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

input

Int[(a + b*ArcTan[c + d*x])^3/(c*e + d*e*x)^4,x]
 

output

(-1/3*(a + b*ArcTan[c + d*x])^3/(c + d*x)^3 + b*(-1/2*(a + b*ArcTan[c + d* 
x])^2/(c + d*x)^2 + ((I/3)*(a + b*ArcTan[c + d*x])^3)/b + b*(-((a + b*ArcT 
an[c + d*x])/(c + d*x)) - (a + b*ArcTan[c + d*x])^2/(2*b) + (b*(Log[(c + d 
*x)^2] - Log[1 + (c + d*x)^2]))/2) - I*((-I)*(a + b*ArcTan[c + d*x])^2*Log 
[2 - 2/(1 - I*(c + d*x))] + (2*I)*b*((I/2)*(a + b*ArcTan[c + d*x])*PolyLog 
[2, -1 + 2/(1 - I*(c + d*x))] - (b*PolyLog[3, -1 + 2/(1 - I*(c + d*x))])/4 
))))/(d*e^4)
 

3.1.20.3.1 Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 47

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c 
 - a*d)   Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d)   Int[1/(c + d*x), x 
], x] /; FreeQ[{a, b, c, d}, x]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 5361

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> 
 Simp[x^(m + 1)*((a + b*ArcTan[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 
1))   Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], 
x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] & 
& IntegerQ[m])) && NeQ[m, -1]
 

rule 5403

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_ 
Symbol] :> Simp[(a + b*ArcTan[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Si 
mp[b*c*(p/d)   Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))]/(1 
 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2* 
d^2 + e^2, 0]
 

rule 5419

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> Simp[(a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, 
c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]
 

rule 5453

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e 
_.)*(x_)^2), x_Symbol] :> Simp[1/d   Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], 
 x] - Simp[e/(d*f^2)   Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/(d + e*x^2) 
), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]
 

rule 5459

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

rule 5527

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

rule 5566

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

rule 7164

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, 
x]}, Simp[w*PolyLog[n + 1, v], x] /;  !FalseQ[w]] /; FreeQ[n, x]
 

3.1.20.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 8.42 (sec) , antiderivative size = 2465, normalized size of antiderivative = 8.59

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2465\)
default	\(\text {Expression too large to display}\)	\(2465\)
parts	\(\text {Expression too large to display}\)	\(2473\)

input

int((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x,method=_RETURNVERBOSE)
 

output

1/d*(-1/3*a^3/e^4/(d*x+c)^3+b^3/e^4*(-1/3/(d*x+c)^3*arctan(d*x+c)^3-1/2/(d 
*x+c)^2*arctan(d*x+c)^2-ln(d*x+c)*arctan(d*x+c)^2+1/2*arctan(d*x+c)^2*ln(1 
+(d*x+c)^2)-arctan(d*x+c)^2*ln((1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+arctan(d 
*x+c)^2*ln((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)-arctan(d*x+c)^2*ln(1+(1+I*(d*x 
+c))/(1+(d*x+c)^2)^(1/2))+2*I*arctan(d*x+c)*polylog(2,-(1+I*(d*x+c))/(1+(d 
*x+c)^2)^(1/2))-2*polylog(3,-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-arctan(d*x 
+c)^2*ln(1-(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))+2*I*arctan(d*x+c)*polylog(2, 
(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))-2*polylog(3,(1+I*(d*x+c))/(1+(d*x+c)^2) 
^(1/2))+1/12*arctan(d*x+c)*(6*I*Pi*arctan(d*x+c)*csgn(I*(1+(1+I*(d*x+c))^2 
/(1+(d*x+c)^2))^2)^2*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))*(d*x+c)+4*I 
*arctan(d*x+c)^2*(d*x+c)-3*I*Pi*arctan(d*x+c)*csgn(I*(1+I*(d*x+c))^2/(1+(d 
*x+c)^2))*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2)/(1+(1+I*(d*x+c))^2/(1+(d*x+ 
c)^2))^2)^2*(d*x+c)-3*I*Pi*arctan(d*x+c)*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x 
+c)^2))^2)*csgn(I*(1+(1+I*(d*x+c))^2/(1+(d*x+c)^2)))^2*(d*x+c)-6*I*Pi*arct 
an(d*x+c)*csgn(I*((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+( 
d*x+c)^2)))*csgn(((1+I*(d*x+c))^2/(1+(d*x+c)^2)-1)/(1+(1+I*(d*x+c))^2/(1+( 
d*x+c)^2)))*(d*x+c)-6*I*Pi*arctan(d*x+c)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c) 
^2))^2*csgn(I*(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))*(d*x+c)-6*I*Pi*arctan(d*x 
+c)*(d*x+c)+3*I*Pi*arctan(d*x+c)*csgn(I*(1+I*(d*x+c))^2/(1+(d*x+c)^2))*csg 
n(I*(1+I*(d*x+c))/(1+(d*x+c)^2)^(1/2))^2*(d*x+c)+3*I*Pi*arctan(d*x+c)*c...
 

