3.1.27 \(\int x^2 (a+b \arctan (c x))^3 \, dx\) [27]

3.1.27.1 Optimal result

Integrand size = 14, antiderivative size = 206 \[ \int x^2 (a+b \arctan (c x))^3 \, dx=\frac {a b^2 x}{c^2}+\frac {b^3 x \arctan (c x)}{c^2}-\frac {b (a+b \arctan (c x))^2}{2 c^3}-\frac {b x^2 (a+b \arctan (c x))^2}{2 c}-\frac {i (a+b \arctan (c x))^3}{3 c^3}+\frac {1}{3} x^3 (a+b \arctan (c x))^3-\frac {b (a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^3}-\frac {b^3 \log \left (1+c^2 x^2\right )}{2 c^3}-\frac {i b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3}-\frac {b^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^3} \]

output

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

3.1.27.2 Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.31 \[ \int x^2 (a+b \arctan (c x))^3 \, dx=\frac {-3 a^2 b c^2 x^2+2 a^3 c^3 x^3+6 a^2 b c^3 x^3 \arctan (c x)+3 a^2 b \log \left (1+c^2 x^2\right )+6 a b^2 \left (c x+\left (i+c^3 x^3\right ) \arctan (c x)^2-\arctan (c x) \left (1+c^2 x^2+2 \log \left (1+e^{2 i \arctan (c x)}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )\right )+b^3 \left (6 c x \arctan (c x)-3 \arctan (c x)^2-3 c^2 x^2 \arctan (c x)^2+2 i \arctan (c x)^3+2 c^3 x^3 \arctan (c x)^3-6 \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-3 \log \left (1+c^2 x^2\right )+6 i \arctan (c x) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-3 \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )}{6 c^3} \]

input

Integrate[x^2*(a + b*ArcTan[c*x])^3,x]
 

output

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

3.1.27.3 Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5361, 5451, 5361, 5451, 2009, 5419, 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.

\(\displaystyle \int x^2 (a+b \arctan (c x))^3 \, dx\)

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5451

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

\(\Big \downarrow \) 5361

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

\(\Big \downarrow \) 5451

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 5419

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

\(\Big \downarrow \) 5455

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

\(\Big \downarrow \) 5379

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

\(\Big \downarrow \) 5529

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

\(\Big \downarrow \) 7164

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

input

Int[x^2*(a + b*ArcTan[c*x])^3,x]
 

output

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

3.1.27.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 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]
 

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

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

Time = 4.19 (sec) , antiderivative size = 1088, normalized size of antiderivative = 5.28

\[\text {Expression too large to display}\]

input

int(x^2*(a+b*arctan(c*x))^3,x)
 

output

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

3.1.27.5 Fricas [F]

\[ \int x^2 (a+b \arctan (c x))^3 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{3} x^{2} \,d x } \]

input

integrate(x^2*(a+b*arctan(c*x))^3,x, algorithm="fricas")
 

output

integral(b^3*x^2*arctan(c*x)^3 + 3*a*b^2*x^2*arctan(c*x)^2 + 3*a^2*b*x^2*a 
rctan(c*x) + a^3*x^2, x)
 

3.1.27.6 Sympy [F]

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

input

integrate(x**2*(a+b*atan(c*x))**3,x)
 

output

Integral(x**2*(a + b*atan(c*x))**3, x)
 

3.1.27.7 Maxima [F]

\[ \int x^2 (a+b \arctan (c x))^3 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{3} x^{2} \,d x } \]

input

integrate(x^2*(a+b*arctan(c*x))^3,x, algorithm="maxima")
 

output

1/24*b^3*x^3*arctan(c*x)^3 - 1/32*b^3*x^3*arctan(c*x)*log(c^2*x^2 + 1)^2 + 
 1/3*a^3*x^3 + 1/2*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4) 
)*a^2*b + integrate(1/32*(4*b^3*c^2*x^4*arctan(c*x)*log(c^2*x^2 + 1) + 28* 
(b^3*c^2*x^4 + b^3*x^2)*arctan(c*x)^3 + 4*(24*a*b^2*c^2*x^4 - b^3*c*x^3 + 
24*a*b^2*x^2)*arctan(c*x)^2 + (b^3*c*x^3 + 3*(b^3*c^2*x^4 + b^3*x^2)*arcta 
n(c*x))*log(c^2*x^2 + 1)^2)/(c^2*x^2 + 1), x)
 

3.1.27.8 Giac [F]

\[ \int x^2 (a+b \arctan (c x))^3 \, dx=\int { {\left (b \arctan \left (c x\right ) + a\right )}^{3} x^{2} \,d x } \]

input

integrate(x^2*(a+b*arctan(c*x))^3,x, algorithm="giac")
 

output

sage0*x
 

3.1.27.9 Mupad [F(-1)]

Timed out. \[ \int x^2 (a+b \arctan (c x))^3 \, dx=\int x^2\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3 \,d x \]

input

int(x^2*(a + b*atan(c*x))^3,x)
 

output

int(x^2*(a + b*atan(c*x))^3, x)