\(\int (a+b \arctan (c+d x))^2 \, dx\) [41]

Optimal result

Integrand size = 12, antiderivative size = 102 \[ \int (a+b \arctan (c+d x))^2 \, dx=\frac {i (a+b \arctan (c+d x))^2}{d}+\frac {(c+d x) (a+b \arctan (c+d x))^2}{d}+\frac {2 b (a+b \arctan (c+d x)) \log \left (\frac {2}{1+i (c+d x)}\right )}{d}+\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.07 \[ \int (a+b \arctan (c+d x))^2 \, dx=\frac {b^2 (-i+c+d x) \arctan (c+d x)^2+2 b \arctan (c+d x) \left (a c+a d x+b \log \left (1+e^{2 i \arctan (c+d x)}\right )\right )+a \left (a c+a d x+2 b \log \left (\frac {1}{\sqrt {1+(c+d x)^2}}\right )\right )-i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c+d x)}\right )}{d} \] Input:

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

Output:

(b^2*(-I + c + d*x)*ArcTan[c + d*x]^2 + 2*b*ArcTan[c + d*x]*(a*c + a*d*x + 
 b*Log[1 + E^((2*I)*ArcTan[c + d*x])]) + a*(a*c + a*d*x + 2*b*Log[1/Sqrt[1 
 + (c + d*x)^2]]) - I*b^2*PolyLog[2, -E^((2*I)*ArcTan[c + d*x])])/d
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5562, 5345, 5455, 5379, 2849, 2752}

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.

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

\(\Big \downarrow \) 5562

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

\(\Big \downarrow \) 5345

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

\(\Big \downarrow \) 5455

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

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 2849

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

\(\Big \downarrow \) 2752

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

Input:

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

Output:

((c + d*x)*(a + b*ArcTan[c + d*x])^2 - 2*b*(((-1/2*I)*(a + b*ArcTan[c + d* 
x])^2)/b - (a + b*ArcTan[c + d*x])*Log[2/(1 + I*(c + d*x))] - (I/2)*b*Poly 
Log[2, 1 - 2/(1 + I*(c + d*x))]))/d
 

Defintions of rubi rules used

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 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 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 5562

