\(\int (a+b \arctan (c+d x))^3 \, dx\) [46]

Optimal result

Integrand size = 12, antiderivative size = 143 \[ \int (a+b \arctan (c+d x))^3 \, dx=\frac {i (a+b \arctan (c+d x))^3}{d}+\frac {(c+d x) (a+b \arctan (c+d x))^3}{d}+\frac {3 b (a+b \arctan (c+d x))^2 \log \left (\frac {2}{1+i (c+d x)}\right )}{d}+\frac {3 i b^2 (a+b \arctan (c+d x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i (c+d x)}\right )}{d}+\frac {3 b^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i (c+d x)}\right )}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.48 \[ \int (a+b \arctan (c+d x))^3 \, dx=\frac {2 a^3 (c+d x)+6 a^2 b (c+d x) \arctan (c+d x)-3 a^2 b \log \left (1+(c+d x)^2\right )+6 a b^2 \left (\arctan (c+d x) \left ((-i+c+d x) \arctan (c+d x)+2 \log \left (1+e^{2 i \arctan (c+d x)}\right )\right )-i \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c+d x)}\right )\right )+2 b^3 \left (\arctan (c+d x)^2 \left ((-i+c+d x) \arctan (c+d x)+3 \log \left (1+e^{2 i \arctan (c+d x)}\right )\right )-3 i \arctan (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c+d x)}\right )+\frac {3}{2} \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c+d x)}\right )\right )}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99, 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, 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.

\(\displaystyle \int (a+b \arctan (c+d x))^3 \, dx\)

\(\Big \downarrow \) 5562

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

\(\Big \downarrow \) 5345

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

\(\Big \downarrow \) 5455

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

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 5529

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

\(\Big \downarrow \) 7164

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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 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]
 

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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (136 ) = 272\).

Time = 0.82 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.95

method	result	size
derivativedivides	\(\frac {\left (d x +c \right ) a^{3}+b^{3} \left (\arctan \left (d x +c \right )^{3} \left (d x +c +i\right )-2 i \arctan \left (d x +c \right )^{3}+3 \arctan \left (d x +c \right )^{2} \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-3 i \arctan \left (d x +c \right ) \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )+\frac {3 \operatorname {polylog}\left (3, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )}{2}\right )+3 a \,b^{2} \left (\arctan \left (d x +c \right )^{2} \left (d x +c +i\right )+2 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-2 i \arctan \left (d x +c \right )^{2}-i \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )\right )+3 a^{2} b \left (\left (d x +c \right ) \arctan \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{d}\)	\(279\)
default	\(\frac {\left (d x +c \right ) a^{3}+b^{3} \left (\arctan \left (d x +c \right )^{3} \left (d x +c +i\right )-2 i \arctan \left (d x +c \right )^{3}+3 \arctan \left (d x +c \right )^{2} \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-3 i \arctan \left (d x +c \right ) \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )+\frac {3 \operatorname {polylog}\left (3, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )}{2}\right )+3 a \,b^{2} \left (\arctan \left (d x +c \right )^{2} \left (d x +c +i\right )+2 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-2 i \arctan \left (d x +c \right )^{2}-i \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )\right )+3 a^{2} b \left (\left (d x +c \right ) \arctan \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{d}\)	\(279\)
parts	\(a^{3} x +\frac {b^{3} \left (\arctan \left (d x +c \right )^{3} \left (d x +c +i\right )-2 i \arctan \left (d x +c \right )^{3}+3 \arctan \left (d x +c \right )^{2} \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-3 i \arctan \left (d x +c \right ) \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )+\frac {3 \operatorname {polylog}\left (3, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )}{2}\right )}{d}+\frac {3 a^{2} b \left (\left (d x +c \right ) \arctan \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {3 a \,b^{2} \left (\arctan \left (d x +c \right )^{2} \left (d x +c +i\right )+2 \arctan \left (d x +c \right ) \ln \left (1+\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )-2 i \arctan \left (d x +c \right )^{2}-i \operatorname {polylog}\left (2, -\frac {\left (1+i \left (d x +c \right )\right )^{2}}{1+\left (d x +c \right )^{2}}\right )\right )}{d}\)	\(280\)

