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

Optimal. Leaf size=144 \[ \frac {3 i b^2 \text {Li}_2\left (1-\frac {2}{i c x^2+1}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{2 c}+\frac {1}{2} x^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right )^3+\frac {i \left (a+b \tan ^{-1}\left (c x^2\right )\right )^3}{2 c}+\frac {3 b \log \left (\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{2 c}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{i c x^2+1}\right )}{4 c} \]

[Out]

1/2*I*(a+b*arctan(c*x^2))^3/c+1/2*x^2*(a+b*arctan(c*x^2))^3+3/2*b*(a+b*arctan(c*x^2))^2*ln(2/(1+I*c*x^2))/c+3/
2*I*b^2*(a+b*arctan(c*x^2))*polylog(2,1-2/(1+I*c*x^2))/c+3/4*b^3*polylog(3,1-2/(1+I*c*x^2))/c

________________________________________________________________________________________

Rubi [B]  time = 2.54, antiderivative size = 545, normalized size of antiderivative = 3.78, number of steps used = 82, number of rules used = 23, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.643, Rules used = {5035, 2454, 2389, 2296, 2295, 6715, 2430, 2416, 2396, 2433, 2374, 6589, 2411, 2346, 2301, 6742, 43, 2394, 2393, 2391, 2375, 2317, 2425} \[ -\frac {3 b^2 \text {PolyLog}\left (2,\frac {1}{2} \left (1-i c x^2\right )\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )}{4 c}-\frac {3 b^3 \text {PolyLog}\left (3,\frac {1}{2} \left (1-i c x^2\right )\right )}{4 c}-\frac {3 b^3 \text {PolyLog}\left (3,\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}+\frac {3 b^3 \log \left (1+i c x^2\right ) \text {PolyLog}\left (2,\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}+\frac {3}{16} i b^2 x^2 \log ^2\left (1+i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )+\frac {3 b^2 \log ^2\left (1+i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )}{16 c}+\frac {3 b \left (1-i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {3 b \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {3 b \log \left (\frac {1}{2} \left (1+i c x^2\right )\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )^2}{8 c}+\frac {3}{16} i b x^2 \log \left (1+i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )^2-\frac {3 b \log \left (1+i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 c}+\frac {b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 c}+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{8 c} \]

Warning: Unable to verify antiderivative.

[In]

Int[x*(a + b*ArcTan[c*x^2])^3,x]

[Out]

(3*b*(1 - I*c*x^2)*((2*I)*a - b*Log[1 - I*c*x^2])^2)/(16*c) + (3*b*(1 - I*c*x^2)*(2*a + I*b*Log[1 - I*c*x^2])^
2)/(16*c) + ((I/16)*(1 - I*c*x^2)*(2*a + I*b*Log[1 - I*c*x^2])^3)/c + (3*b*((2*I)*a - b*Log[1 - I*c*x^2])^2*Lo
g[(1 + I*c*x^2)/2])/(8*c) - (3*b*((2*I)*a - b*Log[1 - I*c*x^2])^2*Log[1 + I*c*x^2])/(16*c) + ((3*I)/16)*b*x^2*
((2*I)*a - b*Log[1 - I*c*x^2])^2*Log[1 + I*c*x^2] + (3*b^3*Log[(1 - I*c*x^2)/2]*Log[1 + I*c*x^2]^2)/(8*c) + (3
*b^2*((2*I)*a - b*Log[1 - I*c*x^2])*Log[1 + I*c*x^2]^2)/(16*c) + ((3*I)/16)*b^2*x^2*((2*I)*a - b*Log[1 - I*c*x
^2])*Log[1 + I*c*x^2]^2 + (b^3*(1 + I*c*x^2)*Log[1 + I*c*x^2]^3)/(16*c) - (3*b^2*((2*I)*a - b*Log[1 - I*c*x^2]
)*PolyLog[2, (1 - I*c*x^2)/2])/(4*c) + (3*b^3*Log[1 + I*c*x^2]*PolyLog[2, (1 + I*c*x^2)/2])/(4*c) - (3*b^3*Pol
yLog[3, (1 - I*c*x^2)/2])/(4*c) - (3*b^3*PolyLog[3, (1 + I*c*x^2)/2])/(4*c)

