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

Optimal result

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

[Out]

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*arc
tan(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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {4946, 5036, 4930, 266, 5004, 5040, 4964, 5114, 6745} \[ \int x^2 (a+b \arctan (c x))^3 \, dx=-\frac {i b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{c^3}-\frac {b (a+b \arctan (c x))^2}{2 c^3}-\frac {i (a+b \arctan (c x))^3}{3 c^3}-\frac {b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c^3}+\frac {1}{3} x^3 (a+b \arctan (c x))^3-\frac {b x^2 (a+b \arctan (c x))^2}{2 c}+\frac {a b^2 x}{c^2}+\frac {b^3 x \arctan (c x)}{c^2}-\frac {b^3 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{2 c^3}-\frac {b^3 \log \left (c^2 x^2+1\right )}{2 c^3} \]

[In]

Int[x^2*(a + b*ArcTan[c*x])^3,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[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 4946

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] - Dist[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 4964

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] + Dist[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 5004

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> 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 5036

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*e*(p + 1))), x] - Dist[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 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x] + Dist[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 6745

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 (a+b \arctan (c x))^3-(b c) \int \frac {x^3 (a+b \arctan (c x))^2}{1+c^2 x^2} \, dx \\ & = \frac {1}{3} x^3 (a+b \arctan (c x))^3-\frac {b \int x (a+b \arctan (c x))^2 \, dx}{c}+\frac {b \int \frac {x (a+b \arctan (c x))^2}{1+c^2 x^2} \, dx}{c} \\ & = -\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+b^2 \int \frac {x^2 (a+b \arctan (c x))}{1+c^2 x^2} \, dx-\frac {b \int \frac {(a+b \arctan (c x))^2}{i-c x} \, dx}{c^2} \\ & = -\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^2 \int (a+b \arctan (c x)) \, dx}{c^2}-\frac {b^2 \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx}{c^2}+\frac {\left (2 b^2\right ) \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2} \\ & = \frac {a b^2 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 {i b^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^3}+\frac {\left (i b^3\right ) \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2}+\frac {b^3 \int \arctan (c x) \, dx}{c^2} \\ & = \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 {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}-\frac {b^3 \int \frac {x}{1+c^2 x^2} \, dx}{c} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(-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*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[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)

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

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

[In]

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

[Out]

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)*csgn(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*arctan(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)))*csg
n(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)-12*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*a
rctan(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))-ln(c*x-I)*ln(-1/2*I*(c*x+I)))-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))-ln(c*x+I)*ln(1/2*I*(c*x-I))))+3*a^2*b*(1/3*c^3*x^3*arctan(c*x)-1/6*c^2*x
^2+1/6*ln(c^2*x^2+1)))

Fricas [F]

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

[In]

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

[Out]

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*arctan(c*x) + a^3*x^2, x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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) + 2
8*(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)*arctan(c*x))*log(c^2*x^2 + 1)^2)/(c^2*x^2 + 1), x)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

int(x^2*(a + b*atan(c*x))^3,x)

[Out]

int(x^2*(a + b*atan(c*x))^3, x)