Input:

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

Output:

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

Fricas [F]

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

integrate((a+b*arctan(d*x+c))^3,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*arcta 
n(d*x + c) + a^3, x)
 

Sympy [F]

\[ \int (a+b \arctan (c+d x))^3 \, dx=\int \left (a + b \operatorname {atan}{\left (c + d x \right )}\right )^{3}\, dx \] Input:

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

Output:

Integral((a + b*atan(c + d*x))**3, x)
 

Maxima [F]

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

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

Output:

7/8*b^3*c^2*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 1/8*b^3*x*arctan 
(d*x + c)^3 + 3*a*b^2*c^2*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 3/ 
32*b^3*x*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)^2 - (3*arctan(d* 
x + c)*arctan((d^2*x + c*d)/d)^2/d - arctan((d^2*x + c*d)/d)^3/d)*a*b^2*c^ 
2 - 7/32*(6*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)^2/d - 4*arctan(d*x + 
 c)*arctan((d^2*x + c*d)/d)^3/d + arctan((d^2*x + c*d)/d)^4/d)*b^3*c^2 + 7 
/8*b^3*arctan(d*x + c)^3*arctan((d^2*x + c*d)/d)/d + 28*b^3*d^2*integrate( 
1/32*x^2*arctan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*d^2*i 
ntegrate(1/32*x^2*arctan(d*x + c)*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) + 96*a*b^2*d^2*integrate(1/32*x^2*arctan(d*x 
+ c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 56*b^3*c*d*integrate(1/32*x*arc 
tan(d*x + c)^3/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^3*d^2*integrate(1/ 
32*x^2*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d^2*x^2 + 2*c*d*x 
 + c^2 + 1), x) + 6*b^3*c*d*integrate(1/32*x*arctan(d*x + c)*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) + 192*a*b^2*c*d*in 
tegrate(1/32*x*arctan(d*x + c)^2/(d^2*x^2 + 2*c*d*x + c^2 + 1), x) + 12*b^ 
3*c*d*integrate(1/32*x*arctan(d*x + c)*log(d^2*x^2 + 2*c*d*x + c^2 + 1)/(d 
^2*x^2 + 2*c*d*x + c^2 + 1), x) + 3*b^3*c^2*integrate(1/32*arctan(d*x + c) 
*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) + 3* 
a*b^2*arctan(d*x + c)^2*arctan((d^2*x + c*d)/d)/d - 12*b^3*d*integrate(...
 

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+b \arctan (c+d x))^3 \, dx=\frac {2 \mathit {atan} \left (d x +c \right )^{3} b^{3} d x +6 \mathit {atan} \left (d x +c \right )^{2} a \,b^{2} d x +6 \mathit {atan} \left (d x +c \right ) a^{2} b c +6 \mathit {atan} \left (d x +c \right ) a^{2} b d x -12 \left (\int \frac {\mathit {atan} \left (d x +c \right ) x}{d^{2} x^{2}+2 c d x +c^{2}+1}d x \right ) a \,b^{2} d^{2}-6 \left (\int \frac {\mathit {atan} \left (d x +c \right )^{2} x}{d^{2} x^{2}+2 c d x +c^{2}+1}d x \right ) b^{3} d^{2}-3 \,\mathrm {log}\left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) a^{2} b +2 a^{3} d x}{2 d} \] Input:

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

Output:

(2*atan(c + d*x)**3*b**3*d*x + 6*atan(c + d*x)**2*a*b**2*d*x + 6*atan(c + 
d*x)*a**2*b*c + 6*atan(c + d*x)*a**2*b*d*x - 12*int((atan(c + d*x)*x)/(c** 
2 + 2*c*d*x + d**2*x**2 + 1),x)*a*b**2*d**2 - 6*int((atan(c + d*x)**2*x)/( 
c**2 + 2*c*d*x + d**2*x**2 + 1),x)*b**3*d**2 - 3*log(c**2 + 2*c*d*x + d**2 
*x**2 + 1)*a**2*b + 2*a**3*d*x)/(2*d)