3.1.96 \(\int \frac {x^2 (a+b \arctan (c x))^2}{d+i c d x} \, dx\) [96]

3.1.96.1 Optimal result

Integrand size = 25, antiderivative size = 277 \[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\frac {i a b x}{c^2 d}+\frac {i b^2 x \arctan (c x)}{c^2 d}+\frac {i (a+b \arctan (c x))^2}{2 c^3 d}+\frac {x (a+b \arctan (c x))^2}{c^2 d}-\frac {i x^2 (a+b \arctan (c x))^2}{2 c d}+\frac {2 b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i b^2 \log \left (1+c^2 x^2\right )}{2 c^3 d}+\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3 d}+\frac {b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^3 d} \]

output

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

3.1.96.2 Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.19 \[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+i c d x} \, dx=-\frac {i \left (6 i a^2 c x-6 a b c x+3 a^2 c^2 x^2-6 i a^2 \arctan (c x)+6 a b \arctan (c x)+12 i a b c x \arctan (c x)-6 b^2 c x \arctan (c x)+6 a b c^2 x^2 \arctan (c x)-12 i a b \arctan (c x)^2+9 b^2 \arctan (c x)^2+6 i b^2 c x \arctan (c x)^2+3 b^2 c^2 x^2 \arctan (c x)^2-4 i b^2 \arctan (c x)^3+12 a b \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+12 i b^2 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+6 b^2 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-3 a^2 \log \left (1+c^2 x^2\right )-6 i a b \log \left (1+c^2 x^2\right )+3 b^2 \log \left (1+c^2 x^2\right )+6 b (-i a+b-i b \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+3 b^2 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )}{6 c^3 d} \]

input

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

output

((-1/6*I)*((6*I)*a^2*c*x - 6*a*b*c*x + 3*a^2*c^2*x^2 - (6*I)*a^2*ArcTan[c* 
x] + 6*a*b*ArcTan[c*x] + (12*I)*a*b*c*x*ArcTan[c*x] - 6*b^2*c*x*ArcTan[c*x 
] + 6*a*b*c^2*x^2*ArcTan[c*x] - (12*I)*a*b*ArcTan[c*x]^2 + 9*b^2*ArcTan[c* 
x]^2 + (6*I)*b^2*c*x*ArcTan[c*x]^2 + 3*b^2*c^2*x^2*ArcTan[c*x]^2 - (4*I)*b 
^2*ArcTan[c*x]^3 + 12*a*b*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + (12 
*I)*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 6*b^2*ArcTan[c*x]^2*L 
og[1 + E^((2*I)*ArcTan[c*x])] - 3*a^2*Log[1 + c^2*x^2] - (6*I)*a*b*Log[1 + 
 c^2*x^2] + 3*b^2*Log[1 + c^2*x^2] + 6*b*((-I)*a + b - I*b*ArcTan[c*x])*Po 
lyLog[2, -E^((2*I)*ArcTan[c*x])] + 3*b^2*PolyLog[3, -E^((2*I)*ArcTan[c*x]) 
]))/(c^3*d)
 

3.1.96.3 Rubi [A] (verified)

Time = 2.45 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.09, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5401, 27, 5361, 5401, 5345, 5379, 5451, 2009, 5419, 5455, 5379, 2849, 2752, 5529, 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.

\(\displaystyle \int \frac {x^2 (a+b \arctan (c x))^2}{d+i c d x} \, dx\)

\(\Big \downarrow \) 5401

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5401

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

\(\Big \downarrow \) 5345

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

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 5451

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5419

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

\(\Big \downarrow \) 5455

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

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

\(\Big \downarrow \) 5529

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

\(\Big \downarrow \) 7164

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

input

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

output

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

3.1.96.3.1 Defintions of rubi rules used

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 2752

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]
 

rule 2849

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]
 

rule 5345

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

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 5379

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] 
 :> Simp[(-(a + b*ArcTan[c*x])^p)*(Log[2/(1 + e*(x/d))]/e), x] + Simp[b*c*( 
p/e)   Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[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 5401

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + ( 
e_.)*(x_)), x_Symbol] :> Simp[f/e   Int[(f*x)^(m - 1)*(a + b*ArcTan[c*x])^p 
, x], x] - Simp[d*(f/e)   Int[(f*x)^(m - 1)*((a + b*ArcTan[c*x])^p/(d + e*x 
)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e 
^2, 0] && GtQ[m, 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 5451

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

rule 5455

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

rule 5529

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 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]
 

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

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

Time = 25.01 (sec) , antiderivative size = 994, normalized size of antiderivative = 3.59