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.36

method	result	size
parts	\(a^{2} x +\frac {b^{2} \left (\arctan \left (d x +c \right )^{2} \left (d x +c +i\right )+2 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-2 i \arctan \left (d x +c \right )^{2}-i \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )\right )}{d}+\frac {2 a b \left (\left (d x +c \right ) \arctan \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{d}\)	\(139\)
derivativedivides	\(\frac {\left (d x +c \right ) a^{2}-i \arctan \left (d x +c \right )^{2} b^{2}+\arctan \left (d x +c \right )^{2} b^{2} \left (d x +c \right )-i \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) b^{2}+2 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) b^{2}+2 a b \left (d x +c \right ) \arctan \left (d x +c \right )-a b \ln \left (1+\left (d x +c \right )^{2}\right )}{d}\)	\(146\)
default	\(\frac {\left (d x +c \right ) a^{2}-i \arctan \left (d x +c \right )^{2} b^{2}+\arctan \left (d x +c \right )^{2} b^{2} \left (d x +c \right )-i \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) b^{2}+2 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right ) b^{2}+2 a b \left (d x +c \right ) \arctan \left (d x +c \right )-a b \ln \left (1+\left (d x +c \right )^{2}\right )}{d}\)	\(146\)
risch	\(a^{2} x +\frac {i \ln \left (-i d x -i c +1\right ) a b c}{d}-\frac {a b \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{2 d}+\frac {a^{2} c}{d}-\frac {i \ln \left (-i d x -i c +1\right )^{2} b^{2}}{4 d}-\frac {i a b \arctan \left (d x +c \right )}{d}-\frac {\ln \left (-i d x -i c +1\right )^{2} b^{2} x}{4}+\frac {i b^{2} \operatorname {dilog}\left (\frac {1}{2}-\frac {1}{2} i d x -\frac {1}{2} i c \right )}{d}-\frac {b^{2} \left (d x +c -i\right ) \ln \left (1+i \left (d x +c \right )\right )^{2}}{4 d}-\frac {\ln \left (-i d x -i c +1\right )^{2} b^{2} c}{4 d}-\frac {\ln \left (-i d x -i c +1\right ) a b}{d}+i \ln \left (-i d x -i c +1\right ) a b x +\left (\frac {b^{2} x \ln \left (1-i \left (d x +c \right )\right )}{2}-\frac {i b \left (2 a x d -b \ln \left (1-i \left (d x +c \right )\right )+i \ln \left (1-i \left (d x +c \right )\right ) b c \right )}{2 d}\right ) \ln \left (1+i \left (d x +c \right )\right )-\frac {i b^{2} \ln \left (\frac {1}{2} i d x +\frac {1}{2} i c +\frac {1}{2}\right ) \ln \left (-i d x -i c +1\right )}{d}-\frac {i a b c \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{2 d}+\frac {a b c \arctan \left (d x +c \right )}{d}+\frac {i a^{2}}{d}+\frac {i b^{2} \ln \left (\frac {1}{2} i d x +\frac {1}{2} i c +\frac {1}{2}\right ) \ln \left (\frac {1}{2}-\frac {1}{2} i d x -\frac {1}{2} i c \right )}{d}\)	\(415\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/16*(12*c^2*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 4*(3*arctan(d*x 
 + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*c^2 + 4*x 
*arctan(d*x + c)^2 + 192*d^2*integrate(1/16*x^2*arctan(d*x + c)^2/(d^2*x^2 
 + 2*c*d*x + c^2 + 1), x) + 16*d^2*integrate(1/16*x^2*log(d^2*x^2 + 2*c*d* 
x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 384*c*d*integrate(1/16* 
x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 64*d^2*integrate(1 
/16*x^2*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) 
 + 32*c*d*integrate(1/16*x*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2 
*c*d*x + c^2 + 1), x) + 64*c*d*integrate(1/16*x*log(d^2*x^2 + 2*c*d*x + c^ 
2 + 1)/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 16*c^2*integrate(1/16*log(d^2*x 
^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) - x*log(d^2*x^ 
2 + 2*c*d*x + c^2 + 1)^2 + 12*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d 
- 12*arctan(d*x + c)*arctan((d^2*x + c*d)/d)^2/d + 4*arctan((d^2*x + c*d)/ 
d)^3/d - 128*d*integrate(1/16*x*arctan(d*x + c)/(d^2*x^2 + 2*c*d*x + c^2 + 
 1), x) + 16*integrate(1/16*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2/(d^2*x^2 + 
2*c*d*x + c^2 + 1), x))*b^2 + a^2*x + (2*(d*x + c)*arctan(d*x + c) - log(( 
d*x + c)^2 + 1))*a*b/d
 

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+b \arctan (c+d x))^2 \, dx=\frac {\mathit {atan} \left (d x +c \right )^{2} b^{2} d x +2 \mathit {atan} \left (d x +c \right ) a b c +2 \mathit {atan} \left (d x +c \right ) a b d x -2 \left (\int \frac {\mathit {atan} \left (d x +c \right ) x}{d^{2} x^{2}+2 c d x +c^{2}+1}d x \right ) b^{2} d^{2}-\mathrm {log}\left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) a b +a^{2} d x}{d} \] Input:

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

Output:

(atan(c + d*x)**2*b**2*d*x + 2*atan(c + d*x)*a*b*c + 2*atan(c + d*x)*a*b*d 
*x - 2*int((atan(c + d*x)*x)/(c**2 + 2*c*d*x + d**2*x**2 + 1),x)*b**2*d**2 
 - log(c**2 + 2*c*d*x + d**2*x**2 + 1)*a*b + a**2*d*x)/d