3.25 \(\int x^4 (a+b \tan ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=271 \[ \frac{3 i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^5}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{10 c^5}+\frac{b^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{10 c^2}-\frac{9 a b^2 x}{10 c^4}+\frac{3 b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{10 c^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{5 c^5}+\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^5}+\frac{3 b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{3 b x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c}-\frac{b^3 x^2}{20 c^3}+\frac{b^3 \log \left (c^2 x^2+1\right )}{2 c^5}-\frac{9 b^3 x \tan ^{-1}(c x)}{10 c^4} \]

[Out]

(-9*a*b^2*x)/(10*c^4) - (b^3*x^2)/(20*c^3) - (9*b^3*x*ArcTan[c*x])/(10*c^4) + (b^2*x^3*(a + b*ArcTan[c*x]))/(1
0*c^2) + (9*b*(a + b*ArcTan[c*x])^2)/(20*c^5) + (3*b*x^2*(a + b*ArcTan[c*x])^2)/(10*c^3) - (3*b*x^4*(a + b*Arc
Tan[c*x])^2)/(20*c) + ((I/5)*(a + b*ArcTan[c*x])^3)/c^5 + (x^5*(a + b*ArcTan[c*x])^3)/5 + (3*b*(a + b*ArcTan[c
*x])^2*Log[2/(1 + I*c*x)])/(5*c^5) + (b^3*Log[1 + c^2*x^2])/(2*c^5) + (((3*I)/5)*b^2*(a + b*ArcTan[c*x])*PolyL
og[2, 1 - 2/(1 + I*c*x)])/c^5 + (3*b^3*PolyLog[3, 1 - 2/(1 + I*c*x)])/(10*c^5)

________________________________________________________________________________________

Rubi [A]  time = 0.755895, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 11, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.786, Rules used = {4852, 4916, 266, 43, 4846, 260, 4884, 4920, 4854, 4994, 6610} \[ \frac{3 i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^5}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{10 c^5}+\frac{b^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{10 c^2}-\frac{9 a b^2 x}{10 c^4}+\frac{3 b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{10 c^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{5 c^5}+\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^5}+\frac{3 b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{3 b x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c}-\frac{b^3 x^2}{20 c^3}+\frac{b^3 \log \left (c^2 x^2+1\right )}{2 c^5}-\frac{9 b^3 x \tan ^{-1}(c x)}{10 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcTan[c*x])^3,x]

[Out]

(-9*a*b^2*x)/(10*c^4) - (b^3*x^2)/(20*c^3) - (9*b^3*x*ArcTan[c*x])/(10*c^4) + (b^2*x^3*(a + b*ArcTan[c*x]))/(1
0*c^2) + (9*b*(a + b*ArcTan[c*x])^2)/(20*c^5) + (3*b*x^2*(a + b*ArcTan[c*x])^2)/(10*c^3) - (3*b*x^4*(a + b*Arc
Tan[c*x])^2)/(20*c) + ((I/5)*(a + b*ArcTan[c*x])^3)/c^5 + (x^5*(a + b*ArcTan[c*x])^3)/5 + (3*b*(a + b*ArcTan[c
*x])^2*Log[2/(1 + I*c*x)])/(5*c^5) + (b^3*Log[1 + c^2*x^2])/(2*c^5) + (((3*I)/5)*b^2*(a + b*ArcTan[c*x])*PolyL
og[2, 1 - 2/(1 + I*c*x)])/c^5 + (3*b^3*PolyLog[3, 1 - 2/(1 + I*c*x)])/(10*c^5)