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2375

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :
> Simp[(Log[d*(e + f*x^m)^r]*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] - Dist[(f*m*r)/(b*n*(p + 1)), Int[(
x^(m - 1)*(a + b*Log[c*x^n])^(p + 1))/(e + f*x^m), x], x] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p,
0] && NeQ[d*e, 1]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2396

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*
(f + g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n])^p)/g, x] - Dist[(b*e*n*p)/g, Int[(Log[(e*(f + g*x))/(e*f -
d*g)]*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*
f - d*g, 0] && IGtQ[p, 1]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2425

Int[(Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)))/(x_), x_Symbol] :> Simp[(Log[
f*x^m]^2*(a + b*Log[c*(d + e*x)^n]))/(2*m), x] - Dist[(b*e*n)/(2*m), Int[Log[f*x^m]^2/(d + e*x), x], x] /; Fre
eQ[{a, b, c, d, e, f, m, n}, x]

Rule 2430

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.)), x_Symbol] :> Simp[x*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]), x] + (-Dist[g*j*m, Int[(x
*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[b*e*n*p, Int[(x*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f
+ g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0]

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 5035

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

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int x \left (a+b \tan ^{-1}\left (c x^2\right )\right )^3 \, dx &=\int \left (\frac {1}{8} x \left (2 a+i b \log \left (1-i c x^2\right )\right )^3+\frac {3}{8} i b x \left (-2 i a+b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )-\frac {3}{8} i b^2 x \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {1}{8} i b^3 x \log ^3\left (1+i c x^2\right )\right ) \, dx\\ &=\frac {1}{8} \int x \left (2 a+i b \log \left (1-i c x^2\right )\right )^3 \, dx+\frac {1}{8} (3 i b) \int x \left (-2 i a+b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right ) \, dx-\frac {1}{8} \left (3 i b^2\right ) \int x \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right ) \, dx+\frac {1}{8} \left (i b^3\right ) \int x \log ^3\left (1+i c x^2\right ) \, dx\\ &=\frac {1}{16} \operatorname {Subst}\left (\int (2 a+i b \log (1-i c x))^3 \, dx,x,x^2\right )+\frac {1}{16} (3 i b) \operatorname {Subst}\left (\int (-2 i a+b \log (1-i c x))^2 \log (1+i c x) \, dx,x,x^2\right )-\frac {1}{16} \left (3 i b^2\right ) \operatorname {Subst}\left (\int (-2 i a+b \log (1-i c x)) \log ^2(1+i c x) \, dx,x,x^2\right )+\frac {1}{16} \left (i b^3\right ) \operatorname {Subst}\left (\int \log ^3(1+i c x) \, dx,x,x^2\right )\\ &=\frac {3}{16} i b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )+\frac {3}{16} i b^2 x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {i \operatorname {Subst}\left (\int (2 a+i b \log (x))^3 \, dx,x,1-i c x^2\right )}{16 c}+\frac {b^3 \operatorname {Subst}\left (\int \log ^3(x) \, dx,x,1+i c x^2\right )}{16 c}+\frac {1}{16} (3 b c) \operatorname {Subst}\left (\int \frac {x (-2 i a+b \log (1-i c x))^2}{1+i c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {x (-2 i a+b \log (1-i c x)) \log (1+i c x)}{1-i c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {x (-2 i a+b \log (1-i c x)) \log (1+i c x)}{1+i c x} \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {x \log ^2(1+i c x)}{1-i c x} \, dx,x,x^2\right )\\ &=\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 c}+\frac {3}{16} i b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )+\frac {3}{16} i b^2 x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 c}+\frac {(3 b) \operatorname {Subst}\left (\int (2 a+i b \log (x))^2 \, dx,x,1-i c x^2\right )}{16 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,1+i c x^2\right )}{16 c}+\frac {1}{16} (3 b c) \operatorname {Subst}\left (\int \left (-\frac {i (-2 i a+b \log (1-i c x))^2}{c}+\frac {(-2 i a+b \log (1-i c x))^2}{c (-i+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \left (-\frac {(2 a+i b \log (1-i c x)) \log (1+i c x)}{c}+\frac {(-2 i a+b \log (1-i c x)) \log (1+i c x)}{c (-i+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \left (\frac {(2 a+i b \log (1-i c x)) \log (1+i c x)}{c}+\frac {(-2 i a+b \log (1-i c x)) \log (1+i c x)}{c (i+c x)}\right ) \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \left (\frac {i \log ^2(1+i c x)}{c}+\frac {\log ^2(1+i c x)}{c (i+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {3 b \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 c}+\frac {3}{16} i b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )-\frac {3 b^3 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b^2 x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 c}-\frac {1}{16} (3 i b) \operatorname {Subst}\left (\int (-2 i a+b \log (1-i c x))^2 \, dx,x,x^2\right )+\frac {1}{16} (3 b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x))^2}{-i+c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log (1+i c x)}{-i+c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log (1+i c x)}{i+c x} \, dx,x,x^2\right )+\frac {1}{16} \left (3 i b^3\right ) \operatorname {Subst}\left (\int \log ^2(1+i c x) \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\log ^2(1+i c x)}{i+c x} \, dx,x,x^2\right )-\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int (2 a+i b \log (x)) \, dx,x,1-i c x^2\right )}{8 c}+\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \log (x) \, dx,x,1+i c x^2\right )}{8 c}\\ &=-\frac {3}{4} a b^2 x^2-\frac {3}{8} i b^3 x^2+\frac {3 b \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+i c x^2\right )\right )}{16 c}+\frac {3 b^3 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{8 c}+\frac {3}{16} i b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )-\frac {3 b^3 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{16 c}+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b^2 x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 c}+\frac {1}{8} \left (3 i b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (1-i c x)) \log \left (\frac {1}{2} i (-i+c x)\right )}{1-i c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 i b^3\right ) \operatorname {Subst}\left (\int \frac {\log (1+i c x) \log \left (-\frac {1}{2} i (i+c x)\right )}{1+i c x} \, dx,x,x^2\right )+\frac {(3 b) \operatorname {Subst}\left (\int (-2 i a+b \log (x))^2 \, dx,x,1-i c x^2\right )}{16 c}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (2-x)) \log (x)}{x} \, dx,x,1+i c x^2\right )}{8 c}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2-x) (-2 i a+b \log (x))}{x} \, dx,x,1-i c x^2\right )}{8 c}+\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,1+i c x^2\right )}{16 c}+\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \log (x) \, dx,x,1-i c x^2\right )}{8 c}\\ &=-\frac {3}{4} a b^2 x^2+\frac {3 b^3 \left (1-i c x^2\right ) \log \left (1-i c x^2\right )}{8 c}+\frac {3 b \left (1-i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {3 b \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+i c x^2\right )\right )}{16 c}+\frac {3 b^3 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{8 c}-\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{16 c}+\frac {3 b^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b^2 x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 c}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {(-2 i a+b \log (x))^2}{2-x} \, dx,x,1-i c x^2\right )}{16 c}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int (-2 i a+b \log (x)) \, dx,x,1-i c x^2\right )}{8 c}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (\frac {1}{2} i (-2 i+i x)\right ) (-2 i a+b \log (x))}{x} \, dx,x,1-i c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\log ^2(x)}{2-x} \, dx,x,1+i c x^2\right )}{16 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \log (x) \, dx,x,1+i c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {1}{2} i (2 i-i x)\right ) \log (x)}{x} \, dx,x,1+i c x^2\right )}{8 c}\\ &=\frac {3}{8} i b^3 x^2+\frac {3 b^3 \left (1-i c x^2\right ) \log \left (1-i c x^2\right )}{8 c}+\frac {3 b \left (1-i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {3 b \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+i c x^2\right )\right )}{8 c}-\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{8 c}+\frac {3 b^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b^2 x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 c}-\frac {3 b^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \text {Li}_2\left (\frac {1}{2} \left (1-i c x^2\right )\right )}{8 c}+\frac {3 b^3 \log \left (1+i c x^2\right ) \text {Li}_2\left (\frac {1}{2} \left (1+i c x^2\right )\right )}{8 c}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right ) (-2 i a+b \log (x))}{x} \, dx,x,1-i c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \log (x) \, dx,x,1-i c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right ) \log (x)}{x} \, dx,x,1+i c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1-i c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1+i c x^2\right )}{8 c}\\ &=\frac {3 b \left (1-i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {3 b \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+i c x^2\right )\right )}{8 c}-\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{8 c}+\frac {3 b^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b^2 x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 c}-\frac {3 b^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \text {Li}_2\left (\frac {1}{2} \left (1-i c x^2\right )\right )}{4 c}+\frac {3 b^3 \log \left (1+i c x^2\right ) \text {Li}_2\left (\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1-i c x^2\right )\right )}{8 c}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1+i c x^2\right )\right )}{8 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1-i c x^2\right )}{8 c}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {x}{2}\right )}{x} \, dx,x,1+i c x^2\right )}{8 c}\\ &=\frac {3 b \left (1-i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {3 b \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{16 c}+\frac {i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^3}{16 c}+\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (\frac {1}{2} \left (1+i c x^2\right )\right )}{8 c}-\frac {3 b \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right )^2 \log \left (1+i c x^2\right )+\frac {3 b^3 \log \left (\frac {1}{2} \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{8 c}+\frac {3 b^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )}{16 c}+\frac {3}{16} i b^2 x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log ^2\left (1+i c x^2\right )+\frac {b^3 \left (1+i c x^2\right ) \log ^3\left (1+i c x^2\right )}{16 c}-\frac {3 b^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \text {Li}_2\left (\frac {1}{2} \left (1-i c x^2\right )\right )}{4 c}+\frac {3 b^3 \log \left (1+i c x^2\right ) \text {Li}_2\left (\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1-i c x^2\right )\right )}{4 c}-\frac {3 b^3 \text {Li}_3\left (\frac {1}{2} \left (1+i c x^2\right )\right )}{4 c}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 224, normalized size = 1.56 \[ \frac {2 a^3 c x^2-3 a^2 b \log \left (c^2 x^4+1\right )+6 a^2 b c x^2 \tan ^{-1}\left (c x^2\right )-6 i b^2 \text {Li}_2\left (-e^{2 i \tan ^{-1}\left (c x^2\right )}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )-6 i a b^2 \tan ^{-1}\left (c x^2\right )^2+6 a b^2 c x^2 \tan ^{-1}\left (c x^2\right )^2+12 a b^2 \tan ^{-1}\left (c x^2\right ) \log \left (1+e^{2 i \tan ^{-1}\left (c x^2\right )}\right )+3 b^3 \text {Li}_3\left (-e^{2 i \tan ^{-1}\left (c x^2\right )}\right )-2 i b^3 \tan ^{-1}\left (c x^2\right )^3+2 b^3 c x^2 \tan ^{-1}\left (c x^2\right )^3+6 b^3 \tan ^{-1}\left (c x^2\right )^2 \log \left (1+e^{2 i \tan ^{-1}\left (c x^2\right )}\right )}{4 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arctan \left (c x^{2}\right ) + a\right )}^{3} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.24, size = 