3.27 \(\int x^2 (a+b \tan ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=206 \[ -\frac {i b^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3}+\frac {a b^2 x}{c^2}-\frac {b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c^3}-\frac {b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3}+\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {b^3 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c^3}+\frac {b^3 x \tan ^{-1}(c x)}{c^2}-\frac {b^3 \log \left (c^2 x^2+1\right )}{2 c^3} \]

[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

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Rubi [A]  time = 0.43, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {4852, 4916, 4846, 260, 4884, 4920, 4854, 4994, 6610} \[ -\frac {i b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3}-\frac {b^3 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^3}+\frac {a b^2 x}{c^2}-\frac {b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c^3}-\frac {b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3}-\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}+\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {b^3 \log \left (c^2 x^2+1\right )}{2 c^3}+\frac {b^3 x \tan ^{-1}(c x)}{c^2} \]

Antiderivative was successfully verified.

[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 260

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 4846

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

Rule 4852

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[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 4884

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 4916

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 4920

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] - 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 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[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 6610

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*} \int x^2 \left (a+b \tan ^{-1}(c x)\right )^3 \, dx &=\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3-(b c) \int \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx\\ &=\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {b \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c}+\frac {b \int \frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{c}\\ &=-\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3+b^2 \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac {b \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{i-c x} \, dx}{c^2}\\ &=-\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3}+\frac {b^2 \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^2}-\frac {b^2 \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^2}+\frac {\left (2 b^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \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 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3}-\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3}-\frac {i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3}+\frac {\left (i b^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2}+\frac {b^3 \int \tan ^{-1}(c x) \, dx}{c^2}\\ &=\frac {a b^2 x}{c^2}+\frac {b^3 x \tan ^{-1}(c x)}{c^2}-\frac {b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3}-\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3}-\frac {i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3}-\frac {b^3 \text {Li}_3\left (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 \tan ^{-1}(c x)}{c^2}-\frac {b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3}-\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{3 c^3}+\frac {1}{3} x^3 \left (a+b \tan ^{-1}(c x)\right )^3-\frac {b \left (a+b \tan ^{-1}(c x)\right )^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 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3}-\frac {b^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 269, normalized size = 1.31 \[ \frac {2 a^3 c^3 x^3+6 a^2 b c^3 x^3 \tan ^{-1}(c x)-3 a^2 b c^2 x^2+3 a^2 b \log \left (c^2 x^2+1\right )+6 a b^2 \left (\left (c^3 x^3+i\right ) \tan ^{-1}(c x)^2-\tan ^{-1}(c x) \left (c^2 x^2+2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+1\right )+i \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+c x\right )+b^3 \left (2 c^3 x^3 \tan ^{-1}(c x)^3-3 \log \left (c^2 x^2+1\right )-3 c^2 x^2 \tan ^{-1}(c x)^2+6 i \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-3 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+2 i \tan ^{-1}(c x)^3-3 \tan ^{-1}(c x)^2+6 c x \tan ^{-1}(c x)-6 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{6 c^3} \]

Warning: Unable to verify antiderivative.