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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 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 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 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 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^4 \left (a+b \tan ^{-1}(c x)\right )^3 \, dx &=\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{1}{5} (3 b c) \int \frac{x^5 \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3-\frac{(3 b) \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{5 c}+\frac{(3 b) \int \frac{x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{5 c}\\ &=-\frac{3 b x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{1}{10} \left (3 b^2\right ) \int \frac{x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx+\frac{(3 b) \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{5 c^3}-\frac{(3 b) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{5 c^3}\\ &=\frac{3 b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{10 c^3}-\frac{3 b x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{(3 b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{i-c x} \, dx}{5 c^4}+\frac{\left (3 b^2\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{10 c^2}-\frac{\left (3 b^2\right ) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{10 c^2}-\frac{\left (3 b^2\right ) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c^2}\\ &=\frac{b^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{10 c^2}+\frac{3 b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{10 c^3}-\frac{3 b x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{5 c^5}-\frac{\left (3 b^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{10 c^4}+\frac{\left (3 b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{10 c^4}-\frac{\left (3 b^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{5 c^4}+\frac{\left (3 b^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{5 c^4}-\frac{\left (6 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}{5 c^4}-\frac{b^3 \int \frac{x^3}{1+c^2 x^2} \, dx}{10 c}\\ &=-\frac{9 a b^2 x}{10 c^4}+\frac{b^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{10 c^2}+\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^5}+\frac{3 b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{10 c^3}-\frac{3 b x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{5 c^5}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{5 c^5}-\frac{\left (3 i b^3\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{5 c^4}-\frac{\left (3 b^3\right ) \int \tan ^{-1}(c x) \, dx}{10 c^4}-\frac{\left (3 b^3\right ) \int \tan ^{-1}(c x) \, dx}{5 c^4}-\frac{b^3 \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )}{20 c}\\ &=-\frac{9 a b^2 x}{10 c^4}-\frac{9 b^3 x \tan ^{-1}(c x)}{10 c^4}+\frac{b^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{10 c^2}+\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^5}+\frac{3 b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{10 c^3}-\frac{3 b x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{5 c^5}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{5 c^5}+\frac{3 b^3 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{10 c^5}+\frac{\left (3 b^3\right ) \int \frac{x}{1+c^2 x^2} \, dx}{10 c^3}+\frac{\left (3 b^3\right ) \int \frac{x}{1+c^2 x^2} \, dx}{5 c^3}-\frac{b^3 \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{20 c}\\ &=-\frac{9 a b^2 x}{10 c^4}-\frac{b^3 x^2}{20 c^3}-\frac{9 b^3 x \tan ^{-1}(c x)}{10 c^4}+\frac{b^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{10 c^2}+\frac{9 b \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^5}+\frac{3 b x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{10 c^3}-\frac{3 b x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{5 c^5}+\frac{1}{5} x^5 \left (a+b \tan ^{-1}(c x)\right )^3+\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{5 c^5}+\frac{b^3 \log \left (1+c^2 x^2\right )}{2 c^5}+\frac{3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{5 c^5}+\frac{3 b^3 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{10 c^5}\\ \end{align*}

Mathematica [A]  time = 0.8301, size = 396, normalized size = 1.46 \[ \frac{-12 i b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right ) \left (a+b \tan ^{-1}(c x)\right )+6 b^3 \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )-3 a^2 b c^4 x^4+6 a^2 b c^2 x^2-6 a^2 b \log \left (c^2 x^2+1\right )+12 a^2 b c^5 x^5 \tan ^{-1}(c x)+4 a^3 c^5 x^5+2 a b^2 c^3 x^3+12 a b^2 c^5 x^5 \tan ^{-1}(c x)^2-6 a b^2 c^4 x^4 \tan ^{-1}(c x)+12 a b^2 c^2 x^2 \tan ^{-1}(c x)-18 a b^2 c x-12 i a b^2 \tan ^{-1}(c x)^2+18 a b^2 \tan ^{-1}(c x)+24 a b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-b^3 c^2 x^2+10 b^3 \log \left (c^2 x^2+1\right )+4 b^3 c^5 x^5 \tan ^{-1}(c x)^3-3 b^3 c^4 x^4 \tan ^{-1}(c x)^2+2 b^3 c^3 x^3 \tan ^{-1}(c x)+6 b^3 c^2 x^2 \tan ^{-1}(c x)^2-18 b^3 c x \tan ^{-1}(c x)-4 i b^3 \tan ^{-1}(c x)^3+9 b^3 \tan ^{-1}(c x)^2+12 b^3 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-b^3}{20 c^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4*(a + b*ArcTan[c*x])^3,x]

[Out]