306, normalized size = 2.12 \[ \frac {x^{2} a^{3}}{2}-\frac {i b^{3} \arctan \left (c \,x^{2}\right )^{3}}{2 c}+\frac {b^{3} x^{2} \arctan \left (c \,x^{2}\right )^{3}}{2}+\frac {3 b^{3} \arctan \left (c \,x^{2}\right )^{2} \ln \left (\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}+1\right )}{2 c}-\frac {3 i b^{3} \arctan \left (c \,x^{2}\right ) \polylog \left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )}{2 c}+\frac {3 b^{3} \polylog \left (3, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right )}{4 c}-\frac {3 i \arctan \left (c \,x^{2}\right )^{2} a \,b^{2}}{2 c}+\frac {3 x^{2} a \,b^{2} \arctan \left (c \,x^{2}\right )^{2}}{2}+\frac {3 \ln \left (\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}+1\right ) \arctan \left (c \,x^{2}\right ) a \,b^{2}}{c}-\frac {3 i \polylog \left (2, -\frac {\left (i c \,x^{2}+1\right )^{2}}{c^{2} x^{4}+1}\right ) a \,b^{2}}{2 c}+\frac {3 x^{2} a^{2} b \arctan \left (c \,x^{2}\right )}{2}-\frac {3 a^{2} b \ln \left (c^{2} x^{4}+1\right )}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*a^3-1/2*I/c*b^3*arctan(c*x^2)^3+1/2*b^3*x^2*arctan(c*x^2)^3+3/2/c*b^3*arctan(c*x^2)^2*ln((1+I*c*x^2)^2
/(c^2*x^4+1)+1)-3/2*I/c*b^3*arctan(c*x^2)*polylog(2,-(1+I*c*x^2)^2/(c^2*x^4+1))+3/4/c*b^3*polylog(3,-(1+I*c*x^
2)^2/(c^2*x^4+1))-3/2*I/c*arctan(c*x^2)^2*a*b^2+3/2*x^2*a*b^2*arctan(c*x^2)^2+3/c*ln((1+I*c*x^2)^2/(c^2*x^4+1)
+1)*arctan(c*x^2)*a*b^2-3/2*I/c*polylog(2,-(1+I*c*x^2)^2/(c^2*x^4+1))*a*b^2+3/2*x^2*a^2*b*arctan(c*x^2)-3/4/c*
a^2*b*ln(c^2*x^4+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, b^{3} x^{2} \arctan \left (c x^{2}\right )^{3} - \frac {3}{64} \, b^{3} x^{2} \arctan \left (c x^{2}\right ) \log \left (c^{2} x^{4} + 1\right )^{2} + \frac {7 \, b^{3} \arctan \left (c x^{2}\right )^{4}}{64 \, c} + 28 \, b^{3} c^{2} \int \frac {x^{5} \arctan \left (c x^{2}\right )^{3}}{32 \, {\left (c^{2} x^{4} + 1\right )}}\,{d x} + 3 \, b^{3} c^{2} \int \frac {x^{5} \arctan \left (c x^{2}\right ) \log \left (c^{2} x^{4} + 1\right )^{2}}{32 \, {\left (c^{2} x^{4} + 1\right )}}\,{d x} + 96 \, a b^{2} c^{2} \int \frac {x^{5} \arctan \left (c x^{2}\right )^{2}}{32 \, {\left (c^{2} x^{4} + 1\right )}}\,{d x} + 12 \, b^{3} c^{2} \int \frac {x^{5} \arctan \left (c x^{2}\right ) \log \left (c^{2} x^{4} + 1\right )}{32 \, {\left (c^{2} x^{4} + 1\right )}}\,{d x} + \frac {1}{2} \, a^{3} x^{2} + \frac {a b^{2} \arctan \left (c x^{2}\right )^{3}}{2 \, c} - 12 \, b^{3} c \int \frac {x^{3} \arctan \left (c x^{2}\right )^{2}}{32 \, {\left (c^{2} x^{4} + 1\right )}}\,{d x} + 3 \, b^{3} c \int \frac {x^{3} \log \left (c^{2} x^{4} + 1\right )^{2}}{32 \, {\left (c^{2} x^{4} + 1\right )}}\,{d x} + 3 \, b^{3} \int \frac {x \arctan \left (c x^{2}\right ) \log \left (c^{2} x^{4} + 1\right )^{2}}{32 \, {\left (c^{2} x^{4} + 1\right )}}\,{d x} + \frac {3 \, {\left (2 \, c x^{2} \arctan \left (c x^{2}\right ) - \log \left (c^{2} x^{4} + 1\right )\right )} a^{2} b}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*b^3*x^2*arctan(c*x^2)^3 - 3/64*b^3*x^2*arctan(c*x^2)*log(c^2*x^4 + 1)^2 + 7/64*b^3*arctan(c*x^2)^4/c + 28
*b^3*c^2*integrate(1/32*x^5*arctan(c*x^2)^3/(c^2*x^4 + 1), x) + 3*b^3*c^2*integrate(1/32*x^5*arctan(c*x^2)*log
(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 96*a*b^2*c^2*integrate(1/32*x^5*arctan(c*x^2)^2/(c^2*x^4 + 1), x) + 12*b^3
*c^2*integrate(1/32*x^5*arctan(c*x^2)*log(c^2*x^4 + 1)/(c^2*x^4 + 1), x) + 1/2*a^3*x^2 + 1/2*a*b^2*arctan(c*x^
2)^3/c - 12*b^3*c*integrate(1/32*x^3*arctan(c*x^2)^2/(c^2*x^4 + 1), x) + 3*b^3*c*integrate(1/32*x^3*log(c^2*x^
4 + 1)^2/(c^2*x^4 + 1), x) + 3*b^3*integrate(1/32*x*arctan(c*x^2)*log(c^2*x^4 + 1)^2/(c^2*x^4 + 1), x) + 3/4*(
2*c*x^2*arctan(c*x^2) - log(c^2*x^4 + 1))*a^2*b/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (a+b\,\mathrm {atan}\left (c\,x^2\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atan(c*x^2))^3,x)

[Out]

int(x*(a + b*atan(c*x^2))^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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