3.1.20.5 Fricas [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{4}} \,d x } \]

input

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="fricas")
 

output

integral((b^3*arctan(d*x + c)^3 + 3*a*b^2*arctan(d*x + c)^2 + 3*a^2*b*arct 
an(d*x + c) + a^3)/(d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4* 
c^3*d*e^4*x + c^4*e^4), x)
 

3.1.20.6 Sympy [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\frac {\int \frac {a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]

input

integrate((a+b*atan(d*x+c))**3/(d*e*x+c*e)**4,x)
 

output

(Integral(a**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d** 
4*x**4), x) + Integral(b**3*atan(c + d*x)**3/(c**4 + 4*c**3*d*x + 6*c**2*d 
**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a*b**2*atan(c + d*x 
)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x 
) + Integral(3*a**2*b*atan(c + d*x)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 
+ 4*c*d**3*x**3 + d**4*x**4), x))/e**4
 

3.1.20.7 Maxima [F]

\[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\int { \frac {{\left (b \arctan \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{4}} \,d x } \]

input

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="maxima")
 

output

-1/2*(d*(1/(d^4*e^4*x^2 + 2*c*d^3*e^4*x + c^2*d^2*e^4) - log(d^2*x^2 + 2*c 
*d*x + c^2 + 1)/(d^2*e^4) + 2*log(d*x + c)/(d^2*e^4)) + 2*arctan(d*x + c)/ 
(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4))*a^2*b - 1/3 
*a^3/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - 1/96* 
(4*b^3*arctan(d*x + c)^3 - 3*b^3*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c 
^2 + 1)^2 - 96*(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^ 
4)*integrate(1/32*(28*(b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2 + b^3)*arctan(d 
*x + c)^3 + 4*(24*a*b^2*d^2*x^2 + 24*a*b^2*c^2 + b^3*c + 24*a*b^2 + (48*a* 
b^2*c + b^3)*d*x)*arctan(d*x + c)^2 - 4*(b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c 
^2)*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1) - (b^3*d*x + b^3*c - 
3*(b^3*d^2*x^2 + 2*b^3*c*d*x + b^3*c^2 + b^3)*arctan(d*x + c))*log(d^2*x^2 
 + 2*c*d*x + c^2 + 1)^2)/(d^6*e^4*x^6 + 6*c*d^5*e^4*x^5 + (15*c^2 + 1)*d^4 
*e^4*x^4 + 4*(5*c^3 + c)*d^3*e^4*x^3 + 3*(5*c^4 + 2*c^2)*d^2*e^4*x^2 + 2*( 
3*c^5 + 2*c^3)*d*e^4*x + (c^6 + c^4)*e^4), x))/(d^4*e^4*x^3 + 3*c*d^3*e^4* 
x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)
 

3.1.20.8 Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\text {Timed out} \]

input

integrate((a+b*arctan(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="giac")
 

output

Timed out
 

3.1.20.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c+d x))^3}{(c e+d e x)^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]

input

int((a + b*atan(c + d*x))^3/(c*e + d*e*x)^4,x)
 

output

int((a + b*atan(c + d*x))^3/(c*e + d*e*x)^4, x)