(-b^3 - 18*a*b^2*c*x + 6*a^2*b*c^2*x^2 - b^3*c^2*x^2 + 2*a*b^2*c^3*x^3 - 3*a^2*b*c^4*x^4 + 4*a^3*c^5*x^5 + 18*
a*b^2*ArcTan[c*x] - 18*b^3*c*x*ArcTan[c*x] + 12*a*b^2*c^2*x^2*ArcTan[c*x] + 2*b^3*c^3*x^3*ArcTan[c*x] - 6*a*b^
2*c^4*x^4*ArcTan[c*x] + 12*a^2*b*c^5*x^5*ArcTan[c*x] - (12*I)*a*b^2*ArcTan[c*x]^2 + 9*b^3*ArcTan[c*x]^2 + 6*b^
3*c^2*x^2*ArcTan[c*x]^2 - 3*b^3*c^4*x^4*ArcTan[c*x]^2 + 12*a*b^2*c^5*x^5*ArcTan[c*x]^2 - (4*I)*b^3*ArcTan[c*x]
^3 + 4*b^3*c^5*x^5*ArcTan[c*x]^3 + 24*a*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 12*b^3*ArcTan[c*x]^2*
Log[1 + E^((2*I)*ArcTan[c*x])] - 6*a^2*b*Log[1 + c^2*x^2] + 10*b^3*Log[1 + c^2*x^2] - (12*I)*b^2*(a + b*ArcTan
[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 6*b^3*PolyLog[3, -E^((2*I)*ArcTan[c*x])])/(20*c^5)

________________________________________________________________________________________

Maple [C]  time = 3.677, size = 3053, normalized size = 11.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctan(c*x))^3,x)

[Out]

-3/80/c^2*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)+I)*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*x^3-3/10/c^5*a^2*b*ln(c^2*x^2+1)+3/5/c^5*b^3*arctan(c*x)^2*ln(2)+I/c^5*b^3*arctan(c*x)+9/
10/c^5*a*b^2*arctan(c*x)+3/5*x^5*a^2*b*arctan(c*x)+3/5*x^5*a*b^2*arctan(c*x)^2-1/5*I/c^5*b^3*arctan(c*x)^3+9/2
0/c^5*b^3*arctan(c*x)^2-1/c^5*b^3*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+3/10/c^5*b^3*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1
))+1/5*x^5*b^3*arctan(c*x)^3+9/160*I/c^3*b^3*arctan(c*x)^2*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*x^2-3/20*I/c^5*b^3*arctan(c*x)^2*Pi*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))^2-3/160/c^2*b^3*arctan(c*x)^2*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)*x^3+3/80/c^2*b^3*arctan(c*x)^2*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*x^3+3/160/c^2*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)+I)^2*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)*x^3+9/160/c^4*b^3*arctan(c*x)^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*c
sgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*x-9/80/c^4*b^3*arctan(c*x)^2*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*x-9/160/c^4*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)+I)^2*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)*x+9/80/c^4*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(
c^2*x^2+1)+I)*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*x+1/5*x^5*a^3-9/10*a*b^2*x/c^4
-9/10*b^3*x*arctan(c*x)/c^4-3/80*I/c^5*b^3*arctan(c*x)^2*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+3/20*I/c^5*b^3*arctan(c*x)^2*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+I*c*x)^2/(c^2*x^2+1)+1)^2)^2+3/20*I/c^5*b^3*arctan(c*x)^2*Pi*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))+3/10*I/c^5*b^3*arctan(c*x)^2*Pi*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))-9/160*I/c^3*b^3*arctan(c*x)^2*Pi*csgn(I*((1+I*c*x)
^2/(c^2*x^2+1)+1)^2)^3*x^2-21/80*I/c^5*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)+I)*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+3/160*I/c^5*b^3*arctan(c*x)^2*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)+21/160*I/c^5*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2
+1)+I)^2*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)+3/5/c^3*a*b^2*x^2*arctan(c*x)-3/5*I/c
^5*b^3*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+3/20*I/c^5*a*b^2*ln(c*x-I)^2+3/10*I/c^5*a*b^2*dilog(-1/
2*I*(c*x+I))-3/20*I/c^5*a*b^2*ln(c*x+I)^2-3/10*I/c^5*a*b^2*dilog(1/2*I*(c*x-I))-1/20*b^3*x^2/c^3-3/5/c^5*a*b^2
*arctan(c*x)*ln(c^2*x^2+1)-3/10/c*a*b^2*x^4*arctan(c*x)+9/80*I/c^3*b^3*arctan(c*x)^2*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*x^2+9/160*I/c^3*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/
(c^2*x^2+1)+I)^2*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)*x^2-9/80*I/c^3*b^3*arctan(c*x
)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)+I)*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*x^2
-3/20*I/c^5*b^3*arctan(c*x)^2*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+I*c*
x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))-9/160*I/c^3*b^3*arctan(c*x)^2*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)*x^2+1/10*a*b^2*x^3/c^2-1/20/c^5*b^3-3/20/c*x^4*a^2*b+3/1
0/c^3*b*a^2*x^2-3/10/c^5*b^3*arctan(c*x)^2*ln(c^2*x^2+1)+3/5/c^5*b^3*arctan(c*x)^2*ln((1+I*c*x)/(c^2*x^2+1)^(1
/2))-3/20/c*b^3*arctan(c*x)^2*x^4+3/10/c^3*b^3*arctan(c*x)^2*x^2+1/10/c^2*b^3*arctan(c*x)*x^3+9/160/c^4*b^3*ar
ctan(c*x)^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*x-3/160/c^2*b^3*arctan(c*x)^2*Pi*csgn(I*((1+I*c*x)^2/(c
^2*x^2+1)+1)^2)^3*x^3+3/160/c^2*b^3*arctan(c*x)^2*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*x^3-9/160/c^4*b^3*arctan(c*x)^2*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*
x-3/20*I/c^5*b^3*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3+21/160*I/c^5*b^3*arctan(c*x)^2*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)^3-3/10*I/c^5*a*b^2*ln(c^2*x^2+1)*ln(c*x-I)+3/10*I/c^5*
a*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))+3/10*I/c^5*a*b^2*ln(c^2*x^2+1)*ln(c*x+I)-3/10*I/c^5*a*b^2*ln(c*x+I)*ln(1/2*
I*(c*x-I))+3/160*I/c^5*b^3*arctan(c*x)^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3-3/20*I/c^5*b^3*arctan(c*x)
^2*Pi*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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{40} \, b^{3} x^{5} \arctan \left (c x\right )^{3} - \frac{3}{160} \, b^{3} x^{5} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2} + \frac{1}{5} \, a^{3} x^{5} + \frac{3}{20} \,{\left (4 \, x^{5} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} a^{2} b + \int \frac{12 \, b^{3} c^{2} x^{6} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right ) + 140 \,{\left (b^{3} c^{2} x^{6} + b^{3} x^{4}\right )} \arctan \left (c x\right )^{3} + 12 \,{\left (40 \, a b^{2} c^{2} x^{6} - b^{3} c x^{5} + 40 \, a b^{2} x^{4}\right )} \arctan \left (c x\right )^{2} + 3 \,{\left (b^{3} c x^{5} + 5 \,{\left (b^{3} c^{2} x^{6} + b^{3} x^{4}\right )} \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{160 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))^3,x, algorithm="maxima")