[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)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{3} x^{2} \arctan \left (c x\right )^{3} + 3 \, a b^{2} x^{2} \arctan \left (c x\right )^{2} + 3 \, a^{2} b x^{2} \arctan \left (c x\right ) + a^{3} x^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [C]  time = 2.53, size = 2020, normalized size = 9.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*x*arctan(c*x)/c^2+a*b^2*x/c^2+1/4*I/c^3*b^3*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*
csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*arctan(c*x)^2*Pi-1/2/c^3*b^3*polylog(3,-
(1+I*c*x)^2/(c^2*x^2+1))-1/2/c^3*b^3*arctan(c*x)^2+1/3*x^3*b^3*arctan(c*x)^3+1/c^3*b^3*ln((1+I*c*x)^2/(c^2*x^2
+1)+1)-1/8/c^2*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*arctan(c*x)^2*x+1/8/c^2*b^3*Pi*csgn(I*(1+I*c*x)^
4/(c^2*x^2+1)^2+2*I*(1+I*c*x)^2/(c^2*x^2+1)+I)^3*arctan(c*x)^2*x+1/4*I/c^3*b^3*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/
((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*arctan(c*x)^2*Pi+1/4*I/c^3*b^3*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*arctan(c*x)^
2*Pi-1/8*I/c^3*b^3*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*arctan(c*x)^2*Pi-1/8*I/c^3*b^3*csgn(I*(1+I*c*x)^4/(
c^2*x^2+1)^2+2*I*(1+I*c*x)^2/(c^2*x^2+1)+I)^3*arctan(c*x)^2*Pi+1/2*I/c^3*a*b^2*ln(c*x-I)*ln(c^2*x^2+1)-1/2*I/c
^3*a*b^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))-1/2*I/c^3*a*b^2*ln(I+c*x)*ln(c^2*x^2+1)+1/2*I/c^3*a*b^2*ln(I+c*x)*ln(1/2
*I*(c*x-I))-1/c*a*b^2*x^2*arctan(c*x)+I/c^3*b^3*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+1/c^3*a*b^2*ar
ctan(c*x)*ln(c^2*x^2+1)-1/4*I/c^3*a*b^2*ln(c*x-I)^2-1/2*I/c^3*a*b^2*dilog(-1/2*I*(I+c*x))+1/4*I/c^3*a*b^2*ln(I
+c*x)^2+1/2*I/c^3*a*b^2*dilog(1/2*I*(c*x-I))-1/8/c^2*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*((1+I
*c*x)^2/(c^2*x^2+1)+1)^2)*arctan(c*x)^2*x+1/4/c^2*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)
^2/(c^2*x^2+1)+1)^2)^2*arctan(c*x)^2*x-1/4/c^2*b^3*Pi*csgn(I*(1+I*c*x)^4/(c^2*x^2+1)^2+2*I*(1+I*c*x)^2/(c^2*x^
2+1)+I)^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)+I)*arctan(c*x)^2*x+1/8/c^2*b^3*Pi*csgn(I*(1+I*c*x)^4/(c^2*x^2+1)^2+2*
I*(1+I*c*x)^2/(c^2*x^2+1)+I)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)+I)^2*arctan(c*x)^2*x-1/4*I/c^3*b^3*csgn(I*(1+I*c*x
)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*arctan(c*x)^2*Pi-1/4*I/c^3*b^
3*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*arctan
(c*x)^2*Pi+1/4*I/c^3*b^3*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*arctan(c*x)^2*P
i-1/2*I/c^3*b^3*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*arctan(c*x)^2*Pi+1/4*I/c
^3*b^3*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2*Pi-1/8*I/c^3*
b^3*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2*Pi+1/4*I/c^3*b^3
*csgn(I*(1+I*c*x)^4/(c^2*x^2+1)^2+2*I*(1+I*c*x)^2/(c^2*x^2+1)+I)^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)+I)*arctan(c*
x)^2*Pi-1/8*I/c^3*b^3*csgn(I*(1+I*c*x)^4/(c^2*x^2+1)^2+2*I*(1+I*c*x)^2/(c^2*x^2+1)+I)*csgn(I*(1+I*c*x)^2/(c^2*
x^2+1)+I)^2*arctan(c*x)^2*Pi-I/c^3*b^3*arctan(c*x)+1/3*I/c^3*b^3*arctan(c*x)^3-1/2/c*b*x^2*a^2+b^2*x^3*a*arcta
n(c*x)^2+a^2*b*x^3*arctan(c*x)-1/c^3*b^3*arctan(c*x)^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2/c^3*b^3*arctan(c*x)
^2*ln(c^2*x^2+1)-1/c^3*a*b^2*arctan(c*x)+1/2/c^3*a^2*b*ln(c^2*x^2+1)-1/c^3*b^3*ln(2)*arctan(c*x)^2-1/2/c*b^3*a
rctan(c*x)^2*x^2+1/3*a^3*x^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{24} \, b^{3} x^{3} \arctan \left (c x\right )^{3} - \frac {1}{32} \, b^{3} x^{3} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2} + \frac {1}{3} \, a^{3} x^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} a^{2} b + \int \frac {4 \, b^{3} c^{2} x^{4} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right ) + 28 \, {\left (b^{3} c^{2} x^{4} + b^{3} x^{2}\right )} \arctan \left (c x\right )^{3} + 4 \, {\left (24 \, a b^{2} c^{2} x^{4} - b^{3} c x^{3} + 24 \, a b^{2} x^{2}\right )} \arctan \left (c x\right )^{2} + {\left (b^{3} c x^{3} + 3 \, {\left (b^{3} c^{2} x^{4} + b^{3} x^{2}\right )} \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{32 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atan(c*x))^3,x)

[Out]

int(x^2*(a + b*atan(c*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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