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(994\)
default	\(\text {Expression too large to display}\)	\(994\)
parts	\(\text {Expression too large to display}\)	\(1038\)

input

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

output

1/c^3*(a^2/d*c*x+2*I/d*a*b*arctan(c*x)*ln(c*x-I)-I/d*a*b*arctan(c*x)*c^2*x 
^2-a^2/d*arctan(c*x)+b^2/d*(arctan(c*x)^2*c*x-1/2*I*arctan(c*x)^2*c^2*x^2- 
2*I*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*dilog(1+I*(1+I*c*x)/(c^2*x^ 
2+1)^(1/2))+I*arctan(c*x)^2*ln(c*x-I)+I*arctan(c*x)*(c*x-I)+2*arctan(c*x)* 
ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^ 
2+1)^(1/2))+Pi*arctan(c*x)^2+I*ln(1+(1+I*c*x)^2/(c^2*x^2+1))-3/2*I*arctan( 
c*x)^2-Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arct 
an(c*x)^2-1/2*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1 
)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-1/2*Pi*csgn((1+I*c*x)^2/(c^ 
2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)^2+1/2*Pi*csgn(I/(1+(1+ 
I*c*x)^2/(c^2*x^2+1)))*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^ 
2+1)))^2*arctan(c*x)^2-1/2*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn((1+ 
I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2 
+1)))*arctan(c*x)^2-1/2*I*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-2/3*arctan(c 
*x)^3-arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-I*arctan(c*x)^2*ln(2 
*I*(1+I*c*x)^2/(c^2*x^2+1)))+2/d*a*b*arctan(c*x)*c*x+1/2*I*a^2/d*ln(c^2*x^ 
2+1)+1/4*I/d*a*b*arctan(1/6*c^3*x^3+7/6*c*x)+1/d*a*b*ln(-1/2*I*(c*x+I))*ln 
(c*x-I)+1/d*a*b*dilog(-1/2*I*(c*x+I))-1/2/d*a*b*ln(c*x-I)^2+1/d*a*b-1/4*I/ 
d*a*b*arctan(1/2*c*x)-1/8/d*a*b*ln(c^4*x^4+10*c^2*x^2+9)+1/2*I/d*a*b*arcta 
n(1/2*c*x-1/2*I)-1/2*I*a^2/d*c^2*x^2+I/d*a*b*c*x-3/4/d*a*b*ln(c^2*x^2+1...
 

3.1.96.5 Fricas [F]

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

input

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

output

integral(1/4*(I*b^2*x^2*log(-(c*x + I)/(c*x - I))^2 + 4*a*b*x^2*log(-(c*x 
+ I)/(c*x - I)) - 4*I*a^2*x^2)/(c*d*x - I*d), x)
 

3.1.96.6 Sympy [F(-1)]

Timed out. \[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.1.96.7 Maxima [F]

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

input

integrate(x^2*(a+b*arctan(c*x))^2/(d+I*c*d*x),x, algorithm="maxima")
 

output

-1/2*a^2*((I*c*x^2 - 2*x)/(c^2*d) - 2*I*log(I*c*x + 1)/(c^3*d)) - 1/96*(16 
*(24*b^2*c^3*integrate(1/16*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^4*d*x^2 + 
c^2*d), x) + 24*b^2*c^3*integrate(1/16*x^3*arctan(c*x)/(c^4*d*x^2 + c^2*d) 
, x) - 72*b^2*c^2*integrate(1/16*x^2*arctan(c*x)^2/(c^4*d*x^2 + c^2*d), x) 
 - 6*b^2*c^2*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^4*d*x^2 + c^2*d), x) 
 - 192*a*b*c^2*integrate(1/16*x^2*arctan(c*x)/(c^4*d*x^2 + c^2*d), x) - 12 
*b^2*c^2*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^4*d*x^2 + c^2*d), x) + 48* 
b^2*c*integrate(1/16*x*arctan(c*x)/(c^4*d*x^2 + c^2*d), x) - 12*b^2*integr 
ate(1/16*log(c^2*x^2 + 1)^2/(c^4*d*x^2 + c^2*d), x) - b^2*arctan(c*x)^3/(c 
^3*d))*c^3*d + 24*b^2*arctan(c*x)^3 + 96*I*c^3*d*integrate(1/16*(4*(3*b^2* 
c^2*x^3 - 2*b^2*x)*arctan(c*x)^2 + (b^2*c^2*x^3 - 2*b^2*x)*log(c^2*x^2 + 1 
)^2 + 4*(8*a*b*c^2*x^3 + b^2*c*x^2)*arctan(c*x) + 2*(b^2*c^2*x^3 + 2*b^2*c 
*x^2*arctan(c*x) + 2*b^2*x)*log(c^2*x^2 + 1))/(c^3*d*x^2 + c*d), x) + 3*I* 
b^2*log(c^2*x^2 + 1)^3 - 12*(-I*b^2*c^2*x^2 + 2*b^2*c*x)*arctan(c*x)^2 + 3 
*(-I*b^2*c^2*x^2 + 2*b^2*c*x + 2*b^2*arctan(c*x))*log(c^2*x^2 + 1)^2 - 12* 
(-I*b^2*arctan(c*x)^2 + (b^2*c^2*x^2 + 2*I*b^2*c*x)*arctan(c*x))*log(c^2*x 
^2 + 1))/(c^3*d)
 

3.1.96.8 Giac [F]

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

input

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

output

sage0*x
 

3.1.96.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int \frac {x^2\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \]

input

int((x^2*(a + b*atan(c*x))^2)/(d + c*d*x*1i),x)
 

output

int((x^2*(a + b*atan(c*x))^2)/(d + c*d*x*1i), x)