[Out]

1/40*b^3*x^5*arctan(c*x)^3 - 3/160*b^3*x^5*arctan(c*x)*log(c^2*x^2 + 1)^2 + 1/5*a^3*x^5 + 3/20*(4*x^5*arctan(c
*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*a^2*b + integrate(1/160*(12*b^3*c^2*x^6*arctan(c*x)*
log(c^2*x^2 + 1) + 140*(b^3*c^2*x^6 + b^3*x^4)*arctan(c*x)^3 + 12*(40*a*b^2*c^2*x^6 - b^3*c*x^5 + 40*a*b^2*x^4
)*arctan(c*x)^2 + 3*(b^3*c*x^5 + 5*(b^3*c^2*x^6 + b^3*x^4)*arctan(c*x))*log(c^2*x^2 + 1)^2)/(c^2*x^2 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{3} x^{4} \arctan \left (c x\right )^{3} + 3 \, a b^{2} x^{4} \arctan \left (c x\right )^{2} + 3 \, a^{2} b x^{4} \arctan \left (c x\right ) + a^{3} x^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))^3,x, algorithm="fricas")

[Out]

integral(b^3*x^4*arctan(c*x)^3 + 3*a*b^2*x^4*arctan(c*x)^2 + 3*a^2*b*x^4*arctan(c*x) + a^3*x^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (a + b \operatorname{atan}{\left (c x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*atan(c*x))**3,x)

[Out]

Integral(x**4*(a + b*atan(c*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arctan \left (c x\right ) + a\right )}^{3} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))^3,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)^3*x^4, x)